Slide 1

School of Engineering
ENGINEERING MECHANICS
CHAPTER 8

Kinetics of Rotational Motion

1

8.1 Brief
Kinetics of rotational motion is the study of rotational
motion caused by external torque.

8.2 Newton’s second law for Rotational Motion
It can be written in mathematical form as:
 = I
 = I

for a single torque
for several torque
where  is the torque (Nm – rotating shaft)
I is the moment of inertia (kgm2)
 is the acceleration (rad/s2)

(Note:- I = mk2, where m is the mass and k is the radius
of gyration.)
2

8.3 Free Body diagram
We note that since the motion is caused by torque
acting on the body, the first step would be to show
all the torque acting on it. This is set out in the form
of a free body diagram indicating clearly all
external torque. A free body diagram is a diagram
showing only the body and the torque acting
directly on it.
8.4 Sign Convention
We are interested to determine the motion of the
body hence we choose the direction of the
resulting rotation as the positive direction.
However, any direction can be selected as
positive but this must be strictly adhered to
throughout the analysis.
3

Example 8.1
A rotor with a mass moment of inertia of 6 kgm2 about
its center of mass has a torque of 90 Nm applied to it.
Determine the angular acceleration of the rotor.

Solution:
By Newton’s law,
 = I

90 Nm

90 = 6
 = 15 rad/s2
4

Example 8.2
The flywheel has a mass of 5 kg and a radius of
gyration of 200 mm. The 20 kg mass, attached to a
rope wrapped around the spindle of diameter 30 mm,
is allowed to fall from rest. Calculate the angular
acceleration of the wheel and the acceleration of the
20 kg mass.
Flywheel
5kg

Spindle

20kg
5

Solution:
For the flywheel,

I = mk2
= 5 x 0.22
= 0.2 kgm2
Torque on flywheel,
 = T x 0.015 Nm
 = I
T x 0.015 = 0.2
T = (0.2 / 0.015) 
---------- (1)

dia =30 mm

T

T

For the mass,
F = ma
20g – T = 20a
a = r
 = a / 0.015

----------- (2)
20g

------------- (3)
6

From (1) & (3),
T = (0.2 / 0.015) x (a / 0.015)
= 0.2a / 0.0152
Sub into (2),
20g – 0.2a / 0.0152 = 20a
a = 0.216 m/s2
From (3),
 = 0.216 / 0.015
= 14.4 rad/s2
7

Example 8.3
A power-driven winch is used to raise a mass of 300
kg with an acceleration of 2 m/s2. The winch drum is
0.5 m in diameter and has a mass moment of inertia
about its center of 8 kgm2. What torque must be
applied to the winch drum?
Winch

300kg

8

T

T

a

For the mass,
F = ma
T - 300g = 300 x 2
T = 3543 N

300g

For the drum,
a = r
2 = 0.25
= 8 rad/s2
 = I
a - 3543 x 0.25 = 8 x 8
a = 950 Nm

9

Example 8.4
A compound pulley having a mass of 4 kg and a
radius of gyration 0.28 m, is connected to two
masses A and B as shown. Assuming no axle
friction, determine the acceleration of the masses.
rB = 0.3 m

rA = 0.1 m

A
10kg

B
20kg

10

TB

TA
TA

10g

TB

20g

I = mk2
= 4 x 0.282
= 0.3136 kgm2
Assume acceleration clockwise:
 = I
TB x 0.3 – TA x 0.1 = 0.3136 --------------- (1)

11

TB

TA
TA

TB

10g

20g

For block A,

F = ma
TA – 10g = 10 aA ---------- (2)

For block B,

F = ma
20g – TB = 20 aB -------- (3)

For pulley,
 = aA / rA = aB / rB

----------- (4)
12

TB

TA
TA

10g

TB

20g

Sub (4) into (2),
TA – 10g = 10 x 0.1
TA =  + 10g

--------------- (5)

Sub (4) into (3),
20g – TB = 20 x 0.3
TB = - 6 + 20g

--------------- (6)
13

TB

TA
TA

10g

TB

20g

Sub (5) & (6) into (1),
(- 6 + 20g) x 0.3 – ( + 10g) x 0.1 = 0.3136
 = 22.154 rad/s2
aA = rA = 0.1 x 22.154 = 2.215 m/s2
aB = rB = 0.3 x 22.154 = 6.646 m/s2
14

Example 8.5
A power driven winch can exert a torque of 1200
Nm. It has a drum of 80 cm diameter and a mass
moment of inertia of 30 kgm2. It is used to raise
block A of mass 200 kg up the slope as shown in
the figure. The coefficient of friction between
block A and the slope is 0.3. Assume the pulley
is frictionless.
(a) Draw the free body diagrams of the winch and
block A.
(b) Determine the acceleration of block A and the
tension in the cable.
Winch

Pulley

Block A 200 kg
60
15

 = 1200 Nm

T
T

200g

Winch

a
N1

60

F1

Block A

Block A:
For the direction perpendicular to the slope,
Fy = 0
N1 – 200gcos60 = 0
N1 = 981
F1 =  N1
= 0.3 x 981
= 294.3 N
16

T
 = 1200 Nm

T

200g

Winch

a
N1

60

F1

Block A

For the direction up the slope,
Fx = ma
T – 294.3 – 200gsin60 = 200a
T = 1993.44 + 200a
---------- (1)
For the drum, a = r
 = (a / 0.4)
17

T

 = 1200 Nm

T

200g

Winch

For the anti-clockwise direction,
N1
 = I
1200 – 0.4T = 30 x (a / 0.4)
1200 – 0.4T – 75a = 0
--------- (2)

a

60

F1

Block A

Substitute (1) into (2)
1200 – 0.4 (1993.44 + 200a) – 75a = 0
a = 2.6 m/s2
Sub. into (1)
T = 1993.44 + 200 x 2.6
= 2513.44 N

End of Chapter 8
18


Slide 2

School of Engineering
ENGINEERING MECHANICS
CHAPTER 8

Kinetics of Rotational Motion

1

8.1 Brief
Kinetics of rotational motion is the study of rotational
motion caused by external torque.

8.2 Newton’s second law for Rotational Motion
It can be written in mathematical form as:
 = I
 = I

for a single torque
for several torque
where  is the torque (Nm – rotating shaft)
I is the moment of inertia (kgm2)
 is the acceleration (rad/s2)

(Note:- I = mk2, where m is the mass and k is the radius
of gyration.)
2

8.3 Free Body diagram
We note that since the motion is caused by torque
acting on the body, the first step would be to show
all the torque acting on it. This is set out in the form
of a free body diagram indicating clearly all
external torque. A free body diagram is a diagram
showing only the body and the torque acting
directly on it.
8.4 Sign Convention
We are interested to determine the motion of the
body hence we choose the direction of the
resulting rotation as the positive direction.
However, any direction can be selected as
positive but this must be strictly adhered to
throughout the analysis.
3

Example 8.1
A rotor with a mass moment of inertia of 6 kgm2 about
its center of mass has a torque of 90 Nm applied to it.
Determine the angular acceleration of the rotor.

Solution:
By Newton’s law,
 = I

90 Nm

90 = 6
 = 15 rad/s2
4

Example 8.2
The flywheel has a mass of 5 kg and a radius of
gyration of 200 mm. The 20 kg mass, attached to a
rope wrapped around the spindle of diameter 30 mm,
is allowed to fall from rest. Calculate the angular
acceleration of the wheel and the acceleration of the
20 kg mass.
Flywheel
5kg

Spindle

20kg
5

Solution:
For the flywheel,

I = mk2
= 5 x 0.22
= 0.2 kgm2
Torque on flywheel,
 = T x 0.015 Nm
 = I
T x 0.015 = 0.2
T = (0.2 / 0.015) 
---------- (1)

dia =30 mm

T

T

For the mass,
F = ma
20g – T = 20a
a = r
 = a / 0.015

----------- (2)
20g

------------- (3)
6

From (1) & (3),
T = (0.2 / 0.015) x (a / 0.015)
= 0.2a / 0.0152
Sub into (2),
20g – 0.2a / 0.0152 = 20a
a = 0.216 m/s2
From (3),
 = 0.216 / 0.015
= 14.4 rad/s2
7

Example 8.3
A power-driven winch is used to raise a mass of 300
kg with an acceleration of 2 m/s2. The winch drum is
0.5 m in diameter and has a mass moment of inertia
about its center of 8 kgm2. What torque must be
applied to the winch drum?
Winch

300kg

8

T

T

a

For the mass,
F = ma
T - 300g = 300 x 2
T = 3543 N

300g

For the drum,
a = r
2 = 0.25
= 8 rad/s2
 = I
a - 3543 x 0.25 = 8 x 8
a = 950 Nm

9

Example 8.4
A compound pulley having a mass of 4 kg and a
radius of gyration 0.28 m, is connected to two
masses A and B as shown. Assuming no axle
friction, determine the acceleration of the masses.
rB = 0.3 m

rA = 0.1 m

A
10kg

B
20kg

10

TB

TA
TA

10g

TB

20g

I = mk2
= 4 x 0.282
= 0.3136 kgm2
Assume acceleration clockwise:
 = I
TB x 0.3 – TA x 0.1 = 0.3136 --------------- (1)

11

TB

TA
TA

TB

10g

20g

For block A,

F = ma
TA – 10g = 10 aA ---------- (2)

For block B,

F = ma
20g – TB = 20 aB -------- (3)

For pulley,
 = aA / rA = aB / rB

----------- (4)
12

TB

TA
TA

10g

TB

20g

Sub (4) into (2),
TA – 10g = 10 x 0.1
TA =  + 10g

--------------- (5)

Sub (4) into (3),
20g – TB = 20 x 0.3
TB = - 6 + 20g

--------------- (6)
13

TB

TA
TA

10g

TB

20g

Sub (5) & (6) into (1),
(- 6 + 20g) x 0.3 – ( + 10g) x 0.1 = 0.3136
 = 22.154 rad/s2
aA = rA = 0.1 x 22.154 = 2.215 m/s2
aB = rB = 0.3 x 22.154 = 6.646 m/s2
14

Example 8.5
A power driven winch can exert a torque of 1200
Nm. It has a drum of 80 cm diameter and a mass
moment of inertia of 30 kgm2. It is used to raise
block A of mass 200 kg up the slope as shown in
the figure. The coefficient of friction between
block A and the slope is 0.3. Assume the pulley
is frictionless.
(a) Draw the free body diagrams of the winch and
block A.
(b) Determine the acceleration of block A and the
tension in the cable.
Winch

Pulley

Block A 200 kg
60
15

 = 1200 Nm

T
T

200g

Winch

a
N1

60

F1

Block A

Block A:
For the direction perpendicular to the slope,
Fy = 0
N1 – 200gcos60 = 0
N1 = 981
F1 =  N1
= 0.3 x 981
= 294.3 N
16

T
 = 1200 Nm

T

200g

Winch

a
N1

60

F1

Block A

For the direction up the slope,
Fx = ma
T – 294.3 – 200gsin60 = 200a
T = 1993.44 + 200a
---------- (1)
For the drum, a = r
 = (a / 0.4)
17

T

 = 1200 Nm

T

200g

Winch

For the anti-clockwise direction,
N1
 = I
1200 – 0.4T = 30 x (a / 0.4)
1200 – 0.4T – 75a = 0
--------- (2)

a

60

F1

Block A

Substitute (1) into (2)
1200 – 0.4 (1993.44 + 200a) – 75a = 0
a = 2.6 m/s2
Sub. into (1)
T = 1993.44 + 200 x 2.6
= 2513.44 N

End of Chapter 8
18


Slide 3

School of Engineering
ENGINEERING MECHANICS
CHAPTER 8

Kinetics of Rotational Motion

1

8.1 Brief
Kinetics of rotational motion is the study of rotational
motion caused by external torque.

8.2 Newton’s second law for Rotational Motion
It can be written in mathematical form as:
 = I
 = I

for a single torque
for several torque
where  is the torque (Nm – rotating shaft)
I is the moment of inertia (kgm2)
 is the acceleration (rad/s2)

(Note:- I = mk2, where m is the mass and k is the radius
of gyration.)
2

8.3 Free Body diagram
We note that since the motion is caused by torque
acting on the body, the first step would be to show
all the torque acting on it. This is set out in the form
of a free body diagram indicating clearly all
external torque. A free body diagram is a diagram
showing only the body and the torque acting
directly on it.
8.4 Sign Convention
We are interested to determine the motion of the
body hence we choose the direction of the
resulting rotation as the positive direction.
However, any direction can be selected as
positive but this must be strictly adhered to
throughout the analysis.
3

Example 8.1
A rotor with a mass moment of inertia of 6 kgm2 about
its center of mass has a torque of 90 Nm applied to it.
Determine the angular acceleration of the rotor.

Solution:
By Newton’s law,
 = I

90 Nm

90 = 6
 = 15 rad/s2
4

Example 8.2
The flywheel has a mass of 5 kg and a radius of
gyration of 200 mm. The 20 kg mass, attached to a
rope wrapped around the spindle of diameter 30 mm,
is allowed to fall from rest. Calculate the angular
acceleration of the wheel and the acceleration of the
20 kg mass.
Flywheel
5kg

Spindle

20kg
5

Solution:
For the flywheel,

I = mk2
= 5 x 0.22
= 0.2 kgm2
Torque on flywheel,
 = T x 0.015 Nm
 = I
T x 0.015 = 0.2
T = (0.2 / 0.015) 
---------- (1)

dia =30 mm

T

T

For the mass,
F = ma
20g – T = 20a
a = r
 = a / 0.015

----------- (2)
20g

------------- (3)
6

From (1) & (3),
T = (0.2 / 0.015) x (a / 0.015)
= 0.2a / 0.0152
Sub into (2),
20g – 0.2a / 0.0152 = 20a
a = 0.216 m/s2
From (3),
 = 0.216 / 0.015
= 14.4 rad/s2
7

Example 8.3
A power-driven winch is used to raise a mass of 300
kg with an acceleration of 2 m/s2. The winch drum is
0.5 m in diameter and has a mass moment of inertia
about its center of 8 kgm2. What torque must be
applied to the winch drum?
Winch

300kg

8

T

T

a

For the mass,
F = ma
T - 300g = 300 x 2
T = 3543 N

300g

For the drum,
a = r
2 = 0.25
= 8 rad/s2
 = I
a - 3543 x 0.25 = 8 x 8
a = 950 Nm

9

Example 8.4
A compound pulley having a mass of 4 kg and a
radius of gyration 0.28 m, is connected to two
masses A and B as shown. Assuming no axle
friction, determine the acceleration of the masses.
rB = 0.3 m

rA = 0.1 m

A
10kg

B
20kg

10

TB

TA
TA

10g

TB

20g

I = mk2
= 4 x 0.282
= 0.3136 kgm2
Assume acceleration clockwise:
 = I
TB x 0.3 – TA x 0.1 = 0.3136 --------------- (1)

11

TB

TA
TA

TB

10g

20g

For block A,

F = ma
TA – 10g = 10 aA ---------- (2)

For block B,

F = ma
20g – TB = 20 aB -------- (3)

For pulley,
 = aA / rA = aB / rB

----------- (4)
12

TB

TA
TA

10g

TB

20g

Sub (4) into (2),
TA – 10g = 10 x 0.1
TA =  + 10g

--------------- (5)

Sub (4) into (3),
20g – TB = 20 x 0.3
TB = - 6 + 20g

--------------- (6)
13

TB

TA
TA

10g

TB

20g

Sub (5) & (6) into (1),
(- 6 + 20g) x 0.3 – ( + 10g) x 0.1 = 0.3136
 = 22.154 rad/s2
aA = rA = 0.1 x 22.154 = 2.215 m/s2
aB = rB = 0.3 x 22.154 = 6.646 m/s2
14

Example 8.5
A power driven winch can exert a torque of 1200
Nm. It has a drum of 80 cm diameter and a mass
moment of inertia of 30 kgm2. It is used to raise
block A of mass 200 kg up the slope as shown in
the figure. The coefficient of friction between
block A and the slope is 0.3. Assume the pulley
is frictionless.
(a) Draw the free body diagrams of the winch and
block A.
(b) Determine the acceleration of block A and the
tension in the cable.
Winch

Pulley

Block A 200 kg
60
15

 = 1200 Nm

T
T

200g

Winch

a
N1

60

F1

Block A

Block A:
For the direction perpendicular to the slope,
Fy = 0
N1 – 200gcos60 = 0
N1 = 981
F1 =  N1
= 0.3 x 981
= 294.3 N
16

T
 = 1200 Nm

T

200g

Winch

a
N1

60

F1

Block A

For the direction up the slope,
Fx = ma
T – 294.3 – 200gsin60 = 200a
T = 1993.44 + 200a
---------- (1)
For the drum, a = r
 = (a / 0.4)
17

T

 = 1200 Nm

T

200g

Winch

For the anti-clockwise direction,
N1
 = I
1200 – 0.4T = 30 x (a / 0.4)
1200 – 0.4T – 75a = 0
--------- (2)

a

60

F1

Block A

Substitute (1) into (2)
1200 – 0.4 (1993.44 + 200a) – 75a = 0
a = 2.6 m/s2
Sub. into (1)
T = 1993.44 + 200 x 2.6
= 2513.44 N

End of Chapter 8
18


Slide 4

School of Engineering
ENGINEERING MECHANICS
CHAPTER 8

Kinetics of Rotational Motion

1

8.1 Brief
Kinetics of rotational motion is the study of rotational
motion caused by external torque.

8.2 Newton’s second law for Rotational Motion
It can be written in mathematical form as:
 = I
 = I

for a single torque
for several torque
where  is the torque (Nm – rotating shaft)
I is the moment of inertia (kgm2)
 is the acceleration (rad/s2)

(Note:- I = mk2, where m is the mass and k is the radius
of gyration.)
2

8.3 Free Body diagram
We note that since the motion is caused by torque
acting on the body, the first step would be to show
all the torque acting on it. This is set out in the form
of a free body diagram indicating clearly all
external torque. A free body diagram is a diagram
showing only the body and the torque acting
directly on it.
8.4 Sign Convention
We are interested to determine the motion of the
body hence we choose the direction of the
resulting rotation as the positive direction.
However, any direction can be selected as
positive but this must be strictly adhered to
throughout the analysis.
3

Example 8.1
A rotor with a mass moment of inertia of 6 kgm2 about
its center of mass has a torque of 90 Nm applied to it.
Determine the angular acceleration of the rotor.

Solution:
By Newton’s law,
 = I

90 Nm

90 = 6
 = 15 rad/s2
4

Example 8.2
The flywheel has a mass of 5 kg and a radius of
gyration of 200 mm. The 20 kg mass, attached to a
rope wrapped around the spindle of diameter 30 mm,
is allowed to fall from rest. Calculate the angular
acceleration of the wheel and the acceleration of the
20 kg mass.
Flywheel
5kg

Spindle

20kg
5

Solution:
For the flywheel,

I = mk2
= 5 x 0.22
= 0.2 kgm2
Torque on flywheel,
 = T x 0.015 Nm
 = I
T x 0.015 = 0.2
T = (0.2 / 0.015) 
---------- (1)

dia =30 mm

T

T

For the mass,
F = ma
20g – T = 20a
a = r
 = a / 0.015

----------- (2)
20g

------------- (3)
6

From (1) & (3),
T = (0.2 / 0.015) x (a / 0.015)
= 0.2a / 0.0152
Sub into (2),
20g – 0.2a / 0.0152 = 20a
a = 0.216 m/s2
From (3),
 = 0.216 / 0.015
= 14.4 rad/s2
7

Example 8.3
A power-driven winch is used to raise a mass of 300
kg with an acceleration of 2 m/s2. The winch drum is
0.5 m in diameter and has a mass moment of inertia
about its center of 8 kgm2. What torque must be
applied to the winch drum?
Winch

300kg

8

T

T

a

For the mass,
F = ma
T - 300g = 300 x 2
T = 3543 N

300g

For the drum,
a = r
2 = 0.25
= 8 rad/s2
 = I
a - 3543 x 0.25 = 8 x 8
a = 950 Nm

9

Example 8.4
A compound pulley having a mass of 4 kg and a
radius of gyration 0.28 m, is connected to two
masses A and B as shown. Assuming no axle
friction, determine the acceleration of the masses.
rB = 0.3 m

rA = 0.1 m

A
10kg

B
20kg

10

TB

TA
TA

10g

TB

20g

I = mk2
= 4 x 0.282
= 0.3136 kgm2
Assume acceleration clockwise:
 = I
TB x 0.3 – TA x 0.1 = 0.3136 --------------- (1)

11

TB

TA
TA

TB

10g

20g

For block A,

F = ma
TA – 10g = 10 aA ---------- (2)

For block B,

F = ma
20g – TB = 20 aB -------- (3)

For pulley,
 = aA / rA = aB / rB

----------- (4)
12

TB

TA
TA

10g

TB

20g

Sub (4) into (2),
TA – 10g = 10 x 0.1
TA =  + 10g

--------------- (5)

Sub (4) into (3),
20g – TB = 20 x 0.3
TB = - 6 + 20g

--------------- (6)
13

TB

TA
TA

10g

TB

20g

Sub (5) & (6) into (1),
(- 6 + 20g) x 0.3 – ( + 10g) x 0.1 = 0.3136
 = 22.154 rad/s2
aA = rA = 0.1 x 22.154 = 2.215 m/s2
aB = rB = 0.3 x 22.154 = 6.646 m/s2
14

Example 8.5
A power driven winch can exert a torque of 1200
Nm. It has a drum of 80 cm diameter and a mass
moment of inertia of 30 kgm2. It is used to raise
block A of mass 200 kg up the slope as shown in
the figure. The coefficient of friction between
block A and the slope is 0.3. Assume the pulley
is frictionless.
(a) Draw the free body diagrams of the winch and
block A.
(b) Determine the acceleration of block A and the
tension in the cable.
Winch

Pulley

Block A 200 kg
60
15

 = 1200 Nm

T
T

200g

Winch

a
N1

60

F1

Block A

Block A:
For the direction perpendicular to the slope,
Fy = 0
N1 – 200gcos60 = 0
N1 = 981
F1 =  N1
= 0.3 x 981
= 294.3 N
16

T
 = 1200 Nm

T

200g

Winch

a
N1

60

F1

Block A

For the direction up the slope,
Fx = ma
T – 294.3 – 200gsin60 = 200a
T = 1993.44 + 200a
---------- (1)
For the drum, a = r
 = (a / 0.4)
17

T

 = 1200 Nm

T

200g

Winch

For the anti-clockwise direction,
N1
 = I
1200 – 0.4T = 30 x (a / 0.4)
1200 – 0.4T – 75a = 0
--------- (2)

a

60

F1

Block A

Substitute (1) into (2)
1200 – 0.4 (1993.44 + 200a) – 75a = 0
a = 2.6 m/s2
Sub. into (1)
T = 1993.44 + 200 x 2.6
= 2513.44 N

End of Chapter 8
18


Slide 5

School of Engineering
ENGINEERING MECHANICS
CHAPTER 8

Kinetics of Rotational Motion

1

8.1 Brief
Kinetics of rotational motion is the study of rotational
motion caused by external torque.

8.2 Newton’s second law for Rotational Motion
It can be written in mathematical form as:
 = I
 = I

for a single torque
for several torque
where  is the torque (Nm – rotating shaft)
I is the moment of inertia (kgm2)
 is the acceleration (rad/s2)

(Note:- I = mk2, where m is the mass and k is the radius
of gyration.)
2

8.3 Free Body diagram
We note that since the motion is caused by torque
acting on the body, the first step would be to show
all the torque acting on it. This is set out in the form
of a free body diagram indicating clearly all
external torque. A free body diagram is a diagram
showing only the body and the torque acting
directly on it.
8.4 Sign Convention
We are interested to determine the motion of the
body hence we choose the direction of the
resulting rotation as the positive direction.
However, any direction can be selected as
positive but this must be strictly adhered to
throughout the analysis.
3

Example 8.1
A rotor with a mass moment of inertia of 6 kgm2 about
its center of mass has a torque of 90 Nm applied to it.
Determine the angular acceleration of the rotor.

Solution:
By Newton’s law,
 = I

90 Nm

90 = 6
 = 15 rad/s2
4

Example 8.2
The flywheel has a mass of 5 kg and a radius of
gyration of 200 mm. The 20 kg mass, attached to a
rope wrapped around the spindle of diameter 30 mm,
is allowed to fall from rest. Calculate the angular
acceleration of the wheel and the acceleration of the
20 kg mass.
Flywheel
5kg

Spindle

20kg
5

Solution:
For the flywheel,

I = mk2
= 5 x 0.22
= 0.2 kgm2
Torque on flywheel,
 = T x 0.015 Nm
 = I
T x 0.015 = 0.2
T = (0.2 / 0.015) 
---------- (1)

dia =30 mm

T

T

For the mass,
F = ma
20g – T = 20a
a = r
 = a / 0.015

----------- (2)
20g

------------- (3)
6

From (1) & (3),
T = (0.2 / 0.015) x (a / 0.015)
= 0.2a / 0.0152
Sub into (2),
20g – 0.2a / 0.0152 = 20a
a = 0.216 m/s2
From (3),
 = 0.216 / 0.015
= 14.4 rad/s2
7

Example 8.3
A power-driven winch is used to raise a mass of 300
kg with an acceleration of 2 m/s2. The winch drum is
0.5 m in diameter and has a mass moment of inertia
about its center of 8 kgm2. What torque must be
applied to the winch drum?
Winch

300kg

8

T

T

a

For the mass,
F = ma
T - 300g = 300 x 2
T = 3543 N

300g

For the drum,
a = r
2 = 0.25
= 8 rad/s2
 = I
a - 3543 x 0.25 = 8 x 8
a = 950 Nm

9

Example 8.4
A compound pulley having a mass of 4 kg and a
radius of gyration 0.28 m, is connected to two
masses A and B as shown. Assuming no axle
friction, determine the acceleration of the masses.
rB = 0.3 m

rA = 0.1 m

A
10kg

B
20kg

10

TB

TA
TA

10g

TB

20g

I = mk2
= 4 x 0.282
= 0.3136 kgm2
Assume acceleration clockwise:
 = I
TB x 0.3 – TA x 0.1 = 0.3136 --------------- (1)

11

TB

TA
TA

TB

10g

20g

For block A,

F = ma
TA – 10g = 10 aA ---------- (2)

For block B,

F = ma
20g – TB = 20 aB -------- (3)

For pulley,
 = aA / rA = aB / rB

----------- (4)
12

TB

TA
TA

10g

TB

20g

Sub (4) into (2),
TA – 10g = 10 x 0.1
TA =  + 10g

--------------- (5)

Sub (4) into (3),
20g – TB = 20 x 0.3
TB = - 6 + 20g

--------------- (6)
13

TB

TA
TA

10g

TB

20g

Sub (5) & (6) into (1),
(- 6 + 20g) x 0.3 – ( + 10g) x 0.1 = 0.3136
 = 22.154 rad/s2
aA = rA = 0.1 x 22.154 = 2.215 m/s2
aB = rB = 0.3 x 22.154 = 6.646 m/s2
14

Example 8.5
A power driven winch can exert a torque of 1200
Nm. It has a drum of 80 cm diameter and a mass
moment of inertia of 30 kgm2. It is used to raise
block A of mass 200 kg up the slope as shown in
the figure. The coefficient of friction between
block A and the slope is 0.3. Assume the pulley
is frictionless.
(a) Draw the free body diagrams of the winch and
block A.
(b) Determine the acceleration of block A and the
tension in the cable.
Winch

Pulley

Block A 200 kg
60
15

 = 1200 Nm

T
T

200g

Winch

a
N1

60

F1

Block A

Block A:
For the direction perpendicular to the slope,
Fy = 0
N1 – 200gcos60 = 0
N1 = 981
F1 =  N1
= 0.3 x 981
= 294.3 N
16

T
 = 1200 Nm

T

200g

Winch

a
N1

60

F1

Block A

For the direction up the slope,
Fx = ma
T – 294.3 – 200gsin60 = 200a
T = 1993.44 + 200a
---------- (1)
For the drum, a = r
 = (a / 0.4)
17

T

 = 1200 Nm

T

200g

Winch

For the anti-clockwise direction,
N1
 = I
1200 – 0.4T = 30 x (a / 0.4)
1200 – 0.4T – 75a = 0
--------- (2)

a

60

F1

Block A

Substitute (1) into (2)
1200 – 0.4 (1993.44 + 200a) – 75a = 0
a = 2.6 m/s2
Sub. into (1)
T = 1993.44 + 200 x 2.6
= 2513.44 N

End of Chapter 8
18


Slide 6

School of Engineering
ENGINEERING MECHANICS
CHAPTER 8

Kinetics of Rotational Motion

1

8.1 Brief
Kinetics of rotational motion is the study of rotational
motion caused by external torque.

8.2 Newton’s second law for Rotational Motion
It can be written in mathematical form as:
 = I
 = I

for a single torque
for several torque
where  is the torque (Nm – rotating shaft)
I is the moment of inertia (kgm2)
 is the acceleration (rad/s2)

(Note:- I = mk2, where m is the mass and k is the radius
of gyration.)
2

8.3 Free Body diagram
We note that since the motion is caused by torque
acting on the body, the first step would be to show
all the torque acting on it. This is set out in the form
of a free body diagram indicating clearly all
external torque. A free body diagram is a diagram
showing only the body and the torque acting
directly on it.
8.4 Sign Convention
We are interested to determine the motion of the
body hence we choose the direction of the
resulting rotation as the positive direction.
However, any direction can be selected as
positive but this must be strictly adhered to
throughout the analysis.
3

Example 8.1
A rotor with a mass moment of inertia of 6 kgm2 about
its center of mass has a torque of 90 Nm applied to it.
Determine the angular acceleration of the rotor.

Solution:
By Newton’s law,
 = I

90 Nm

90 = 6
 = 15 rad/s2
4

Example 8.2
The flywheel has a mass of 5 kg and a radius of
gyration of 200 mm. The 20 kg mass, attached to a
rope wrapped around the spindle of diameter 30 mm,
is allowed to fall from rest. Calculate the angular
acceleration of the wheel and the acceleration of the
20 kg mass.
Flywheel
5kg

Spindle

20kg
5

Solution:
For the flywheel,

I = mk2
= 5 x 0.22
= 0.2 kgm2
Torque on flywheel,
 = T x 0.015 Nm
 = I
T x 0.015 = 0.2
T = (0.2 / 0.015) 
---------- (1)

dia =30 mm

T

T

For the mass,
F = ma
20g – T = 20a
a = r
 = a / 0.015

----------- (2)
20g

------------- (3)
6

From (1) & (3),
T = (0.2 / 0.015) x (a / 0.015)
= 0.2a / 0.0152
Sub into (2),
20g – 0.2a / 0.0152 = 20a
a = 0.216 m/s2
From (3),
 = 0.216 / 0.015
= 14.4 rad/s2
7

Example 8.3
A power-driven winch is used to raise a mass of 300
kg with an acceleration of 2 m/s2. The winch drum is
0.5 m in diameter and has a mass moment of inertia
about its center of 8 kgm2. What torque must be
applied to the winch drum?
Winch

300kg

8

T

T

a

For the mass,
F = ma
T - 300g = 300 x 2
T = 3543 N

300g

For the drum,
a = r
2 = 0.25
= 8 rad/s2
 = I
a - 3543 x 0.25 = 8 x 8
a = 950 Nm

9

Example 8.4
A compound pulley having a mass of 4 kg and a
radius of gyration 0.28 m, is connected to two
masses A and B as shown. Assuming no axle
friction, determine the acceleration of the masses.
rB = 0.3 m

rA = 0.1 m

A
10kg

B
20kg

10

TB

TA
TA

10g

TB

20g

I = mk2
= 4 x 0.282
= 0.3136 kgm2
Assume acceleration clockwise:
 = I
TB x 0.3 – TA x 0.1 = 0.3136 --------------- (1)

11

TB

TA
TA

TB

10g

20g

For block A,

F = ma
TA – 10g = 10 aA ---------- (2)

For block B,

F = ma
20g – TB = 20 aB -------- (3)

For pulley,
 = aA / rA = aB / rB

----------- (4)
12

TB

TA
TA

10g

TB

20g

Sub (4) into (2),
TA – 10g = 10 x 0.1
TA =  + 10g

--------------- (5)

Sub (4) into (3),
20g – TB = 20 x 0.3
TB = - 6 + 20g

--------------- (6)
13

TB

TA
TA

10g

TB

20g

Sub (5) & (6) into (1),
(- 6 + 20g) x 0.3 – ( + 10g) x 0.1 = 0.3136
 = 22.154 rad/s2
aA = rA = 0.1 x 22.154 = 2.215 m/s2
aB = rB = 0.3 x 22.154 = 6.646 m/s2
14

Example 8.5
A power driven winch can exert a torque of 1200
Nm. It has a drum of 80 cm diameter and a mass
moment of inertia of 30 kgm2. It is used to raise
block A of mass 200 kg up the slope as shown in
the figure. The coefficient of friction between
block A and the slope is 0.3. Assume the pulley
is frictionless.
(a) Draw the free body diagrams of the winch and
block A.
(b) Determine the acceleration of block A and the
tension in the cable.
Winch

Pulley

Block A 200 kg
60
15

 = 1200 Nm

T
T

200g

Winch

a
N1

60

F1

Block A

Block A:
For the direction perpendicular to the slope,
Fy = 0
N1 – 200gcos60 = 0
N1 = 981
F1 =  N1
= 0.3 x 981
= 294.3 N
16

T
 = 1200 Nm

T

200g

Winch

a
N1

60

F1

Block A

For the direction up the slope,
Fx = ma
T – 294.3 – 200gsin60 = 200a
T = 1993.44 + 200a
---------- (1)
For the drum, a = r
 = (a / 0.4)
17

T

 = 1200 Nm

T

200g

Winch

For the anti-clockwise direction,
N1
 = I
1200 – 0.4T = 30 x (a / 0.4)
1200 – 0.4T – 75a = 0
--------- (2)

a

60

F1

Block A

Substitute (1) into (2)
1200 – 0.4 (1993.44 + 200a) – 75a = 0
a = 2.6 m/s2
Sub. into (1)
T = 1993.44 + 200 x 2.6
= 2513.44 N

End of Chapter 8
18


Slide 7

School of Engineering
ENGINEERING MECHANICS
CHAPTER 8

Kinetics of Rotational Motion

1

8.1 Brief
Kinetics of rotational motion is the study of rotational
motion caused by external torque.

8.2 Newton’s second law for Rotational Motion
It can be written in mathematical form as:
 = I
 = I

for a single torque
for several torque
where  is the torque (Nm – rotating shaft)
I is the moment of inertia (kgm2)
 is the acceleration (rad/s2)

(Note:- I = mk2, where m is the mass and k is the radius
of gyration.)
2

8.3 Free Body diagram
We note that since the motion is caused by torque
acting on the body, the first step would be to show
all the torque acting on it. This is set out in the form
of a free body diagram indicating clearly all
external torque. A free body diagram is a diagram
showing only the body and the torque acting
directly on it.
8.4 Sign Convention
We are interested to determine the motion of the
body hence we choose the direction of the
resulting rotation as the positive direction.
However, any direction can be selected as
positive but this must be strictly adhered to
throughout the analysis.
3

Example 8.1
A rotor with a mass moment of inertia of 6 kgm2 about
its center of mass has a torque of 90 Nm applied to it.
Determine the angular acceleration of the rotor.

Solution:
By Newton’s law,
 = I

90 Nm

90 = 6
 = 15 rad/s2
4

Example 8.2
The flywheel has a mass of 5 kg and a radius of
gyration of 200 mm. The 20 kg mass, attached to a
rope wrapped around the spindle of diameter 30 mm,
is allowed to fall from rest. Calculate the angular
acceleration of the wheel and the acceleration of the
20 kg mass.
Flywheel
5kg

Spindle

20kg
5

Solution:
For the flywheel,

I = mk2
= 5 x 0.22
= 0.2 kgm2
Torque on flywheel,
 = T x 0.015 Nm
 = I
T x 0.015 = 0.2
T = (0.2 / 0.015) 
---------- (1)

dia =30 mm

T

T

For the mass,
F = ma
20g – T = 20a
a = r
 = a / 0.015

----------- (2)
20g

------------- (3)
6

From (1) & (3),
T = (0.2 / 0.015) x (a / 0.015)
= 0.2a / 0.0152
Sub into (2),
20g – 0.2a / 0.0152 = 20a
a = 0.216 m/s2
From (3),
 = 0.216 / 0.015
= 14.4 rad/s2
7

Example 8.3
A power-driven winch is used to raise a mass of 300
kg with an acceleration of 2 m/s2. The winch drum is
0.5 m in diameter and has a mass moment of inertia
about its center of 8 kgm2. What torque must be
applied to the winch drum?
Winch

300kg

8

T

T

a

For the mass,
F = ma
T - 300g = 300 x 2
T = 3543 N

300g

For the drum,
a = r
2 = 0.25
= 8 rad/s2
 = I
a - 3543 x 0.25 = 8 x 8
a = 950 Nm

9

Example 8.4
A compound pulley having a mass of 4 kg and a
radius of gyration 0.28 m, is connected to two
masses A and B as shown. Assuming no axle
friction, determine the acceleration of the masses.
rB = 0.3 m

rA = 0.1 m

A
10kg

B
20kg

10

TB

TA
TA

10g

TB

20g

I = mk2
= 4 x 0.282
= 0.3136 kgm2
Assume acceleration clockwise:
 = I
TB x 0.3 – TA x 0.1 = 0.3136 --------------- (1)

11

TB

TA
TA

TB

10g

20g

For block A,

F = ma
TA – 10g = 10 aA ---------- (2)

For block B,

F = ma
20g – TB = 20 aB -------- (3)

For pulley,
 = aA / rA = aB / rB

----------- (4)
12

TB

TA
TA

10g

TB

20g

Sub (4) into (2),
TA – 10g = 10 x 0.1
TA =  + 10g

--------------- (5)

Sub (4) into (3),
20g – TB = 20 x 0.3
TB = - 6 + 20g

--------------- (6)
13

TB

TA
TA

10g

TB

20g

Sub (5) & (6) into (1),
(- 6 + 20g) x 0.3 – ( + 10g) x 0.1 = 0.3136
 = 22.154 rad/s2
aA = rA = 0.1 x 22.154 = 2.215 m/s2
aB = rB = 0.3 x 22.154 = 6.646 m/s2
14

Example 8.5
A power driven winch can exert a torque of 1200
Nm. It has a drum of 80 cm diameter and a mass
moment of inertia of 30 kgm2. It is used to raise
block A of mass 200 kg up the slope as shown in
the figure. The coefficient of friction between
block A and the slope is 0.3. Assume the pulley
is frictionless.
(a) Draw the free body diagrams of the winch and
block A.
(b) Determine the acceleration of block A and the
tension in the cable.
Winch

Pulley

Block A 200 kg
60
15

 = 1200 Nm

T
T

200g

Winch

a
N1

60

F1

Block A

Block A:
For the direction perpendicular to the slope,
Fy = 0
N1 – 200gcos60 = 0
N1 = 981
F1 =  N1
= 0.3 x 981
= 294.3 N
16

T
 = 1200 Nm

T

200g

Winch

a
N1

60

F1

Block A

For the direction up the slope,
Fx = ma
T – 294.3 – 200gsin60 = 200a
T = 1993.44 + 200a
---------- (1)
For the drum, a = r
 = (a / 0.4)
17

T

 = 1200 Nm

T

200g

Winch

For the anti-clockwise direction,
N1
 = I
1200 – 0.4T = 30 x (a / 0.4)
1200 – 0.4T – 75a = 0
--------- (2)

a

60

F1

Block A

Substitute (1) into (2)
1200 – 0.4 (1993.44 + 200a) – 75a = 0
a = 2.6 m/s2
Sub. into (1)
T = 1993.44 + 200 x 2.6
= 2513.44 N

End of Chapter 8
18


Slide 8

School of Engineering
ENGINEERING MECHANICS
CHAPTER 8

Kinetics of Rotational Motion

1

8.1 Brief
Kinetics of rotational motion is the study of rotational
motion caused by external torque.

8.2 Newton’s second law for Rotational Motion
It can be written in mathematical form as:
 = I
 = I

for a single torque
for several torque
where  is the torque (Nm – rotating shaft)
I is the moment of inertia (kgm2)
 is the acceleration (rad/s2)

(Note:- I = mk2, where m is the mass and k is the radius
of gyration.)
2

8.3 Free Body diagram
We note that since the motion is caused by torque
acting on the body, the first step would be to show
all the torque acting on it. This is set out in the form
of a free body diagram indicating clearly all
external torque. A free body diagram is a diagram
showing only the body and the torque acting
directly on it.
8.4 Sign Convention
We are interested to determine the motion of the
body hence we choose the direction of the
resulting rotation as the positive direction.
However, any direction can be selected as
positive but this must be strictly adhered to
throughout the analysis.
3

Example 8.1
A rotor with a mass moment of inertia of 6 kgm2 about
its center of mass has a torque of 90 Nm applied to it.
Determine the angular acceleration of the rotor.

Solution:
By Newton’s law,
 = I

90 Nm

90 = 6
 = 15 rad/s2
4

Example 8.2
The flywheel has a mass of 5 kg and a radius of
gyration of 200 mm. The 20 kg mass, attached to a
rope wrapped around the spindle of diameter 30 mm,
is allowed to fall from rest. Calculate the angular
acceleration of the wheel and the acceleration of the
20 kg mass.
Flywheel
5kg

Spindle

20kg
5

Solution:
For the flywheel,

I = mk2
= 5 x 0.22
= 0.2 kgm2
Torque on flywheel,
 = T x 0.015 Nm
 = I
T x 0.015 = 0.2
T = (0.2 / 0.015) 
---------- (1)

dia =30 mm

T

T

For the mass,
F = ma
20g – T = 20a
a = r
 = a / 0.015

----------- (2)
20g

------------- (3)
6

From (1) & (3),
T = (0.2 / 0.015) x (a / 0.015)
= 0.2a / 0.0152
Sub into (2),
20g – 0.2a / 0.0152 = 20a
a = 0.216 m/s2
From (3),
 = 0.216 / 0.015
= 14.4 rad/s2
7

Example 8.3
A power-driven winch is used to raise a mass of 300
kg with an acceleration of 2 m/s2. The winch drum is
0.5 m in diameter and has a mass moment of inertia
about its center of 8 kgm2. What torque must be
applied to the winch drum?
Winch

300kg

8

T

T

a

For the mass,
F = ma
T - 300g = 300 x 2
T = 3543 N

300g

For the drum,
a = r
2 = 0.25
= 8 rad/s2
 = I
a - 3543 x 0.25 = 8 x 8
a = 950 Nm

9

Example 8.4
A compound pulley having a mass of 4 kg and a
radius of gyration 0.28 m, is connected to two
masses A and B as shown. Assuming no axle
friction, determine the acceleration of the masses.
rB = 0.3 m

rA = 0.1 m

A
10kg

B
20kg

10

TB

TA
TA

10g

TB

20g

I = mk2
= 4 x 0.282
= 0.3136 kgm2
Assume acceleration clockwise:
 = I
TB x 0.3 – TA x 0.1 = 0.3136 --------------- (1)

11

TB

TA
TA

TB

10g

20g

For block A,

F = ma
TA – 10g = 10 aA ---------- (2)

For block B,

F = ma
20g – TB = 20 aB -------- (3)

For pulley,
 = aA / rA = aB / rB

----------- (4)
12

TB

TA
TA

10g

TB

20g

Sub (4) into (2),
TA – 10g = 10 x 0.1
TA =  + 10g

--------------- (5)

Sub (4) into (3),
20g – TB = 20 x 0.3
TB = - 6 + 20g

--------------- (6)
13

TB

TA
TA

10g

TB

20g

Sub (5) & (6) into (1),
(- 6 + 20g) x 0.3 – ( + 10g) x 0.1 = 0.3136
 = 22.154 rad/s2
aA = rA = 0.1 x 22.154 = 2.215 m/s2
aB = rB = 0.3 x 22.154 = 6.646 m/s2
14

Example 8.5
A power driven winch can exert a torque of 1200
Nm. It has a drum of 80 cm diameter and a mass
moment of inertia of 30 kgm2. It is used to raise
block A of mass 200 kg up the slope as shown in
the figure. The coefficient of friction between
block A and the slope is 0.3. Assume the pulley
is frictionless.
(a) Draw the free body diagrams of the winch and
block A.
(b) Determine the acceleration of block A and the
tension in the cable.
Winch

Pulley

Block A 200 kg
60
15

 = 1200 Nm

T
T

200g

Winch

a
N1

60

F1

Block A

Block A:
For the direction perpendicular to the slope,
Fy = 0
N1 – 200gcos60 = 0
N1 = 981
F1 =  N1
= 0.3 x 981
= 294.3 N
16

T
 = 1200 Nm

T

200g

Winch

a
N1

60

F1

Block A

For the direction up the slope,
Fx = ma
T – 294.3 – 200gsin60 = 200a
T = 1993.44 + 200a
---------- (1)
For the drum, a = r
 = (a / 0.4)
17

T

 = 1200 Nm

T

200g

Winch

For the anti-clockwise direction,
N1
 = I
1200 – 0.4T = 30 x (a / 0.4)
1200 – 0.4T – 75a = 0
--------- (2)

a

60

F1

Block A

Substitute (1) into (2)
1200 – 0.4 (1993.44 + 200a) – 75a = 0
a = 2.6 m/s2
Sub. into (1)
T = 1993.44 + 200 x 2.6
= 2513.44 N

End of Chapter 8
18


Slide 9

School of Engineering
ENGINEERING MECHANICS
CHAPTER 8

Kinetics of Rotational Motion

1

8.1 Brief
Kinetics of rotational motion is the study of rotational
motion caused by external torque.

8.2 Newton’s second law for Rotational Motion
It can be written in mathematical form as:
 = I
 = I

for a single torque
for several torque
where  is the torque (Nm – rotating shaft)
I is the moment of inertia (kgm2)
 is the acceleration (rad/s2)

(Note:- I = mk2, where m is the mass and k is the radius
of gyration.)
2

8.3 Free Body diagram
We note that since the motion is caused by torque
acting on the body, the first step would be to show
all the torque acting on it. This is set out in the form
of a free body diagram indicating clearly all
external torque. A free body diagram is a diagram
showing only the body and the torque acting
directly on it.
8.4 Sign Convention
We are interested to determine the motion of the
body hence we choose the direction of the
resulting rotation as the positive direction.
However, any direction can be selected as
positive but this must be strictly adhered to
throughout the analysis.
3

Example 8.1
A rotor with a mass moment of inertia of 6 kgm2 about
its center of mass has a torque of 90 Nm applied to it.
Determine the angular acceleration of the rotor.

Solution:
By Newton’s law,
 = I

90 Nm

90 = 6
 = 15 rad/s2
4

Example 8.2
The flywheel has a mass of 5 kg and a radius of
gyration of 200 mm. The 20 kg mass, attached to a
rope wrapped around the spindle of diameter 30 mm,
is allowed to fall from rest. Calculate the angular
acceleration of the wheel and the acceleration of the
20 kg mass.
Flywheel
5kg

Spindle

20kg
5

Solution:
For the flywheel,

I = mk2
= 5 x 0.22
= 0.2 kgm2
Torque on flywheel,
 = T x 0.015 Nm
 = I
T x 0.015 = 0.2
T = (0.2 / 0.015) 
---------- (1)

dia =30 mm

T

T

For the mass,
F = ma
20g – T = 20a
a = r
 = a / 0.015

----------- (2)
20g

------------- (3)
6

From (1) & (3),
T = (0.2 / 0.015) x (a / 0.015)
= 0.2a / 0.0152
Sub into (2),
20g – 0.2a / 0.0152 = 20a
a = 0.216 m/s2
From (3),
 = 0.216 / 0.015
= 14.4 rad/s2
7

Example 8.3
A power-driven winch is used to raise a mass of 300
kg with an acceleration of 2 m/s2. The winch drum is
0.5 m in diameter and has a mass moment of inertia
about its center of 8 kgm2. What torque must be
applied to the winch drum?
Winch

300kg

8

T

T

a

For the mass,
F = ma
T - 300g = 300 x 2
T = 3543 N

300g

For the drum,
a = r
2 = 0.25
= 8 rad/s2
 = I
a - 3543 x 0.25 = 8 x 8
a = 950 Nm

9

Example 8.4
A compound pulley having a mass of 4 kg and a
radius of gyration 0.28 m, is connected to two
masses A and B as shown. Assuming no axle
friction, determine the acceleration of the masses.
rB = 0.3 m

rA = 0.1 m

A
10kg

B
20kg

10

TB

TA
TA

10g

TB

20g

I = mk2
= 4 x 0.282
= 0.3136 kgm2
Assume acceleration clockwise:
 = I
TB x 0.3 – TA x 0.1 = 0.3136 --------------- (1)

11

TB

TA
TA

TB

10g

20g

For block A,

F = ma
TA – 10g = 10 aA ---------- (2)

For block B,

F = ma
20g – TB = 20 aB -------- (3)

For pulley,
 = aA / rA = aB / rB

----------- (4)
12

TB

TA
TA

10g

TB

20g

Sub (4) into (2),
TA – 10g = 10 x 0.1
TA =  + 10g

--------------- (5)

Sub (4) into (3),
20g – TB = 20 x 0.3
TB = - 6 + 20g

--------------- (6)
13

TB

TA
TA

10g

TB

20g

Sub (5) & (6) into (1),
(- 6 + 20g) x 0.3 – ( + 10g) x 0.1 = 0.3136
 = 22.154 rad/s2
aA = rA = 0.1 x 22.154 = 2.215 m/s2
aB = rB = 0.3 x 22.154 = 6.646 m/s2
14

Example 8.5
A power driven winch can exert a torque of 1200
Nm. It has a drum of 80 cm diameter and a mass
moment of inertia of 30 kgm2. It is used to raise
block A of mass 200 kg up the slope as shown in
the figure. The coefficient of friction between
block A and the slope is 0.3. Assume the pulley
is frictionless.
(a) Draw the free body diagrams of the winch and
block A.
(b) Determine the acceleration of block A and the
tension in the cable.
Winch

Pulley

Block A 200 kg
60
15

 = 1200 Nm

T
T

200g

Winch

a
N1

60

F1

Block A

Block A:
For the direction perpendicular to the slope,
Fy = 0
N1 – 200gcos60 = 0
N1 = 981
F1 =  N1
= 0.3 x 981
= 294.3 N
16

T
 = 1200 Nm

T

200g

Winch

a
N1

60

F1

Block A

For the direction up the slope,
Fx = ma
T – 294.3 – 200gsin60 = 200a
T = 1993.44 + 200a
---------- (1)
For the drum, a = r
 = (a / 0.4)
17

T

 = 1200 Nm

T

200g

Winch

For the anti-clockwise direction,
N1
 = I
1200 – 0.4T = 30 x (a / 0.4)
1200 – 0.4T – 75a = 0
--------- (2)

a

60

F1

Block A

Substitute (1) into (2)
1200 – 0.4 (1993.44 + 200a) – 75a = 0
a = 2.6 m/s2
Sub. into (1)
T = 1993.44 + 200 x 2.6
= 2513.44 N

End of Chapter 8
18


Slide 10

School of Engineering
ENGINEERING MECHANICS
CHAPTER 8

Kinetics of Rotational Motion

1

8.1 Brief
Kinetics of rotational motion is the study of rotational
motion caused by external torque.

8.2 Newton’s second law for Rotational Motion
It can be written in mathematical form as:
 = I
 = I

for a single torque
for several torque
where  is the torque (Nm – rotating shaft)
I is the moment of inertia (kgm2)
 is the acceleration (rad/s2)

(Note:- I = mk2, where m is the mass and k is the radius
of gyration.)
2

8.3 Free Body diagram
We note that since the motion is caused by torque
acting on the body, the first step would be to show
all the torque acting on it. This is set out in the form
of a free body diagram indicating clearly all
external torque. A free body diagram is a diagram
showing only the body and the torque acting
directly on it.
8.4 Sign Convention
We are interested to determine the motion of the
body hence we choose the direction of the
resulting rotation as the positive direction.
However, any direction can be selected as
positive but this must be strictly adhered to
throughout the analysis.
3

Example 8.1
A rotor with a mass moment of inertia of 6 kgm2 about
its center of mass has a torque of 90 Nm applied to it.
Determine the angular acceleration of the rotor.

Solution:
By Newton’s law,
 = I

90 Nm

90 = 6
 = 15 rad/s2
4

Example 8.2
The flywheel has a mass of 5 kg and a radius of
gyration of 200 mm. The 20 kg mass, attached to a
rope wrapped around the spindle of diameter 30 mm,
is allowed to fall from rest. Calculate the angular
acceleration of the wheel and the acceleration of the
20 kg mass.
Flywheel
5kg

Spindle

20kg
5

Solution:
For the flywheel,

I = mk2
= 5 x 0.22
= 0.2 kgm2
Torque on flywheel,
 = T x 0.015 Nm
 = I
T x 0.015 = 0.2
T = (0.2 / 0.015) 
---------- (1)

dia =30 mm

T

T

For the mass,
F = ma
20g – T = 20a
a = r
 = a / 0.015

----------- (2)
20g

------------- (3)
6

From (1) & (3),
T = (0.2 / 0.015) x (a / 0.015)
= 0.2a / 0.0152
Sub into (2),
20g – 0.2a / 0.0152 = 20a
a = 0.216 m/s2
From (3),
 = 0.216 / 0.015
= 14.4 rad/s2
7

Example 8.3
A power-driven winch is used to raise a mass of 300
kg with an acceleration of 2 m/s2. The winch drum is
0.5 m in diameter and has a mass moment of inertia
about its center of 8 kgm2. What torque must be
applied to the winch drum?
Winch

300kg

8

T

T

a

For the mass,
F = ma
T - 300g = 300 x 2
T = 3543 N

300g

For the drum,
a = r
2 = 0.25
= 8 rad/s2
 = I
a - 3543 x 0.25 = 8 x 8
a = 950 Nm

9

Example 8.4
A compound pulley having a mass of 4 kg and a
radius of gyration 0.28 m, is connected to two
masses A and B as shown. Assuming no axle
friction, determine the acceleration of the masses.
rB = 0.3 m

rA = 0.1 m

A
10kg

B
20kg

10

TB

TA
TA

10g

TB

20g

I = mk2
= 4 x 0.282
= 0.3136 kgm2
Assume acceleration clockwise:
 = I
TB x 0.3 – TA x 0.1 = 0.3136 --------------- (1)

11

TB

TA
TA

TB

10g

20g

For block A,

F = ma
TA – 10g = 10 aA ---------- (2)

For block B,

F = ma
20g – TB = 20 aB -------- (3)

For pulley,
 = aA / rA = aB / rB

----------- (4)
12

TB

TA
TA

10g

TB

20g

Sub (4) into (2),
TA – 10g = 10 x 0.1
TA =  + 10g

--------------- (5)

Sub (4) into (3),
20g – TB = 20 x 0.3
TB = - 6 + 20g

--------------- (6)
13

TB

TA
TA

10g

TB

20g

Sub (5) & (6) into (1),
(- 6 + 20g) x 0.3 – ( + 10g) x 0.1 = 0.3136
 = 22.154 rad/s2
aA = rA = 0.1 x 22.154 = 2.215 m/s2
aB = rB = 0.3 x 22.154 = 6.646 m/s2
14

Example 8.5
A power driven winch can exert a torque of 1200
Nm. It has a drum of 80 cm diameter and a mass
moment of inertia of 30 kgm2. It is used to raise
block A of mass 200 kg up the slope as shown in
the figure. The coefficient of friction between
block A and the slope is 0.3. Assume the pulley
is frictionless.
(a) Draw the free body diagrams of the winch and
block A.
(b) Determine the acceleration of block A and the
tension in the cable.
Winch

Pulley

Block A 200 kg
60
15

 = 1200 Nm

T
T

200g

Winch

a
N1

60

F1

Block A

Block A:
For the direction perpendicular to the slope,
Fy = 0
N1 – 200gcos60 = 0
N1 = 981
F1 =  N1
= 0.3 x 981
= 294.3 N
16

T
 = 1200 Nm

T

200g

Winch

a
N1

60

F1

Block A

For the direction up the slope,
Fx = ma
T – 294.3 – 200gsin60 = 200a
T = 1993.44 + 200a
---------- (1)
For the drum, a = r
 = (a / 0.4)
17

T

 = 1200 Nm

T

200g

Winch

For the anti-clockwise direction,
N1
 = I
1200 – 0.4T = 30 x (a / 0.4)
1200 – 0.4T – 75a = 0
--------- (2)

a

60

F1

Block A

Substitute (1) into (2)
1200 – 0.4 (1993.44 + 200a) – 75a = 0
a = 2.6 m/s2
Sub. into (1)
T = 1993.44 + 200 x 2.6
= 2513.44 N

End of Chapter 8
18


Slide 11

School of Engineering
ENGINEERING MECHANICS
CHAPTER 8

Kinetics of Rotational Motion

1

8.1 Brief
Kinetics of rotational motion is the study of rotational
motion caused by external torque.

8.2 Newton’s second law for Rotational Motion
It can be written in mathematical form as:
 = I
 = I

for a single torque
for several torque
where  is the torque (Nm – rotating shaft)
I is the moment of inertia (kgm2)
 is the acceleration (rad/s2)

(Note:- I = mk2, where m is the mass and k is the radius
of gyration.)
2

8.3 Free Body diagram
We note that since the motion is caused by torque
acting on the body, the first step would be to show
all the torque acting on it. This is set out in the form
of a free body diagram indicating clearly all
external torque. A free body diagram is a diagram
showing only the body and the torque acting
directly on it.
8.4 Sign Convention
We are interested to determine the motion of the
body hence we choose the direction of the
resulting rotation as the positive direction.
However, any direction can be selected as
positive but this must be strictly adhered to
throughout the analysis.
3

Example 8.1
A rotor with a mass moment of inertia of 6 kgm2 about
its center of mass has a torque of 90 Nm applied to it.
Determine the angular acceleration of the rotor.

Solution:
By Newton’s law,
 = I

90 Nm

90 = 6
 = 15 rad/s2
4

Example 8.2
The flywheel has a mass of 5 kg and a radius of
gyration of 200 mm. The 20 kg mass, attached to a
rope wrapped around the spindle of diameter 30 mm,
is allowed to fall from rest. Calculate the angular
acceleration of the wheel and the acceleration of the
20 kg mass.
Flywheel
5kg

Spindle

20kg
5

Solution:
For the flywheel,

I = mk2
= 5 x 0.22
= 0.2 kgm2
Torque on flywheel,
 = T x 0.015 Nm
 = I
T x 0.015 = 0.2
T = (0.2 / 0.015) 
---------- (1)

dia =30 mm

T

T

For the mass,
F = ma
20g – T = 20a
a = r
 = a / 0.015

----------- (2)
20g

------------- (3)
6

From (1) & (3),
T = (0.2 / 0.015) x (a / 0.015)
= 0.2a / 0.0152
Sub into (2),
20g – 0.2a / 0.0152 = 20a
a = 0.216 m/s2
From (3),
 = 0.216 / 0.015
= 14.4 rad/s2
7

Example 8.3
A power-driven winch is used to raise a mass of 300
kg with an acceleration of 2 m/s2. The winch drum is
0.5 m in diameter and has a mass moment of inertia
about its center of 8 kgm2. What torque must be
applied to the winch drum?
Winch

300kg

8

T

T

a

For the mass,
F = ma
T - 300g = 300 x 2
T = 3543 N

300g

For the drum,
a = r
2 = 0.25
= 8 rad/s2
 = I
a - 3543 x 0.25 = 8 x 8
a = 950 Nm

9

Example 8.4
A compound pulley having a mass of 4 kg and a
radius of gyration 0.28 m, is connected to two
masses A and B as shown. Assuming no axle
friction, determine the acceleration of the masses.
rB = 0.3 m

rA = 0.1 m

A
10kg

B
20kg

10

TB

TA
TA

10g

TB

20g

I = mk2
= 4 x 0.282
= 0.3136 kgm2
Assume acceleration clockwise:
 = I
TB x 0.3 – TA x 0.1 = 0.3136 --------------- (1)

11

TB

TA
TA

TB

10g

20g

For block A,

F = ma
TA – 10g = 10 aA ---------- (2)

For block B,

F = ma
20g – TB = 20 aB -------- (3)

For pulley,
 = aA / rA = aB / rB

----------- (4)
12

TB

TA
TA

10g

TB

20g

Sub (4) into (2),
TA – 10g = 10 x 0.1
TA =  + 10g

--------------- (5)

Sub (4) into (3),
20g – TB = 20 x 0.3
TB = - 6 + 20g

--------------- (6)
13

TB

TA
TA

10g

TB

20g

Sub (5) & (6) into (1),
(- 6 + 20g) x 0.3 – ( + 10g) x 0.1 = 0.3136
 = 22.154 rad/s2
aA = rA = 0.1 x 22.154 = 2.215 m/s2
aB = rB = 0.3 x 22.154 = 6.646 m/s2
14

Example 8.5
A power driven winch can exert a torque of 1200
Nm. It has a drum of 80 cm diameter and a mass
moment of inertia of 30 kgm2. It is used to raise
block A of mass 200 kg up the slope as shown in
the figure. The coefficient of friction between
block A and the slope is 0.3. Assume the pulley
is frictionless.
(a) Draw the free body diagrams of the winch and
block A.
(b) Determine the acceleration of block A and the
tension in the cable.
Winch

Pulley

Block A 200 kg
60
15

 = 1200 Nm

T
T

200g

Winch

a
N1

60

F1

Block A

Block A:
For the direction perpendicular to the slope,
Fy = 0
N1 – 200gcos60 = 0
N1 = 981
F1 =  N1
= 0.3 x 981
= 294.3 N
16

T
 = 1200 Nm

T

200g

Winch

a
N1

60

F1

Block A

For the direction up the slope,
Fx = ma
T – 294.3 – 200gsin60 = 200a
T = 1993.44 + 200a
---------- (1)
For the drum, a = r
 = (a / 0.4)
17

T

 = 1200 Nm

T

200g

Winch

For the anti-clockwise direction,
N1
 = I
1200 – 0.4T = 30 x (a / 0.4)
1200 – 0.4T – 75a = 0
--------- (2)

a

60

F1

Block A

Substitute (1) into (2)
1200 – 0.4 (1993.44 + 200a) – 75a = 0
a = 2.6 m/s2
Sub. into (1)
T = 1993.44 + 200 x 2.6
= 2513.44 N

End of Chapter 8
18


Slide 12

School of Engineering
ENGINEERING MECHANICS
CHAPTER 8

Kinetics of Rotational Motion

1

8.1 Brief
Kinetics of rotational motion is the study of rotational
motion caused by external torque.

8.2 Newton’s second law for Rotational Motion
It can be written in mathematical form as:
 = I
 = I

for a single torque
for several torque
where  is the torque (Nm – rotating shaft)
I is the moment of inertia (kgm2)
 is the acceleration (rad/s2)

(Note:- I = mk2, where m is the mass and k is the radius
of gyration.)
2

8.3 Free Body diagram
We note that since the motion is caused by torque
acting on the body, the first step would be to show
all the torque acting on it. This is set out in the form
of a free body diagram indicating clearly all
external torque. A free body diagram is a diagram
showing only the body and the torque acting
directly on it.
8.4 Sign Convention
We are interested to determine the motion of the
body hence we choose the direction of the
resulting rotation as the positive direction.
However, any direction can be selected as
positive but this must be strictly adhered to
throughout the analysis.
3

Example 8.1
A rotor with a mass moment of inertia of 6 kgm2 about
its center of mass has a torque of 90 Nm applied to it.
Determine the angular acceleration of the rotor.

Solution:
By Newton’s law,
 = I

90 Nm

90 = 6
 = 15 rad/s2
4

Example 8.2
The flywheel has a mass of 5 kg and a radius of
gyration of 200 mm. The 20 kg mass, attached to a
rope wrapped around the spindle of diameter 30 mm,
is allowed to fall from rest. Calculate the angular
acceleration of the wheel and the acceleration of the
20 kg mass.
Flywheel
5kg

Spindle

20kg
5

Solution:
For the flywheel,

I = mk2
= 5 x 0.22
= 0.2 kgm2
Torque on flywheel,
 = T x 0.015 Nm
 = I
T x 0.015 = 0.2
T = (0.2 / 0.015) 
---------- (1)

dia =30 mm

T

T

For the mass,
F = ma
20g – T = 20a
a = r
 = a / 0.015

----------- (2)
20g

------------- (3)
6

From (1) & (3),
T = (0.2 / 0.015) x (a / 0.015)
= 0.2a / 0.0152
Sub into (2),
20g – 0.2a / 0.0152 = 20a
a = 0.216 m/s2
From (3),
 = 0.216 / 0.015
= 14.4 rad/s2
7

Example 8.3
A power-driven winch is used to raise a mass of 300
kg with an acceleration of 2 m/s2. The winch drum is
0.5 m in diameter and has a mass moment of inertia
about its center of 8 kgm2. What torque must be
applied to the winch drum?
Winch

300kg

8

T

T

a

For the mass,
F = ma
T - 300g = 300 x 2
T = 3543 N

300g

For the drum,
a = r
2 = 0.25
= 8 rad/s2
 = I
a - 3543 x 0.25 = 8 x 8
a = 950 Nm

9

Example 8.4
A compound pulley having a mass of 4 kg and a
radius of gyration 0.28 m, is connected to two
masses A and B as shown. Assuming no axle
friction, determine the acceleration of the masses.
rB = 0.3 m

rA = 0.1 m

A
10kg

B
20kg

10

TB

TA
TA

10g

TB

20g

I = mk2
= 4 x 0.282
= 0.3136 kgm2
Assume acceleration clockwise:
 = I
TB x 0.3 – TA x 0.1 = 0.3136 --------------- (1)

11

TB

TA
TA

TB

10g

20g

For block A,

F = ma
TA – 10g = 10 aA ---------- (2)

For block B,

F = ma
20g – TB = 20 aB -------- (3)

For pulley,
 = aA / rA = aB / rB

----------- (4)
12

TB

TA
TA

10g

TB

20g

Sub (4) into (2),
TA – 10g = 10 x 0.1
TA =  + 10g

--------------- (5)

Sub (4) into (3),
20g – TB = 20 x 0.3
TB = - 6 + 20g

--------------- (6)
13

TB

TA
TA

10g

TB

20g

Sub (5) & (6) into (1),
(- 6 + 20g) x 0.3 – ( + 10g) x 0.1 = 0.3136
 = 22.154 rad/s2
aA = rA = 0.1 x 22.154 = 2.215 m/s2
aB = rB = 0.3 x 22.154 = 6.646 m/s2
14

Example 8.5
A power driven winch can exert a torque of 1200
Nm. It has a drum of 80 cm diameter and a mass
moment of inertia of 30 kgm2. It is used to raise
block A of mass 200 kg up the slope as shown in
the figure. The coefficient of friction between
block A and the slope is 0.3. Assume the pulley
is frictionless.
(a) Draw the free body diagrams of the winch and
block A.
(b) Determine the acceleration of block A and the
tension in the cable.
Winch

Pulley

Block A 200 kg
60
15

 = 1200 Nm

T
T

200g

Winch

a
N1

60

F1

Block A

Block A:
For the direction perpendicular to the slope,
Fy = 0
N1 – 200gcos60 = 0
N1 = 981
F1 =  N1
= 0.3 x 981
= 294.3 N
16

T
 = 1200 Nm

T

200g

Winch

a
N1

60

F1

Block A

For the direction up the slope,
Fx = ma
T – 294.3 – 200gsin60 = 200a
T = 1993.44 + 200a
---------- (1)
For the drum, a = r
 = (a / 0.4)
17

T

 = 1200 Nm

T

200g

Winch

For the anti-clockwise direction,
N1
 = I
1200 – 0.4T = 30 x (a / 0.4)
1200 – 0.4T – 75a = 0
--------- (2)

a

60

F1

Block A

Substitute (1) into (2)
1200 – 0.4 (1993.44 + 200a) – 75a = 0
a = 2.6 m/s2
Sub. into (1)
T = 1993.44 + 200 x 2.6
= 2513.44 N

End of Chapter 8
18


Slide 13

School of Engineering
ENGINEERING MECHANICS
CHAPTER 8

Kinetics of Rotational Motion

1

8.1 Brief
Kinetics of rotational motion is the study of rotational
motion caused by external torque.

8.2 Newton’s second law for Rotational Motion
It can be written in mathematical form as:
 = I
 = I

for a single torque
for several torque
where  is the torque (Nm – rotating shaft)
I is the moment of inertia (kgm2)
 is the acceleration (rad/s2)

(Note:- I = mk2, where m is the mass and k is the radius
of gyration.)
2

8.3 Free Body diagram
We note that since the motion is caused by torque
acting on the body, the first step would be to show
all the torque acting on it. This is set out in the form
of a free body diagram indicating clearly all
external torque. A free body diagram is a diagram
showing only the body and the torque acting
directly on it.
8.4 Sign Convention
We are interested to determine the motion of the
body hence we choose the direction of the
resulting rotation as the positive direction.
However, any direction can be selected as
positive but this must be strictly adhered to
throughout the analysis.
3

Example 8.1
A rotor with a mass moment of inertia of 6 kgm2 about
its center of mass has a torque of 90 Nm applied to it.
Determine the angular acceleration of the rotor.

Solution:
By Newton’s law,
 = I

90 Nm

90 = 6
 = 15 rad/s2
4

Example 8.2
The flywheel has a mass of 5 kg and a radius of
gyration of 200 mm. The 20 kg mass, attached to a
rope wrapped around the spindle of diameter 30 mm,
is allowed to fall from rest. Calculate the angular
acceleration of the wheel and the acceleration of the
20 kg mass.
Flywheel
5kg

Spindle

20kg
5

Solution:
For the flywheel,

I = mk2
= 5 x 0.22
= 0.2 kgm2
Torque on flywheel,
 = T x 0.015 Nm
 = I
T x 0.015 = 0.2
T = (0.2 / 0.015) 
---------- (1)

dia =30 mm

T

T

For the mass,
F = ma
20g – T = 20a
a = r
 = a / 0.015

----------- (2)
20g

------------- (3)
6

From (1) & (3),
T = (0.2 / 0.015) x (a / 0.015)
= 0.2a / 0.0152
Sub into (2),
20g – 0.2a / 0.0152 = 20a
a = 0.216 m/s2
From (3),
 = 0.216 / 0.015
= 14.4 rad/s2
7

Example 8.3
A power-driven winch is used to raise a mass of 300
kg with an acceleration of 2 m/s2. The winch drum is
0.5 m in diameter and has a mass moment of inertia
about its center of 8 kgm2. What torque must be
applied to the winch drum?
Winch

300kg

8

T

T

a

For the mass,
F = ma
T - 300g = 300 x 2
T = 3543 N

300g

For the drum,
a = r
2 = 0.25
= 8 rad/s2
 = I
a - 3543 x 0.25 = 8 x 8
a = 950 Nm

9

Example 8.4
A compound pulley having a mass of 4 kg and a
radius of gyration 0.28 m, is connected to two
masses A and B as shown. Assuming no axle
friction, determine the acceleration of the masses.
rB = 0.3 m

rA = 0.1 m

A
10kg

B
20kg

10

TB

TA
TA

10g

TB

20g

I = mk2
= 4 x 0.282
= 0.3136 kgm2
Assume acceleration clockwise:
 = I
TB x 0.3 – TA x 0.1 = 0.3136 --------------- (1)

11

TB

TA
TA

TB

10g

20g

For block A,

F = ma
TA – 10g = 10 aA ---------- (2)

For block B,

F = ma
20g – TB = 20 aB -------- (3)

For pulley,
 = aA / rA = aB / rB

----------- (4)
12

TB

TA
TA

10g

TB

20g

Sub (4) into (2),
TA – 10g = 10 x 0.1
TA =  + 10g

--------------- (5)

Sub (4) into (3),
20g – TB = 20 x 0.3
TB = - 6 + 20g

--------------- (6)
13

TB

TA
TA

10g

TB

20g

Sub (5) & (6) into (1),
(- 6 + 20g) x 0.3 – ( + 10g) x 0.1 = 0.3136
 = 22.154 rad/s2
aA = rA = 0.1 x 22.154 = 2.215 m/s2
aB = rB = 0.3 x 22.154 = 6.646 m/s2
14

Example 8.5
A power driven winch can exert a torque of 1200
Nm. It has a drum of 80 cm diameter and a mass
moment of inertia of 30 kgm2. It is used to raise
block A of mass 200 kg up the slope as shown in
the figure. The coefficient of friction between
block A and the slope is 0.3. Assume the pulley
is frictionless.
(a) Draw the free body diagrams of the winch and
block A.
(b) Determine the acceleration of block A and the
tension in the cable.
Winch

Pulley

Block A 200 kg
60
15

 = 1200 Nm

T
T

200g

Winch

a
N1

60

F1

Block A

Block A:
For the direction perpendicular to the slope,
Fy = 0
N1 – 200gcos60 = 0
N1 = 981
F1 =  N1
= 0.3 x 981
= 294.3 N
16

T
 = 1200 Nm

T

200g

Winch

a
N1

60

F1

Block A

For the direction up the slope,
Fx = ma
T – 294.3 – 200gsin60 = 200a
T = 1993.44 + 200a
---------- (1)
For the drum, a = r
 = (a / 0.4)
17

T

 = 1200 Nm

T

200g

Winch

For the anti-clockwise direction,
N1
 = I
1200 – 0.4T = 30 x (a / 0.4)
1200 – 0.4T – 75a = 0
--------- (2)

a

60

F1

Block A

Substitute (1) into (2)
1200 – 0.4 (1993.44 + 200a) – 75a = 0
a = 2.6 m/s2
Sub. into (1)
T = 1993.44 + 200 x 2.6
= 2513.44 N

End of Chapter 8
18


Slide 14

School of Engineering
ENGINEERING MECHANICS
CHAPTER 8

Kinetics of Rotational Motion

1

8.1 Brief
Kinetics of rotational motion is the study of rotational
motion caused by external torque.

8.2 Newton’s second law for Rotational Motion
It can be written in mathematical form as:
 = I
 = I

for a single torque
for several torque
where  is the torque (Nm – rotating shaft)
I is the moment of inertia (kgm2)
 is the acceleration (rad/s2)

(Note:- I = mk2, where m is the mass and k is the radius
of gyration.)
2

8.3 Free Body diagram
We note that since the motion is caused by torque
acting on the body, the first step would be to show
all the torque acting on it. This is set out in the form
of a free body diagram indicating clearly all
external torque. A free body diagram is a diagram
showing only the body and the torque acting
directly on it.
8.4 Sign Convention
We are interested to determine the motion of the
body hence we choose the direction of the
resulting rotation as the positive direction.
However, any direction can be selected as
positive but this must be strictly adhered to
throughout the analysis.
3

Example 8.1
A rotor with a mass moment of inertia of 6 kgm2 about
its center of mass has a torque of 90 Nm applied to it.
Determine the angular acceleration of the rotor.

Solution:
By Newton’s law,
 = I

90 Nm

90 = 6
 = 15 rad/s2
4

Example 8.2
The flywheel has a mass of 5 kg and a radius of
gyration of 200 mm. The 20 kg mass, attached to a
rope wrapped around the spindle of diameter 30 mm,
is allowed to fall from rest. Calculate the angular
acceleration of the wheel and the acceleration of the
20 kg mass.
Flywheel
5kg

Spindle

20kg
5

Solution:
For the flywheel,

I = mk2
= 5 x 0.22
= 0.2 kgm2
Torque on flywheel,
 = T x 0.015 Nm
 = I
T x 0.015 = 0.2
T = (0.2 / 0.015) 
---------- (1)

dia =30 mm

T

T

For the mass,
F = ma
20g – T = 20a
a = r
 = a / 0.015

----------- (2)
20g

------------- (3)
6

From (1) & (3),
T = (0.2 / 0.015) x (a / 0.015)
= 0.2a / 0.0152
Sub into (2),
20g – 0.2a / 0.0152 = 20a
a = 0.216 m/s2
From (3),
 = 0.216 / 0.015
= 14.4 rad/s2
7

Example 8.3
A power-driven winch is used to raise a mass of 300
kg with an acceleration of 2 m/s2. The winch drum is
0.5 m in diameter and has a mass moment of inertia
about its center of 8 kgm2. What torque must be
applied to the winch drum?
Winch

300kg

8

T

T

a

For the mass,
F = ma
T - 300g = 300 x 2
T = 3543 N

300g

For the drum,
a = r
2 = 0.25
= 8 rad/s2
 = I
a - 3543 x 0.25 = 8 x 8
a = 950 Nm

9

Example 8.4
A compound pulley having a mass of 4 kg and a
radius of gyration 0.28 m, is connected to two
masses A and B as shown. Assuming no axle
friction, determine the acceleration of the masses.
rB = 0.3 m

rA = 0.1 m

A
10kg

B
20kg

10

TB

TA
TA

10g

TB

20g

I = mk2
= 4 x 0.282
= 0.3136 kgm2
Assume acceleration clockwise:
 = I
TB x 0.3 – TA x 0.1 = 0.3136 --------------- (1)

11

TB

TA
TA

TB

10g

20g

For block A,

F = ma
TA – 10g = 10 aA ---------- (2)

For block B,

F = ma
20g – TB = 20 aB -------- (3)

For pulley,
 = aA / rA = aB / rB

----------- (4)
12

TB

TA
TA

10g

TB

20g

Sub (4) into (2),
TA – 10g = 10 x 0.1
TA =  + 10g

--------------- (5)

Sub (4) into (3),
20g – TB = 20 x 0.3
TB = - 6 + 20g

--------------- (6)
13

TB

TA
TA

10g

TB

20g

Sub (5) & (6) into (1),
(- 6 + 20g) x 0.3 – ( + 10g) x 0.1 = 0.3136
 = 22.154 rad/s2
aA = rA = 0.1 x 22.154 = 2.215 m/s2
aB = rB = 0.3 x 22.154 = 6.646 m/s2
14

Example 8.5
A power driven winch can exert a torque of 1200
Nm. It has a drum of 80 cm diameter and a mass
moment of inertia of 30 kgm2. It is used to raise
block A of mass 200 kg up the slope as shown in
the figure. The coefficient of friction between
block A and the slope is 0.3. Assume the pulley
is frictionless.
(a) Draw the free body diagrams of the winch and
block A.
(b) Determine the acceleration of block A and the
tension in the cable.
Winch

Pulley

Block A 200 kg
60
15

 = 1200 Nm

T
T

200g

Winch

a
N1

60

F1

Block A

Block A:
For the direction perpendicular to the slope,
Fy = 0
N1 – 200gcos60 = 0
N1 = 981
F1 =  N1
= 0.3 x 981
= 294.3 N
16

T
 = 1200 Nm

T

200g

Winch

a
N1

60

F1

Block A

For the direction up the slope,
Fx = ma
T – 294.3 – 200gsin60 = 200a
T = 1993.44 + 200a
---------- (1)
For the drum, a = r
 = (a / 0.4)
17

T

 = 1200 Nm

T

200g

Winch

For the anti-clockwise direction,
N1
 = I
1200 – 0.4T = 30 x (a / 0.4)
1200 – 0.4T – 75a = 0
--------- (2)

a

60

F1

Block A

Substitute (1) into (2)
1200 – 0.4 (1993.44 + 200a) – 75a = 0
a = 2.6 m/s2
Sub. into (1)
T = 1993.44 + 200 x 2.6
= 2513.44 N

End of Chapter 8
18


Slide 15

School of Engineering
ENGINEERING MECHANICS
CHAPTER 8

Kinetics of Rotational Motion

1

8.1 Brief
Kinetics of rotational motion is the study of rotational
motion caused by external torque.

8.2 Newton’s second law for Rotational Motion
It can be written in mathematical form as:
 = I
 = I

for a single torque
for several torque
where  is the torque (Nm – rotating shaft)
I is the moment of inertia (kgm2)
 is the acceleration (rad/s2)

(Note:- I = mk2, where m is the mass and k is the radius
of gyration.)
2

8.3 Free Body diagram
We note that since the motion is caused by torque
acting on the body, the first step would be to show
all the torque acting on it. This is set out in the form
of a free body diagram indicating clearly all
external torque. A free body diagram is a diagram
showing only the body and the torque acting
directly on it.
8.4 Sign Convention
We are interested to determine the motion of the
body hence we choose the direction of the
resulting rotation as the positive direction.
However, any direction can be selected as
positive but this must be strictly adhered to
throughout the analysis.
3

Example 8.1
A rotor with a mass moment of inertia of 6 kgm2 about
its center of mass has a torque of 90 Nm applied to it.
Determine the angular acceleration of the rotor.

Solution:
By Newton’s law,
 = I

90 Nm

90 = 6
 = 15 rad/s2
4

Example 8.2
The flywheel has a mass of 5 kg and a radius of
gyration of 200 mm. The 20 kg mass, attached to a
rope wrapped around the spindle of diameter 30 mm,
is allowed to fall from rest. Calculate the angular
acceleration of the wheel and the acceleration of the
20 kg mass.
Flywheel
5kg

Spindle

20kg
5

Solution:
For the flywheel,

I = mk2
= 5 x 0.22
= 0.2 kgm2
Torque on flywheel,
 = T x 0.015 Nm
 = I
T x 0.015 = 0.2
T = (0.2 / 0.015) 
---------- (1)

dia =30 mm

T

T

For the mass,
F = ma
20g – T = 20a
a = r
 = a / 0.015

----------- (2)
20g

------------- (3)
6

From (1) & (3),
T = (0.2 / 0.015) x (a / 0.015)
= 0.2a / 0.0152
Sub into (2),
20g – 0.2a / 0.0152 = 20a
a = 0.216 m/s2
From (3),
 = 0.216 / 0.015
= 14.4 rad/s2
7

Example 8.3
A power-driven winch is used to raise a mass of 300
kg with an acceleration of 2 m/s2. The winch drum is
0.5 m in diameter and has a mass moment of inertia
about its center of 8 kgm2. What torque must be
applied to the winch drum?
Winch

300kg

8

T

T

a

For the mass,
F = ma
T - 300g = 300 x 2
T = 3543 N

300g

For the drum,
a = r
2 = 0.25
= 8 rad/s2
 = I
a - 3543 x 0.25 = 8 x 8
a = 950 Nm

9

Example 8.4
A compound pulley having a mass of 4 kg and a
radius of gyration 0.28 m, is connected to two
masses A and B as shown. Assuming no axle
friction, determine the acceleration of the masses.
rB = 0.3 m

rA = 0.1 m

A
10kg

B
20kg

10

TB

TA
TA

10g

TB

20g

I = mk2
= 4 x 0.282
= 0.3136 kgm2
Assume acceleration clockwise:
 = I
TB x 0.3 – TA x 0.1 = 0.3136 --------------- (1)

11

TB

TA
TA

TB

10g

20g

For block A,

F = ma
TA – 10g = 10 aA ---------- (2)

For block B,

F = ma
20g – TB = 20 aB -------- (3)

For pulley,
 = aA / rA = aB / rB

----------- (4)
12

TB

TA
TA

10g

TB

20g

Sub (4) into (2),
TA – 10g = 10 x 0.1
TA =  + 10g

--------------- (5)

Sub (4) into (3),
20g – TB = 20 x 0.3
TB = - 6 + 20g

--------------- (6)
13

TB

TA
TA

10g

TB

20g

Sub (5) & (6) into (1),
(- 6 + 20g) x 0.3 – ( + 10g) x 0.1 = 0.3136
 = 22.154 rad/s2
aA = rA = 0.1 x 22.154 = 2.215 m/s2
aB = rB = 0.3 x 22.154 = 6.646 m/s2
14

Example 8.5
A power driven winch can exert a torque of 1200
Nm. It has a drum of 80 cm diameter and a mass
moment of inertia of 30 kgm2. It is used to raise
block A of mass 200 kg up the slope as shown in
the figure. The coefficient of friction between
block A and the slope is 0.3. Assume the pulley
is frictionless.
(a) Draw the free body diagrams of the winch and
block A.
(b) Determine the acceleration of block A and the
tension in the cable.
Winch

Pulley

Block A 200 kg
60
15

 = 1200 Nm

T
T

200g

Winch

a
N1

60

F1

Block A

Block A:
For the direction perpendicular to the slope,
Fy = 0
N1 – 200gcos60 = 0
N1 = 981
F1 =  N1
= 0.3 x 981
= 294.3 N
16

T
 = 1200 Nm

T

200g

Winch

a
N1

60

F1

Block A

For the direction up the slope,
Fx = ma
T – 294.3 – 200gsin60 = 200a
T = 1993.44 + 200a
---------- (1)
For the drum, a = r
 = (a / 0.4)
17

T

 = 1200 Nm

T

200g

Winch

For the anti-clockwise direction,
N1
 = I
1200 – 0.4T = 30 x (a / 0.4)
1200 – 0.4T – 75a = 0
--------- (2)

a

60

F1

Block A

Substitute (1) into (2)
1200 – 0.4 (1993.44 + 200a) – 75a = 0
a = 2.6 m/s2
Sub. into (1)
T = 1993.44 + 200 x 2.6
= 2513.44 N

End of Chapter 8
18


Slide 16

School of Engineering
ENGINEERING MECHANICS
CHAPTER 8

Kinetics of Rotational Motion

1

8.1 Brief
Kinetics of rotational motion is the study of rotational
motion caused by external torque.

8.2 Newton’s second law for Rotational Motion
It can be written in mathematical form as:
 = I
 = I

for a single torque
for several torque
where  is the torque (Nm – rotating shaft)
I is the moment of inertia (kgm2)
 is the acceleration (rad/s2)

(Note:- I = mk2, where m is the mass and k is the radius
of gyration.)
2

8.3 Free Body diagram
We note that since the motion is caused by torque
acting on the body, the first step would be to show
all the torque acting on it. This is set out in the form
of a free body diagram indicating clearly all
external torque. A free body diagram is a diagram
showing only the body and the torque acting
directly on it.
8.4 Sign Convention
We are interested to determine the motion of the
body hence we choose the direction of the
resulting rotation as the positive direction.
However, any direction can be selected as
positive but this must be strictly adhered to
throughout the analysis.
3

Example 8.1
A rotor with a mass moment of inertia of 6 kgm2 about
its center of mass has a torque of 90 Nm applied to it.
Determine the angular acceleration of the rotor.

Solution:
By Newton’s law,
 = I

90 Nm

90 = 6
 = 15 rad/s2
4

Example 8.2
The flywheel has a mass of 5 kg and a radius of
gyration of 200 mm. The 20 kg mass, attached to a
rope wrapped around the spindle of diameter 30 mm,
is allowed to fall from rest. Calculate the angular
acceleration of the wheel and the acceleration of the
20 kg mass.
Flywheel
5kg

Spindle

20kg
5

Solution:
For the flywheel,

I = mk2
= 5 x 0.22
= 0.2 kgm2
Torque on flywheel,
 = T x 0.015 Nm
 = I
T x 0.015 = 0.2
T = (0.2 / 0.015) 
---------- (1)

dia =30 mm

T

T

For the mass,
F = ma
20g – T = 20a
a = r
 = a / 0.015

----------- (2)
20g

------------- (3)
6

From (1) & (3),
T = (0.2 / 0.015) x (a / 0.015)
= 0.2a / 0.0152
Sub into (2),
20g – 0.2a / 0.0152 = 20a
a = 0.216 m/s2
From (3),
 = 0.216 / 0.015
= 14.4 rad/s2
7

Example 8.3
A power-driven winch is used to raise a mass of 300
kg with an acceleration of 2 m/s2. The winch drum is
0.5 m in diameter and has a mass moment of inertia
about its center of 8 kgm2. What torque must be
applied to the winch drum?
Winch

300kg

8

T

T

a

For the mass,
F = ma
T - 300g = 300 x 2
T = 3543 N

300g

For the drum,
a = r
2 = 0.25
= 8 rad/s2
 = I
a - 3543 x 0.25 = 8 x 8
a = 950 Nm

9

Example 8.4
A compound pulley having a mass of 4 kg and a
radius of gyration 0.28 m, is connected to two
masses A and B as shown. Assuming no axle
friction, determine the acceleration of the masses.
rB = 0.3 m

rA = 0.1 m

A
10kg

B
20kg

10

TB

TA
TA

10g

TB

20g

I = mk2
= 4 x 0.282
= 0.3136 kgm2
Assume acceleration clockwise:
 = I
TB x 0.3 – TA x 0.1 = 0.3136 --------------- (1)

11

TB

TA
TA

TB

10g

20g

For block A,

F = ma
TA – 10g = 10 aA ---------- (2)

For block B,

F = ma
20g – TB = 20 aB -------- (3)

For pulley,
 = aA / rA = aB / rB

----------- (4)
12

TB

TA
TA

10g

TB

20g

Sub (4) into (2),
TA – 10g = 10 x 0.1
TA =  + 10g

--------------- (5)

Sub (4) into (3),
20g – TB = 20 x 0.3
TB = - 6 + 20g

--------------- (6)
13

TB

TA
TA

10g

TB

20g

Sub (5) & (6) into (1),
(- 6 + 20g) x 0.3 – ( + 10g) x 0.1 = 0.3136
 = 22.154 rad/s2
aA = rA = 0.1 x 22.154 = 2.215 m/s2
aB = rB = 0.3 x 22.154 = 6.646 m/s2
14

Example 8.5
A power driven winch can exert a torque of 1200
Nm. It has a drum of 80 cm diameter and a mass
moment of inertia of 30 kgm2. It is used to raise
block A of mass 200 kg up the slope as shown in
the figure. The coefficient of friction between
block A and the slope is 0.3. Assume the pulley
is frictionless.
(a) Draw the free body diagrams of the winch and
block A.
(b) Determine the acceleration of block A and the
tension in the cable.
Winch

Pulley

Block A 200 kg
60
15

 = 1200 Nm

T
T

200g

Winch

a
N1

60

F1

Block A

Block A:
For the direction perpendicular to the slope,
Fy = 0
N1 – 200gcos60 = 0
N1 = 981
F1 =  N1
= 0.3 x 981
= 294.3 N
16

T
 = 1200 Nm

T

200g

Winch

a
N1

60

F1

Block A

For the direction up the slope,
Fx = ma
T – 294.3 – 200gsin60 = 200a
T = 1993.44 + 200a
---------- (1)
For the drum, a = r
 = (a / 0.4)
17

T

 = 1200 Nm

T

200g

Winch

For the anti-clockwise direction,
N1
 = I
1200 – 0.4T = 30 x (a / 0.4)
1200 – 0.4T – 75a = 0
--------- (2)

a

60

F1

Block A

Substitute (1) into (2)
1200 – 0.4 (1993.44 + 200a) – 75a = 0
a = 2.6 m/s2
Sub. into (1)
T = 1993.44 + 200 x 2.6
= 2513.44 N

End of Chapter 8
18


Slide 17

School of Engineering
ENGINEERING MECHANICS
CHAPTER 8

Kinetics of Rotational Motion

1

8.1 Brief
Kinetics of rotational motion is the study of rotational
motion caused by external torque.

8.2 Newton’s second law for Rotational Motion
It can be written in mathematical form as:
 = I
 = I

for a single torque
for several torque
where  is the torque (Nm – rotating shaft)
I is the moment of inertia (kgm2)
 is the acceleration (rad/s2)

(Note:- I = mk2, where m is the mass and k is the radius
of gyration.)
2

8.3 Free Body diagram
We note that since the motion is caused by torque
acting on the body, the first step would be to show
all the torque acting on it. This is set out in the form
of a free body diagram indicating clearly all
external torque. A free body diagram is a diagram
showing only the body and the torque acting
directly on it.
8.4 Sign Convention
We are interested to determine the motion of the
body hence we choose the direction of the
resulting rotation as the positive direction.
However, any direction can be selected as
positive but this must be strictly adhered to
throughout the analysis.
3

Example 8.1
A rotor with a mass moment of inertia of 6 kgm2 about
its center of mass has a torque of 90 Nm applied to it.
Determine the angular acceleration of the rotor.

Solution:
By Newton’s law,
 = I

90 Nm

90 = 6
 = 15 rad/s2
4

Example 8.2
The flywheel has a mass of 5 kg and a radius of
gyration of 200 mm. The 20 kg mass, attached to a
rope wrapped around the spindle of diameter 30 mm,
is allowed to fall from rest. Calculate the angular
acceleration of the wheel and the acceleration of the
20 kg mass.
Flywheel
5kg

Spindle

20kg
5

Solution:
For the flywheel,

I = mk2
= 5 x 0.22
= 0.2 kgm2
Torque on flywheel,
 = T x 0.015 Nm
 = I
T x 0.015 = 0.2
T = (0.2 / 0.015) 
---------- (1)

dia =30 mm

T

T

For the mass,
F = ma
20g – T = 20a
a = r
 = a / 0.015

----------- (2)
20g

------------- (3)
6

From (1) & (3),
T = (0.2 / 0.015) x (a / 0.015)
= 0.2a / 0.0152
Sub into (2),
20g – 0.2a / 0.0152 = 20a
a = 0.216 m/s2
From (3),
 = 0.216 / 0.015
= 14.4 rad/s2
7

Example 8.3
A power-driven winch is used to raise a mass of 300
kg with an acceleration of 2 m/s2. The winch drum is
0.5 m in diameter and has a mass moment of inertia
about its center of 8 kgm2. What torque must be
applied to the winch drum?
Winch

300kg

8

T

T

a

For the mass,
F = ma
T - 300g = 300 x 2
T = 3543 N

300g

For the drum,
a = r
2 = 0.25
= 8 rad/s2
 = I
a - 3543 x 0.25 = 8 x 8
a = 950 Nm

9

Example 8.4
A compound pulley having a mass of 4 kg and a
radius of gyration 0.28 m, is connected to two
masses A and B as shown. Assuming no axle
friction, determine the acceleration of the masses.
rB = 0.3 m

rA = 0.1 m

A
10kg

B
20kg

10

TB

TA
TA

10g

TB

20g

I = mk2
= 4 x 0.282
= 0.3136 kgm2
Assume acceleration clockwise:
 = I
TB x 0.3 – TA x 0.1 = 0.3136 --------------- (1)

11

TB

TA
TA

TB

10g

20g

For block A,

F = ma
TA – 10g = 10 aA ---------- (2)

For block B,

F = ma
20g – TB = 20 aB -------- (3)

For pulley,
 = aA / rA = aB / rB

----------- (4)
12

TB

TA
TA

10g

TB

20g

Sub (4) into (2),
TA – 10g = 10 x 0.1
TA =  + 10g

--------------- (5)

Sub (4) into (3),
20g – TB = 20 x 0.3
TB = - 6 + 20g

--------------- (6)
13

TB

TA
TA

10g

TB

20g

Sub (5) & (6) into (1),
(- 6 + 20g) x 0.3 – ( + 10g) x 0.1 = 0.3136
 = 22.154 rad/s2
aA = rA = 0.1 x 22.154 = 2.215 m/s2
aB = rB = 0.3 x 22.154 = 6.646 m/s2
14

Example 8.5
A power driven winch can exert a torque of 1200
Nm. It has a drum of 80 cm diameter and a mass
moment of inertia of 30 kgm2. It is used to raise
block A of mass 200 kg up the slope as shown in
the figure. The coefficient of friction between
block A and the slope is 0.3. Assume the pulley
is frictionless.
(a) Draw the free body diagrams of the winch and
block A.
(b) Determine the acceleration of block A and the
tension in the cable.
Winch

Pulley

Block A 200 kg
60
15

 = 1200 Nm

T
T

200g

Winch

a
N1

60

F1

Block A

Block A:
For the direction perpendicular to the slope,
Fy = 0
N1 – 200gcos60 = 0
N1 = 981
F1 =  N1
= 0.3 x 981
= 294.3 N
16

T
 = 1200 Nm

T

200g

Winch

a
N1

60

F1

Block A

For the direction up the slope,
Fx = ma
T – 294.3 – 200gsin60 = 200a
T = 1993.44 + 200a
---------- (1)
For the drum, a = r
 = (a / 0.4)
17

T

 = 1200 Nm

T

200g

Winch

For the anti-clockwise direction,
N1
 = I
1200 – 0.4T = 30 x (a / 0.4)
1200 – 0.4T – 75a = 0
--------- (2)

a

60

F1

Block A

Substitute (1) into (2)
1200 – 0.4 (1993.44 + 200a) – 75a = 0
a = 2.6 m/s2
Sub. into (1)
T = 1993.44 + 200 x 2.6
= 2513.44 N

End of Chapter 8
18


Slide 18

School of Engineering
ENGINEERING MECHANICS
CHAPTER 8

Kinetics of Rotational Motion

1

8.1 Brief
Kinetics of rotational motion is the study of rotational
motion caused by external torque.

8.2 Newton’s second law for Rotational Motion
It can be written in mathematical form as:
 = I
 = I

for a single torque
for several torque
where  is the torque (Nm – rotating shaft)
I is the moment of inertia (kgm2)
 is the acceleration (rad/s2)

(Note:- I = mk2, where m is the mass and k is the radius
of gyration.)
2

8.3 Free Body diagram
We note that since the motion is caused by torque
acting on the body, the first step would be to show
all the torque acting on it. This is set out in the form
of a free body diagram indicating clearly all
external torque. A free body diagram is a diagram
showing only the body and the torque acting
directly on it.
8.4 Sign Convention
We are interested to determine the motion of the
body hence we choose the direction of the
resulting rotation as the positive direction.
However, any direction can be selected as
positive but this must be strictly adhered to
throughout the analysis.
3

Example 8.1
A rotor with a mass moment of inertia of 6 kgm2 about
its center of mass has a torque of 90 Nm applied to it.
Determine the angular acceleration of the rotor.

Solution:
By Newton’s law,
 = I

90 Nm

90 = 6
 = 15 rad/s2
4

Example 8.2
The flywheel has a mass of 5 kg and a radius of
gyration of 200 mm. The 20 kg mass, attached to a
rope wrapped around the spindle of diameter 30 mm,
is allowed to fall from rest. Calculate the angular
acceleration of the wheel and the acceleration of the
20 kg mass.
Flywheel
5kg

Spindle

20kg
5

Solution:
For the flywheel,

I = mk2
= 5 x 0.22
= 0.2 kgm2
Torque on flywheel,
 = T x 0.015 Nm
 = I
T x 0.015 = 0.2
T = (0.2 / 0.015) 
---------- (1)

dia =30 mm

T

T

For the mass,
F = ma
20g – T = 20a
a = r
 = a / 0.015

----------- (2)
20g

------------- (3)
6

From (1) & (3),
T = (0.2 / 0.015) x (a / 0.015)
= 0.2a / 0.0152
Sub into (2),
20g – 0.2a / 0.0152 = 20a
a = 0.216 m/s2
From (3),
 = 0.216 / 0.015
= 14.4 rad/s2
7

Example 8.3
A power-driven winch is used to raise a mass of 300
kg with an acceleration of 2 m/s2. The winch drum is
0.5 m in diameter and has a mass moment of inertia
about its center of 8 kgm2. What torque must be
applied to the winch drum?
Winch

300kg

8

T

T

a

For the mass,
F = ma
T - 300g = 300 x 2
T = 3543 N

300g

For the drum,
a = r
2 = 0.25
= 8 rad/s2
 = I
a - 3543 x 0.25 = 8 x 8
a = 950 Nm

9

Example 8.4
A compound pulley having a mass of 4 kg and a
radius of gyration 0.28 m, is connected to two
masses A and B as shown. Assuming no axle
friction, determine the acceleration of the masses.
rB = 0.3 m

rA = 0.1 m

A
10kg

B
20kg

10

TB

TA
TA

10g

TB

20g

I = mk2
= 4 x 0.282
= 0.3136 kgm2
Assume acceleration clockwise:
 = I
TB x 0.3 – TA x 0.1 = 0.3136 --------------- (1)

11

TB

TA
TA

TB

10g

20g

For block A,

F = ma
TA – 10g = 10 aA ---------- (2)

For block B,

F = ma
20g – TB = 20 aB -------- (3)

For pulley,
 = aA / rA = aB / rB

----------- (4)
12

TB

TA
TA

10g

TB

20g

Sub (4) into (2),
TA – 10g = 10 x 0.1
TA =  + 10g

--------------- (5)

Sub (4) into (3),
20g – TB = 20 x 0.3
TB = - 6 + 20g

--------------- (6)
13

TB

TA
TA

10g

TB

20g

Sub (5) & (6) into (1),
(- 6 + 20g) x 0.3 – ( + 10g) x 0.1 = 0.3136
 = 22.154 rad/s2
aA = rA = 0.1 x 22.154 = 2.215 m/s2
aB = rB = 0.3 x 22.154 = 6.646 m/s2
14

Example 8.5
A power driven winch can exert a torque of 1200
Nm. It has a drum of 80 cm diameter and a mass
moment of inertia of 30 kgm2. It is used to raise
block A of mass 200 kg up the slope as shown in
the figure. The coefficient of friction between
block A and the slope is 0.3. Assume the pulley
is frictionless.
(a) Draw the free body diagrams of the winch and
block A.
(b) Determine the acceleration of block A and the
tension in the cable.
Winch

Pulley

Block A 200 kg
60
15

 = 1200 Nm

T
T

200g

Winch

a
N1

60

F1

Block A

Block A:
For the direction perpendicular to the slope,
Fy = 0
N1 – 200gcos60 = 0
N1 = 981
F1 =  N1
= 0.3 x 981
= 294.3 N
16

T
 = 1200 Nm

T

200g

Winch

a
N1

60

F1

Block A

For the direction up the slope,
Fx = ma
T – 294.3 – 200gsin60 = 200a
T = 1993.44 + 200a
---------- (1)
For the drum, a = r
 = (a / 0.4)
17

T

 = 1200 Nm

T

200g

Winch

For the anti-clockwise direction,
N1
 = I
1200 – 0.4T = 30 x (a / 0.4)
1200 – 0.4T – 75a = 0
--------- (2)

a

60

F1

Block A

Substitute (1) into (2)
1200 – 0.4 (1993.44 + 200a) – 75a = 0
a = 2.6 m/s2
Sub. into (1)
T = 1993.44 + 200 x 2.6
= 2513.44 N

End of Chapter 8
18