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MM203
Mechanics of
Machines: Part 1
1
Module
• Lectures
• Tutorials
• Labs
• Why study dynamics?
• Problem solving
2
Vectors
PQ  R
R
R
Q
Q
PQ  QP
P
P
R
P  R  Q  R   Q
5 2 7
3  1  4
     
5  5
    
3  3
−Q
P
P  Q  R  P  Q  R
3
Unit vectors - components
 4
1 0
3  40  31
 
   
v  vx i  v y j  vz k
v  v  vx  v y  vz
2
2
2
4
3  ?
 
 4
 4  
 3   3
  0 
 
4
Direction cosines
• l, m, and n – direction cosines between v and
x-, y-, and z-axes
vy
vx
vz
l , m , n
v
v
v
v  vli  mj  nk  , l  m  n  1
2
– Calculate 3 direction cosines for
2
2
 3
4
 
6 
8 
 
5
Dot (or scalar) product
P Q  PxQx  PyQy  PzQz
P Q  PQ cos
Q

P
• Component of Q in P
direction
P Q
QP  Q cos  
P
P
Q
QP
6
Angle between vectors
P  Q PxQx  Py Qy  Pz Qz
cos 

PQ
PQ
 lPlQ  mP mQ  nP nQ
i i  ?
ij ?
PP  P
2
7
Dot product
• Commutative and distributive
P Q  Q P
P  Q  R   P  Q  P  R
8
Particle kinematics
• What is kinematics?
• What is a particle?
•
•
•
•
Rectilinear motion - review
Plane curvilinear motion - review
Relative motion
Space curvilinear motion
9
Rectilinear motion
ds
v
dt
dv
a
dt
• Combining gives
v dv  a ds
• What do dv and ds represent?
10
Example
• The acceleration of a particle is a = 4t − 30
(where a is in m/s2 and t is in seconds).
Determine the velocity and displacement in
terms of time. (Problem 2/5, M&K)
11
Vector calculus
d uP 
 uP  uP
dt
• Vectors can vary both in length and in
direction
  P i  P j  P k
P
x
y
z
12
Plane curvilinear motion
• Choice of coordinate system (axes)
– Depends on problem – how information is given
and/or what simplifies solution
– Practice
13
Plane curvilinear motion
• Rectangular coordinates
• Position vector - r
y
j
r  xi  yj
v  r  xi  y j  v x i  v y j
i
x
a  v  r  xi  yj  a x i  a y j
• e.g. projectile motion
• ENSURE CONSISTENCY IN DIRECTIONS
14
Plane curvilinear motion
• Normal and tangential coordinates
• Instantaneous radius of curvature – r
ds  r db
ds
db
v v
r
dt
dt
y
Note that dr/dt can be ignored in this case
– see M&K.
• What is direction of v?
r
db
ds
x
15
Plane curvilinear motion
v  vet  rbet
d ve t 
a  v 
 ve t  ve t
dt
e t  ben
an  vb  rb 
2
v
2
r
en
et
r
at  v  s
16
Example
• A test car starts from rest on a horizontal
circular track of 80 m radius and increases its
speed at a uniform rate to reach 100 km/h in
10 seconds. Determine the magnitude of the
acceleration of the car 8 seconds after the
start. (Answer: a = 6.77 m/s2). (Problem 2/97, M&K)
17
Example
• To simulate a condition of “weightlessness” in its
cabin, an aircraft travelling at 800 km/h moves an a
sustained curve as shown. At what rate in degrees
per second should the pilot drop his longitudinal line
of sight to effect the desired condition? Use g = 9.79
m/s2. (Answer: db/dt = 2.52 deg/s). (Problem 2/111, M&K)
b
18
Example
• A ball is thrown horizontally at 15 m/s
from the top of a cliff as shown and
lands at point C. The ball has a
horizontal
acceleration
in
the
negative x-direction due to wind.
Determine the radius of curvature of
the path at B where its trajectory
makes an angle of 45° with the
horizontal. Neglect air resistance in
the vertical direction. (Answer: r =
41.8 m). (Problem 2/125, M&K)
A
B
50 m
C
40 m
x
19
Plane curvilinear motion
• Polar coordinates
r  re r
e r  e , e  er
v  r  re r  re r
v  re  re
r

y
e
er
r

x
20
Plane curvilinear motion
a  ar e r  a e
where
2

ar  r  r
 
1 d 2



a  r  2r 
r
r dt
21
Example
• An aircraft flies over an observer with a constant
speed in a straight line as shown. Determine the
signs (i.e. +ve, -ve, or 0) for
r , r , r, , , and
y
• for positions
• A, B, and C.
•
(Problem 2/134, M&K)
C
B
v
A
r

x
22
Example
• At the bottom of a loop at point P as shown, an aircraft has
a horizontal velocity of 600 km/h and no horizontal
acceleration. The radius of curvature of the loop is 1200 m.
For the radar tracking station shown, determine the
recorded values of d2r/dt2 and d2/dt2 for this instant.
(Answer: d2r/dt2 = 12.5 m/s2, d2/dt2 = 0.0365 rad/s2). (Problem
2/141, M&K)
P
r

400 m
600 km/h
1000 m
23
Relative motion
• Absolute (fixed axes)
• Relative (translating axes)
• Used when measurements are taken from a
moving observation point, or where use of
moving axes simplifies solution of problem.
• Motion of moving coordinate system may be
specified w.r.t. fixed system.
24
Relative motion
• Set of translating
axes (x-y) attached to Y
particle B (arbitrarily).
The position of A
relative to the frame
x-y (i.e. relative to B)
is
O
y
rA
A
rA/B
B
x
rB
X
rA/ B  xi  yj
25
Relative motion
• Absolute positions of points A and B (w.r.t.
fixed axes X-Y) are related by
rA  rB  rA / B
or
rB  rA  rB / A
where
rB / A  rA / B
26
Relative motion
• Differentiating w.r.t. time gives
rA  rB  rA / B
v A  v B  v A/ B
a A  aB  a A/ B
v A/ B  v B / A
a A / B  a B / A
• Coordinate systems may be rectangular,
tangential and normal, polar, etc.
27
Inertial systems
• A translating reference system with no
acceleration is known as an inertial system. If
aB = 0 then
a A  a A/ B
• Replacing a fixed reference system with an
inertial system does not affect calculations (or
measurements) of accelerations (or forces).
28
Example
• A yacht moving in the direction
shown is tacking windward
against a north wind. The log
registers a hull speed of 6.5
knots. A “telltale” (a string tied to
the rigging) indicates that the
direction of the apparent wind is
35° from the centerline of the
boat. What is the true wind
velocity? (Answer: vw = 14.40
knots). (Problem 2/191, M&K)
vw
50°
35°
29
Example
• To increase his speed, the water skier A cuts across the
wake of the boat B which has a velocity of 60 km/h as
shown. At the instant when  = 30°, the actual path of the
skier makes an angle b = 50° with the tow rope. For this
position, determine the velocity vA of the skier and the value
of d/dt. (Answer: vA = 80.8 km/h, d/dt = 0.887 rad/s). (Problem
2/193, M&K)
A
b
B
m
0
1

vB
30
Example
• Car A is travelling at a constant
speed of 60 km/h as it rounds a
circular curve of 300 m radius. At the
instant shown it is at  = 45°. Car B is
passing the centre of the circle at the
same instant. Car A is located
relative to B using polar coordinates
with the pole moving with B. For this
instant, determine vA/B and the values
fo d/dt and dr/dt as measured by an
observer in car B. (Answer: vA/B =
36.0 m/s, d/dt = 0.1079 rad/s, dr/dt =
−15.71 m/s). (Problem 2/201, M&K)
A

r
B
31
Space curvilinear motion
•
•
•
•
Rectangular coordinates (x, y, z)
Cylindrical coordinates (r, , z)
Spherical coordinates (R, , f)
Coordinate transformations – not covered
• Tangential and normal system not used due to
complexity involved.
32
Space curvilinear motion
• Rectangular coordinates (x, y, z) – similar to
2D
R  xi  yj  zk
v  xi  y j  zk
a  xi  yj  zk
33
Space curvilinear motion
• Cylindrical coordinates (r, , z)
R  re r  zk
z
R

y
r

v  re r  re  zk

z

x

2

a  r  r er  r  2r e  zk
34
Space curvilinear motion
• Spherical coordinates (R, , f)
v  R e R  R cosfe  Rfef
z
a  aR e R  a e  af ef
where
  Rf 2  R 2 cos2 f
aR  R
cosf d 2 
a 
R   2 Rf sin f
R dt
1 d 2
af 
R f  R 2 sin f cosf
R dt
 
 
R
y
f
R

x
35
Example
• A section of a roller-coaster is a horizontal cylindrical helix.
The velocity of the cars as they pass point A is 15 m/s. The
effective radius of the cylindrical helix is 5 m and the helix
angle is 40°. The tangential acceleration at A is gcosg.
Compute the magnitude of the acceleration of the
passengers as they pass A. (Answer: a = 27.5 m/s2). (Problem
2/171, M&K)
5m
A
A
g = 40°
36
Example
• The robot shown rotates about a
fixed vertical axis while its arm
extends and elevates. At a given
instant, f = 30°, df/dt = 10 deg/s =
constant, l = 0.5 m, dl/dt = 0.2 m/s,
d2l/dt2 = −0.3 m/s2, and W = 20
deg/s = constant. Determine the
magnitudes of the velocity and
acceleration of the gripped part P.
(Answer: v = 0.480 m/s, a = 0.474
m/s2). (Problem 2/177, M&K)
y
5
0.7
O
l
W
P
m
f
x
37
Particle kinetics
• Newton’s laws
F  ma
• Applied and reactive forces must be
considered – free body diagrams
• Forces required to produce motion
• Motion due to forces
38
Particle kinetics
• Constrained and unconstrained motion
• Degrees of freedom
• Rectilinear motion – covered
• Curvilinear motion
39
Rectilinear motion - example
 F  ma
• The 10 Mg truck hauls a 20 Mg trailer. If the unit starts from
rest on a level road with a tractive force of 20 kN between the
driving wheels and the road, compute the tension T in the
horizontal drawbar and the acceleration a of the rig. (Answer:
T = 13.33 kN, a = 0.667 m/s2). (Problem 3/5, M&K)
20 Mg
10 Mg
40
Example
• The motorized drum turns at a constant speed causing the
vertical cable to have a constant downwards velocity v.
Determine the tension in the cable in terms of y. Neglect the
diameter and mass of the small pulleys. (Problem 3/48, M&K)
2b
y
v
• Answer:
2 2

m
b
v 
2
2

T
b  y  g 
3 
2y
4y 

m
41
Curvilinear motion
• Rectangular coordinates
F
x
 max ,
F
y
 may
• Normal and tangential coordinates
F
n
 man ,
• Polar coordinates
F
r
 mar ,
 F  ma
t
t
 F  ma
42
00
10
• A pilot flies an airplane at a
constant speed of 600 km/h
in a vertical circle of radius
1000 m. Calculate the force
exerted by the seat on the
90 kg pilot at point A and at
point A. (Answer: RA = 3380
N, RB = 1617 N). (Problem 3/63,
m
Example
M&K)
600 km/h
43
Example
• The 30 Mg aircraft is climbing at an angle of 15° under a jet
thrust T of 180 kN. At the instant shown, its speed is 300
km/h and is increasing at a rate of 1.96 m/s2. Also  is
decreasing as the aircraft begins to level off. If the radius of
curvature at this instant is 20 km, compute the lift L and the
drag D. (Lift and drag are the aerodynamic forces normal to
and opposite to the flight direction, respectively). (Answer: D
= 45.0 kN, L = 274 kN). (Problem 3/69, M&K)

T
44
Example
• A child's slide has a quarter circle shape as shown. Assuming
that friction is negligible, determine the velocity of the child at
the end of the slide ( = 90°) in terms of the radius of
curvature r and the initial angle 0.
• Answer

v  2 gr1  sin  0 
45
Slide
• Does it matter what profile slide has?
• What if friction added?
46
Example
• A flat circular discs rotates about a
vertical axis through the centre point
at a slowly increasing angular
velocity w. With w = 0, the position of
the two 0.5 kg sliders is x = 25 mm.
Each spring has a stiffness of 400
x
N/m. Determine the value of x for w =
240 rev/min and the normal force
exerted by the side of the slot on the
block. Neglect any friction and the
mass of the springs. (Answer: x =
118.8 mm, N = 25.3 N). (Problem 3/83,
w
x
80 mm
80 mm
M&K)
47
Work and energy
• Work/energy analysis – don’t need to calculate
accelerations
A′
• Work done by force F
F
dr
A
a
dU  F  dr  F ds cosa
where
ds  dr
R+dr
r
O
• Integration of F = ma w.r.t. displacement gives
equations for work and energy
48
Work and energy
• Active forces and reactive forces (constraint
forces that do no work)
• Total work done by force
U   F  dr   Fx dx  Fy dy  Fz dz
or
U   Ft ds
• where Ft = tangential force component
49
Work and energy
• If displacement is in same direction as force
then work is +ve (otherwise –ve)
• Ignore reactive forces
• Kinetic energy
T  mv
1
2
2
• Gravitational potential energy
Vg  mgh
50
Example
• A small vehicle enters the top of
a circular path with a horizontal
velocity v0 and gathers speed as
it moves down the path.
Determine the angle b (in terms
of v0) at which it leaves the path
and becomes a projectile.
Neglect friction and treat the
vehicle as a particle. (Problem 3/87,
M&K)
• Answer:
v0
R
b
2

v0 
1 2

b  cos  

3
3
gR


51
Example
• The small slider of mass m is
released from point A and
slides without friction to point
D. From point D onwards the
coefficient of kinetic friction
between the slider and the
slide is mk. Determine the
distance s travelled by the
slider up the incline beyond D.
m
A
2R
s
B
R
D
30°
(Problem 3/125, M&K)
• Answer:
s
4R
C
1  mk 3
52
Example
• A rope of length pr/2 and
mass per unit length r is
released with  = 0 in a
smooth vertical channel and
falls through a hole in the
supporting
surface.
Determine the velocity v of
the chain as the last part of it
leaves the slot. (Problem 3/173,
M&K)
• Answer:
r

p 4 
v  gr  
2 p
53
Linear impulse and momentum
• Integration of F = ma w.r.t. time gives
equations of impulse and momentum.
• Useful where time over which force acts is
very short (e.g. impact) or where force acts
over specified length of time.
54
Linear impulse and momentum
d




F

m
v

m
v

G

dt
• If mass m is constant then sum of forces =
time rate of change of linear momentum
G  mv
• Linear momentum of particle
• Units – kg·m/s or N·s
• Scalar form:
F  G ,
F  G ,
F  G

x
x

y
y

z
z
55
Linear impulse and momentum
• Integrate over time
  F dt  G
t2
t1
2
 G 1  G
or
G1  
t2
t1
 F dt  G
2
• Product of force and time is called linear
impulse
t
• Scalar form
t  Fx dt  mvx 2  mvx 1 , etc.
2
1
56
Linear impulse and momentum
• Note that all forces must be included (i.e. both
active and reactive)
57
Linear impulse and momentum
?
m 2 v2
F
m 1 v1
−F
?
• If there are no unbalanced forces acting on a system
then the total linear momentum of the system will
remain constant (principle of conservation of linear
momentum)
58
Impact
• How to determine velocities after impact?
• Forces normal to contact
surface. Fd is force during
deformation period while Fr is
force during recovery period.
• The ratio of the restoration
impulse to the deformation
impulse is called the coefficiente 
of restitution
(V2)n
m2
(V1)n
m1


t
t0
t0
0
Fr dt
Fd dt
59
Impact
• For particle 1, (v0)n being the intermediate normal
velocity component (of both particles) and (v1)′n being
normal velocity component after collision

e

t
t0
t0
0
m1 v1 n  m1 v0 n

Fd dt m1 v0 n  m1 v1 n
Fr dt
• Similarly for particle 2
m2 v2 n  m2 v0 n
e
m2 v0 n  m2 v2 n
60
Impact
• Combining gives
v2 n  v1 n
e
v1 n  v2 n
• e = 0 for plastic impact, e = 1 for elastic impact
• Note that tangential velocities are not affected
by impact
61
Example
• A 75 g projectile traveling at 600 m/ strikes and becomes
embedded in the 50 kg block which is initially stationary.
Compute the energy lost during the impact. Express your
answer as an absolute value and as a percentage of the
original energy of the system. (Problem 3/180, M&K)
75 g
600 m/s
50 kg
62
Example
d/2
d/2
• The pool ball shown must
be hit so as to travel into
the side pocket as shown.
Specify the location x of the
cushion impact if e = 0.8.
(Answer: x = 0.268d) (Problem
3/251, M&K)
x
d
63
Example
• The vertical motion of the 3 kg
load is controlled by the forces P
applied to the end rollers of the
framework shown. If the upward
velocity of the cylinder is
increased from 2 m/s to 4 m/s in 2
seconds, calculate the average
force Rav under each of the two
rollers during the 2 s interval.
Neglect the small mass of the
frame. (Answer: Rav = 16.22 N)
v
P
3 kg
P
(Problem 3/199, M&K)
64
Example
• A 1000 kg spacecraft is
traveling in deep space with a
speed vs = 2000 m/s when
struck at its mass centre by a
10 kg meteor with velocity vm
of magnitude 5000 m/s. The
meteor becomes embedded in
the satellite. Determine the
final velocity of the spacecraft.
(Answer: v = 36.9i + 1951j –
14.76k m/s) (Problem 3/201, M&K)
z
vm
4
x
2
5
vs
y
65
Cross (or vector) product
• Magnitude of cross-product
P Q  PQsin 
Q P  PQsin   PQsin 
Q

P
• Direction of cross-product governed by righthand rule
66
Right-hand rule
PQ  R
• Middle finger in direction of R if thumb in
direction of P and index finger in direction of
Q.
• Use right-handed reference frame for x,y, and
z.
67
Cross (or vector) product
i  j  k , etc.
P×Q
i  i  0 , etc.
Q
P
Q×P=−P×Q
• Distributive
P  Q  R  P  Q  P  R
68
Cross (or vector) product
P Q
y
z
i
j
P  Q  Px
Qx
Py
Qy
k
Pz 
Qz
 Pz Qy i  PxQz  Pz Qx  j  PxQy  Py Qx k
• Derivative
d P  Q 
  P  Q
 PQ
dt
69
Angular impulse and momentum
• The angular momentum of a particle about
any point is the moment of the linear
momentum about that point.
• Units are kg·m/s·m or N·m·s
70
Angular impulse and momentum
• Planar motion
• There are 3
components of the
angular momentum
of P about arbitrary
point O: i.e. about
x-,y-, and z-axes.
y
mv
P
r
O
x
71
Angular impulse and momentum
• Since P is coplanar
with x- and y-axes, it
has no moment about
these axes. It only has
a moment about the zaxis.
y
mv
P
r
O
x
72
Angular impulse and momentum
• Is angular momentum of
P about O positive or
negative? – governed by
right-hand rule
y
mv
P
r
O
x
73
Right-hand rule
• Curl fingers in. Rotation indicated by fingers is
in direction of thumb. Is this positive or
negative in this case?
74
Angular impulse and momentum
p

H O  m vr cos     m vr sin 
2

y
• Direction of
component about zaxis is in z-direction
HO  mvr sin  k
mv

P
r
O
x
75
Angular impulse and momentum
HO  x mvy  y mvx k
mvy
mv
y
P
mvx
r
O
x
x
76
Angular impulse and momentum
HO  r  mv
• Note
r  mv  mv r
H x  my vz  z vy , etc.
77
Angular impulse and momentum
• The resultant moment of all forces about O is
M
O
 r  F
• From Newton’s 2nd law
M
 r  mv
• Differentiate w.r.t. time
HO  r  mv
  v  mv  r  mv
H
O
• Now v  mv  0
• so

O
M
O
 HO
78
Angular impulse and momentum
• The moment of all forces on the particle about a fixed point O
equals the time rate of change of the angular momentum
about that point.

M

H
 O x O x , etc.
• If moment about O is zero then angular momentum is
constant (principle of conservation of angular momentum).
• If moment about any axis is zero then component of angular
momentum about that axis is constant.
79
Angular impulse and momentum
• Particle following circular
path at constant angular
velocity. Is angular
momentum about O varying
with time?
• Is angular momentum about
O′ varying with time?
• Is component about z-axis
varying with time?
z
O′
O
y
x
w m
80
Angular impulse and momentum

M

H
 O O
 M
t2
t1
O
dt  H O 2  H O1
• i.e. change in angular momentum is equal to
total angular impulse
 M
t2
t1
Ox
dt  H O x 2  H O x 1 , etc.
81
Angular impulse and momentum
• Example – ice skater
w1
w2
82
Example
• Calculate HO, the angular
momentum of the particle
shown about O (a) using the
vector definition and (b) using
a geometrical approach. The
centre of the particle lies in
the x-y plane. (Answer: HO =
128.7k N·m·s) (Problem 3/221,
M&K)
y
2 kg
7 m/s
30°
8m
O
6m
x
83
Example
• A particle of mass m moves with negligible
friction across a horizontal surface and is
connected by a light spring fastened at point
O. The velocity at A is as shown. Determine
the velocity at B. (Problem 3/226, M&K)
A
350
mm
vA = 4 m/s
54°
O
230
mm
B
65°
vB
84
Example
r
D
2r
w0
• Each of 4 spheres of mass m is treated
as a particle. Spheres A and B are
mounted on a light rod and are rotating
initially with an angular velocity w0 about
a vertical axis through O. The other two
spheres are similarly (but independently)
mounted and have no initial velocity.
When assembly AB reaches the position
indicated it latches with CD and the two
move with a common angular velocity w.
Neglect friction. Determine w and n the
percentage loss of kinetic energy.
(Answer w = w0/5, n = 80%). (Problem 3/227,
r
O
B
A
2r
M&K)
C
85
Example
• The particle of mass m is launched from point O with a
horizontal velocity u at time t = 0. Determine its angular
momentum about O as a function of t. (Answer H0 =
−½mgut2k). (Problem 3/233, M&K)
m
u
O
y
x
86
Relative motion
• Fixed reference frame X-Y
• Moving reference frame x-y
a A  a B  a rel
y
A
  F  ma A  ma B  a rel 
Y
  F  m a rel
rA/B = rrel
rA
x
B
rB
O
X
87
Relative motion
• Special case – inertial system or “Newtonian frame
of reference” with zero acceleration
• Note that work-energy and impulse momentum
equations are equally valid in inertial system – but
relative momentum/relative energy etc. will, in
general, be different to those measured relative to
fixed frame of reference.
88
Example
• The ball A of mass 10 kg is attached to the light rod of length
l = 0.8 m. The rod is attached to a carriage of mass 250 kg
which moves on rails with an acceleration aO as shown. The
rod is free to rotate horizontally about O. If d/dt = 3 rad/s
when  = 90°, find the kinetic energy T of the system if the
carriage has a velocity of 0.8 m/s. Treat the ball as a particle.
(Answer: T = 112 J). (Problem 3/311, M&K)
A
O
l

aO
89
Example
• The small slider A moves with negligible friction down the
tapered block, which moves to the right with constant speed
v = v0. Use the principle of work-energy to determine the
magnitude vA of the absolute velocity of the slider as it
passes point C if it is released at point B with no velocity
relative to the block. (Problem 3/316, M&K)
• Answer:
2
v A  v0  2 gl sin   2v0 cos 2 gl sin 
B
v
A
l
C

90