Transcript Chapter 10 Rotational motion and Energy

Chapter 10
Rotational motion
and Energy
Rotational Motion

Up until now we have been looking at the
kinematics and dynamics of translational
motion – that is, motion without rotation. Now
we will widen our view of the natural world to
include objects that both rotate and translate.

We will develop descriptions (equations) that
describe rotational motion

Now we can look at motion of bicycle wheels
and even more!
I.
Rotational variables
- Angular position, displacement, velocity, acceleration
II. Rotation with constant angular acceleration
III. Relation between linear and angular variables
- Position, speed, acceleration
IV. Kinetic energy of rotation
V. Rotational inertia
VI. Torque
VII. Newton’s second law for rotation
VIII. Work and rotational kinetic energy
Rotational kinematics
 In
the kinematics of rotation we encounter
new kinematic quantities
 Angular displacement
q
 Angular speed
w
 Angular acceleration
a
 Rotational Inertia
I
 Torque
t
 All these quantities are defined relative to
an axis of rotation
Angular displacement

Measured in radians or degrees
 There is no dimension
Dq = qf – qi CW
qi
Dq
Axis of rotation
2
Dq  q f  q i 
rad
3
qf
2r

1 rev  360 
 2 rad
r

1 rad  57.3  0.159 rev
Note: we do not reset θ to
zero with each complete
rotation of the reference line
about the rotation axis. 2
turns  θ =4π
Translation: body’s movement described by x(t).
Rotation: body’s movement given by θ(t) = angular position
of the body’s reference line as function of time.
Angular displacement: body’s rotation about its axis
changing the angular position from θ1 to θ2.
Dq  q2  q1
Clockwise rotation  negative
Counterclockwise rotation  positive
Angular displacement and arc length

Arc length depends
on the distance it is
measured away
from the axis of
rotation
2
Dq  q f  q i 
rad
3
S p  Dq  rp
r
P
S q  Dq  rq
s
Dq 
r
qi
Q
sp
sq
Axis of rotation
qf
Angular Speed

Angular speed is the rate of change of
angular position
Dq
w
Dt

We can also define the instantaneous
angular speed
Dq
w  lim
Dt 0 Dt
Average angular velocity and
tangential speed
Recall that speed is distance divided by time
elapsed
 Tangential speed is arc length divided by time
s
elapsed
v 

t
Dt
s
 And because Dq 
r
we can write
Dq  s 1 vT
w 


Dt
r Dt r
Average Angular Acceleration
 Rate
of change of angular velocity
Dw w f  w i
a

Dt
Dt
a
 Instantaneous
Dw w 2  w1

Dt
Dt
angular acceleration
Dw
a  lim
Dt  0 D t
Angular acceleration and tangential
acceleration
 We
can find a link between tangential
acceleration at and angular acceleration α
vf
vi

w

w
Dw
aT
f
i
r
r
a



Dt
Dt
Dt
r
 So
aT
a
r
Centripetal acceleration
 We
have that
vT
ac 
r
2
we also know that vT2  w 2 r 2
 So we can also say
 But
ac  w r
2
Example: Rotation

A dryer rotates at 120 rpm. What distance do
your clothes travel during one half hour of drying
time in a 70 cm diameter dryer? What angle is
swept out?
 Distance: s = Dqr and w = Dq/Dt so s = wDtr
 s = 120 /min x 0.5 h x 60 min/h x 0.35 m
= 1.3 km
 Angle: DqwDt = 120 r/min x 0.5 x 60 min
= 120x2r /min x 0.5 h x 60 min/h = 2.3 x
104 r
Rotational motion with constant
angular acceleration

We will consider cases where a is constant

Definitions of rotational and translational
quantities look similar

The kinematic equations describing rotational
motion also look similar

Each of the translational kinematic equations
has a rotational analogue
Rotational and Translational
Kinematic Equations
v f  v i  at
Dx  v i t  at
1
2
2
v f  v  2aDx
2
2
i
vi  vf
v
2
Constant a motion
What is the angular acceleration of a car’s wheels
(radius 25 cm) when a car accelerates from 2 m/s to
5 m/s in 8 seconds?
vi 2m / s
wi  
 8rad / s
r 0.25m
v f 5m / s
wf  
 20rad / s
r 0.25m
w f  wi (20  8)rad / s
a

 1.5rad / s
Dt
8s
Example: Centripetal Acceleration

A 1000 kg car goes around a bend that has a
radius of 100 m, travelling at 50 km/h. What is the
centripetal force? What keeps the car on the bend?
[What keeps the skater in the arc?]
2
 km 1000 
1000kg   50


2
mv
h 3600 

Fc 

 1929 N
r
100m
Friction keeps the car and skater on the bend
Car rounding a bend

Frictional force of road on tires supplies
centripetal force
 If ms between road and tires is lowered then
frictional force may not be enough to provide
centripetal force…car will slide
 Locking wheels makes things worse as
mk < ms
 Banking of roads at corners reduces the risk
of skidding…
Car rounding a bend

Horizontal component of the normal force of
the road on the car can provide the
centripetal force
Ncosq
 If
N
mv
N sin q 
r
2
then no friction is required
Nsinq
q
Fg
Rotational Dynamics

Easier to move door at A than at B using the
same force F
hinge
A

B
More torque is exerted at A than at B
Torque

Torque is the rotational analogue of Force

Torque, t, is defined to be
t = Fr
Where F is the force applied tangent to the
rotation and r is the distance from the axis of
rotation
F
r
Torque

A general definition of torque is
F
q
t = Fsinq r

r
Units of torque are Nm

Sign convention used with torque

Torque is positive if object tends to rotate CCW

Torque is negative if object tends to rotate CW
Condition for Equilibrium

We know that if an object is in (translational)
equilibrium then it does not accelerate. We
can say that SF = 0
 An object in rotational equilibrium does not
change its rotational speed. In this case we
can say that there is no net torque or in other
words that:
St = 0
Torque and angular acceleration

An unbalanced torque (t) gives rise to an
angular acceleration (a)
 We can find an expression analogous to
F = ma that relates t and a
 We can see that
Ft = mat
 and Ftr = matr = mr2a
Ft
m
(since at = ra)
r
 Therefore
t = mr2a
Torque and Angular Acceleration

Angular acceleration is directly proportional to
the net torque, but the constant of proportionality
has to do with both the mass of the object and
the distance of the object from the axis of rotation
– in this case the constant is mr2

This constant is called the moment of inertia.
Its symbol is I, and its units are kgm2

I depends on the arrangement of the rotating
system. It might be different when the same
mass is rotating about a different axis
Newton’s Second Law for Rotation

Now we have
t = Ia
 Where
I is a constant related to the
distribution of mass in the rotating
system
 This is a new version of Newton’s
second law that applies to rotation
Angular Acceleration and I
The angular acceleration reached by a
rotating object depends on, M, r, (their
distribution) and T
T
When objects are rolling under the influence
of gravity, only the mass distribution and the
radius are important
Moments of Inertia for Rotating
Objects
The total torque on a rotating system is the
sum of the torques acting on all particles of
the system about the axis of rotation –
and since a is the same for all particles:
I  Smr2 = m1r12+ m2r22+ m3r32+…
Axis of rotation
Continuous Objects
To calculate the moment of inertia for
continuous objects, we imagine the object to
consist of a continuum of very small mass
elements dm. Thus the finite sum Σmi r2i
becomes the integral
I   r dm
2
Moment of Inertia of a Uniform Rod
Lets find the moment of inertia of a uniform
rod of length L and mass M about an axis
perpendicular to the rod and through one end.
Assume that the rod has negligible thickness.
L
Moment of Inertia of a Uniform Rod
We choose a mass element dm at a distance x
from the axes. The mass per unit length (linear
mass density) is λ = M / L
Moment of Inertia of a Uniform Rod
dm = λ dx
M
dm 
dx
L
L
L
L
M
M 2
I   x dm   x
dx 
x dx

L
L 0
0
0
2
M 1

x
L 3
2
3 L
0
3
M L
1
2

 ML
L 3
3
Example:Moment of Inertia of a Dumbbell
A dumbbell consist of point masses 2kg and 1kg
attached by a rigid massless rod of length 0.6m.
Calculate the rotational inertia of the dumbbell
(a) about the axis going through the center of the
mass and (b) going through the 2kg mass.
Example:Moment of Inertia of a Dumbbell
m1 x1  m2 x2 (2kg)(0)  (1kg)(0.6m)
X 

 0.2m
m1  m2
2kg  1kg
(a)
I CM  m1( x1  X CM )2  m2 ( x2  X CM )2
 (2kg)(0m  0.2m)2  (1kg)(0.6m  0.2m)2  0.24kgm2
Example:Moment of Inertia of a Dumbbell
(b)
I  m2 L2  (1kg)(0.6m) 2  0.4kgm2
Moment of Inertia of a Uniform Hoop
dm
All mass of the hoop M
is at distance r = R from
the axis
R
I   r dm   R dm  R
2
2
2
 dm  MR
2
Moment of Inertia of a Uniform Disc
We expect that I will
be smaller than MR2
since the mass is
uniformly distributed
from r = 0 to r = R
rather than being
concentrated at r=R
as it is in the hoop.
dr
r
R
Each mass element is a hoop of radius r and
thickness dr. Mass per unit area
σ = M / A = M /πR2
Moment of Inertia of a Uniform
Disc
dm  dA 
dr
R
R
0
0
I   r 2 dm   r 2 2rdr  2  r 3dr

r
M

A
2M R
M 4 1
2
I

R  MR
2
A 4 2R
2
4
R
Moments of inertia I for Different Mass
Arrangements
Moments of inertia I for Different Mass
Arrangements
Torque
Torque:
Twist  “Turning action of force F ”.
Units: Nm
Radial component, Fr : does not
cause rotation  pulling a door
parallel to door’s plane.
Tangential component, Ft: does cause rotation  pulling a
door perpendicular to its plane. Ft= F sinφ
t  r  ( F  sin  )  r  Ft  (r sin  ) F  r F
r┴ : Moment arm of F
Vector quantity
r : Moment arm of Ft
Sign: Torque >0 if body rotates counterclockwise.
Torque <0 if clockwise rotation.
Superposition principle: When several torques act on
a body, the net torque is the sum of the individual
torques
Newton’s second law for rotation
F  ma  t  Ia
Proof:
Particle can move only along the
circular path  only the tangential
component of the force Ft (tangent
to the circular path) can accelerate
the particle along the path.
F t  mat
t  Ft  r  mat  r  m(a  r )r  (mr 2 )a  Ia
t net  Ia
Kinetic energy of rotation
Reminder: Angular velocity, ω
particles within the rotating body.
is the same for all
Linear velocity, v of a particle within the rigid body
depends on the particle’s distance to the rotation axis (r).
1
2 1
2 1
K  m1v1  m2v2  m3v32  ... 
2
2
2

1
1
1

mi vi2 
mi (w  ri ) 2  
2
2
2
i
i




i

mi ri2 w 2


Moment of Inertia
Rotational Kinetic Energy
We must rewrite our statements of conservation of
mechanical energy to include KEr
Must now allow that (in general):
½ mv2+mgh+ ½ Iw2 = constant
Could also add in e.g. spring PE
VII. Work and Rotational kinetic energy
Translation
DK  K f  K i 
1 2 1 2
mv f  mvi  W
2
2
Rotation
DK  K f  K i 
xf
1 2 1 2
Iw f  Iwi  W
2
2
qf
W   Fdx
W   t  dq
xi
Work-kinetic energy
Theorem
Work, rotation about fixed axis
qi
W  t (q f  q i )
W  F d
dW
P
 F v
dt
P
Work, constant torque
dW
 t w
dt
Power, rotation about
fixed axis
Proof:
W  DK  K f  K i 
1 2 1 2 1
1
1
1
1
1
mv f  mvi  m(w f r ) 2  m(wi r ) 2  (mr 2 )w 2f  (mr 2 )wi2  Iw 2f  Iwi2
2
2
2
2
2
2
2
2
qf
dW  Ft ds  Ft  r  dq  t  dq  W   t  dq
qi
P
dW t  dq

 t w
dt
dt