Rotational Kinematics
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,
roundabouts and divers.
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
qi
Dq
Axis of rotation
2
Dq  q f  q i 
rad
3
qf
CW
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 
r
3
S p  Dq  rp
r
S q  Dq  rq
s
Dq 
r
qi
P
Q
sp
sq
Axis of rotation
qf
CW
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
elapsed
s
vt 
Dt
s
• And because Dq  we can write
r
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
• Instantaneous 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
• But we also know that
• So we can also say
2
v w r
2
T
ac  w r
2
2 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?
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?
A rotating wheel requires 3.00 s to rotate through
37.0 revolutions. Its angular speed at the end of the
3.00-s interval is 98.0 rad/s. What is the constant
angular acceleration of the wheel?
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  F r
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
(since at = ra)
• Therefore
t = mr2a
r
m
Ft
Torque and Angular Acceleration
t = mr2a
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
• We now have that
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
A fish takes a line and pulls it with a tension of 15 N for 20
seconds. The spool has a radius of 7.5 cm. If the moment
of inertia of the reel is 10 kgm2, through how many rotations
does the reel spin? (Assume there is no friction)
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
I for a small mass m rotating about a point a
distance r away is mr2
What is the moment of inertia for an object that
is rotating – such as a rolling object?
Disc?
Sphere?
Hoop?
Cylinder?
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
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
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
(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

M
I  r dm 
2

0
M
R dm  R
2
2

0
dm  R
2
M
m0
 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
M
dm  dA 
2rdr
A
R
dr
R
I   r dm   r  2rdr  2  r dr
2
2
3
0
0
r
R
2M R
M
1
4
2
I 

R  MR
2
A 4
2R
2
4
Moments of inertia I for Different Mass Arrangements
Moments of inertia I for Different Mass Arrangements
Which one will win?
A hoop, disc and sphere are all rolled
down an inclined plane. Which one will win?
Which one will win?
A hoop, disc and sphere are all rolled
down an inclined plane. Which one will win?
1. Hoop I = MR2
2. Disc
I = ½MR2
3. Sphere I = 2/5MR2
a = t/ I
1. a1 = t/ MR2
2. α2= 2(t/ MR2)
3.a3 = 2.5(t/ MR2)
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
Kinetic energy of a body in pure rotation
1
2
K  MvCOM
2
Kinetic energy of a body in pure translation
1 2
K  Iw
2
VII. Work and Rotational kinetic energy
Translation
DK  K f  K i 
1 2 1 2
mv f  mvi  W
2
2
xf
W   Fdx
Rotation
DK  K f  K i 
1 2 1 2
Iw f  Iwi  W
2
2
qf
W   t  dq
xi
qi
W  F d
W  t (q f  q i )
P
dW
 F v
dt
Work-kinetic energy
Theorem
Work, rotation about fixed axis
Work, constant torque
Power, rotation about
fixed axis
dW
P
 t w
dt
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
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
Example - Rotational KE
• What is the linear speed of a ball with radius 1 cm
when it reaches the end of a 2.0 m high 30o
incline?
2m
mgh+ ½ mv2+ ½ Iw2 = constant
• Is there enough information?
Example - Rotational KE
mgh+ ½ mv2+ ½ Iω2 = constant
2
I Sphere  MR 2
5
1
1 2
2
So we
Mghi  Mv f   MR 2w 2f
2
2 5
have that:
1 2 1 2 2
ghi  v f  R w f
2
5
The velocity of the centre of mass and the
tangential speed of the sphere are the same,
so we can say that:
vt v
w 
R R
1 2 1 2
2
ghi  v f  v f  0.7v f
2
5
Rearranging for vf:
ghi
9.8  2
vf 

 5.3m / s
0.7
0.7
Example: Conservation of KEr
A boy of mass 30 kg is flung off the edge of a roundabout
(m = 400 kg, r = 1 m) that is travelling at 2 rpm. What is
the speed of the roundabout after he falls off?
During a certain period of time, the angular position of a
swinging door is described by
θ= 5.00 + 10.0t + 2.00t2
where θ is in radians and t is in seconds. Determine the
angular position, angular speed, and angular
acceleration of the door (a) at t = 0 and (b) at t = 3.00 s.
The four particles are connected by
rigid rods of negligible mass. The
origin is at the center of the
rectangle. If the system rotates in
the xy plane about the z axis with
an angular speed of 6.00 rad/s,
calculate (a) the moment of inertia
of the system about the z axis and
(b) the rotational kinetic energy of
the system.
Parallel axis theorem
R
Rotational inertia about a given axis =
Rotational Inertia about a parallel axis
that extends trough body’s Center of
Mass + Mh2
I  I COM  Mh
2
h = perpendicular distance between the given axis and axis
through COM.
Proof:

( x  a)  ( y  b) dm 


 ( x  y )dm  2a xdm  2b ydm  (a


 
I  r 2dm 
2
2
2
2
2
 b 2 )dm
I   R 2 dm  2aMxCOM  2bMyCOM  Mh 2  I COM  Mh 2
Many machines employ cams for various purposes, such as
opening and closing valves. In Figure, the cam is a circular
disk rotating on a shaft that does not pass through the center
of the disk. In the manufacture of the cam, a uniform solid
cylinder of radius R is first machined. Then an off-center hole
of radius R/2 is drilled, parallel to the axis of the cylinder, and
centered at a point a distance R/2 from the center of the
cylinder. The cam, of mass M, is then slipped onto the circular
shaft and welded into place. What is the kinetic energy of the
cam when it is rotating with angular speed ω about the axis of
the shaft?
Find the net torque on the wheel in Figure about the axle
through O if a = 10.0 cm and b = 25.0 cm.
A block of mass m1 = 2.00 kg
and a block of mass m2 =
6.00 kg are connected by a
massless string over a pulley
in the shape of a solid disk
having radius R = 0.250m
and mass M = 10.0 kg.
These blocks are allowed to move on a fixed block-wedge
of angle = 30.0 as in Figure. The coefficient of kinetic
friction is 0.360 for both blocks. Draw free-body diagrams
of both blocks and of the pulley. Determine (a) the
acceleration of the two blocks, and (b) the tensions in the
string on both sides of the pulley.
A uniform rod 1.1 m long with mass 0.7 kg is pivoted at
one end, as shown in Fig., and released from a horizontal
position. Find the torque about the pivot exerted by the
force of gravity as a function of the angle that the rod
makes with the horizontal direction.
A seesaw pivots as shown in Fig. (a) What is the net torque
about the pivot point? (b) Give an example for which the
application of three different forces and their points of
application will balance the seesaw. Two of the forces must
point down and the other one up.
Four small spheres are fastened to the corners of a frame of
negligible mass lying in the xy plane (Fig. 10.7). Two of the
spheres have mass m = 3.1kg and are a distance a = 1.7 m
from the origin and the other two have mass M = 1.4 kg and
are a distance a = 1.5 m from the origin.
(a) If the rotation of the system occurs about the y axis, as in
Figure a, with an angular speed ω = 5.1rad/s, find the
moment of inertia Iy about the y axis and the rotational kinetic
energy about this axis.
Suppose the system rotates in the xy plane about an axis
(the z axis) through O (Fig. b). Calculate the moment of inertia
about the z axis and the rotational energy about this axis.
The reel shown in Figure has
radius R and moment of
inertia I. One end of the block
of mass m is connected to a
spring of force constant k, and
the other end is fastened to a
cord wrapped around the reel.
The reel axle and the incline
are frictionless. The reel is
wound counterclockwise so
that the spring stretches a
distance d from its unstretched position and is then released
from rest. (a) Find the angular speed of the reel when the
spring is again unstretched. (b) Evaluate the angular speed
numerically at this point if I = 1.00 kg·m2, R = 0.300 m,
k = 50.0 N/m, m = 0.500 kg, d = 0.200 m, and θ= 37.0°.
A tennis ball is a hollow sphere with a thin wall. It is set
rolling without slipping at 4.03 m/s on a horizontal section of
a track, as shown in Figure. It rolls around the inside of a
vertical circular loop 90.0 cm in diameter, and finally leaves
the track at a point 20.0 cm below the horizontal section.
(a) Find the speed of the ball at the top of the loop.
Demonstrate that it will not fall from the track. (b) Find its
speed as it leaves the track. (c) Suppose that static friction
between ball and track were negligible, so that the ball slid
instead of rolling. Would its
speed then be higher, lower,
or the same at the top of the
loop? Explain.