Physics 101: Lecture 16
Rotational Kinematics

Today’s lecture will cover Textbook Chapter 8
Physics 101: Lecture 16, Pg 1
Rotational Kinematics

The motion of a rigid body about a fixed axis is
described by using the same concepts as for linear
motion (see C&J Chapter 2):
Displacement (Dq), Velocity (w), Acceleration (a)
Angular Displacement:
Identify the axis of rotation and choose a line
perpendicular to this axis. Observe the motion of a point
on this line. How can one define the change of position of
this point during rotation about an axis ?
Answer:
Change of angle the line makes with a reference line: Dq
Physics 101: Lecture 16, Pg 2
Angular Displacement

Follow point P on line perpendicular to the rotation axis:
t=t0:
reference
P
t=tf:
q0=0
q
f
line
counter clockwise
Angular displacement:
-
clockwise
Dq = qf - q0 = qf
SI unit: radian (rad)
qf
Dq = -qf
1 rad = 360 degrees/(2 p)
Dq (in rad) = arc length / radius = s/r
Physics 101: Lecture 16, Pg 3
Angular Velocity and Acceleration
With the concept of displacement in place we can now
define angular velocity and acceleration to describe the
motion of a rigid body rotating about an axis:

Average angular velocity = angular displacement/elapsed time
wave = Dq / Dt = (qf-q0)/(t-t0)
Average angular acceleration = change in velocity /elapsed time
aave = Dw /Dt = (wf-w0)/(t-t0)
Physics 101: Lecture 16, Pg 4
See text: chapter 8
Rotational Kinematics
(with comparison to 1-D linear kinematics)
Angular
Linear
a  constant
a  constant
ω  ω0  αt
v  v 0  at
1
q  q0  w 0 t  at 2
2
x  x 0  v 0t 
w 2  w 02  2aΔθ
1 2
at
2
v2 = v02 +2 a Dx
And for a point at a distance R from the rotation axis:
s = Rq
vT= wR
aT = aR
See Table 8.1
Physics 101: Lecture 16, Pg 5
Angular and Tangential Variables

A point on a line perpendicular to the rotation axis at distance R
from the rotational axis moves with a tangential speed
(t0=0s,q0=0)
vT = s/t = q R/t = w R (q in rad and w in rad/s)
and an average tangential acceleration
aT = (vT0 – vTf)/t = R (w0-wf)/t = a R
How does this relate to the case of uniform circular motion we
discussed before ?
Uniform motion = constant tangential speed and aT = 0
The change of direction of vT, however, results into a centripetal
acceleration:
ac = vT2/R = w2 R (=constant)
Nonuniform motion = increasing/decreasing tangential speed
Magnitude of the total acceleration is then given by
a = (aT2 + ac2)1/2
Physics 101: Lecture 16, Pg 6
Conceptual Question
You and a friend are playing on the merry-go-round. You stand at the
outer edge of the merry-go-round and your friend stands halfway
between the outer edge and the center. Assume the rotation rate of
the merry-go-round is constant.
Who has the greatest angular velocity?
1. You do
2. Your friend does
3. Same
CORRECT
Since the angular displacement is the same in both
cases.
Physics 101: Lecture 16, Pg 7
you
Conceptual Question
you
Who has the greatest tangential velocity?
1. You do
2. Your friend does
3. Same
CORRECT
This is like the example of the "crack-the-whip."
The person farthest from the pivot has the hardest
job. The skater has to cover more distance than
anyone else. To accomplish this, the skater must
skate faster to keep the line straight.
vT = w R (w is the same but R is larger)
Physics 101: Lecture 16, Pg 8
Concept Question
you
Who has the greatest centripetal acceleration?
1. You do
2. Your friend does
3. Same
CORRECT
Things toward the outer edge want to "fly off" more
than things toward the middle. Force is greater on you
because you want to fly off more.
Centripetal acceleration is ac = R w2 and you have the
largest radius.
Physics 101: Lecture 16, Pg 9