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Summary Lecture 12
Rotational Motion
10.8
Torque
10.9
Newton 2 for rotation
10.10 Work and Power
11.2
Today
Rolling motion
20-minute test
on material in
lectures 1-7
at end of lecture
Problems:Chap. 10:
21, 28 , 33, 39, 49, 54, 67
Some Rotational Inertia
Parallel-axis Theorem
The rotational inertia of a body about any parallel axis, is
equal to its
R.I. about an axis through its CM,
PLUS
R.I. of its CM about a parallel axis
through the point of rotation
CM
h
Axis of Rotation
I = ICM + Mh2
Proof of Parallel-axis Theorem
One rotation about yellow axis involves
one rotation of CM about this axis
plus one rotation of body about CM.
I = Icm + Mh2
What is it about
here?
Example
R
RI of ring of mass M
about CM is
RI of CM about
suspension point, distance
R away is
MR2
MR2.
So total RI is
2MR2
The Story so far...
, , 
Rotational Variables
relation to linear variables
vector nature
Rotational kinematics
with const. 
Analogue equations to
linear motion
Rotation and
Kinetic Energy
Rotational Inertia
Torque
…is the “turning ability” of a force
Where would
you put the door
knob?
here?
r
r
F
The magnitude of the
torque is Fr,
and this is greater here!
or here?
F
Torque
F
Axis
r
If F and r are perpendicular
 = rF
(Unit: N m)
The same F at larger r has bigger turning effect.
In General
Torque is a vector
F
Axis
r

Fsin
=rxF
 = r F sin
 = r x perp component of F
Direction of :
Perpendicular to r and F
Sense:
Right-hand screw rule (out of screen)
(Hint: Which way would it accelerate the body?
Sense is same as change in ang vel. .)
Newton 2 for Rotational motion
For Translational motion we had:
 F  ma
ext
For Rotational motion we expect:
   I
ext
Example


Apply a force of 3200 N
Distance r = 2 m from axis
Observe ang accel of 1.2 rad s-2
 = r x F = 6400 N m
F
 = I (Newton 2)
r
 I = /
 I = 6400/1.2
What is I for
rotating object?
 I = 5334 kg m2
Work
Work done by a torque 
which rotates a body through
an angle D:


Dw = .D
D
D
R

w  .d
cf
If  is constant w =  
Power
dw τ.dθ
p

dt
dt

w  F.ds
P = .
cf P = F.v
Example
A car engine has a
power of 100 HP
at 1800 RPM
What is the torque provided by the
engine?
1 HP = 746 W
P = .
P = 74600 W
  = P/ 
 = 74600/ 188
1800 RPM
1800 x 2

rad s 1
60
  188 rad s 1
 = 397 N m
Rolling Motion
R
2R
vcm t = 2R
But in turning one revolution (2 radian) in
time t,  = 2/t
So that t = 2/
vcm 2/= 2R
vcm = R
Rolling Motion
Object of
radius R
Cons. Energy says
PEinitial = KEfinal
mgh
h
= Ktrans + Krot
= ½ mvcm2 + ½ I2
so mgh = ½ mv2 + ½ Iv2/R2
vcm= R
 = vcm/R
total energy (KE) = ½ mv2 + ½ v2bmR2/R2
= ½ mv2 + ½ mv2b
The kinetic energy is shared
= ½ mv2(1 + b)
between translational and
rotational motion.
b is the coefficient in the expression
remember
or
for the Rotational Inertia I
I = bmR2
mgh = Ktotal = Ktrans + KRot
= ½ mvcm2 + ½ mvcm2b
h h
= ½ mvcm2(1 + b)
2gh
2
v cm 
(1  β)
The kinetic energy is shared
The larger b, the more of the
between translational and available energy goes into
rotational motion.
rotational energy, and the smaller
the centre of mass velocity
The fraction of KE that
is translational is
2
1 mv
cm
2
2
1 mv
cm (1  b )
2

1
1b

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