


{Fnet = ∑ Fi = ma}
€

τ net =
∑
→
→
→ →
 → → →

τ i = r × F = r × m a = m r × (r × α ) = Iα
€
Work and
Rotational Kinetic Energy
NET Work done
ON system
Rotational work,
fixed axis rotation
€
(if torque is const)
Power,
€
fixed axis rotation
€
W = ΔKE trans



2
2
1
 = 2 m (v f − vi )
W = ΔKE rot
= 12 I (ω 2f − ω i2 )
θ2
W = ∫ τ net dθ
€
θ1
= τ (θ f − θ i )
dW d
P=
= (τθ ) = τω
dt dt
€
€

W =

x2
∫ Fdx
x1

1 − D motion

 

dW d  
= ( F • x ) = F • v
P =


dt dt
We can compare linear variables with rotational variables
x
v
a
Δt
F
m
θ
ω
α
Δt
τ
I
s = rθ
v T = rω
aT = rα
The same can be done for work and energy:
€
€
€
For translational
systems For rotational systems
W = τ ⋅θ
1
KE = Iω 2
2
W =F⋅x
1
KE = mv 2
2
€
€
All points on wheel move
with same ω. All points on
outer rim move with same
linear speed v = vcom.
  
v =ω ×r
Note at point P:
at point T:
€
All points on wheel
move to the right with same
linear velocity vcom as center
of wheel
Combination of “pure
rotation” and “pure
translation”
vector sum of velocity = 0
(point of stationary contact)
vector sum of velocity = 2vcom (top moves twice as fast as com)
1
2
2
1
2
I comω +
Mv
2
com
= KE tot
Note: rotation about COM and translation of COM combine for total KE
€
Remember: vcom=ωr
€
Example: Rolling down a ramp with no fric6on
Object, with mass m and radius r, roles
from top of incline plane to bottom.
What is v, a, and Δt at bottom
ΔE mech = 0
ΔKE tot = −ΔU
AT BOTTOM
(KE
rot +trans,COM
1
2
)
2
mvCOM
€
h
[
L sin θ = h
θ
− (0) init = − (0) final − ( mgh ) init
€
+ 12 I comω 2 = mgL sin θ
final
]
Using 1-D kinematics
v 2 = v02 + 2aL
v2
g sin θ
a=
=
I 
2L 
1 +

2
 mr 
AND
2
1
2
 v
2
mvCOM
+ 12 I COM   = mgL sin θ
 r
I COM 

2
vCOM
m
+
= 2mgL sin θ

2 

r
vCOM =
2gL sin θ

I 
1 +

 mr 2 
2L 
I 
t=
1 +

2
g sin θ  mr 
AT BOTTOM
€
Compare Different Objects
Assuming same work done (same change in U),
objects with larger rotational inertial have larger
KErot and during rolling, their KEtrans is smaller.
KE tot = KE trans + KE rot
vCOM =

I com 
= KE trans 1 +

 mr 2 
2gL sin θ

I com 
1 +

 mr 2 
Roll a hoop, disk, and solid sphere down a ramp - what wins?
€
Rotational
Fraction of Energy in
Object
Inertia, Icom
Translation
Rotation
€
Moment of inertia
large → small
Hoop
1mr 2
0.5
0.5
Disk
1
2
mr 2
0.67
0.33
Sphere
2
5
mr 2
0.71
0.29
sliding block
(no friction)
€
0
€
1
0
slowest
Δt bottom =
fastest
2L 
I 
1 +

g sin θ  mr 2 
Question
A ring and a solid disc, both with radius r and mass m, are
released from rest at the top of a ramp. Which one gets to
the bottom first?
1. Solid disc
2. Ring (hoop)
3. both reach bo9om at
same ;me
Question #2
Two solid disks of equal mass, but different radii, are
released from rest at the top of a ramp. Which one arrives
at the bottom first?
1. The smaller radius disk.
2. The larger radius disk.
3. Both arrive at the same time.
Using 1-D kinematics
v 2 = v02 + 2aL
v2
g sin θ
a=
=
I 
2L 
1 +

2
 mr 
AND
The equation for the speed of the a disk at the bottom of
4
the ramp is
gl sin θ
3
Notice, it does not depend on the radius or the mass of
the disk!!
2L 
I 
t=
1 +

2
g sin θ  mr 
€
Rolling
Down
Ramp with a Frictional Force
Rolling
Down aa Ramp
Consider a round uniform body of mass M and radius R
rolling down an inclined plane of angle θ . We will
acom
calculate the acceleration acom of the center of mass
along the x-axis using Newton's second law for the
translational and rotational motion.
Newton's second law for motion along the x-axis: f s − Mg sin θ = Macom (eq. 1)
Newton's second law for rotation about the center of mass: τ = Rf s = I comα
α=−
acom
R
f s = − I com
− I com
acom
R
2
We substitute α in the second equation and get Rf s = − I com
acom
R
2
acom
R
→
(eq. 2). We substitute f s from equation 2 into equation 1 →
− Mg sin θ = Macom
acom = −
g sin θ
I com
1+
MR 2
g sin θ
| acom |=
I
1 + com2
MR
acom
Cylinder
MR 2
I1 =
2
g sin θ
a1 =
1 + I1 / MR 2
g sin θ
a1 =
1 + MR 2 / 2 MR 2
g sin θ
a1 =
1 + 1/ 2
2 g sin θ
a1 =
= (0.67) g sin θ
3
Hoop
I 2 = MR 2
a2 =
g sin θ
1 + I 2 / MR 2
g sin θ
a2 =
1 + MR 2 / MR 2
g sin θ
a2 =
1+1
g sin θ
a2 =
= (0.5) g sin θ
2
Sample Problem
A solid cylinder starts from rest at the upper end of the
track as shown. What is the angular speed of the
cylinder about its center when it is at the top of the
loop?
Sample Problem #2
A solid cylinder of radius 10 cm and mass 12 kg starts
from rest and rolls without slipping a distance of 6 m
down a house roof that is inclined at 30º.
Where does it hit?
A
B
Forces of Rolling
1) If object is rolling with acom=0 (i.e. no net forces), then vcom=ωR = constant (smooth roll)
…if constant speed, it has no tendency to slide at point of contact - no frictional forces
acom = 0
α= 0
acom = αR
2)
τ =0
Icomα = τnet
f=0
τnet = |R||f|sinθ
If object is rolling with acom≠ 0 (i.e. there are net forces) and no slipping occurs,
then α ≠ 0 ⇒ τ ≠ 0
… static friction needed to supply torque !
kinetic friction - during sliding (acom ≠ αR)
static friction -
smooth rolling (acom = αR)
Rolling Down a Ramp: Accelera2on by fs
- Acceleration
downwards
Yo-Yo
- Friction provides
torque
- Static friction points
upwards
Newton’s 2nd Law
Linear version
xˆ : f s − Fg sin θ = −ma com
yˆ : N − Fg cos θ = 0
- Tension provides
torque
- Here θ = 90°
- Axle: R ⇒ R0
Angular version
zˆ : I com α = τ = Rf s
a com
→ a com = αR
α=
€
€
R 2 ( mg sin θ − ma com )
=
I com
a com Rf s
=
R
I com
€
a com =
€
€
a com =
g sin θ
1 + I com mR 2
€
g
1 + I com mR02
Problem
A yo-yo has a rotational inertia of Icom and mass of m. Its axle
radius is R0 and string’s length is h. The yo-yo is thrown so
that its initial speed down the string is v0.
a) How long does it take to reach the end of the string?
1-D kinematics given acom
−h = Δy = −v0 t − 12 a com t 2 ⇒ solve for t (quadradic equation)
b)
As it reaches the end of the string, what is its total KE?
Conservation of mechanical energy
2
 vcom,0 
2
1
1
KE f = KE i + U = 2 mvcom,0 + 2 I com 
 + mgh
€
 R0 
c) As it reaches the end of the string, what is its linear speed?
1-D kinematics given acom
− vcom = −v0 − a com t ⇒ solve for vcom
€
d)
As it reaches the end of the string, what is its translational KE?
Knowing |vcom|
1 2
KE trans = mvcom
2
€
a com =
g
1 + I com mR02
downwards
e)
As it reaches the end of the string, what is its angular speed?
Knowing |vcom|
v
ω=
€
f)
com
R
As it reaches the end of the string, what is its rotational KE?
1
2
=
I
ω
or KE rot = K Ef ,tot − KE trans
com
€ 2
Two ways: KE
rot
Which way will it roll??
1
1
3
Problem 11‐13
NON-smooth rolling motion
A bowler throws a bowling ball of radius R along a lane. The ball slides on the lane, with initial speed
vcom,0 and initial angular speed ω0 = 0. The coefficient of kinetic friction between the ball and the
lane is µk. The kinetic frictional force fk acting on the ball while producing a torque that causes an
angular acceleration of the ball. When the speed vcom has decreased enough and the angular speed
ω has increased enough, the ball stops sliding and then rolls smoothly.
a) [After it stops sliding] What is the vcom in terms of ω ?
Smooth rolling means vcom =Rω
b) During the sliding, what is the ball’s linear acceleration?
fk = µ k N
a com = − f k m
From 2nd law: xˆ : − f k = ma com
But
So
= µ k mg
= −µ k g
(linear)
yˆ : N − mg = 0
c)
d)
During the sliding, what is the ball’s angular acceleration?

fk = µ k N
From 2€nd law: τ = Rf k (−zˆ)
So€
€But

Iα (− zˆ ) = τ = Rf k (−zˆ)
= µ k mg
(angular)
α = Rµ k mg I
What is the speed of the ball when smooth rolling begins?
vcom€= v0 + a com t
vcom = v€0 − µ k gt
When does
€ vcom =Rω ?
From kinematics: ω = ω 0 + αt
t = ω = Iω
α
e)
Iα = Rf k = R(µ k mg)
vcom
Rµ k mg
How long does the ball slide?
€
t=
v0 − vcom
µkg
vcom
€
€
 vcom  

 I
R 
= v0 − µ k g

R
µ
mg
k




v0
=
(1 + I mR 2 )