Physics 207: Lecture 24
Announcements
• No labs next week, May 2 – 5
• Exam 3 review session: Wed, May 4 from 8:00 – 9:30 pm; here
Today’s Agenda
z
z
z
Recap:
ÍRotational dynamics and torque
Work and energy with example
Many body dynamics examples
ÍWeight and massive pulley
ÍRolling and sliding examples
ÍRotation around a moving axis: Puck on ice
ÍRolling down an incline
ÍBowling ball: sliding to rolling
ÍAtwood’s Machine with a massive pulley
1
Review: Torque and Angular Acceleration
τNET = Iα
z
z
z
This is the rotational analogue
of FNET = ma
Torque is the rotational analogue of force:
ÍThe amount of “twist” provided by a force.
Moment of inertia I is the rotational analogue of mass
ÍIf I is big, more torque is required to achieve a given angular
acceleration.
2
Page 1
Torque
z
Recall the definition of torque:
τ = rFθ
= r F sin φ
Fr φ F
= r sin φ F
z
φ
τ = rpF
Fθ
rp = “distance of closest approach”
r
φ
Equivalent definitions!
rp
3
Torque
τ = r Fsin φ
F
z
So if φ = 0o, then τ = 0
z
And if φ = 90o, then τ = maximum
r
F
r
4
Page 2
Lecture 23, Act 3
Torque
z
In which of the cases shown below is the torque provided by the
applied force about the rotation axis biggest? In both cases the
magnitude and direction of the applied force is the same.
(a) case 1
L
(b) case 2
(c) same
F
F
L
axis
case 1
case 2
5
Lecture 23, Act 3
Solution
z
Torque = F x (distance of closest approach)
ÍThe applied force is the same.
ÍThe distance of closest approach is the same.
Torque is the same!
F
F
L
L
case 1
case 2
6
Page 3
Torque and the
Right Hand Rule:
z
The right hand rule can tell you the direction of torque:
ÍPoint your hand along the direction from the axis to the
point where the force is applied.
ÍCurl your fingers in the direction of the force.
ÍYour thumb will point in the direction
of the torque.
F
y
r
x
τ
z
7
The Cross Product
z
We can describe the vectorial nature of torque in a compact
form by introducing the “cross product”.
ÍThe cross product of two vectors is a third vector:
B
AXB=C
z
z
φ
The length of C is given by:
C = AB sin φ
The direction of C is perpendicular to
the plane defined by A and B, and in
the direction defined by the right hand
rule.
A
C
8
Page 4
The Cross Product
z
Cartesian components of the cross product:
C=AXB
B
CX = AY BZ - BY AZ
CY = AZ BX - BZ AX
A
C
CZ = AX BY - BX AY
Note: B X A = - A X B
9
Torque & the Cross Product:
z
So we can define torque as:
τ =rXF
= rF sin φ
τX = rY FZ - FY rZ = y FZ - FY z
τY = rZ FX - FZ rX = z FX - FZ x
τZ = rX FY - FX rY = x FY - FX y
F
τ
r
y
z
x
10
Page 5
Comment on τ = Iα
z
When we write τ = Iα we are really talking about the z
component of a more general vector equation. (Recall that
we normally choose the z-axis to be the the rotation axis.)
τz = Izαz
τz
Iz
z
z
We usually omit the
z subscript for simplicity.
αz
11
Example
z
To loosen a stuck nut, a (stupid) man pulls at an angle of
45o on the end of a 50 cm wrench with a force of 200 N.
ÍWhat is the magnitude of the torque on the nut?
ÍIf the nut suddenly turns freely, what is the angular
acceleration of the wrench? (The wrench
has a mass of 3 kg, and its shape
45o
is that of a thin rod).
F = 200 N
L = 0.5 m
12
Page 6
Example
z
Torque τ = LFsin φ = (0.5 m)(200 N)(sin 45)
z
If the nut turns freely, τ = Iα
Í We know τ and we want α, so we need to figure out I.
1
1
I = ML2 = (3 kg)(0.5 m)2 = 0.25 kgm2
3
3
= 70.7 Nm
45o
F = 200 N
L = 0.5m
So α = τ / I = (70.7 Nm) / (0.25 kgm2)
α
α = 283 rad/s2
13
Work
z
Consider the work done by a force F acting on an object
constrained to move around a fixed axis. For an
infinitesimal angular displacement dθ:
β
.
ÍdW = F dr = FR dθ cos(β)
= FR dθ cos(90-φ)
= FR dθ sin(φ)
= FR sin(φ) dθ
dW = τ dθ
z
z
z
F
φ
R
dθ
dr = R dθ
axis
We can integrate this to find: W = τθ
Analogue of W = F •∆r
W will be negative if τ and θ have opposite signs!
14
Page 7
Work & Power
z
The work done by a torque τ acting through a displacement θ is
given by:
W = τθ
z
The power provided by a constant torque is therefore given by:
P=
dθ
dW
=τ
= τω
dt
dt
15
Work & Kinetic Energy:
∆K = WNET
z
Recall the Work/Kinetic Energy Theorem:
z
This is true in general, and hence applies to rotational
motion as well as linear motion.
z
So for an object that rotates about a fixed axis:
∆K =
(
)
1 2
I ω − ω i2 = WNET
2 f
16
Page 8
Example: Disk & String
z
A massless string is wrapped 10 times around a disk of
mass M = 40 g and radius R = 10 cm. The disk is
constrained to rotate without friction about a fixed axis
though its center. The string is pulled with a force F = 10 N
until it has unwound. (Assume the string does not slip, and
that the disk is initially not spinning).
ÍHow fast is the disk spinning after the string has
unwound?
R
M
F
17
Disk & String...
z
The work done is W = τ θ
ÍThe torque is τ = RF (since φ = 90o)
ÍThe angular displacement θ is
2π rad/rev x 10 rev.
R
z
So W = (.1 m)(10 N)(20π rad) = 62.8 J
1442443 1
424
3
τ
θ
M
F
18
Page 9
Disk & String...
WNET = W = 62.8 J = ∆K =
1 2
Iω
2
Recall that I for a disk about
its central axis is given by:
1
MR 2
2
1 1
∆K = ⎛⎜ MR 2 ⎞⎟ ω 2 = W
⎠
2 ⎝2
R
I=
So
ω=
4W
=
MR 2
M
ω
4 (62 .8 J )
ω = 792.5 rad/s
(.04 kg )(.1)2
19
Lecture 23, Act 4
Work & Energy
z
Strings are wrapped around the circumference of two solid disks
and pulled with identical forces for the same distance.
Disk 1 has a bigger radius, but both have the same moment of
inertia. Both disks rotate freely around axes though their
centers, and start at rest.
ÍWhich disk has the biggest angular velocity after the pull ?
ω2
ω1
(a) disk 1
(b) disk 2
(c) same
F
F
20
Page 10
Lecture 23, Act 4
Solution
z
The work done on both disks is the same!
ÍW = Fd
z
The change in kinetic energy of each will
therefore also be the same since W = ∆K.
But we know ∆K =
1
I ω2
2
ω2
ω1
So since I1 = I2
ω1 = ω2
F
F
d
21
Lecture 24, Act 1
Rotations
z
z
Two wheels can rotate freely about fixed axles through their centers.
The wheels have the same mass, but one has twice the radius of the
other.
Forces F1 and F2 are applied as shown. What is F2 / F1 if the angular
acceleration of the wheels is the same?
(a) 1
F2
(b) 2
(c) 4
F1
22
Page 11
Lecture 24, Act 1
Solution
We know
but
τ= Iα
τ = FR
so FR = mR 2α
F = mR α
Since R2 = 2 R1
I = mR 2
and
F2 mR 2α R2
=
=
F1 mR1α R1
F2
=2
F1
F2
F1
23
Falling weight & pulley
z
A mass m is hung by a string that is
wrapped around a pulley of radius R
attached to a heavy flywheel. The moment
of inertia of the pulley + flywheel is I. The
string does not slip on the pulley.
α
I
R
T
ÍStarting at rest, how long does it take
for the mass to fall a distance L.
m
a
mg
L
24
Page 12
Falling weight & pulley...
z
z
For the hanging mass use F = ma
Ímg - T = ma
For the pulley + flywheel use τ = Iα
Íτ = TR = Iα
TR = I
z
Realize that a = αR
z
Now solve for a using the above
equations.
α
I
R
a
R
T
m
mg
a
L
⎛ mR 2 ⎞
a=⎜
⎟g
⎝ mR 2 + I ⎠
25
Falling weight & pulley...
z
Using 1-D kinematics (Lecture 2) we
can solve for the time required for the
weight to fall a distance L:
α
I
R
L=
1 2
at
2
where
t=
2L
a
⎛ mR 2 ⎞
a=⎜
⎟g
⎝ mR 2 + I ⎠
T
m
a
mg
L
26
Page 13
Rotation around a moving axis.
z
A string is wound around a puck (disk) of mass M and radius R.
The puck is initially lying at rest on a frictionless horizontal
surface. The string is pulled with a force F and does not slip as it
unwinds.
ÍWhat length of string L has unwound after the puck has moved
a distance D?
M
R
F
Top view
27
Rotation around a moving axis...
A=
z
The CM moves according to F = MA
z
The distance moved by the CM is thus D =
z
z
F
M
1 2
F 2
t
At =
2
2M
2F
τ
RF
=
=
MR
I 1
2
MR
2
The disk will rotate about
its CM according to τ = Iα
α=
So the angular displacement is θ =
F 2
1 2
αt =
t
2
MR
1
I = MR 2
2
α
M
A
R
F
Page 14
28
Rotation around a moving axis...
z
So we know both the distance moved by the CM and the angle of
rotation about the CM as a function of time:
D=
F 2
t
2M
θ=
(a)
θ 2
=
D R
Divide (b) by (a):
F 2
t
MR
(b)
The length of string
pulled out is L = Rθ:
Rθ = 2 D
L = 2D
θ
F
F
D
L
29
Comments on CM acceleration:
z
We just used τ = Iα for rotation about an axis through the CM even
though the CM was accelerating!
ÍThe CM is not an inertial reference frame! Is this OK??
(After all, we can only use F = ma in an inertial reference frame).
z
YES! We can always write τ = Iα for an axis through the CM.
ÍThis is true even if the CM is accelerating.
ÍWe will prove this when we discuss angular momentum!
α
M
A
R
F
Page 15
30
Rolling
z
An object with mass M, radius R, and moment of inertia I
rolls without slipping down a plane inclined at an angle θ
with respect to horizontal. What is its acceleration?
z
Consider CM motion and rotation about
the CM separately when solving this
problem (like we did with the last
problem)...
I
M
R
θ
31
Rolling...
z
z
z
Static friction f causes rolling. It is an unknown, so we must solve
for it.
First consider the free body diagram of the object and use FNET =
MACM :
In the x direction Mg sin θ - f = MA
Now consider rotation about the CM
and use τ = Iα realizing that
τ = Rf and A = αR
M
y
R
x
A
Rf = I
R
f =I
θ
A
f
Mg
R2
32
Page 16
Rolling...
Mg sin θ - f = mA
z
We have two equations:
z
We can combine these to eliminate f:
A=g
f =I
A
R2
MR 2 sin θ
MR 2 + I
I
A
For a sphere:
A=g
MR 2 sin θ
2
MR 2 + MR 2
5
=
5
g sin θ
7
M
R
θ
33
Lecture 24, Act 2
Rotations
z
Two uniform cylinders are machined out of solid aluminum. One has
twice the radius of the other.
ÍIf both are placed at the top of the same ramp and released, which
is moving faster at the bottom?
(a) bigger one
(b) smaller one
(c) same
34
Page 17
Lecture 24, Act 2
Solution
z
Consider one of them. Say it has radius R, mass M and falls a
height H.
Energy conservation: - ∆U = ∆K
but
I=
1
MR 2
2
and
MgH =
ω=
1
1
I ω 2 + MV 2
2
2
V
R
MgH =
2
1 ⎛1
1
2
2⎞ V
⎜ MR ⎟ 2 + MV
⎠R
2 ⎝2
2
MgH =
1
1
3
MV 2 + MV 2 = MV 2
4
2
4
H
35
Lecture 24, Act 2
Solution
So:
MgH =
3
MV 2
4
gH =
V =
3 2
V
4
4
gH
3
So, (c) does not depend on size,
as long as the shape is the same!!
H
36
Page 18
Sliding to Rolling
z
A bowling ball of mass M and radius R is thrown with initial
velocity v0. It is initially not rotating. After sliding with
kinetic friction along the lane for a distance D it finally rolls
without slipping and has a new velocity vf. The coefficient of
kinetic friction between the ball and the lane is µ.
ÍWhat is the final velocity, vf, of the ball?
vf= ωR
ω
v0
f = µMg
D
37
Sliding to Rolling...
z
z
z
While sliding, the force of friction will accelerate the ball in
the -x direction: F = -µMg = Ma so a = -µg
The speed of the ball is therefore v = v0 - µgt (a)
Friction also provides a torque about the CM of the ball.
Using τ = Iα and remembering that I = 2/5MR2 for a solid
sphere about an axis through its CM:
τ = µMgR =
x
2
MR 2 α
5
v f= ωR
α=
5µg
2R
ω = ω0 + αt =
5 µg
t (b)
2R
ω
v0
D
Page 19
f = µMg
38
Sliding to Rolling...
v = v 0 − µgt (a)
ω=
5µg
t (b)
2R
z
We have two equations:
z
Using (b) we can solve for t as a function of ω:
z
Plugging this into (a) and using vf = ωR (the condition for
rolling without slipping):
5
Doesn’t depend
vf = v0
7
on µ, M, g!!
2 Rω
5 µg
ω
vf= ωR
x
t=
v0
f = µMg
D
39
Lecture 24, Act 3
Rotations
z
A bowling ball (uniform solid sphere) rolls along the floor without
slipping.
ÍWhat is the ratio of its rotational kinetic energy to its translational
kinetic energy?
(a)
1
5
(b)
2
5
(c)
1
2
Recall that I = 2 MR 2 for a solid sphere about
5
an axis through its CM:
40
Page 20
Lecture 24, Act 3
Solution
z
The total kinetic energy is partly due to rotation and partly due to
translation (CM motion).
K=
1
1
I ω 2 + MV 2
2
2
rotational
translational
K
K
41
Lecture 24, Act 3
Solution
K=
1
1
I ω 2 + MV 2
2
2
Since it rolls without slipping: ω =
rotational
Translational
K
K
K ROT
K TRANS
V
R
2
1 2
⎛ 2 MR 2 ⎞ V
Iω
⎟ 2
⎜
5
⎠R = 2
= 2
=⎝
2
1
2
5
MV
MV
2
42
Page 21
Atwoods Machine with Massive Pulley:
z
z
z
y
A pair of masses are hung over a
massive disk-shaped pulley as shown.
ÍFind the acceleration of the blocks.
For the hanging masses use F = ma
Í -m1g + T1 = -m1a
Í -m2g + T2 = m2a
For the pulley use τ = Iα = I
Í T1R - T2R= I
x
α
R
T2
T1
a
R
a 1
= MRa
R 2
1
(Since I = MR 2 for a disk)
2
M
a
m2
m1
a
m2g
m1g
43
Atwoods Machine with Massive Pulley...
z
y
We have three equations and three
unknowns (T1, T2, a). Solve for a.
x
-m1g + T1 = -m1a
(1)
-m2g + T2 = m2a
(2)
T1 - T2 =
1
Ma
2
(3)
⎛
⎞
m1 − m 2
a=⎜
⎟g
⎝ m1 + m 2 + M 2 ⎠
M
α
R
T2
T1
a
m2
m1
a
m1g
m2g
44
Page 22