PHYS 1443 – Section 001
Lecture #16
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Wednesday, July 6, 2011
Dr. Jaehoon Yu
Calculation of Moment of Inertia
Torque and Angular Acceleration
Rolling Motion & Rotational Kinetic Energy
Work, Power and Energy in Rotation
Angular Momentum & Its Conservation
Equilibrium
The final homework is homework #9, due 10pm, Saturday, July 9!!
Wednesday, July 6, 2011
PHYS 1443-001, Summer 2011 Dr.
Jaehoon Yu
1
• Quiz #3 results
Announcements
– Class average: 20.1/35
• Equivalent to 59.7/100
• Extremely consistent: 62.5 and 61.4
– Top score: 33/35
• Quiz #4 tomorrow, Thursday, July 7
– Beginning of the class
– Covers CH10.1 through what we learn today (CH12.2?)
• Please to not forget the planetarium special credit sheet
submission tomorrow
• Final Comprehensive Exam
– 8 – 10am, Monday, July 11 in SH103
– Covers CH1.1 through what we learn Thursday, July 7
– Mixture of multiple choice and free response problems
• Bring your two special projects during the intermission
Wednesday, July 6, 2011
PHYS 1443-001, Summer 2011 Dr.
Jaehoon Yu
2
Moment of Inertia
Rotational Inertia:
For a group
of particles
Measure of resistance of an object to
changes in its rotational motion.
Equivalent to mass in linear motion.
I   mi ri2
i
What are the dimension and
unit of Moment of Inertia?
For a rigid
body
I   r 2 dm
 ML  kg  m
2
2
Determining Moment of Inertia is extremely important for
computing equilibrium of a rigid body, such as a building.
Dependent on the axis of rotation!!!
Wednesday, July 6, 2011
PHYS 1443-001, Summer 2011 Dr.
Jaehoon Yu
3
Calculation of Moments of Inertia
Moments of inertia for large objects can be computed, if we assume
the object consists of small volume elements with mass, mi.
2
2
I

lim
r
m

r

i
i
The moment of inertia for the large rigid object is
 dm
m 0
i
i
It is sometimes easier to compute moments of inertia in terms
How can we do this?
of volume of the objects rather than their mass
Using the volume density,  , replace   dm
dm  dV The moments of
2
I


r
dV
 dV
dm in the above equation with dV.
inertia becomes
Example: Find the moment of inertia of a uniform hoop of mass M and radius R
about an axis perpendicular to the plane of the hoop and passing through its center.
y
O
The moment
of inertia is
dm
R
x
Wednesday, July 6, 2011
What do you notice
from this result?
I   r 2 dm  R2  dm  MR 2
The moment of inertia for this
object is the same as that of a
point of mass M at the distance R.
PHYS 1443-001, Summer 2011 Dr.
Jaehoon Yu
4
Ex.10 – 11 Rigid Body Moment of Inertia
Calculate the moment of inertia of a uniform rigid rod of length L and mass M about an
axis perpendicular to the rod and passing through its center of mass.
M


The line density of the rod is
y
L
M
so the masslet is
dm  dx  dx
L
dx
x
x
L
The moment
of inertia is
M  L 
 L

     
 2
3L  2 
3
What is the moment of inertia
when the rotational axis is at
one end of the rod.
Will this be the same as the above.
Why or why not?
Wednesday, July 6, 2011
L /2
M 1 3
x2 M

x 
I   r dm  
dx

 L /2
L  3  L / 2
L
L /2
2
I
3
 M  L3  ML2

 4  
3L
12

L
M 1 3
x2 M
2

x 
dx
 r dm  0

L  3 0
L
2
M
M 3
ML
3
 L   0  

L 


3L
3L
3

L
 
Since the moment of inertia is resistance to motion, it makes perfect sense
for it to be harder to move when it is rotating about the axis at one end.
PHYS 1443-001, Summer 2011 Dr.
Jaehoon Yu
5
Check out
Figure 10 – 20
for moment of
inertia for
various shaped
objects
Wednesday, July 6, 2011
PHYS 1443-001, Summer 2011 Dr.
Jaehoon Yu
6
Torque & Angular Acceleration
Ft
r F
r
Let’s consider a point object with mass m rotating on a circle.
What forces do you see in this motion?
m
The tangential force Ft and the radial force Fr
Ft  mat  mr
The torque due to tangential force Ft is   F r  ma r  mr 2  I
t
t
The tangential force Ft is
What do you see from the above relationship?
What does this mean?
  I
Torque acting on a particle is proportional to the angular acceleration.
What law do you see from this relationship?
Analogs to Newton’s 2nd law of motion in rotation.
How about a rigid object?
The external tangential force δFt is d Ft  d mat  d mr
dFt
2
r
dm 
d
F
r

d

The
torque
due
to
tangential
force
F
is
t
t
δm
The total torque is   lim  d   d   lim  r 2d m    r 2 dm  I
d 0
d m0
r
Contribution from radial force is 0, because its
What is the contribution due
line of action passes through the pivoting
O
to
radial
force
and
why?
Wednesday, July 6, 2011
PHYS 1443-001, Summer 2011 point,
Dr. making the moment arm 0.
7

Jaehoon Yu

Rolling Motion of a Rigid Body
What is a rolling motion?
A more generalized case of a motion where the
rotational axis moves together with an object
A rotational motion about a moving axis
To simplify the discussion, let’s
make a few assumptions
1.
2.
Limit our discussion on very symmetric
objects, such as cylinders, spheres, etc
The object rolls on a flat surface
Let’s consider a cylinder rolling on a flat surface, without slipping.
Under what condition does this “Pure Rolling” happen?
The total linear distance the CM of the cylinder moved is sR
R  s
s=Rθ
Wednesday, July 6, 2011
Thus the linear
speed of the CM is
ds d
vCM  R R

dt
dt
The condition for a “Pure Rolling motion”
PHYS 1443-001, Summer 2011 Dr.
Jaehoon Yu
8
More Rolling Motion of a Rigid Body
The magnitude of the linear acceleration of the CM is
P’
CM
dvCM
d
aCM 
R
R

dt
dt
As we learned in rotational motion, all points in a rigid body
2vCM moves at the same angular speed but at different linear speeds.
vCM
CM is moving at the same speed at all times.
At any given time, the point that comes to P has 0 linear
speed while the point at P’ has twice the speed of CM
P
Why??
A rolling motion can be interpreted as the sum of Translation and Rotation
P’
CM
P
vCM
P’
vCM
CM
v=0
vCM
Wednesday, July 6, 2011
+
v=Rω
v=Rω
2vCM
P’
=
P
PHYS 1443-001, Summer 2011 Dr.
Jaehoon Yu
CM
vCM
P
9
Rotational Kinetic Energy
y
vi
mi
ri

O
x
What do you think the kinetic energy of a rigid object
that is undergoing a circular motion is?
1
1
Kinetic energy of a masslet, mi, K i  mi vi2  mi ri 2 
2
2
moving at a tangential speed, vi, is
Since a rigid body is a collection of masslets, the total kinetic energy of the
rigid object is
1
1
2 
2 
K R   Ki   mi ri     mi ri 
2 i
2 i

i
Since moment of Inertia, I, is defined as
I   mi ri2
i
The above expression is simplified as
Wednesday, July 6, 2011
1 
K R  I
2
PHYS 1443-001, Summer 2011 Dr.
Jaehoon Yu
10
Kinetic Energy of a Rolling Sphere
R

h
θ
vCM
Since vCM=Rω
What is the speed of the
CM in terms of known
quantities and how do you
find this out?
Wednesday, July 6, 2011
Let’s consider a sphere with radius R
rolling down the hill without slipping.
1 2
1
2
KE  KERotation  KELinear  2 ICM 2MvCM
2
1 2
1 v
C
M
 ICM   MvCM
2  R 2
1
I
CM 2
 2 
M
v

CM
2
R


Since the kinetic energy at the bottom of the hill must
be equal to the potential energy at the top of the hill
1
I
CM 2
Mgh

M
v
 2

K
CM
2
R

2
gh
v

CM
2
1

I
/
MR
CM
PHYS 1443-001, Summer 2011 Dr.
11
Jaehoon Yu
Ex. 10 – 16: Rolling Kinetic Energy
For solid sphere as shown in the figure, calculate the linear speed of the CM at the
bottom of the hill and the magnitude of linear acceleration of the CM. Solve this problem
using Newton’s second law, the dynamic method.
What are the forces involved in this motion?
Gravitational Force, Frictional Force, Normal Force
Newton’s second law applied to the CM gives
n
f
M
h
Mg
Mg
sin


fMa
F 
n

Mg
cos

0
F 
θ
x
CM
y
Since the forces Mg and n go through the CM, their moment arm is 0


CM
and do not contribute to torque, while the static friction f causes torque
We know that
2 2
ICM
 MR
5
a

R

CM
We
obtain
2 2
MR
ICM
aCM 2Ma
f 
5
CM
R  R  R 5


Substituting f in
dynamic equations
Wednesday, July 6, 2011
fR ICM

5
7
a

g
sin

Mg
sin
Ma
CM CM
7
5
PHYS 1443-001, Summer 2011 Dr.
Jaehoon Yu
12
Work, Power, and Energy in Rotation

Δs
Δθ
r
O
Let’s consider the motion of a rigid body with a single external
force F exerting on the point P, moving the object by Δs.
The work done by the force F as the object rotates through
the infinitesimalurdistance
r Δs=rΔθ is
W  F  s  F sin  r
What is Fsinϕ?
What is the work done by
radial component Fcosϕ?
Since the magnitude of torque is rFsinϕ,
The rate of work, or power, becomes
The rotational work done by an external force
equals the change in rotational Kinetic energy.
The work put in by the external force then
The tangential component of the force F.
Zero, because it is perpendicular to the
displacement.
W  rF sin    
P
How was the power
W  
  defined in linear motion?


t
t
  


I

I




 t 
f
1
2
1
2
W   I d  I 2f  I i2
i
Wednesday, July 6, 2011
  I
PHYS 1443-001, Summer 2011
Dr.
Work-kinetic
Jaehoon Yu
Energy Theorem
13
Angular Momentum of a Particle
If you grab onto a pole while running, your body will rotate about the pole, gaining
angular momentum. We’ve used the linear momentum to solve physical problems
with linear motions, the angular momentum will do the same for rotational motions.
z
y
Let’s consider a point-like object ( particle) with mass m located
at the vector location r and moving with linear velocity v
The angular momentum L of this
particle relative to the origin O is
What is the unit and dimension of angular momentum?
x
Note that L depends on origin O. Why?
2
k
gms
 2/ [M
L
T1]
Because r changes
What else do you learn? The direction of L is +z.

mvr
sin
Since p is mv, the magnitude of L becomes L
 mr 2 sin  v r  I

What do you learn from this?
Wednesday, July 6, 2011
If the direction of linear velocity points to the origin of
rotation, the particle does not have any angular momentum.
If the linear velocity is perpendicular to position vector, the
particle moves exactly the same way as a point on a14rim.
PHYS 1443-001, Summer 2011 Dr.
Jaehoon Yu
Example for Rigid Body Angular Momentum
A rigid rod of mass M and length l is pivoted without friction at its center. Two particles of mass
m1 and m2 are attached to either end of the rod. The combination rotates on a vertical plane with
an angular speed of ω. Find an expression for the magnitude of the angular momentum.
y
m2
l

m2 g
O
m1
m1 g
x
The moment of inertia of this system is
1 2 1 2 1
2

I

I

I

M
l

ml

m
l
I
rod
m
m
1
2
1
2
12
4
4
2
l
1
l2
1


L

I

M

m

m
 M

m

m



1
2
1
2
4
3


4
3




Find an expression for the magnitude of the angular acceleration of the
system when the rod makes an angle θ with the horizon.
If m1 = m2, no angular
momentum because the net
torque is 0.
If θ//, at equilibrium
so no angular momentum.
Wednesday, July 6, 2011
First compute the
net external torque
l
l


m
g
cos




m
g
cos


1
2
2
2


gl
cos

m

m






ext 2
2
1
2
2
1
lcos
m
2
m
c
o
sg
1m
2g
m
1
2
2


Thus α

ext
 2

l
l 1
 1M
m
m
I
M

m

m

1
2

1
2
becomes
PHYS 1443-001, Summer 2011 4
Dr.3
15 
 3
Jaehoon Yu
Conservation of Angular Momentum
Remember under what condition the linear momentum is conserved?
u
r
u
r
dp
F

0

Linear momentum is conserved when the net external force is 0. 
d
t
u
r
By the same token, the angular momentum of a system
is constant in both magnitude and direction, if the
resultant external torque acting on the system is 0.
What does this mean?
p

c
o
n
s
t
ur
r
dL


e
x
t
 dt  0
u
r
o
n
s
t
L c
Angular momentum of the system before and
after a certain change is the same.
r
r
c
o
n
s
t
a
n
t
L i  Lf 
Mechanical Energy

U

K
U
Three important conservation laws K
i
i
f
f
r r
for isolated system that does not get p
Linear Momentum
i p
f
affected by external forces
r r
L
Angular Momentum
i L
f
Wednesday, July 6, 2011
PHYS 1443-001, Summer 2011 Dr.
Jaehoon Yu
16
Ex. 11 – 3 Neutron Star
A star rotates with a period of 30 days about an axis through its center. After the star
undergoes a supernova explosion, the stellar core, which had a radius of 1.0x104km, collapses
into a neutron star of radius 3.0km. Determine the period of rotation of the neutron star.
What is your guess about the answer?
Let’s make some assumptions:
The period will be significantly shorter,
because its radius got smaller.
1. There is no external torque acting on it
2. The shape remains spherical
3. Its mass remains constant
L
L
i
f
Using angular momentum
conservation
I

I
i

f
f
The angular speed of the star with the period T is
2
I i
mr
i 2
f  I  2 
Thus
f
Tf 
2

f
2

T
mr
f T
i
2
 rf2 
3
.
0



6
  2 Ti 

0
.
23
s

2
.
7

10
days

30
days


4
r 
1
.
0

10


 i 
Wednesday, July 6, 2011
PHYS 1443-001, Summer 2011 Dr.
Jaehoon Yu
17
Similarity Between Linear and Rotational Motions
All physical quantities in linear and rotational motions show striking similarity.
Quantities
Mass
Length of motion
Speed
Acceleration
Force
Work
Power
Momentum
Kinetic Energy
Wednesday, July 6, 2011
Linear
Mass
M
Distance
r
t
v
a
t
u
r
L
v
Rotational
Moment of Inertia
I
m
r2
Angle  (Radian)

t


t

r
r u
r
Force Fm
a Torque I
rr



Work WF
 d Work W
u
rr
P


P

F
v
u
r
r
pm
v
Kinetic
1 2
K mv
2
PHYS 1443-001, Summer 2011 Dr.
Jaehoon Yu
u
r
u
r
LI
Rotational
1 2
K

R I
2
18
Conditions for Equilibrium
What do you think the term “An object is at its equilibrium” means?
The object is either at rest (Static Equilibrium) or its center of mass
is moving at a constant velocity (Dynamic Equilibrium).
When do you think an object is at its equilibrium?
ur
Translational Equilibrium: Equilibrium in linear motion  F  0
Is this it?
The above condition is sufficient for a point-like object to be at its
translational equilibrium. For an object with size, however, this is
not sufficient. One more condition is needed. What is it?
Let’s consider two forces equal in magnitude but in opposite direction acting
on a rigid object as shown in the figure. What do you think will happen?
F
d
d
CM
-F

r
The object will rotate about the CM. The net torque
 0
acting on the object about any axis must be 0.
For an object to be at its static equilibrium, the object should not
have linear or angular speed. vCM  0   0
Wednesday, July 6, 2011
PHYS 1443-001, Summer 2011 Dr.
Jaehoon Yu
19
More on Conditions for Equilibrium
To simplify the problem, we will only deal with forces acting on x-y plane, giving torque
only along z-axis. What do you think the conditions for equilibrium be in this case?
The six possible equations from the two vector equations turns to three equations.
ur
F 0
F
F
x
y
0
0
AND
r
  0

z
0
What happens if there are many forces exerting on an object?
r’
r5 O O’
Wednesday, July 6, 2011
If an object is at its translational static equilibrium, and
if the net torque acting on the object is 0 about one
axis, the net torque must be 0 about any arbitrary axis.
Why is this true?
Because the object is not moving, no matter what the
rotational axis is, there should not be any motion. It is
simply a matter of mathematical manipulation.
PHYS 1443-001, Summer 2011 Dr.
Jaehoon Yu
20
How do we solve equilibrium problems?
1.
2.
3.
4.
5.
6.
Identify all the forces and their directions and locations
Draw a free-body diagram with forces indicated on it with
their directions and locations properly noted
Write down force equation for each x and y component with
proper signs
Select a rotational axis for torque calculations  Selecting
the axis such that the torque of one of the unknown forces
become 0 makes the problem easier to solve
Write down the torque equation with proper signs
Solve the equations for unknown quantities
Wednesday, July 6, 2011
PHYS 1443-001, Summer 2011 Dr.
Jaehoon Yu
21
Ex. 12 – 3: Seesaw Balancing
A uniform 40.0 N board supports the father and the daughter each weighing 800 N and
350 N, respectively, and is not moving. If the support (or fulcrum) is under the center of
gravity of the board, and the father is 1.00 m from CoG, what is the magnitude of the
normal force n exerted on the board by the support?
1m
F
MFg
x
n
MBg
Since there is no linear motion, this system
is in its translational equilibrium
D
F
MDg
x
F
0
 n M B g M F g M D g  0
n  40.0  800  350  1190N
y
Therefore the magnitude of the normal force
Determine where the child should sit to balance the system.
The net torque about the fulcrum
by the three forces are
Therefore to balance the system
the daughter must sit
Wednesday, July 6, 2011
  M B g  0  n  0  M F g 1.00  M D g  x  0
x

MFg
800
1.00m 
1.00m  2.29m
MDg
350
PHYS 1443-001, Summer 2011 Dr.
Jaehoon Yu
22
Seesaw Example Cont’d
Determine the position of the child to balance the
system for different position of axis of rotation.
Rotational axis
1m
F
MFg

x
n
x/2
D
MFg
MBg
The net torque about the axis of
rotation by all the forces are
 M B g  x / 2  M F g  1.00  x / 2  n x / 2  M D g  x / 2  0
n  MBg  MF g  MDg
  M B g  x / 2  M F g  1.00  x / 2
 M B g  M F g  M D g  x / 2  M D g  x / 2
Since the normal force is
The net torque can
be rewritten
 M F g 1.00  M D g  x  0
Therefore
x
Wednesday, July 6, 2011
MFg
800

1.00m 
1.00m  2.29m
MDg
350
PHYS 1443-001, Summer 2011 Dr.
Jaehoon Yu
What do we learn?
No matter where the
rotation axis is, net effect of
the torque is identical.
23
Ex. 12.4 for Mechanical Equilibrium
A person holds a 50.0N sphere in his hand. The forearm is horizontal. The biceps
muscle is attached 3.00 cm from the joint, and the sphere is 35.0cm from the joint. Find
the upward force exerted by the biceps on the forearm and the downward force exerted
by the upper arm on the forearm and acting at the joint. Neglect the weight of forearm.
FB
Since the system is in equilibrium, from
the translational equilibrium condition
F  0
O
l
mg
 F  F  F  mg  0
F
From the rotational equilibrium condition   F  0  F  d  mg  l  0
d
x
U
y
B
U
U
B
FB  d  mg  l
mg  l 50.0  35.0

 583N
FB 
3.00
d
Force exerted by the upper arm is
FU  FB  mg  583  50.0  533N
Thus, the force exerted by
the biceps muscle is
Wednesday, July 6, 2011
PHYS 1443-001, Summer 2011 Dr.
Jaehoon Yu
24
Example 12 – 6
A 5.0 m long ladder leans against a wall at a point 4.0m above the ground. The ladder is
uniform and has mass 12.0kg. Assuming the wall is frictionless (but ground is not),
determine the forces exerted on the ladder by the ground and the wall.
FW
FBD
mg
FGy
O
FGx
First the translational equilibrium,
using components
 Fx FGx  FW  0
 F  mg  F
y
0
Gy
Thus, the y component of the force by the ground is
FGy  mg  12.0  9.8N  118N
The length x0 is, from Pythagorian theorem
x0  5.02  4.02  3.0m
Wednesday, July 6, 2011
PHYS 1443-001, Summer 2011 Dr.
Jaehoon Yu
25
Example 12 – 6 cont’d
From the rotational equilibrium

O
 mg x0 2  FW 4.0  0
Thus the force exerted on the ladder by the wall is
mg x0 2 118 1.5

 44 N
4.0
4.0
The x component of the force by the ground is
FW 
F
x
 FGx  FW  0
Solve for FGx
FGx  FW  44 N
Thus the force exerted on the ladder by the ground is
FG  FGx2  FGy2  442  1182  130N
The angle between the  tan 1  FGy 
1  118 
o

tan

70





ground force to the floor
 44 
 FGx 
Wednesday, July 6, 2011
PHYS 1443-001, Summer 2011 Dr.
Jaehoon Yu
26