Transcript phys1443-fall04-111004

PHYS 1443 – Section 003
Lecture #19
Wednesday, Nov. 10, 2004
Dr. Jaehoon Yu
1.
2.
3.
4.
5.
6.
Moment of Inertia
Parallel Axis Theorem
Torque and Angular Acceleration
Rotational Kinetic Energy
Work, Power and Energy in Rotation
Angular Momentum & Its Conservation
Today’s homework is HW #10, due 1pm next Wednesday!!
Wednesday, Nov. 10, 2004
PHYS 1443-003, Fall 2004
Dr. Jaehoon Yu
1
Moment of Inertia
Measure of resistance of an object to
changes in its rotational motion.
Equivalent to mass in linear motion.
Rotational Inertia:
For a group
of particles
I   mi ri
2
i
What are the dimension and
unit of Moment of Inertia?
For a rigid
body
I   r 2 dm
ML 
2
kg m
2
Determining Moment of Inertia is extremely important for
computing equilibrium of a rigid body, such as a building.
Wednesday, Nov. 10, 2004
PHYS 1443-003, Fall 2004
Dr. Jaehoon Yu
2
Example for Moment of Inertia
In a system of four small spheres as shown in the figure, assuming the radii are negligible
and the rods connecting the particles are massless, compute the moment of inertia and
the rotational kinetic energy when the system rotates about the y-axis at angular speed w.
y
m
Since the rotation is about y axis, the moment of
inertia about y axis, Iy, is
b
l
M
O
m
l
M
x
b
I   mi ri2  Ml2 Ml 2 m  02  m  02  2Ml 2
i
This is because the rotation is done about y axis,
and the radii of the spheres are negligible.
1 2 1
K R  Iw  2 Ml 2 w 2  Ml 2w 2
2
2
Why are some 0s?

Thus, the rotational kinetic energy is

Find the moment of inertia and rotational kinetic energy when the system rotates on
the x-y plane about the z-axis that goes through the origin O.

2
2
2
I   mi ri 2  Ml 
Ml 2 mb2 mb 2  2 Ml  mb
i
Wednesday, Nov. 10, 2004

1
1
K R  Iw 2  2 Ml 2  2mb2 w 2  Ml 2  mb2 w 2
2
2
PHYS 1443-003, Fall 2004
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, Dmi.
2
2
I

lim
r
D
m

r

i
i
The moment of inertia for the large rigid object is
 dm
Dm 0
i
It is sometimes easier to compute moments of inertia in terms
of volume of the elements rather than their mass
Using the volume density, r, replace
dm in the above equation with dV.
r
i
How can we do this?
dm
The moments of
dm  rdV inertia becomes
dV
I   rr 2 dV
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, Nov. 10, 2004
I   r 2 dm  R 2  dm  MR 2
What do you notice
from this result?
PHYS 1443-003, Fall 2004
Dr. Jaehoon Yu
The moment of inertia for this
object is the same as that of a
point of mass M at the distance R.
4
Example for 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
so the masslet is
dm  dx  M dx
L
dx
x
x
L
The moment
of inertia is
M  L   L 

     
3L  2   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, Nov. 10, 2004
L/2
2
M 1 3 
x
M
x 
I   r dm  
dx 

L / 2
L  3  L / 2
L
L/2
2
3
 M  L3  ML2
  

3
L
12

 4
L
M 1
I  r 2 dm   x M dx   x3 
0
L 3 0
L
M
M 3
ML2
3
L   0  L  

3L
3L
3

2
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 1443-003, Fall 2004
Dr. Jaehoon Yu
5
Parallel Axis Theorem
Moments of inertia for highly symmetric object is easy to compute if the
rotational axis is the same as the axis of symmetry. However if the axis of
rotation does not coincide with axis of symmetry, the calculation can still be
2
done in simple manner using parallel-axis theorem. I  I CM  MD
y


2
2
2

x

y
dm (1)
I

r
dm
Moment of inertia is defined


(x,y)
x  xCM  x' y  yCM  y'
Since x and y are
r
yCM
y
y’
One can substitute x and y in Eq. 1 to obtain
CM
(xCM,yCM)
D
xCM
x’
x


I   xCM  x'   yCM  y ' dm

2

2


2
2
2
2
 xCM
 yCM
dm

2
x
x
'
dm

2
y
y
'
dm

x
'

y
'
dm
CM
CM




Since the x’ and y’ are the
 x' dm  0  y' dm  0
x distance from CM, by definition
Therefore, the parallel-axis theorem
2
2
 dm   x'2  y'2 dm  MD 2  I CM
I  xCM
 yCM
What does this
Moment of inertia of any object about any arbitrary axis are the same as
theorem
tell you?
of inertia
for a rotation about the CM and that6of
Wednesday, Nov. 10, 2004 the sum of moment
PHYS 1443-003,
Fall 2004
Dr. Jaehoon
Yu axis.
the CM about the
rotation
Example for Parallel Axis Theorem
Calculate the moment of inertia of a uniform rigid rod of length L and mass M about an
axis that goes through one end of the rod, using parallel-axis theorem.
The line density of the rod is  
y
so the masslet is
CM
dm  dx 
dx
x
L
x The moment of
inertia about
the CM
M
L
M
dx
L
L/2
2
M 1 3 
x
M
I CM   r dm  
x 
dx 

L / 2
L
3
L

 L / 2
3
3
M  L   L   M  L3  ML2
  

       
3L  2   2   3L  4  12
2
L/2
2
Using the parallel axis theorem
I  I CM
ML2  L 
ML2 ML2 ML2
2
D M 
  M 


12  2 
12
4
3
The result is the same as using the definition of moment of inertia.
Parallel-axis theorem is useful to compute moment of inertia of a rotation of a rigid
object with complicated shape about an arbitrary axis
Wednesday, Nov. 10, 2004
PHYS 1443-003, Fall 2004
Dr. Jaehoon Yu
7
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 radial force Fr
Ft  mat  mr
2
The torque due to tangential force Ft is   Ft r  mat r  mr   I
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 dFt is dFt  dmat  dmr
dFt
d dFt r  r 2 dm
The torque due to tangential force Ft is
dm
The total torque is     r 2 dm I
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, Nov. 10, 2004
PHYS 1443-003, Fall 2004 point, making the moment arm 0.
8
Dr. Jaehoon Yu
Example for Torque and Angular Acceleration
A uniform rod of length L and mass M is attached at one end to a frictionless pivot and is
free to rotate about the pivot in the vertical plane. The rod is released from rest in the
horizontal position. What are the initial angular acceleration of the rod and the initial linear
acceleration of its right end?
The only force generating torque is the gravitational force Mg
L/2
  Fd  F
Mg
L
L
 Mg  I
2
2
L
Since the moment of inertia of the rod I  r 2 dm 
0
0
when it rotates about one end
L
We obtain
  MgL 
2I
MgL 3 g

2ML2 2 L
3
Wednesday, Nov. 10, 2004
L
3


M
x
ML2


2
x dx      
3
 L  3  0
Using the relationship between tangential and
angular acceleration
3g
at  L 
2
PHYS 1443-003, Fall 2004
Dr. Jaehoon Yu
What does this mean?
The tip of the rod falls faster than
an object undergoing a free fall.
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Rotational Kinetic Energy
y
vi
What do you think the kinetic energy of a rigid object
that is undergoing a circular motion is?
1
1
2
2 

m
v
Kinetic energy of a masslet, mi,
Ki

m
r
i i
i i w
2
2
moving at a tangential speed, vi, is
mi
ri
q
O
x
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 w    mi ri w
2 i
2 i

i
Since moment of Inertia, I, is defined as
I   mi ri2
i
The above expression is simplified as
Wednesday, Nov. 10, 2004
1 
K R  Iw
2
PHYS 1443-003, Fall 2004
Dr. Jaehoon Yu
10
Total Kinetic Energy of a Rolling Body
What do you think the total kinetic
energy of the rolling cylinder is?
P’
CM
Since it is a rotational motion about the point
P, we can write the total kinetic energy
1
K  I Pw 2
2
2vCM
vCM
Where, IP, is the moment of
inertia about the point P.
Using the parallel axis theorem, we can rewrite

P

1
1
1
1
2
2
2
2
2 2
K  I Pw  I CM  MR w  I CM w  MR w
2
2
2
2
1
1
Since vCM=Rw, the above
2
2
K  I CM w  MvCM
relationship can be rewritten as
2
2
What does this equation mean?
Rotational kinetic
energy about the CM
Total kinetic energy of a rolling motion is the sum
of the rotational kinetic energy about the CM
Wednesday, Nov. 10, 2004
PHYS 1443-003, Fall 2004
Dr. Jaehoon Yu
Translational Kinetic
energy of the CM
And the translational
kinetic of the CM
11
Kinetic Energy of a Rolling Sphere
Let’s consider a sphere with radius R
rolling down a hill without slipping.
R
w
h
q
vCM
Since vCM=Rw
What is the speed of the
CM in terms of known
quantities and how do you
find this out?
1
1
2
2 2
K  I CM w  MR w
2
2
2
1
 vCM   1 Mv 2
 I CM 
CM

2
2
 R 
1I
 2
  CM2  M vCM
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


M
 2
vCM  Mgh
K
2 R

vCM 
Wednesday, Nov. 10, 2004
PHYS 1443-003, Fall 2004
Dr. Jaehoon Yu
2 gh
1  I CM / MR 2
12
Example for 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
F
F
q
x
 Mg sin q  f  MaCM
y
 n  Mg cosq  0
Since the forces Mg and n go through the CM, their moment arm is 0

and do not contribute to torque, while the static friction f causes torque CM
We know that
I CM 
2
MR 2
5
aCM  R
We
obtain
2
MR 2
I CM 
2
f 
 aCM   MaCM
5
R  R  R  5


Substituting f in
dynamic equations
Wednesday, Nov. 10, 2004
 fR  I CM 
7
Mg sin q  MaCM
5
PHYS 1443-003, Fall 2004
Dr. Jaehoon Yu
aCM 
5
g sin q
7
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