Transcript Thursday, June 22, 2006

PHYS 1443 – Section 001
Lecture #13
Thursday, June 22, 2006
Dr. Jaehoon Yu
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Thursday, June 22, 2006
CM and the Center of Gravity
Fundamentals on Rotational Motion
Rotational Kinematics
Relationship between angular and linear quantities
Rolling Motion of a Rigid Body
Torque
Torque and Vector Product
Moment of Inertia
PHYS 1443-001, Summer 2006
Dr. Jaehoon Yu
1
Announcements
• Reading assignments
– CH. 11.6, 11.8, 11.9 and 11.10
• Last quiz next Wednesday
– Early in the class
– Covers Ch. 10 – what we cover next Tuesday
Thursday, June 22, 2006
PHYS 1443-001, Summer 2006
Dr. Jaehoon Yu
2
Center of Mass and Center of Gravity
The center of mass of any symmetric object lies on an
axis of symmetry and on any plane of symmetry, if
object’s mass is evenly distributed throughout the body.
How do you think you can
determine the CM of
objects that are not
symmetric?
Center of Gravity
Dmi
Axis of
One can use gravity to locate CM.
symmetry
1. Hang the object by one point and draw a vertical line
following a plum-bob.
2. Hang the object by another point and do the same.
3. The point where the two lines meet is the CM.
Since a rigid object can be considered as a collection
of small masses, one can see the total gravitational
force exerted on the object as
Fg   Fi   Dmi g  Mg
i
Dmig
CM
What does this
equation tell you?
Thursday, June 22, 2006
i
The net effect of these small gravitational
forces is equivalent to a single force acting on
a point (Center of Gravity) with mass M.
PHYS 1443-001, Summer 2006
3
The CoG is the point in an object
asDr.ifJaehoon
all the
gravitational
force
is
acting
on!
Yu
Motion of a Group of Particles
We’ve learned that the CM of a system can represent the motion of a system.
Therefore, for an isolated system of many particles in which the total mass
M is preserved, the velocity, total momentum, acceleration of the system are
d 1
drCM
 
dt  M
dt
Velocity of the system
vCM 
Total Momentum
of the system
pCM  MvCM
Acceleration of
the system
aCM
External force exerting
on the system
If net external force is 0
Thursday, June 22, 2006

M
1

m
r

 i i  M
i i
i i
1

m
v

 i i  M
 Fext  MaCM   miai 
dptot
dt
dri
mi vi


dt
M
 m v  m v   p  p
M
d 1
dvCM
 
dt  M
dt
 Fext  0 
 mi
dptot
dt
ptot  const
PHYS 1443-001, Summer 2006
Dr. Jaehoon Yu
tot
 mi
dvi  mi ai

dt
M
What about the
internal forces?
System’s momentum
is conserved.
4
Fundamentals on Rotation
Linear motions can be described as the motion of the center of
mass with all the mass of the object concentrated on it.
Is this still true for
rotational motions?
No, because different parts of the object have
different linear velocities and accelerations.
Consider a motion of a rigid body – an object that
does not change its shape – rotating about the axis
protruding out of the slide.
The arc length is l Rq
l
Therefore the angle, q, is q  R . And the unit of
the angle is in radian. It is dimensionless!!
One radian is the angle swept by an arc length equal to the radius of the arc.
Since the circumference of a circle is 2pr,
360  2pr / r  2p
The relationship between radian and degrees is 1 rad  360  / 2p  180 / p
Thursday, June 22, 2006
PHYS 1443-001, Summer 2006
Dr. Jaehoon Yu
 180o 3.14  57.3o
5
Example 10 – 1
A particular bird’s eyes can barely distinguish objects that subtend an angle no
smaller than about 3x10-4 rad. (a) How many degrees is this? (b) How small an
object can the bird just distinguish when flying at a height of 100m?
(a) One radian is 360o/2p. Thus


4
3  10 rad  3 10 rad 
4
 360
o

2p rad  0.017
(b) Since l=rq and for small angle
arc length is approximately the
same as the chord length.
l  rq 
4
100m  3 10 rad 
2
Thursday, June 22, 2006
3 10 m  3cm
PHYS 1443-001, Summer 2006
Dr. Jaehoon Yu
6
o
Rotational Kinematics
The first type of motion we have learned in linear kinematics was
under a constant acceleration. We will learn about the rotational
motion under constant angular acceleration, because these are the
simplest motions in both cases.
Just like the case in linear motion, one can obtain
Angular Speed under constant
angular acceleration:
Angular displacement under
constant angular acceleration:
One can also obtain
Thursday, June 22, 2006
 f  i   t
1 2
q f  qi  it   t
2
    2 q f  qi 
2
f
PHYS 1443-001, Summer 2006
Dr. Jaehoon Yu
2
i
7
Angular Displacement, Velocity, and Acceleration
Using what we have learned in the previous slide, how
Dq q f  q i
would you define the angular displacement?
q f  qi
How about the average angular speed?
Unit? rad/s

And the instantaneous angular speed?
Unit? rad/s
  lim
By the same token, the average angular
acceleration
Unit? rad/s2
And the instantaneous angular
acceleration? Unit? rad/s2

t f  ti
Dt 0
t f  ti
Dt 0
Dq
Dt
Dq dq

Dt
dt
 f  i
  lim


qf
qi
D
Dt
D d

dt
Dt
When rotating about a fixed axis, every particle on a rigid object rotates through
the same angle and has the same angular speed and angular acceleration.
Thursday, June 22, 2006
PHYS 1443-001, Summer 2006
Dr. Jaehoon Yu
8
Example for Rotational Kinematics
A wheel rotates with a constant angular acceleration of 3.50 rad/s2. If
the angular speed of the wheel is 2.00 rad/s at ti=0, a) through what
angle does the wheel rotate in 2.00s?
Using the angular displacement formula in the previous slide, one gets
q f qi
1 2
 t   t
2
1
2
 2.00  2.00  3.50  2.00
2
11.0

rev.  1.75rev.
2p
Thursday, June 22, 2006
PHYS 1443-001, Summer 2006
Dr. Jaehoon Yu
 11.0rad
9
Example for Rotational Kinematics cnt’d
What is the angular speed at t=2.00s?
Using the angular speed and acceleration relationship
 f   i  t
 2.00  3.50  2.00  9.00rad / s
Find the angle through which the wheel rotates between t=2.00 s and
t=3.00 s.
Using the angular kinematic formula
qf
1 2
 q i   t  t
 11.0rad 2
1
At t=2.00s qt 2  2.00  2.00  2 3.50  2.00
1
2
q

2.00  3.00 3.50   3.00   21.8rad
t 3
At t=3.00s
2
Angular
displacement
Thursday, June 22, 2006
Dq  q   q 2
10.8
rev.  1.72rev.
 10.8rad 
2p
PHYS 1443-001, Summer 2006
Dr. Jaehoon Yu
10
Relationship Between Angular and Linear Quantities
What do we know about a rigid object that rotates
about a fixed axis of rotation?
Every particle (or masslet) in the object moves in a
circle centered at the axis of rotation.
When a point rotates, it has both the linear and angular
components in its motion.
What is the linear component of the motion you see?
Linear velocity along the tangential direction.
The
direction
of 
follows a
right-hand
rule.
How do we related this linear component of the motion
with angular component?
dl
d
dq
The arc-length is l  Rq So the tangential speed v is v    rq   r
dt
dt dt
 r
What does this relationship tell you about Although every particle in the object has the same
the tangential speed of the points in the angular speed, its tangential speed differs
object and their angular speed?:
proportional to its distance from the axis of rotation.
Thursday, June 22, 2006
The farther
away2006
the particle is from the center of
PHYS 1443-001,
Summer
Dr.
Jaehoon
rotation,
theYuhigher the tangential speed.
11
Is the lion faster than the horse?
A rotating carousel has one child sitting on a horse near the outer edge
and another child on a lion halfway out from the center. (a) Which child has
the greater linear speed? (b) Which child has the greater angular speed?
(a) Linear speed is the distance traveled
divided by the time interval. So the child
sitting at the outer edge travels more
distance within the given time than the child
sitting closer to the center. Thus, the horse
is faster than the lion.
(b) Angular speed is the angle traveled divided by the time interval. The
angle both the children travel in the given time interval is the same.
Thus, both the horse and the lion have the same angular speed.
Thursday, June 22, 2006
PHYS 1443-001, Summer 2006
Dr. Jaehoon Yu
12
How about the acceleration?
How many different linear accelerations do you see
in a circular motion and what are they? Two
Tangential, at, and the radial acceleration, ar.
Since the tangential speed v is
v  r
The magnitude of tangential a  dv  d r  r d  r
 
t
acceleration at is
dt
dt
dt
What does this
relationship tell you?
Although every particle in the object has the same angular
acceleration, its tangential acceleration differs proportional to its
distance from the axis of rotation.
2
v2


r

The radial or centripetal acceleration ar is ar 

r
r
 r 2
What does The father away the particle is from the rotation axis, the more radial
this tell you? acceleration it receives. In other words, it receives more centripetal force.
Total linear acceleration is
Thursday, June 22, 2006
a  at2  ar2

r 
2
 r
PHYS 1443-001, Summer 2006
Dr. Jaehoon Yu

2 2
 r  2 4
13
Example
(a) What is the linear speed of a child seated 1.2m from the center of
a steadily rotating merry-go-around that makes one complete
revolution in 4.0s? (b) What is her total linear acceleration?
First, figure out what the angular
speed of the merry-go-around is.
Using the formula for linear speed
1rev

 2p 1.6rad / s
4.0s
4.0 s
v  r  1.2m 1.6rad / s  1.9m / s
Since the angular speed is constant, there is no angular acceleration.
Tangential acceleration is
Radial acceleration is
Thus the total
acceleration is
Thursday, June 22, 2006
at  r  1.2m  0rad / s 2  0m / s 2
2
2
ar  r  1.2m  1.6rad / s   3.1m / s 2
a  a  a  0  3.1  3.1m / s 2
2
t
2
r
PHYS 1443-001, Summer 2006
Dr. Jaehoon Yu
2
14
Example for Rotational Motion
Audio information on compact discs are transmitted digitally through the readout system consisting of laser
and lenses. The digital information on the disc are stored by the pits and flat areas on the track. Since
the speed of readout system is constant, it reads out the same number of pits and flats in the same time
interval. In other words, the linear speed is the same no matter which track is played. a) Assuming the
linear speed is 1.3 m/s, find the angular speed of the disc in revolutions per minute when the inner most
(r=23mm) and outer most tracks (r=58mm) are read.
Using the relationship
between angular and
tangential speed v  r
r  23mm

v 1.3m / s
1.3


 56.5rad / s
r 23mm 23 10 3
 9.00rev / s  5.4 10 2 rev / min
r  58mm  
1.3m / s
1.3

 22.4rad / s
58mm 58 10 3

b) The maximum playing time of a standard music
CD is 74 minutes and 33 seconds. How many
revolutions does the disk make during that time?
qf
     540  210rev / min

i
f
2
2
Thursday, June 22, 2006

 375rev / min
375
 q i   t  0  60 rev / s  4473s  2.8 104 rev
l  vt Dt
c) What is the total length of the track past through the readout mechanism?
d) What is the angular acceleration of the CD over
the 4473s time interval, assuming constant ?
 2.110 2 rev / min



f
 i 
Dt
PHYS 1443-001, Summer 2006
Dr. Jaehoon Yu

 1.3m / s  4473s
 5.8  103 m
22.4  56.5rad / s
4473s
 7.6 10 3 rad / s 2
15
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 the object
A rotational motion about the 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 without slipping on a flat surface
Under what condition does this “Pure Rolling” happen?
The total linear distance the CM of the cylinder moved is
s  Rq
R q s
s=Rq
Thursday, June 22, 2006
Thus the linear
speed of the CM is
vCM 
Condition
PHYS 1443-001, Summer 2006
Dr. Jaehoon Yu
ds
dq
R
 R
dt
dt
for “Pure Rolling”
16
More Rolling Motion of a Rigid Body
The magnitude of the linear acceleration of the CM is
P’
CM
aCM
dvCM
d

R
 R
dt
dt
As we learned in the rotational motion, all points in a rigid body
2vCM moves at the same angular speed but at a different linear speed.
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
Thursday, June 22, 2006
+
v=R
v=R
2vCM
P’
=
P
PHYS 1443-001, Summer 2006
Dr. Jaehoon Yu
CM
vCM
P
17
Torque
Torque is the tendency of a force to rotate an object about an axis.
Torque, t, is a vector quantity.
F
f
r
P
Line of
Action
d2
d
Moment
arm
F2
Consider an object pivoting about the point P
by the force F being exerted at a distance r.
The line that extends out of the tail of the force
vector is called the line of action.
The perpendicular distance from the pivoting point
P to the line of action is called Moment arm.
Magnitude of torque is defined as the product of the force
exerted on the object to rotate it and the moment arm.
When there are more than one force being exerted on certain
points of the object, one can sum up the torque generated by each
force vectorially. The convention for sign of the torque is positive if
rotation is in counter-clockwise and negative if clockwise.
Thursday, June 22, 2006
PHYS 1443-001, Summer 2006
Dr. Jaehoon Yu
t  rF sin f  Fd
t  t
1
t 2
 F1d1  F2 d2
18
Example for Torque
A one piece cylinder is shaped as in the figure with core section protruding from the larger
drum. The cylinder is free to rotate around the central axis shown in the picture. A rope
wrapped around the drum whose radius is R1 exerts force F1 to the right on the cylinder,
and another force exerts F2 on the core whose radius is R2 downward on the cylinder. A)
What is the net torque acting on the cylinder about the rotation axis?
R1
F1
The torque due to F1
t1  R1F1 and due to F2
So the total torque acting on
the system by the forces is
R2
t  t
1
t 2  R2 F2
 t 2  R1F1  R2 F2
F2
Suppose F1=5.0 N, R1=1.0 m, F2= 15.0 N, and R2=0.50 m. What is the net torque
about the rotation axis and which way does the cylinder rotate from the rest?
Using the
above result
Thursday, June 22, 2006
t 
 R1F1  R2 F2
The cylinder rotates in
 5.0 1.0 15.0  0.50  2.5N  m counter-clockwise.
PHYS 1443-001, Summer 2006
Dr. Jaehoon Yu
19
Torque and Vector Product
z
O
r
trxF
Let’s consider a disk fixed onto the origin O and
the force F exerts on the point p. What happens?
p
The disk will start rotating counter clockwise about the Z axis
y
The magnitude of torque given to the disk by the force F is
q
t  Fr sin f
F
x
But torque is a vector quantity, what is the direction?
How is torque expressed mathematically?
What is the direction?
The direction of the torque follows the right-hand rule!!
The above operation is called
Vector product or Cross product
What is the result of a vector product?
Another vector
Thursday, June 22, 2006
t  rF
C  A B
C  A  B  A B sin q
What is another vector operation we’ve learned?
Scalar product
PHYS 1443-001, Summer 2006
Dr. Jaehoon Yu
C  A  B  A B cosq
Result? A scalar
20
Properties of Vector Product
Vector Product is Non-commutative
What does this mean?
If the order of operation changes the result changes
Following the right-hand rule, the direction changes
Vector Product of two parallel vectors is 0.
 A B sin 0  0
C  A  B  A B sin q
A B  B  A
A  B  B  A
Thus,
A A  0
If two vectors are perpendicular to each other
A B  A B sin q  A B sin 90  A B  AB
Vector product follows distribution law


A B  C  A B  A C
The derivative of a Vector product with respect to a scalar variable is


d A B
dA
dB

 B  A
dt
dt
dt
Thursday, June 22, 2006
PHYS 1443-001, Summer 2006
Dr. Jaehoon Yu
21
More Properties of Vector Product
The relationship between
unit vectors, i, j and k
i i  j  j  k  k  0
i  j   j i  k
j  k  k  j  i
k  i  i  k  j
Vector product of two vectors can be expressed in the following determinant form
i
A  B  Ax
Bx
j
k
Ay
Az  i
By
Bz
Ay
Az
By
Bz
Ax
 j
Bx
Ax
Az
k
Bx
Bz
Ay
By
 Ay Bz  Az B y  i   Ax Bz  Az Bx  j  Ax B y  Ay Bx k
Thursday, June 22, 2006
PHYS 1443-001, Summer 2006
Dr. Jaehoon Yu
22
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.
Thursday, June 22, 2006
PHYS 1443-001, Summer 2006
Dr. Jaehoon Yu
23
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 .
y
m
Since the rotation is about y axis, the moment of
inertia about y axis, Iy, is
b
l
M
O
l
M
x
b
m
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  I  2 Ml 2  2  Ml 2 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
Thursday, June 22, 2006

1
1
K R  I 2  2 Ml 2  2mb2  2  Ml 2  mb2  2
2
2
PHYS 1443-001, Summer 2006
Dr. Jaehoon Yu
24
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
Thursday, June 22, 2006
What do you notice
from this result?
I   r 2 dm  R 2  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 2006
Dr. Jaehoon Yu
25
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?
Thursday, June 22, 2006
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 1443-001, Summer 2006
Dr. Jaehoon Yu
26
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 t  Ft r  mat r  mr   I
The tangential force Ft is
What do you see from the above relationship?
What does this mean?
t  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
dt dFt r  r 2 dm
The torque due to tangential force Ft is
dm
The total torque is t    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?
point, making the moment arm 0.
Thursday, June 22, 2006
PHYS 1443-001, Summer 2006
27
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
t  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
L
We obtain
  MgL 
2I
MgL 3 g

2ML2 2 L
3
Thursday, June 22, 2006
3


M
x
ML2


2
x dx      
3
 L  3  0
Using the relationship between tangential and
angular acceleration
3g
at  L 
2
What does this mean?
The tip of the rod falls faster than
an object undergoing a free fall.
PHYS 1443-001, Summer 2006
Dr. Jaehoon Yu
28
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 
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     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
Thursday, June 22, 2006
1 
K R  I
2
PHYS 1443-001, Summer 2006
Dr. Jaehoon Yu
29
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 P 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 P  I CM  MR   I CM   MR 
2
2
2
2
1
1
Since vCM=R, the above
2
2
K  I CM   MvCM
relationship can be rewritten as
2
2
What does this equation mean?
Rotational kinetic
energy about the CM
Translational Kinetic
energy of the CM
Total kinetic energy of a rolling motion is the sum
of the rotational kinetic energy about the CM
Thursday, June 22, 2006
And the translational
PHYS 1443-001, Summer 2006
kinetic of the CM
Dr. Jaehoon Yu
30
Kinetic Energy of a Rolling Sphere
R

h
q
vCM
Since vCM=R
What is the speed of the
CM in terms of known
quantities and how do you
find this out?
Let’s consider a sphere with radius R
rolling down a hill without slipping.
1
1
2
2 2
K  I CM   MR 
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 
Thursday, June 22, 2006
PHYS 1443-001, Summer 2006
Dr. Jaehoon Yu
2 gh
1  I CM / MR 2
31
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
t
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
Thursday, June 22, 2006
 fR  I CM 
7
Mg sin q  MaCM
5
PHYS 1443-001, Summer 2006
Dr. Jaehoon Yu
aCM 
5
g sin q
7
32