Download Rotational & Translational Motion Rolling without Slipping ... and more Exercises Acting in PDF only on Docsity!
Rigid Bodies:
Rotational & Translational Motion
Rolling without Slipping
W11D
Announcements
Sunday Tutoring in 26-152 from 1-5 pm
Problem Set 8 due Week 11 Tuesday at 9 pm in box outside 26-
No Math Review Week 11
Exam 3 Tuesday Nov 26 7:30 to 9:30 pm
Conflict Exam 3 Wednesday Nov 27 8 am to 10 am, 10 am to 12
noon
Nov 27 Drop Date
Demo: Rotation and Translation
of a Rigid Body
Thrown Rigid Rod
Translational Motion: the gravitational external force
acts on center-of-mass
Rotational Motion: object rotates about center-of-
mass. Note that the center-of-mass may be
accelerating
sy
ext total cm total
cm
s d d
m m
dt dt
= = =
p V
F A
! ! !!
Rotation about a moving axis
For straight line motion, the bicycle wheel
rotates about a fixed direction and center of
mass is translating
Demo: Bicycle Wheel
Constraint conditions and rolling without slipping
Center of Mass Reference Frame
Frame O: At rest with respect to
ground
Frame O cm
: Origin located at center
of mass
Position vectors in different frames:
Relative velocity between the two
reference frames
Law of addition of velocities:
V
cm
= d
R
cm
/ dt
A
cm
= d
V
cm
/ dt =
r
i
r
cm, i
R
cm
r
cm, i
r
i
R
cm
v
i
v
cm, i
V
cm
v
cm, i
v
i
V
cm
Rolling Without Slipping
The velocity of the point on the rim that is in contact with
the ground is zero in the reference frame fixed to the ground.
Kinetic Energy of
Rotation and Translation
Kinetic energy of rotation about center-of-mass
Translational kinetic energy
Kinetic energy is sum
K
rot
I
cm
cm
2
K = K
trans
+ K
rot
mv
cm
2
I
cm
! cm
2
K
trans
mv
cm
2
Concept Question: Rolling
Without Slipping
If a wheel of radius R rolls without slipping through an
angle θ, what is the relationship between the distance
the wheel rolls, x, and the product Rθ?
- x > Rθ.
- x = Rθ.
- x < Rθ.
Concept Q. Answer :
Rolling Without Slipping
Answer 2. Rolling without slipping condition, x = Rθ.
Table Problem: Bicycle Wheel
A bicycle wheel of radius R is rolling without slipping along a
horizontal surface. The center of mass of the bicycle in
moving with a constant speed V in the positive x -direction. A
bead is lodged on the rim of the wheel. Assume that at t = 0,
the bead is located at the top of the wheel at x ( t = 0) = x 0 and
y ( t = 0) = 2 R. What are the x - and y -components of the
position of the bead as a function of time according to an
observer fixed to the ground?
Cycloid
Courtesy Wikipedia and Wolfram
x(t) = R(t − sin t) and y(t) = R(1 − cos t)
curve traced by a point on the rim of the circular wheel as it rolls along a straight line
Earth s Spin Angular Momentum
about center of mass of earth
- Period and angular velocity
- Magnitude
L
cm
spin = I cm
spin
m e
R
e
2 ! spin
n ˆ
spin
=
T spin
= 7.29 # 10
$ 5 rad % s
$ 1
!
L cm
spin = 7.09! 10
33 kg " m
2 " s
1
ˆ n
Earth s Angular Momentum
For a body undergoing orbital motion like the earth orbiting the sun, the two
terms can be thought of as an orbital angular momentum about the center-of-
mass of the earth-sun system, denoted by S ,
Spin angular momentum about center-of-mass of earth
Total angular momentum about S
sys
,cm ,cm ,
ˆ S S s e e cm L = R! p = r m v k
!! !
spin 2
cm cm spin spin
e e
L = I = m R! n
L
S
total = r s , e
m e
v cm
k +
m e
R
e
2 ! spin
ˆ n
Demo B 113: Rolling Cylinders
Concept Question:
Cylinder Race
Two cylinders of the same size and mass roll down an
incline, starting from rest. Cylinder A has most of its
mass concentrated at the rim, while cylinder B has most
of its mass concentrated at the center. Which reaches
the bottom first?
1. A
2. B
- Both at the same time
Concept Q. Answer :
Cylinder Race
Answer 2: Because the moment of inertia of
cylinder B is smaller, more of the mechanical
energy will go into the translational kinetic energy
hence B will have a greater center of mass speed
and hence reach the bottom first.
Concept Question:
Cylinder race with different masses
Two cylinders of the same size but different masses
roll down an incline, starting from rest. Cylinder A has a
greater mass. Which reaches the bottom first?
1. A
2. B
- Both at the same time
Announcements
Sections 1-4 No Class Week 11 Monday
Sunday Tutoring in 26-152 from 1-5 pm
Problem Set 8 due Week 11 Tuesday at 9 pm in box outside 26-
No Math Review Week 11
Exam 3 Tuesday Nov 26 7:30-9:30 pm
Conflict Exam 3 Wednesday Nov 27 8-10 am, 10-12 noon
Nov 27 Drop Date
Angular Momentum and Torque
About any fixed point S
Decomposition:
L
S
L
S
orbital
L
cm
spin
r
s , cm
! m
T
v
cm
L
cm
spin
S
S , i
ext
d
L
S
dt
i
"
cm
ext
d
L
cm
spin
dt
S , cm
ext
d
L
S
orbital
dt
Worked Example: Descending and
Ascending Yo-Yo
A Yo-Yo of mass m has an axle
of radius b and a spool of radius
R. It s moment of inertia about
the center of mass can be taken
to be I = (1/2)mR
2 and the
thickness of the string can be
neglected. The Yo-Yo is released
from rest. What is the
acceleration of the Yo-Yo as it
descends.
30
Worked Example: Descending
and Ascending Yo-Yo
31
! cm
r cm, T
T = # b
i " # T
j = bT
k
Torque about cm:
Torque equation:
Newton s Second Law:
Constraint:
Tension:
Acceleration:
bT = I!
mg! T = ma
y
a y
= b!
z
T = Ia y
/ b
2
a y
mb
2
( mb
2
g
Demo B107: Descending and
Ascending Yo-Yo
wheel+axle
M = 435 g
outer
R! 6. 3 cm
inner
R! 4. 9 cm
2 2
cm outer inner
4 2
1. 385 10 g cm
I! M R + R
Demonstration:
Pulling a Yo-Yo
Concept Question: Pulling a Yo-Yo 2
Concept Q. Ans.: Pulling a Yo-Yo 2
Answer 2. When the string is pulled up, the only
horizontal force is static friction and it points to the left
so the yo-yo accelerates to the left. Therefore
somewhere between A and B the direction of rotation
changes.
Table Problem: Pulling a Yo-Yo
A Yo-Yo of mass m has an axle of radius b and a spool of
radius R. It s moment of inertia about the center of mass
can be taken to be I = (1/2)mR
2 and the thickness of the
string can be neglected. The Yo-Yo is placed upright on a
table and the string is pulled with a horizontal force to the
right as shown in the figure. The coefficient of static friction
between the Yo-Yo and the table is. What is the
maximum magnitude of the pulling force, F, for which the
Yo-Yo will roll without slipping?
μ s
Concept Question: Cylinder Rolling
Down Inclined Plane
A cylinder is rolling without slipping down an inclined plane.
The friction at the contact point P is
- Static and points up the inclined plane.
- Static and points down the inclined plane.
- Kinetic and points up the inclined plane.
- Kinetic and points down the inclined plane.
- Zero because it is rolling without slipping.
40
Concept Q. Ans.: Cylinder Rolling
Down Inclined Plane
Answer 1. The friction at the contact point P is static and
points up the inclined plane. This friction produces a torque
about the center of mass that points into the plane of the
figure. This torque produces an angular acceleration into
the plane, increasing the angular speed as the cylinder rolls
down.
41
Table Problem: Cylinder on Inclined
Plane Torque About Center of Mass
A hollow cylinder of outer radius R and mass m with moment of inertia
I cm
about the center of mass starts from rest and moves down an
incline tilted at an angle from the horizontal. The center of mass of
the cylinder has dropped a vertical distance h when it reaches the
bottom of the incline. Let g denote the gravitational constant. The
coefficient of static friction between the cylinder and the surface is s
.
The cylinder rolls without slipping down the incline. Using the torque
method about the center of mass, calculate the velocity of the center of
mass of the cylinder when it reaches the bottom of the incline.
Concept Question: Angular Collisions
A long narrow uniform stick lies motionless on ice
(assume the ice provides a frictionless surface).
The center of mass of the stick is the same as the
geometric center (at the midpoint of the stick). A
puck (with putty on one side) slides without spinning
on the ice toward the stick, hits one end of the stick,
and attaches to it.
Which quantities are constant?
- Angular momentum of puck about center of mass of
stick.
- Momentum of stick and ball.
- Angular momentum of stick and ball about any
point.
- Mechanical energy of stick and ball.
- None of the above 1-4.
- Three of the above 1.
- Two of the above 1-4.
Concept Q. Ans.: Angular Collisions
Answer: 7
(2) and (3) are correct. There are no external
forces acting on this system so the momentum of
the center of mass is constant (1). There are no
external torques acting on the system so the
angular momentum of the system about any point
is constant (3). However there is a collision force
acting on the puck, so the torque about the center
of the mass of the stick on the puck is non-zero,
hence the angular momentum of puck about
center of mass of stick is not constant. The
mechanical energy is not constant because the
collision between the puck and stick is inelastic.
Table Problem: Angular Collision
A long narrow uniform stick of length l and mass m lies
motionless on a frictionless ). The moment of inertia
of the stick about its center of mass is l cm
. A puck (with
putty on one side) has the same mass m as the stick.
The puck slides without spinning on the ice with a
speed of v 0
toward the stick, hits one end of the stick,
and attaches to it. (You may assume that the radius of
the puck is much less than the length of the stick so that
the moment of inertia of the puck about its center of
mass is negligible compared to l cm .) What is the angular
velocity of the stick plus puck after the collision?
