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Monday, Apr. 14, 2008
PHYS 1441002, Spring 2008
Dr. Jaehoon Yu
1
PHYS 1441 – Section 002
Lecture #21
Monday, Apr. 14, 2008
Dr.
Jae
hoon
Yu
•
Rolling Motion
•
Rotational Dynamics
–
Torque
–
Equilibrium
–
Moment of Inertia
–
Torque and Angular Acceleration
•
Rotational Kinetic Energy
Today’s homework is HW #11, due 9pm, Wednesday, Apr. 23!!
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PHYS 1441002, Spring 2008
Dr. Jaehoon Yu
2
Announcements
•3
rd
term exam
– Next Monday, Apr. 21, in class
– Covers: Ch. 6.7 – what we complete this Wednesday, Apr.
16
– This is the final term exam in the semester
– The worst of the three term exams will be dropped from
the final grading
• Quiz results
– Class average: 2/6
• Equivalent to 33/100
• How did you do before?: 50, 44 and 47
– Top score: 4.2/6
Monday, Apr. 14, 2008
PHYS 1441002, Spring 2008
Dr. Jaehoon Yu
3
Rolling Motion of a Rigid Body
What is a rolling motion?
To simplify the discussion, let’s
make a few assumptions
Let’s consider a cylinder rolling on a flat surface, without slipping.
A more generalized case of a motion where the
rotational axis moves together with an object
Under what condition does this “Pure Rolling” happen?
The total linear distance the CM of the cylinder moved is
Thus the linear
speed of the CM is
A rotational motion about a moving axis
1.
Limit our discussion on very symmetric
objects, such as cylinders, spheres, etc
2.
The object rolls on a flat surface
R
θ
s
s=R
θ
s
=
CM
s
v
t
Δ
=
Δ
The condition for a “Pure Rolling motion”
R
t
θ
Δ
=
Δ
ω
R
=
R
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PHYS 1441002, Spring 2008
Dr. Jaehoon Yu
4
More Rolling Motion of a Rigid Body
As we learned in rotational motion, all points in a rigid body
moves at the same angular speed but at different linear speeds.
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
The magnitude of the linear acceleration of the CM is
A rolling motion can be interpreted as the sum of Translation and Rotation
CM
a
Why??
P
P’
CM
v
CM
2v
CM
CM is moving at
the same speed at all times.
P
P’
CM
v
CM
v
CM
v
CM
+
P
P’
CM
v=R
ω
v=0
v=R
ω
=
P
P’
CM
2v
CM
v
CM
CM
v
t
Δ
=
Δ
R
t
ω
Δ
=
Δ
α
R
=
Monday, Apr. 14, 2008
PHYS 1441002, Spring 2008
Dr. Jaehoon Yu
5
Starting from rest, the car accelerates for 20.0 s
with a constant linear acceleration of 0.800 m/s
2
.
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This note was uploaded on 10/21/2009 for the course PHYS phys1441 taught by Professor Dr.jhaehoonyou during the Spring '08 term at UT Arlington.
 Spring '08
 Dr.JhaehoonYou
 Acceleration, Energy, Inertia, Kinetic Energy

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