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J. P. Student
Multiple-choice answer sheets:
Physics 1D03
Rotational Dynamics
Examples involving t
Parallel-axis theorem
=Ia
Text sections 10.5, 10.7
Physics 1D03
Moments of inertia for some familiar objects: (see Table 10.2 in the text
for more):
I=MR2
L
I = 1/12 ML2
I = ½ MR2
L
I = 1/3 ML2
You do not have to know how to derive any of these !
Physics 1D03
QUIZ: If m1 > m2, the magnitudes of the accelerations of the
masses obey the relation:
a) a1 > a2
b) a1 = a2
R2
R1
c) a1 < a2
d) not enough info
m1
Physics 1D03
m2
Quiz:
Atwood’s Machine, again
Two masses, m1 > m2, are attached to the end of a light string which
passes over a pulley. The pulley rotates (ie: there is friction
between the pulley and string) on a frictionless horizontal axis and
has mass M. How do the tensions in two sections of the string
compare?
A) T1 = T2
B) T1 < T2
T2
T1
C) T1 > T2
m1
m2
Physics 1D03
Atwood’s Machine
m1 = 3 kg, m2 = 2 kg, R = 10 cm.
Find the accelerations, tensions if
R
1) the string is massless
2) the pulley has moment of inertia I = 0.04 kg
m2
.
T2
T1
m1
a =1.09 m/s2, T1 = 26.1 N, T2 = 21.8 N
Physics 1D03
m2
Parallel Axis Theorem
ICM
I
I = ICM + MD2
D
ICM : for an axis through the centre of mass
I : for another axis, parallel to the first
Physics 1D03
Example:
Uniform thin rod
L/2
I = (1/3) ML2
ICM = (1/12) ML2
Since: I = ICM + MD2
Physics 1D03
Example:
Uniform thin hoop (mass M, radius R); axis perpendicular to hoop
P
CM
ICM = MR2
(why?)
P
CM
IP = ICM + MD2 (here “D” = R)
= 2 MR2
Physics 1D03
Example – 2 ways to solve a problem
The metre stick is pivoted at the 25-cm mark. What is
its angular acceleration when it is released?
Physics 1D03
Summary
Newton’s 2nd law for rotation about a fixed axis:
t
external
= Ia
Parallel-axis Theorem: I=ICM+MD2
Practice: Look over Examples in Chapter 10
Physics 1D03