(from wikipedia.com) Mass Moments of Inertia (from wikipedia.com)

Mass Moments of Inertia (from wikipedia.com)
Rotational Kinetic Energy
Ball spinning around in a circle at a constant speed:
KE =
Replace v with _______
KE =
Mass moment of inertia: _____
More than one point masses: ________
Distributed mass: _________
EF 151 Spring, 2011 Lecture 4-2
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Mass Moments of Inertia (from wikipedia.com)
EF 151 Spring, 2011 Lecture 4-2
EF 151 Spring, 2011 Lecture 4-2
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Mass Moments of Inertia (from wikipedia.com)
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EF 151 Spring, 2011 Lecture 4-2
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Mass Moments of Inertia (from wikipedia.com)
Example: Hurricane
Estimate the rotational energy stored in a
hurricane. Model the hurricane as a uniform
cylinder 300 km in diameter, 5 km high, and a
rotational speed of 0.0004 rad/s (outer edge
speed of 60m/s). The density of air is 1.3kg/
m3.
An average house uses about 10x1010 J of
energy in a year. If we could harness the
rotational energy of a hurricane, how
many houses could we run for a year?
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Parallel Axis Theorem
Hurricane Katrina
Kg-m2/s2 is also a unit of:
A.  N/m
B.  lb/sec
C.  Joule
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Example: Yardstick
Given: A 0.15 lb yardstick is hung so it pivots about a point
3 inches down from the top.
Required: Mass moment of inertia about the pivot point
I = I CM + Md 2
I=
ICM =
M=
d = ____________ distance between axis of interest and
axis ___________ to this through center of mass
EF 151 Spring, 2011 Lecture 4-2
Icm is:
A.  (1/12)ML2
B.  (1/3)ML2
C.  (1/2)ML2
D.  ML2
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EF 151 Spring, 2011 Lecture 4-2
d is:
A.  3 in.
B.  15 in.
C.  18 in.
D.  36 in.
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Given: The yardstick is held horizontally, and then released. The lower
end strikes a ping pong ball at the bottom of the stroke.
Required: Speed with which yardstick strikes ping pong ball.
Given: To make the ping pong ball go further, a 0.13 lb, 2.5
in. diameter tennis ball is attached to the end of the yardstick.
Treat the tennis ball as a thin-walled hollow sphere.
Required: Mass moment of inertia about the pivot point
3 in.
Example: Yardstick
33 in.
Example: Yardstick, cont.
h in mgh is:
A.  3 in.
B.  15 in.
C.  18 in.
D.  36 in.
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Example: Mass Moment of Inertia
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R = P !Q
Vector Cross-Product
A person holding a set of dumbbells can be modeled as:
body: solid cylinder, m=60 kg, r=15 cm
arm: solid cylinder, m=5 kg, r=3 cm, l = 40 cm
dumbbell: point mass, m=3 kg
Determine the mass moment of inertia about a vertical
axis through the body of the person holding the
dumbbells with their arms extended horizontally.
Magnitude: _____________
R
Direction: ________________
to plane containing P and Q;
determined using ______ rule
Q
Commutative? _____
P
Q! P " P !Q
Special Cases:
! =0
! = 90!
!
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P !Q =
P !Q =
EF 151 Spring, 2011 Lecture 4-2
iˆ ! iˆ =
ˆj ! kˆ =
iˆ ! ˆj =
ˆj ! iˆ =
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Vector Cross Product: General Solution
iˆ
px
qx
ˆj
py
qy
P !Q
Example: Cross Product
Given: A = 2iˆ + 3 ˆj
B = 4iˆ ! ˆj + 5kˆ
kˆ
pz
qz
EF 151 Spring, 2011 Lecture 4-2
Required:
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A!B
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