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Unformatted text preview: celeration; i.e., the point on the rim has the bigger tangential
acceleration.
(f) The angular acceleration is α, and is the same for both points.
(g) The centripetal acceleration is acent = v 2 /r = (rω )2 /r = rω 2 . Since ω is the same
for both points, the point on the rim has the bigger centripetal acceleration, since
it has the bigger value of r.
1 [ 2
1
I = 2 5 (0.500 kg )(0.0500 m ) + (0.500 kg )(0.200 m ) + 12 (0.0600 kg )(0.300 m )
2 2 2 = 0.0415 kg ⋅ m 2
he methane molecule (CH4 ) has four
2.TThepercent difference between I and I app is:
hydrogen atoms located at the vertices of
a regular tetrahedron of edge length 0.18
I − I app 0.0415 kg ⋅ m 2 − 0.0400 kg ⋅ m 2
nm, with the carbon atom at the center
=
= 3.6 %
2
I
0.0415
of the tetrahedron. Find the moment of kg ⋅ m
inertia of this molecule for rotation about
an axis that passes through the centers of
(b) The rotationaland one would hydrogen because Icm of a hollow sphere is greater
inertia of the increase
the carbon...
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This document was uploaded on 03/14/2014 for the course PHYS 18 at UC Merced.
 Spring '10
 collins
 Physics, Work

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