HW 8 Solutions

G the centripetal acceleration is acent v 2 r r 2 r

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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.

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