ch11-p015

# ch11-p015 - F N = 2 Mg ) that the center of mass speed...

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15. The physics of a rolling object usually requires a separate and very careful discussion (above and beyond the basics of rotation discussed in chapter 10); this is done in the first three sections of chapter 11. Also, the normal force on something (which is here the center of mass of the ball) following a circular trajectory is discussed in section 6-6 (see particularly sample problem 6-7). Adapting Eq. 6-19 to the consideration of forces at the bottom of an arc, we have F N Mg = Mv 2 /r which tells us (since we are given
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Unformatted text preview: F N = 2 Mg ) that the center of mass speed (squared) is v 2 = gr , where r is the arc radius (0.48 m) Thus, the ball’s angular speed (squared) is ω 2 = v 2 /R 2 = gr/R 2 , where R is the ball’s radius. Plugging this into Eq. 10-5 and solving for the rotational inertia (about the center of mass), we find I com = 2 MhR 2 /r – MR 2 = MR 2 [2(0.36/0.48) – 1] . Thus, using the β notation suggested in the problem, we find β = 2(0.36/0.48) – 1 = 0.50....
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## This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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