ch11-p016

# ch11-p016 - rotational inertia(about the center of mass we...

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as t = 2 h/g . Then Eq. 4-21 (squared, and using d for the horizontal displacement) gives v 2 = gd 2 /2 h . Plugging this into our expression for β gives 2 g ( H h )/ v 2 – 1 = 4 h ( H h )/ d 2 1 Therefore, with the values given in the problem, we find β = 0.25. 16. The physics of a rolling object usually requires a separate and very careful discussion (above and beyond the basics of rotation discussed in chapter 11); this is done in the first three sections of Chapter 11. Using energy conservation with Eq. 11-5 and solving for the
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Unformatted text preview: rotational inertia (about the center of mass), we find I com = 2 MhR 2 /r – MR 2 = MR 2 [2 g ( H – h )/ v 2 – 1] . Thus, using the β notation suggested in the problem, we find β = 2 g ( H – h )/ v 2 – 1. To proceed further, we need to find the center of mass speed v, which we do using the projectile motion equations of Chapter 4. With v o y = 0, Eq. 4-22 gives the time-of-flight...
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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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