ch11-p031

# ch11-p031 - v o = 40.0 m/s. Then with y = 1 2 y max , Eq....

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31. (a) Since the speed is (momentarily) zero when it reaches maximum height, the angular momentum is zero then. (b) With the convention (used in several places in the book) that clockwise sense is to be associated with the negative sign, we have L = r mv where r = 2.00 m, m = 0.400 kg, and v is given by free-fall considerations (as in chapter 2). Specifically, y max is determined by Eq. 2-16 with the speed at max height set to zero; we find y max = v o 2 /2 g where
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Unformatted text preview: v o = 40.0 m/s. Then with y = 1 2 y max , Eq. 2-16 can be used to give v = v o / 2 . In this way we arrive at L = 22.6 kg . m 2 /s. (c) As mentioned in the previous part, we use the minus sign in writing = r F with the force F being equal (in magnitude) to mg . Thus, = 7.84 N . m. (d) Due to the way r is defined it does not matter how far up the ball is. The answer is the same as in part (c), = 7.84 N . m....
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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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