Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 8 - Rotational Motion - General Problems - Page 227: 88

Answer

$v_2 = 4.0~m/s$

Work Step by Step

Since the string pulls at a $90^{\circ}$ angle to the motion of the mass, there is no net external torque on the system. Therefore angular momentum is conserved: $I_2~\omega_2 = I_1~\omega_1$ $(m~r_2^2)(\frac{v_2}{r_2}) = (m~r_1^2)(\frac{v_1}{r_1})$ $(r_2)(v_2) = (r_1)(v_1)$ $v_2 = \frac{ (r_1)(v_1)}{v_2} = \frac{(0.80~m)(2.4~m/s)}{0.48~m}$ $v_2 = 4.0~m/s$
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