University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 10 - Dynamics of Rotational Motion - Problems - Exercises - Page 335: 10.77

Answer

(a) h=1.87m

Work Step by Step

Mass of the two balls (m) = 5kg Mass of the rod (M) = 8kg Length of the rod (l) = 4m Height from where the ball is dropped (H) = 12m Let the height reached by the second ball be h. Applying energy conservation, mgH=\frac{1}{2}mv^{2} v is the velocity of ball just before it strikes the rod v=\sqrt (2gh) Let \omega be the angular velocity of the system just after the ball strikes the rod Conserving angular momentum mv\frac{l}{2} = (\frac{Mx^{l}}{12} + 2\timesx^{\frac{l}{2}})\omega Now substituting value of v from above eqn we get \omega = 3.03 rad/sec v = 6.06 m/sec Now applying energy conservation mgh = \frac{1}{2} mx^{2} h = x^{2}/2g h = 1.87m
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