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 - Problems - Page 225: 59

Answer

7.27 m/s.

Work Step by Step

The lower end of the pole does not slip on the ground, the friction does no work. Mechanical energy is conserved. The initial energy is the potential energy, treating all the mass as though it were at the center of mass. The final energy is rotational kinetic energy, and the pole is rotating about its point of contact with the ground. The linear velocity of the falling tip of the pole is its angular velocity multiplied by the length. $$PE_i=KE_f$$ $$mgh=\frac{1}{2}I \omega^2$$ $$mg(L/2)=\frac{1}{2}(\frac{1}{3}mL^2)(v_{tip}/L)^2$$ $$ v_{tip}=\sqrt{3gL}=7.27m/s$$
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