Engineering Mechanics: Statics & Dynamics (14th Edition)

Published by Pearson
ISBN 10: 0133915425
ISBN 13: 978-0-13391-542-6

Chapter 15 - Kinetics of a Particle: Impulse and Momentum - Section 15.7 - Principle of Angular Impulse and Momentum - Problems - Page 294: 110


$4.03m/s$, $725J$

Work Step by Step

We can determine the required speed and the work done as follows: According to the principle of conservation of angular momentum $mr_1v_1=mrv_t$ $\implies v_t=\frac{r_1v_1}{r}$ $\implies v_t=\frac{8(3)}{6}=4m/s$ Now, $v=\sqrt{v_t^2+v_r^2}$ We plug in the known values to obtain: $v=\sqrt{16+0.25}=4.03m/s$ The work done can be calculated as $W=\frac{mv_2^2}{2}-\frac{mv_1^2}{2}$ $\implies W=\frac{m}{2}(v_2^2-v_1^2)$ We plug in the known values to obtain: $W=\frac{200}{2}(16.25-9)$ This simplifies to: $W=725J$
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