College Physics (7th Edition)

Published by Pearson
ISBN 10: 0-32160-183-1
ISBN 13: 978-0-32160-183-4

Chapter 8 - Rotational Motion and Equilibrium - Learning Path Questions and Exercises - Exercises - Page 303: 1

Answer

The mud is at 9 o'clock position. The velocity of this point will be the resultant of velocity towards the center and $r\omega$. So, $v=\sqrt (v_{cm}^{2}+r\omega^{2})$ For pure rolling, $v_{cm}=r\omega$, So, $v=\sqrt 2v_{cm}$ Hence, $\theta=tan^{-1}\frac{r\omega}{v_{cm}}=45^{\circ}$ Hence the mud on the wheel will fly off at an angle of $45^{\circ}$ with horizontal and with a speed equal to $\sqrt 2v_{cm}$.

Work Step by Step

The mud is at 9 o'clock position. The velocity of this point will be the resultant of velocity towards the center and $r\omega$. So, $v=\sqrt (v_{cm}^{2}+r\omega^{2})$ For pure rolling, $v_{cm}=r\omega$, So, $v=\sqrt 2v_{cm}$ Hence, $\theta=tan^{-1}\frac{r\omega}{v_{cm}}=45^{\circ}$ Hence the mud on the wheel will fly off at an angle of $45^{\circ}$ with horizontal and with a speed equal to $\sqrt 2v_{cm}$.
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