Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 8 - Dynamics II: Motion in a Plane - Exercises and Problems - Page 215: 64

Answer

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Work Step by Step

a) From the figure below, we can apply Newton's second law to find the ball's angular velocity. $$\sum F_x=F_n\cos\theta=ma_r=m\dfrac{v^2}{r}$$ where $\cos\theta=\dfrac{r}{R}$; thus $$ F_n\dfrac{r}{R}=m\dfrac{v^2}{r}$$ where $v=\omega r$ $$ F_n\dfrac{\color{blue}{\bf\not} r}{R}=m\dfrac{\omega^2 \color{blue}{\bf\not} r^{\color{red}{\bf\not} 2}}{\color{red}{\bf\not} r}$$ Hence, $$\omega=\sqrt{\dfrac{F_n}{mR}}\tag 1$$ Now we need to find the normal force; $$\sum F_y=F_n\sin\theta-mg=ma_y=m(0)=0$$ Thus, $$F_n=\dfrac{mg}{\sin\theta}$$ where $\sin\theta=\dfrac{R-y}{R}$ So that $$F_n=\dfrac{mgR}{R-y}$$ Plugging into (1); $$\omega=\sqrt{\dfrac{ \color{red}{\bf\not} m \color{red}{\bf\not} R g}{ \color{red}{\bf\not} m \color{red}{\bf\not} R(R-y)}}$$ $$\boxed{\omega=\sqrt{\dfrac{ g}{ R-y }}}$$ ________________________________________________ b) The minimum value of $\omega$ to move on a circle is approach as $R-y$ approaches $R$. In other words, we have the minimum value of $\omega$ when $R-y\approx R$ and hence, $$\boxed{\omega_{mini}=\sqrt{\dfrac{g}{R}}}$$ ________________________________________________ c) $y=\frac{1}{2}R$ since the ball is halfway up. Plugging the given into the boxed formula in part (a). $$ \omega=\sqrt{\dfrac{ 9.8}{ R-\frac{1}{2}R }}=\sqrt{\dfrac{ 9.8}{ \frac{1}{2}\times0.2 }} =\bf 9.9\;\rm rad/s$$ $$ \omega=\rm \dfrac{9.9\;rad}{1\;s}\times\dfrac{60\;s}{1\;min}\times \dfrac{1\;rev}{2\pi \;rad}$$ $$ \omega=\color{red}{\bf 94.5}\rm rpm $$
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