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Q86P

Expert-verifiedFound in: Page 333

Book edition
9th Edition

Author(s)
Raymond A. Serway, John W. Jewett

Pages
1624 pages

ISBN
9781133947271

**Review. A clown balances a small spherical grape at the top of his bald head, which also has the shape of a sphere. After drawing sufficient applause, the grape starts from rest and rolls down without slipping. It will leave contact with the clown’s scalp when the radial line joining it to the center of curvature makes what angle with the vertical?**

Hence, the required angle is $54.{0}^{\circ}$

Consider the free body diagram shown below for the given situation.

The energy conservation between the apex and the point where the grape leaves meet the surface.

$\begin{array}{l}mg\Delta y=\frac{1}{2}m{v}_{f}^{2}+\frac{1}{2}I{\omega}_{f}^{2}\\ mgR(1-\mathrm{cos}\theta )=\frac{1}{2}m{v}_{f}^{2}+\frac{1}{2}\left(\frac{2}{5}m{R}^{2}\right){\left(\frac{{v}_{f}}{R}\right)}^{2}\end{array}$

From this, we get

$g(1-\mathrm{cos}\theta )=\frac{7}{10}\left(\frac{{v}_{f}^{2}}{R}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{.......}\left(1\right)$

Take a look at the radial forces at work on the grape.

$mg\mathrm{cos}\theta -n=\frac{m{v}_{f}^{2}}{R}$

We $n\to 0$obtain at the moment where the grape leaves the surface

$\begin{array}{c}mg\mathrm{cos}\theta =\frac{m{v}_{f}^{2}}{R}\\ \frac{{v}_{f}^{2}}{R}=g\mathrm{cos}\theta \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{.........}\text{\hspace{0.17em}}\left(2\right)\end{array}$

We get by substituting equation into$\left(2\right)$ equation . $\left(1\right)$

$\begin{array}{c}g-g\mathrm{cos}\theta =\frac{7}{10}g\mathrm{cos}\theta \\ \mathrm{cos}\theta =\frac{10}{17}\end{array}$

As a result, the needed angle is

$\begin{array}{c}\theta ={\mathrm{cos}}^{-1}\left(\frac{10}{17}\right)\\ ={54.0}^{\circ}\end{array}$

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