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Expert-verified Found in: Page 333 ### Physics For Scientists & Engineers

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 $54.{0}^{\circ }$

See the step by step solution

## Step 1: Free body diagram

Consider the free body diagram shown below for the given situation. ## Step 2: Energy conservation

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

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

From this, we get

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

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

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

## Step 3: Substitution of the values

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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