Question

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?

Short answer

Hence, the required angle is$54.0^{\circ }$

Step-by-step solution

Free body diagram

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

Energy conservation

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

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

From this, we get

$g(1-\cos \theta )=\frac{7}{10}\left(\frac{v_f^2}{R}\right)\mathrm{.......}\left(1\right)$

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

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

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

$\begin{array}{c}mg\cos \theta =\frac{m{v}_f^2}{R}\\ \frac{v_f^2}{R}=g\cos \theta \mathrm{.........}\left(2\right)\end{array}$

Substitution of the values

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

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

As a result, the needed angle is

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