Question

A solid hemisphere of radius $R$ is placed on an inclined plane of inclination $\theta .$ What will be the maximum value of $\theta$ for which the hemisphere will not topple? (Assume that the solid hemisphere will not slide)

• A. $\theta ={\tan }^{-1}\left(\frac{2R}{\pi }\right)$
• B. $\theta ={\tan }^{-1}\left(\frac{3\pi }{4}\right)$
• C. $\theta ={\tan }^{-1}\left(2\right)$
• D. $\theta ={\tan }^{-1}\left(\frac{8}{3}\right)$

Answer 1

The correct option is D $\theta ={\tan }^{-1}\left(\frac{8}{3}\right)$

When the hemisphere is going to topple, the normal will shift towards point $\text{A}$.
Hence, FBD of hemisphere when it's just about to topple.


Here, $y_c=\frac{3R}{8}$ { $COM$ of the solid hemisphere}

Force $mg$ will act at the $COM$ of the sphere.

From torque equlibrium at point $\text{A}$:
$mg\sin \theta \times \left(\frac{3R}{8}\right)=mg\cos \theta \times R$
$\Rightarrow \tan \theta =\frac{8}{3}$
$\therefore \theta ={\tan }^{-1}\left(\frac{8}{3}\right)$