Question

A small mass of 10 g lies in a hemispherical bowl of radius 0.5 m at a height of 0.2 m from the bottom of the bowl. The mass will be in equilibrium if the bowl rotates at an angular speed of $\left(g=10m{s}^{-2}\right)$

• A. $\frac{10}{\sqrt{3}}rad/s$
• B. $10\sqrt{3}rad/s$
• C. $10rad/s$
• D. $\sqrt{20}rad/s$

Answer 1

Answer: (A)

$x=\sqrt{0.5^2-0.3^2}=0.4m$
$\sin \theta =\frac{0.3}{0.5}=\frac{3}{5}$
$\Rightarrow \cos \theta =\frac{4}{5}$
F.B.D. of $P$
$N\sin \theta =mg$
$N\cos \theta =m{\omega }^2\cdot x$
$\Rightarrow \tan \theta =\frac{g}{{\omega }^2\cdot x}$

$\omega =\sqrt{\frac{g}{x\tan \theta }}=\sqrt{\frac{10}{(0.4)\times \frac{3}{4}}}$
$=\frac{10}{\sqrt{3}}rad{s}^{-1}$