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 $\left(g=10 ms ^{-2}\right)$

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

Answer 1

Answer: (A)

$x=\sqrt{0.5^{2}-0.3^{2}}=0.4 \,m$
$\sin \theta=\frac{0.3}{0.5}=\frac{3}{5}$
$ \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}}}$
$=\frac{10}{\sqrt{3}} \,rad\,s^{-1}$