Question

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

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

Answer 1

The correct option is A $\frac{10}{\sqrt{3}}rad./s$

Let the angular velocity be $w$.

From figure, $tan\theta =\frac{0.3}{0.4}=\frac{3}{4}$

As the mass is in equilibrium, thus $Nsin\theta =mg$ ..........(1)

Also $Ncos\theta =mr{w}^2$ where $r=0.4$ m .............(2)

Dividing (1) and (2) we get $tan\theta =\frac{g}{r{w}^2}$

$\therefore$ $w^2=\frac{g}{rtan\theta }=\frac{10}{0.4\times \frac{3}{4}}=\frac{100}{3}$ $⟹w=\frac{10}{\sqrt{3}}$ rad/s