Question

A particle $P$ will be equilibrium inside a hemispherical bowl of radius $0.5m$ at a height $0.2m$ from the bottom when the bow is rotated at an angular speed ($g=10m/se{c}^2$)

• A. $10/\sqrt{3}rad/sec$
• B. $104\sqrt{3}rad/sec$
• C. $10rad/sec$
• D. $\sqrt{20}rad/sec$

Answer 1

The correct option is A $10/\sqrt{3}rad/sec$

$N\sin \theta =mg$

$m{w}^{2}r=N\cos \theta$

$m{w}^{2}r=\frac{mg}{\sin \theta }\cos \theta$

$w^2=\frac{g}{r}\frac{1}{\tan \theta }$

$=\frac{10\times 10}{4}\times \frac{1}{3}\times 4$

$=\frac{100}{3}$

$\therefore w=\frac{10}{\sqrt{3}}$rad/sec