Question

A particle moving with constant speed, $u$, inside a fixed smooth hemispherical bowl of radius, $a$, describes a horizontal circle at a distance $\frac{a}{2}$ below its centre. Then

• A. The radius of the circular motion is $\frac{a\sqrt{5}}{2}$.
• B. The value of u is $\sqrt{\frac{3ag}{2}}$.
• C. The normal reaction of the spherical surface on the particle is $\frac{mg}{2}$.
• D. None of these.

Answer 1

The correct option is B The value of u is $\sqrt{\frac{3ag}{2}}$.

$r=a\sin {60}^o=\frac{a\sqrt{3}}{2}$

$N\sin {60}^o=\frac{{mu}^2}{r}$

$N\cos {60}^o=mg$

From the above two equations, we get $u=\sqrt{\frac{3ag}{2}}$