Question

A particle moving with constant speed $u$ inside a fixed smooth spherical 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 $\frac{a\sqrt{3}}{2}$
• 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. the magnitude of the resultant force acting on the particle is zero, in an inertial frame.

Answer 1

The correct option is A the radius of the circular motion is $\frac{a\sqrt{3}}{2}$

Since the body is given just enough speed to complete the vertical circle

$u^2=5gR$

$v^2=u^2-2gR$

$=5gR-2gR$

$=3gR$

When the velocity is vertical, the body is acted by two acceleration $a_T{=}^{\prime }{g}^{\prime }$ tangentially and centripetal acceleration $a_c=\frac{v^2}{R}$ radially.

$a_c=\frac{v^2}{R}=\frac{3gR}{R}$

$=3g$

Total acceleration $=\sqrt{a_T^2+a_c^2}$

$=\sqrt{g^2+(3g{)}^2}$

$=g\sqrt{10}$

Hence a is the correct answer.