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 $\frac{a\sqrt{3}}{2}$
• 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 value of u is $\sqrt{\frac{5ag}{2}}$

Answer 1

Answer: (A), (B)

The correct options are
A The value of u is $\sqrt{\frac{3ag}{2}}$
B The radius of the circular motion is $\frac{a\sqrt{3}}{2}$

The forces on the particle are $mg$ ,downward $Normal$ towards the centre and

The $Centripetalforce$ Towards the centre

Balancing the Net force in horizontal and Vertical Direction

$Nsin30=mgandNcos30=\frac{m{v}^2}{r}$

Dividing both the equations

$tan30=\frac{gr}{v^2}$

and

$tan60=\frac{2r}{a}\sqrt{3}=\frac{2r}{a}r=\frac{\sqrt{3}a}{2}$

so $u=\sqrt{\frac{3ag}{2}}$