Question

A particle of mass m is sliding on a fixed frictionless concave surface of radius q as shown in the diagram. The particle is released from rest from point A (at height h<<a) from rest, then the kinetic energy of a body varies with $\theta$ as

• A. $mgh-mga(1-cos\theta )$
• B. $mga-mgh(1-cos\theta )$
• C. $mga(1-cos\theta )$
• D. $mgh(1-cos\theta )$

Answer 1

The correct option is A $mgh-mga(1-cos\theta )$

$mgh=\frac{1}{2}m{v}^2+mg(a-a\cos \theta )$

$\Rightarrow \frac{1}{2}m{v}^2=mgh-mga(1-\cos \theta )$

$\Rightarrow K.B.=mgh-mga(1-\cos \theta )$