Question

The figure below shows a small spherical ball of mass m rolling down the loop track. The ball is released on the linear portion at a vertical height H from the lowest point. The circular part shown has a radius R. The tangential acceleration of the centre when the ball is at A is (neglect the radius of mass 'm')

• A. $-\frac{5}{7}gcos\theta$
• B. $-\frac{5}{7}gsin\theta$
• C. $-\frac{5}{9}gcos\theta$
• D. $-\frac{5}{9}gsin\theta$

Answer 1

The correct option is A $-\frac{5}{7}gcos\theta$

Using conservation of mechanical energy between top of the track and point A

$mgH=mg[R+Rsin\theta ]+\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2$

$I=\frac{2}{5}m{R}^2\&v=\omega r$ (Pure rolling)

$mgH=mgR[1+sin\theta ]+\frac{7}{10}m{v}^2$

Differentiating w.r.t. time

$0=mgR[0+cos\theta \frac{d\theta }{dt}]+\frac{7}{10}m\times 2v\frac{dv}{dt}$

$R\times \frac{d\theta }{dt}=v$

$\frac{dv}{dt}=-\frac{5}{7}gcos\theta$