Question

A uniform disc of mass m and radius r is pivoted at point p and is free to rotate in vertical plane. The centre C of disc is initially in horizontal position with p. if it is released from this position then it's angular acceleration when the line PC is inclined to the horizontal at an angle q

• A. $\alpha =\frac{2g\cos \theta }{6R}$
• B. $\alpha =\frac{5g\cos \theta }{5R}$
• C. $\alpha =\frac{4g\cos \theta }{3R}$
• D. $\alpha =\frac{2g\cos \theta }{3R}$

Answer 1

The correct option is D $\alpha =\frac{2g\cos \theta }{3R}$

we need to find the torque on the disc about its point P when it is at angle with the horizontal.

now by the equation we can say

$\tau =F\times R\times \sin \left(90-\theta \right)$

plug in values in it

$\tau =mg\times R\times \cos \theta$

now we can use

$\tau =I\alpha$

now the moment of inertia of disc about point $P$

$I=\frac{3}{2}m{R}^2$

now plug in all values in above equation

$mgR\cos \theta =\frac{3}{2}m{R}^2\times a$

so angular acceleration is given as

$\alpha =\frac{2g\cos \theta }{3R}$

so above is the angular acceleration of disc at given position

Hence,

option $\left(D\right)$ is correct answer.