Question

A uniform disk with a mass of 5.0 kg and diameter 30 cm rotates on a frictionless fixed axis through its center and perpendicular to the disk faces. A uniform force of 4.0 N is applied tangentially to the rim of the disk. What is the angular acceleration of the disk?

Answer 1

Answer:

Angular acceleration of the disk will be $\alpha =10.714rad/sec^2$

Explanation:

We have given mass of the disk m = 5 kg

Diameter of the disk d = 30 cm = 0.3 m

So radius $r=\frac{d}{2}=\frac{0.3}{2}=0.15m$

Moment of inertia of disk is given by $I=\frac{1}{2}mr^2=\frac{1}{2}\times 5\times 0.15^2=0.056kgm^2$

Force is given by F=4 N

Torque is given as $\tau =Fr=4\times 0.15=0.6N-m$

We also know that torque is given by $\tau =I\alpha$

$0.6=0.056\times \alpha$

$\alpha =10.714rad/sec^2$