Question

A cord is wound a round the circumfererence of a wheel of mass 50Kg and diameter 0.3m. The axis of the wheel is horizonatal. A mass of 0.5Kg is attached at the end of the cord. Find the angular acceleration of the wheel?

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

Here the weight of the 0.5Kg applies a torque on the wheel about its axis.
$\tau =rFsin\theta$
Moment of Inertia $I=\frac{M{R}^2}{2}But\tau =I\alpha \therefore \alpha =\frac{\tau }{I}=\frac{rFsin\theta \times 2}{M{R}^2}Herer=R=0.15mF=mg=(0.5\times 9.8)N\theta ={90}^{\circ }M=50KgSubstituting,\alpha =\frac{0.15\times 0.5\times 9.8\times sin{90}^{\circ }\times 2}{50\times 0.15\times 0.15}=1.3rad/s^2$