Question

A ball of mass $1kg$ and radius $0.5m,$ starting from test rolls down on a $30^{\circ }$ inclined plane. The torque acting on the ball at the distance of the $7m$ from the starting point is close to (Take acceleration due to gravity as $10m/s^2$)

• A. 0.25 N-m
• B. 0.7 N-m
• C. 0.5 N-m
• D. 0.4 N-m
• E. 1.4 N-m

Answer 1

Answer: (B)

Given, ball of mass $M=1kg$,
radius of ball $R=0.5m$, angle of inclination $\theta =30^{\circ }$ A ball placed on a inclined plane is given in figure.

Acceleration of ball
$a=\frac{g\sin \theta }{1+\frac{I}{M{R}^2}}$
where, $I=$ moment of inertia of ball
$\Rightarrow a=\frac{10\times \sin 30^{\circ }}{1+\frac{2}{5}\frac{M{R}^2}{M{R}^2}}=\frac{5\times 5}{7}$
Torque on the ball $\tau =I\alpha =I\frac{a}{R}$
$\Rightarrow \tau =\frac{2}{5}M{R}^2\left[\frac{5\times 5}{7}\right]\frac{1}{R}$
$=\frac{2}{5}\times 1\times 0.5\times \frac{5\times 5}{7}$
$=\frac{5}{7}=0.71N-m$