Question

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

• 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 $M=1\, kg$,
radius of ball $R=0.5\,m$, 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} $
$=\frac{5}{7}=0.71\, N - m$