Question

A solid sphere of mass $M$ and radius $R$ is released from the top of an inclined plane of inclination $\theta$. The minimum coefficient of friction between the plane and the sphere so that it rolls down the plane without sliding is given by:

• A. $\mu =tan\theta$
• B. $\mu =\frac{2}{3}tan\theta$
• C. $\mu =\frac{2}{5}tan\theta$
• D. $\mu =\frac{2}{7}tan\theta$

Answer 1

Answer: (D)

$\mu =\frac{2}{7}tan\theta$

Total kinetic energy of the rolling balls- $KE=\frac{1}{2}M{v}^2+\frac{1}{2}I{\omega }^2$
= $KE=\frac{1}{2}M{v}^2+\frac{1}{2}kM{R}^2(v/R{)}^2$
=$\frac{1}{2}(1+k)M{v}^2$
Equating this to the potential energy we get-
$\frac{1}{2}(1+k)M{v}^2=Mgy$
Thus we get-
$v^2=\frac{2gy}{1+k}$
We write the distance y as $Ssin\theta$.
Now by taking the derivative we get-
$2v\frac{dv}{dt}=\frac{2gsin\theta }{1+k}\frac{dS}{dt}$
or
$2v\frac{dv}{dt}=\frac{2gvsin\theta }{1+k}$
or
$a=\frac{gsin\theta }{1+k}$
Now as the frictional force is normal to the radius of the axis of rotation and it is applied at the surface of the ball at a distance R from the axis, so friction makes a torque on the ball equal to $\mu RMgcos\theta$
In order for the ball not to slip, the torque on the ball from friction can not be less than the total torque on the ball when it rolls, therefore-
thus we get $\mu RMgcos\theta \ge \frac{I}{R}a$
or $\mu \ge k\frac{a}{gcos\theta }$ (we have $I=kM{R}^2$)
Substituting the acceleration we got earlier in this equation we get-
$\mu \ge k\frac{gsin\theta /(1+k)}{gcos\theta }$
or
$\mu =\frac{k}{1+k}tan\theta =\frac{1}{1+\frac{M{R}^2}{I}}tan\theta$
Substituting moment of inertia I of the ball as$\frac{2}{5}M{R}^2$ we get-
$\mu \ge \frac{2}{7}tan\theta$