Question

A solid sphere of mass $m$ and radius $R$ is rigidly enclosed in a solid hollow shell of same mass $m$ & radius $R$. This body is rolling down a rough incline of inclination $\dfrac{\pi }{4}$. If the minimum value of coefficient of static friction is $\dfrac{x}{{23}}$. Find the value of $x$?

Answer 1

Hint: For the given question we have to formulate an equation from the relation between the angle of inclination and coefficient of friction where we could use moment of inertia. Then by substituting the values we will find the coefficient of friction, then by comparing with the given answer of the question, we will find the value of $x$.

Complete step by step answer:
It is given in the question that both the hollow shell and the solid sphere is rolling through an inclined plane at angle $\dfrac{\pi }{4}$ to the horizontal.
Let the coefficient of static friction be $\mu $.
When an object is in rolling motion at angle with the inclined plane then the relation between the angle of inclination and coefficient of friction is,
$\mu \geqslant \dfrac{{\tan \theta }}{{1 + \dfrac{{{R^2}}}{{{K^2}}}}} - - - - - \left( 1 \right)$
The variables are defined as,
$\mu = $ coefficient of friction
$\theta = $ angle of inclination
$R = $ radius
$K = $ radius of gyration
Multiplying the numerator and denominator of the right side by $m$ of equation $\left( 1 \right)$ we get,
$\mu \geqslant \dfrac{{\tan \theta }}{{1 + \dfrac{{M{R^2}}}{{M{K^2}}}}} - - - - - \left( 2 \right)$
We know that $I = M{K^2}$ where $I$ is the moment of inertia.
Hence the formula arises as,
$\mu \geqslant \dfrac{{\tan \theta }}{{1 + \dfrac{{M{R^2}}}{I}}} - - - - - \left( 3 \right)$
The net moment of inertia $I = \dfrac{2}{3}m{R^2} + \dfrac{2}{5}m{R^2} = \dfrac{{16}}{{15}}m{R^2}$
The total mass of the two bodies are $M = 2m$
And the angle of inclination is given as, $\theta = \dfrac{\pi }{4}$
Again, we know, $\tan \dfrac{\pi }{4} = 1$
Substituting all the values in equation $\left( 3 \right)$ we get,
$\mu \geqslant \dfrac{1}{{1 + \dfrac{{2m{R^2}}}{{\dfrac{{16}}{{15}}m{R^2}}}}}$
Eliminating $m{R^2}$ we get,
$\mu \geqslant \dfrac{1}{{1 + \dfrac{2}{{\dfrac{{16}}{{15}}}}}} \geqslant \dfrac{1}{{1 + \dfrac{{15}}{8}}} \geqslant \dfrac{8}{{23}}$
So, the minimum value of coefficient of friction is $\mu = \dfrac{8}{{23}}$.
In the given question the minimum value of coefficient of friction is given as $\dfrac{x}{{23}}$.
Comparing both we get,
$\dfrac{x}{{23}} = \dfrac{8}{{23}}$
$ \Rightarrow x = 8$
Thus, the value of $x = 8$.

Note: It must be noted that the total moment of inertia of the bodies is the algebraic sum of the moment of inertia of the hollow shell and the solid sphere. Moment of inertia of an object is generally the mass of the object when it is rotating.