Question

A circular rigid body of mass m, radius R and radius of gyration (k) rolls without slipping on an inclined plane of a inclination $\theta$. Find the linear acceleration of the rigid body and force of friction on it. What must be the minimum value of coefficient of friction so that rigid body may roll without sliding?

Answer 1

if a is the acceleration of the centre of mass of the rigid body and f the force of friction between sphere and the plane, the equation of translatory and rotatory motion of the rigid body will be.

mg sin $\theta$ - f = ma (Translatory motion) fR = I $\alpha$ (Rotatory motion)

$f=\frac{I\alpha }{R}$

I = $m{k}^2$, due to pure rolling a = $\alpha R$
$mgsin\theta -\frac{I\alpha }{R}=m\alpha R$

Mg sin $\theta =m\alpha R+\frac{I\alpha }{R}mgsin\theta R+\frac{m{k}^2\alpha }{R}$

Mg sin $\theta =ma+\frac{m{k}^2\alpha }{R}mgsin\theta =a\left[\frac{R^2+k^2}{R^2}\right]$

$a=\frac{gsin\theta }{\left[\frac{R^2+k^2}{R^2}\right]}a=\frac{gsin\theta }{(1+\frac{k^2}{R^2})}f=\frac{I\alpha }{R}f=\frac{m{k}^{2}a}{R^2}\Rightarrow \frac{mg{k}^{2}sin\theta }{R^2+k^2}$

$fleq\cup N\frac{m{k}^2}{k^2}a\le \cup \le mgcos\theta$

$R^2\frac{k^2}{R^2}\times \frac{gsin\theta }{(k^2+R^2)}\ne \cup gcos\theta \cup \ge \frac{tan\theta }{[1+\frac{R^2}{k^2}]}{\mu }_{min}=\frac{tan\theta }{[1+\frac{R^2}{k^2}]}$