Question

A wheel of radius $R=10\, cm$ and moment of inertia $I=0.05 \, kg\, m ^{2}$ is rotating about a fixed horizontal axis $O$ with angular velocity $\omega_{0}=10 \, rad / \sec$. A uniform rigid rod of mass $m=3\, kg$ and length $l=50\, cm$ is hinged at one end $A$ such that it can rotate about end $A$ in a vertical plane. End $B$ of the rod is tied with a thread as shown in figure such that the rod is horizontal and is just in contact with the surface of rotating wheel. Horizontal distance between axis of rotation $O$ of cylinder and $A$ is equal to $a$ $=30\, cm$. The wheel stops rotating after one second after the thread has burnt. If the coefficient of friction $\mu$ between the rod and the surface of the wheel is equal to $\frac{x}{5}$. Then the value of $x$ is _____$.\left(g=10\,ms ^{-2}\right)$

Answer 1

When thread is burnt, rod rests over the wheel, therefore, a normal reaction comes into existence, due to which friction is developed. This friction produces a retarding moment on the wheel and the wheel ultimately stops rotating.
To calculate retarding moment produced by the friction, first consider rotational motion of the wheel,
$\omega_{0}=10\, rad / \sec \,\omega=0, t=1 $ second, $ a=?$
Using $\omega=\omega_{0}+$ at, $\alpha=-10\, rad / \sec ^{2}$
Retardation moment, $\tau=I|\alpha|=0.5 \,Nm$
Let friction between rod and wheel be, $f$ then, $\tau=f R$ or $\tau=f R $
$\Rightarrow f=\frac{\tau}{R}=\frac{0.5}{0.1}=5\, N$
Let normal reaction exerted by wheel on rod be $N$.
Considering free body diagram of the rod.

Taking moments about $A$,
$N . a=m g \frac{l}{2}$
or $N \cdot(0.3)=3 \times 10 \times \frac{0.5}{2}$
Hence $N=25$ newton
But friction, $f=\mu N $
$\Rightarrow \mu=\frac{f}{N}=\frac{5}{25}=\frac{1}{5}$
Hence, $\mu=1 / 5$