Question

A wheel of radius $R=10cm$ and moment of inertia $I=0.05kg{m}^2$ is rotating about a fixed horizontal axis $O$ with angular velocity ${\omega }_0=10rad/\sec$. A uniform rigid rod of mass $m=3kg$ and length $l=50cm$ 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 $a$ $=30cm$. 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 _____$.\left(g=10m{s}^{-2}\right)$

Answer 1

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,
${\omega }_0=10rad/\sec \omega =0,t=1$ second, $a=?$
Using $\omega ={\omega }_0+$ at, $\alpha =-10rad/{\sec }^2$
Retardation moment, $\tau =I|\alpha |=0.5Nm$
Let friction between rod and wheel be, $f$ then, $\tau =fR$ or $\tau =fR$
$\Rightarrow f=\frac{\tau }{R}=\frac{0.5}{0.1}=5N$
Let normal reaction exerted by wheel on rod be $N$.
Considering free body diagram of the rod.

Taking moments about $A$,
$N.a=mg\frac{l}{2}$
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, $\mu =1/5$