A wheel of radius `R=10 cm` and moment of inertia `I=0.05kg m^(2)` is rotating about a fixed horizontal axis `O` with angular velocity `omega_(0)=10rads^(-1)`. A uniform riigid rod of mass `m=3 kg` and length `l=50 cm` is hinged at one end `A` such that it can rotate at 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`.
If the wheel stops rotating after one second after the thread has burnt, calculate coefficient of friction ,`mu` between the rod and the surface of the wheel. `(g = 10 ms^(-2))`

Answer

`0.2`

Explanation

When thread is burnt, rod rest 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//s^(-1) omega=0, t=1` second , `a=?`
Using `omega=omega_(0)+at, alpha=-10rad s^-(12)`
Retardtion moment `tau=I|alpha|=0.5Nm`
Let friction between rod and wheel be `F`, then `tau=FR` or `F=2` newton.
Let normal reacton exerted by wheel on rod be `N`.
Considering free body diagram of the rod.

Taking moment about `A`
`N.a=mgl/2` or `N=25newton `
But friction `F=muN`
`mu=0.2`