A uniform cylinder of radius 1 m, mass 1 kg spins about its axis with an angular velocity 20 rad/s. At certain moment, the cylinder is placed into a corner as shown in the figure. The coefficient of friction between the horizontal wall and the cylinder is `mu` , whereas the vertical wall is frictionless. If the number of rounds made by the cylinder is 5 before it stops, then the value of `mu` is (acceleration due to gravity, `g = 10 m//s^2` )

Options

A. `3/pi`
B. `2/pi`
C. `1/pi`
D. `(0.4)/(pi)`

Answer

C

Explanation

Free body diagram of cylinder is

We have following equation
`N_2 = mg ` …(i)
`mu N_2 = N_1 " " ….(ii)`
Torque about centre of cylinder is
`mu N_2R = (mR^2)/(2) alpha " " .....(iiii)`
From eqs, (i) and (iii) , we get
` mu mg R = (mR^2)/(2) alpha rArr alpha = (2mu g)/(R)`
Now using
` omega^2 = omega_0^2 - 2 alpha theta`
we have , `0 = (20)^2 - 2 ((2 xx mu xx 10)/(1)) theta`
Here , ` theta = 5 xx 2pi`
so , `(20)^2 = 40 xx 10pi (mu) rArr mu = 1/pi`