Suppose we consider friction between string and the pulley while still considering the string to be massless. So, in such a theoretical case, the tension in the string in contant with the pulley will not be constant and its variation is calculated as follows.
If the string is just on the verge of slipping on the pulley, then friction acting is limiting and if `T_(2) gt T_(1)` then friction will act towards `T_(1)` in tangential direction as shown.

`d N = (T + df) sin (d theta//2) + (T + dT) sin (d theta//2)`
`dN = T sin (d theta//2) + df sin (d del//2) + T sin (d theta//2) + d T sin (d theta//2)`
`dN ~~ 2T sin (d theta//2)`
`d N ~~ 2T (d theta)/(2)`
`d N = T d theta` ........(1)
Alsso, `(T + df) cos ((d theta)/(2)) = (T + d T) cos ((d theta)/(2))`
`T cos ((d theta)/(2)) + df cos ((d theta)/(2)) = T cos ((d theta)/(2)) 1 dT cos ((d theta)/(2))`
`df = dT = mu d N`.........(2)
from (1) & (2)
`dT = mu T d theta`
`int_(T_1)^(T_2) (dT)/(T) = mu int_(0)^(theta) d theta`
`ln ((T_(2))/(T_(1))) = mu theta`
`T_(2) = T_(1) e^(mu theta)`, where `theta` is the angle of contact between the string and the pulley Based on above information, answer the following questions.
Figure shows a cylinder mounted on an horizontal axle. A massless string is wound on it with two and a half turns and connected to two masses m and 2m. If the system is in limiting equilibrium, find the coefficient of friction between the string and the pulley surface.

Options

A. `(1)/(5pi) 1 n (2)`
B. `(2)/(5 pi) 1 n (2)`
C. `(1)/(10 pi) 1 n (2)`
D. `(1)/(3 pi) 1 n (2)`

Answer

A

Explanation

If the system is in equilibrium, the tensions in two hanging parts of the string will be mg and 2mg respectively. From equation, we have `2 mg = mg e^(mu (5 pi))`
[As the angle of contact `= 5 pi` ]
`mu = (1)/(5 pi) I n (2)`