Question

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 contact 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  T₂ > T₁ then friction will act towards T₁ in tangential direction as shown.

dN = (T + df) sin (dθ/2) + (T + dT) sin (dθ/2)
dN = T sin (dθ/2) + df sin (dθ/2) + T sin (dθ/2) + dT sin (dθ/2)
dN ≈ 2T sin (dθ/2)
dN ≈ 2T
dN = Tdθ    ... (1)
Also,    (T + df) cos$\left(\frac{\mathrm{d\theta }}{2}\right)$ = (T + dT) cos $\left(\frac{\mathrm{d\theta }}{2}\right)$
T cos $\left(\frac{\mathrm{d\theta }}{2}\right)$ + df cos $\left(\frac{\mathrm{d\theta }}{2}\right)$ = T cos $\left(\frac{\mathrm{d\theta }}{2}\right)$ + dT cos $\left(\frac{\mathrm{d\theta }}{2}\right)$
df = dT = µdN    ... (2)
from (1) & (2),
dT = µTdθ 
$\underset{{\mathrm{T}}_1}{\overset{{\mathrm{T}}_2}{\int }}\frac{\mathrm{dT}}{\mathrm{T}}=\mathrm{\mu }\underset{0}{\overset{\mathrm{\theta }}{\int }}\mathrm{d\theta }$
$\mathrm{\ell n}\left(\frac{{\mathrm{T}}_2}{{\mathrm{T}}_1}\right)=\mathrm{\mu \theta }$
T₂ = T₁e^µθ , where θ is the angle of contact between the string and the pulley. 

Based on above information, answer the following questions.