A fixed pulley carries a weightless thread with masses `m_(1)` and `m_(2)` at its ends. There is friction between the thread and the pulley. It is such that the thread starts slipping when the ratio `m_(2)//m_(1)=eta_(0)` find :
(a) The friction coefficient
(b) The acceleration of the mass when `m_(2)//m_(1)=eta gt eta_(0)`

Explanation

Let, us consider a small element of the thread and draw free body diagram for this element.
(a) Applying Newton's second law of motion in projection form, `F_n=mw_n` for this element,
`(T+dT)sin(d theta//2)+Tsin (d theta//2)-dN=dm omega^2R=0`
or, `2Tsin(d theta//2)=dN`, [neglecting the term ( `dTsin theta//2`)]
or, `T d theta =dN`, as `sin (d theta)/(2)=(d theta)/(2)` (1)
Also, `d f r =kdN=(T+dT)-T=dT` (2)
From Eqs. (1) and (2),
`kTd theta=dT` or `(dT)/(T)=kd theta`
In this case `Q=pi` so,
or, or, `1n(T_2)/(T_1)=kpi` (3)
So, `k=1/pi1nT_2/T_1=1/pieta_0`
as `T_2/T_1=(m_2g)/(m_1g)=m_2/m_1=eta_0`
(b) When `m_2/m_1=eta`, which is greater than `eta_0`, the blocks will move with same value of acceleration. (say w) and clearly `m_2` moves downward. From Newton's second law in projection form (downward for `m_2` and upward for `m_1`) we get:
`m_2g-T_2=m_2w` (4)
and `T_1-m_1g=m_1w` (5)
Also `T_2/T_1=eta_0` (6)
Simultaneous solution of Eqs. (4), (5) and (6) yields:
`w=((m_2-eta_0m_1)g)/((m_2+eta_0m_1))=((eta-eta_0))/((eta+eta_0))g` (as `m_2/m_1=eta`)