If we consider a system comprising a massless string over a frictionless pulley,then we write the torque equation as $(T_2-T_1)R=I a$ ,where $T_2$ and $T_1$ are tensions on either side of the pulley.
The tangential force acting on the pulley is the friction $F$ between the pulley and the string. How is that the torque applied by friction is equal to the torque applied by the difference in tensions ? In other words how is the friction equal to the difference in the tensions ?
If we consider the pulley and the string over it as one system such that the string does not slip then the net force acting is the difference in the tensions and net torque $(T_2-T_1)R$.
But when we consider pulley in isolation then the force which applies torque is the friction between the string and pulley.
Could someone help me understand mathematically how do we calculate net torque on the pulley by considering pulley as the system i.e how is friction $F =T_2-T_1$ ?
Best Answer
**The previous answer I had posted was completely wrong
Consider an elemental length of string wrapped around the pulley. We know that the string is massless(light - its an approximation). There is a tension acting on the string $T+dT$ from one side and $T$ from the other side.(There is a small change in tension because we have considered a small part of the string only). There is a small amount of friction $df$ acting on this string.
The force equation of this string is as follows.
$$T+dT-T-df=dm\cdot a$$ We know that $dm=0$ so, $$dT=df$$ Integrate both sides and you get $$f_{\mathrm{total}}=\Delta T=T_2-T_1$$