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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 $F =T_2-T_1$ ?

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If the pulley is frictionless, where is the torque of friction coming from? Do you mean massless pulley, because that's when torque of friction equals torque due to tension. –  udiboy1209 Aug 23 '13 at 17:04
    
You are contradicting yourself, since in your first sentence you are stating that the pulley is frictionless. Or do you mean between the pulley and the bearing? –  fibonatic Aug 23 '13 at 17:10
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A frictionless pulley means friction is absent between the pulley and the axle(bearings).There is sufficient friction present between the pulley and the string such that no slipping occurs. –  Tanya Sharma Aug 23 '13 at 17:26
    
I am pretty surprised that my question has been downvoted.I am curious to know who has given it a negative vote.Is there a way to know which member has downvoted or upvoted you ? –  Tanya Sharma Aug 24 '13 at 6:52
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1 Answer 1

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 ?

**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 $$f_{\mathrm{total}}=\Delta T=T_2-T_1$$

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The torque equation you have written is incorrect.My simple question is how does torque due to friction between string and pulley resulting in a value (T2-T1)R ?It is the friction which causes a torque on the pulley ,not tensions .I hope I have made myself more clear. –  Tanya Sharma Aug 24 '13 at 6:49
    
What, according to you is the correct torque equation? –  udiboy1209 Aug 24 '13 at 8:04
    
Iα = (T2-T1)R - fr Where T1 and T2 are the tensions, R is the pulley radius, r is the axle radius and f is the frictional force at the axle. –  Tanya Sharma Aug 25 '13 at 16:07
    
@TanyaSharma, I changed my answer –  udiboy1209 Aug 25 '13 at 16:23
    
Nice! In the same fashion can you explain how is the downward force on the pulley T1+T2 ? If I consider pulley with the string then I am convinced that the net downward force on the pulley is T1+T2 but what if we see pulley in isolation .The tangential force on the pulley is by friction and the normal force from the string acts radially .How is their net resultant in vertical direction equal to T1+T2 ? I hope I am clear with my question :) –  Tanya Sharma Aug 25 '13 at 16:34
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