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[Post New]15 Feb 2008 00:02:11 IST
    Subject: Re:Try dis plz. Rates assured
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karthik2007 (3733)

Blazing goIITian

Olaaa!! Perrrfect answer. 655  [884 rates]
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Here is my method.

Here there is no direct way of finding the frictional torque, so consider an annular ring. Its area = 2rdr.

mass = m/r2 x m' = 2mdr/R.

Normal rxn = 2mgdr/R, so force due to friction = 2umgdr/R.

Frictional torque about center = 2umgdr x r

Hence, torque for the disc = [0 ][r ] 2u(2pir)rgdr = 2uMgr/3

Hence we get the desired result.
Will nip in at times to solve problems :)
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