A uniform cylinder of mass `m` and radius `R` starts descending at a moment `t=0` due to gravity, Neglecting the mass of the thread, find
(a) the tension of each thread and the angular acceleration of the cylinder,
(b) the time dependence of the instantaneous power developed by the gravitational force.

Explanation

Linear motion: `mg-2T=ma` (`i`)
Rotational motion: `2TcdotR=I_(c.m.)alpha=(1)/(2)mR^(2)alpha` (`ii`)
`a=Ralpha` (`iii`)
From (`ii`), `T=(1)/(4)mRalpha=(1)/(4)maimplies ma=4T` in (`i`)
`mg-2T=4Timplies T=(mg)/(6)`
`T=(1)/(4)mRalphaimplies(mg)/(6)=(1)/(4)mRalphaimplies (2g)/(3R)`
`a=Ralpha=(2g)/(3)`
Linear velocity of cylinder after time `t`
`v=at=(2"gt")/(3)`
Power delivered by gravitational force
`P=m vecg cdot vecv=mgv=(2mg^(2)t)/(3)`
Tension`=(mg)/(6)`
Angular acceleration `=(2g)/(3R)`
Power `=(2mg^(2)t)/(3)`
OR
Here the cylinder rolls without slipping on vertical surface
`a=(gsintheta)/(1+(k^(2))/(R^(2)))=(gsin90^(@))/(1+(1)/(2))=(2)/(3)g`