Question

A force `F` acts tangentially at the highest point of a solid cylinder of mass `m` kept on a rough horizontal plane. If the cylinder rolls without slipping, find the acceleration of the center of cylinder.

Answer 1

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`