Find the tension in the tape and the linear acceleration of the cylinder up the incline, assuming there is no slipping. (`Take M = 4m` and `g =10 m//s^(2)`)

Explanation

Since the cylinder does not slip, the displacement, velocity and acceleration of the hanging mass are, respectively, equal to the displacement, velocity and acceleration of the upper point `B` of the cylinder as the string leaves the upper point tangentially,
Now `v`(velocity of the upper point) `=v_(m)=omegar=2v_(CM)`
(`:.omegar=v_(m)`)
From conservation of energy, when `m` lowers through `h`
`mgh=1/2mv^(2)+(1/2mv_(CM)^(2)+1/2Iomega^(2))`

( `:.` loss in `PE=` gain i `KE`)
`=1/2mv^(2)+1/2(MV^(2))/4+1/2(1/2Mr^(2))(v_(CM)^(2))/(r^(2))`
(`:.I=1/2Mr^(2)`)
(`:'I=1/2Mr^(2)`)
`=1/2mv^(2)+1/8MV^(2)+1/4M(v^(2))/4`
`=1/2mv^(2)+3/16Mv^(2)`
`16mgh=8mv^(2)+3Mv^(2)`
`impliesv^(2)=(16mg)/(8m+3M)`
From `v^(2)=2ah`
`impliesa=(8mg)/(8m+3M)=g/(1+3M//8m)=4m//s^(2)`
Considering downwardd motion `m`
`mg-T=ma`
or `T=mg-ma=2(10-4)=12N`