In the above question what is the acceleration of the cylinder?

Options

A. g
B. `g//3`
C. `2g//3`
D. `(3g)/(4)`

Answer

C

Explanation

Let the cylinder fall through a height .h. in time t, then the loss in P.E. is equal to gain in K.E. thus
`Mgh=(1)/(2)Mv^(2)+(1)/(2)Iomega^(2)`
or `Mgh=(1)/(2)Mv^(2)+(1)/(2)xx(1)/(2)MR^(2)xx(v^(2))/(R^(2))`
`=(1)/(2)Mv^(2)+(1)/(4)Mv^(2)=(3)/(4)Mv^(2)`
or `gh=3//4v^(2),"But "v^(2)=2ah`
`gh=(3)/(4)xx2ah`
`therefore a=(2g)/(3)`