A string is wrapped several time on a cylinder of mass `M` and radius `R`, the cylinder is pivoted about its axis of symmetery . A block mass of `m` tied to the string rests on support so that the string is black. The block is lifted upto a height `h` and the support is removed. (shown in figure)

When the string experience a jerk, a large impulsive force is generated for a short duration, so that contribution of weight `mg` can be neglected suring this duration. then what will be speed of blocks `m`, just after string has becomes taut

Options

A. `(sqrt(2gh))/([1+M/m])`
B. `(sqrt(2gh))/([1+M/(2m)])`
C. `(sqrt(gh))/([1+M/m])`
D. `(sqrt(gh))/([1+M/(2m)])`

Answer

B

Explanation

When the string experiences a jerk, the large impulse developed is of very short duration so that the contribution of weight `mg` can be neglected during this time interval. The angular momentum of the system is conserved, as the tension is internal force for the system. thus we have `vec(L)_(i)=vec(L)_(f)`
`mv_(1)R+1/2MR^(2)omega_(1)=mv_(0)R=msqrt(2gh)R`
The string is inextensible, so `v_(1)=Romega_(1)`. on solving for `omega_(1)` we get `omega_(1)=(sqrt(2gh))/(R[1+(M//2m)])`
`v_(1)=Romega_(1)=(sqrt(2gh))/([1+(M//2m)])`
the final kinetic energy `K_(1)` is given by
`K_(1)=1/2mv_(1)^(2)+1/2Iomega_(1)^(2)=1/2mv_(1)^(2)+1/2(1/2MR^(2))((v_(1)^(2))/(R^(2)))`
`=1/2(m+M/2)v_(1)^(2)=1/2[(mv_(0)^(2))/(1+(M//2m))]`
`=(K_(0))/(1+(M//2m))`,`{ `:'` K_(o)=1/2mv_(0)^(2)}`