A block of mass `M` is situation on a smooth horizontal frictionless table. A thread tied to the block passes through a hole in the table and carries a mass `m` at its other end if the length of the thread above the table is l, what should be the value of `m` so that it may remain suspended at a constant height and the block `M` moves in a circular p-ath with an angular velocity `omega` on the table?

Options

A. `(Mlomega^(2))/(g)`
B. `(Mlomega^(2))/(3g)`
C. `(Mlomega^(2))/(5g)`
D. `(2Mlomega^(2))/(g)`

Answer

A

Explanation

The situation is shown in figure. For the mass `m` to be stationary, the tension in the string should provide the necessary centripetal force on the rotating mass `M` . Now
`T=mg` and `T=Mlomega^(2)` or `mg=mlomega^(2)`
or `m=(Mlomega^(2))/(g)` .