A rope is wound round a hollow cylinder of mass 2 kg and radius 0.5 m. If the rope is pulled with a force of 80 N, the angular acceleration of the cylinder will be -

Options

A. 20 rad/`s^(2)`
B. 40 rad/`s^2 `
C. 60 rad/`s^2 `
D. 80 rad/`s^2 `

Answer

D

Explanation

`I=Mr^(2)`
Torque acting on the cylinder by pulling the rope with a force F is
`T=r F sin theta `
`=F.(r sin theta )=F.r _(1) `
`=80xx0.5 = 40n-m`
`and 1=Mr^(2) = 2xx(0.5)^(2) = 0.5 kg m^(2)`
`therefore alpha =( tau)/(I) (40N -m)/(0.5 kg - m^(2))=80 rad //s^(2)`