A rope is wound around a hollow cylinder of mass 3 kg and radius 40 cm . What is the angular acceleration of the cylinder if the rope is pulled with a force of 30 N ?

Options

A. 0.25 rad `s^(-2)`
B. 25 rad `s^(-2)`
C. 5 m `s^(-2)`
D. `25 m s^(-2)`

Answer

B

Explanation

`m = 3` kg , r= 40 cm = `40 xx 10^(-2) m , F = 30` N
Moment of inertia of hollow cylinder about its axis = `mr^(2) = 3 kg xx (0.4)^(2) m^(2) = 0.48 kg m^(2)`
The torque is given by t = `I alpha`
where I = moment of inertia , `alpha`= angular acceleration
In the given case , `tau = r F` , as the force is acting perpendicularly to the radial vector .
`therefore alpha = (tau)/(I) = (Fr)/(mr^(2)) = (F)/(mr) = (30)/(3 xx 40 xx 10^(-2)) = (30 xx 100)/(3 xx 40)`
`alpha = 25 rad s^(-2)`