A rope is wound round a hollow cylinder of mass `3 kg and radius 40 cm`. If the rope is pulled with a force of `30 N`, what is the angular acceleration of the cylinder ?

Options

A. `15 rad s^(-2)`
B. `20 rad s^(-2)`
C. `25 rad s^(-2)`
D. `30 rad s^(-2)`

Answer

C

Explanation

Here, M = 3kg
`R=40cm = 40 xx 10^(-2)`m
Force applied, `F=30N`
Torque, `tau=FR = (30N)(40 xx 10^(-2)m) =12Nm`
Moment of inertia of hollow cylinder about its axis is
`I=MR^(2)=(3)kg(40 xx 10^(-2)m)^(2)=0.48 kg m^(2)`
Let `alpha` is the angular acceleration produced.
As `tau=Ialpha`
`therefore alpha = tau/I = (12N m)/(0.48 kg m^(2)) = 25 rad s^(-2)`