A rope of negligible mass is wound around a hollow cylinder of mass 4 kg and radius 40 cm. What is the angular acceleration of the cylinder, if the rope is pulled with a force of 4N? Assume that there is no splipping.

Options

A. `"2 rad/s"^(2)`
B. `"1.5 rad/s"^(2)`
C. `"2.5 rad/s"^(2)`
D. `"3 rad/s"^(2)`

Answer

C

Explanation

The M.I. of the hollow cylinder about its geometric axis is given by `I=MR^(2)=4xx(0.4)^(2)="0.64 kg m"^(2)`
When the rope is pulled with a force `F=4N,` a torque `tau=FxxR=4xx0.4=1.6Nm` is produced and it produces and angular acceleration `alpha." "(because tau=Ialpha)`
`therefore" "alpha=(tau)/(I)=(Fxxr)/(I)`
`therefore" "alpha=(4xx0.4)/(0.64)=(1.6)/(0.64)="2.5 rad/s"^(2)`