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 ?

A rope is wound around a hollow cylinder of mass 3 kg and radius 40 cm

1.	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 ?
Answer» 0.25 rad `s^(-2)` 25 rad `s^(-2)` 5 m `s^(-2)` `25 m s^(-2)` Solution :`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 BYT = `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)`