Answer :

We need to find angular acceleration of cylinder when force was applied tangential to hollow cylinder (by pulling rope wound in it).


Torque produced by action of force on a body,


τ = F r


Where


F is force applied on body


r is perpendicular distance of point of application of force with axis of rotation.


Given


F = 30 N


r = 40 Cm = .4 m


τ = 30 × .4


= 12 Nm


On action of this torque (or tangential force), body gains an angular acceleration, say, α. In terms of α,


Torque, τ = I α


Where


I is moment of inertia of body


α is angular acceleration


Moment of inertia of hollow cylinder,


I = Mr2


Where


M is mass of cylinder


r is radius of cylinder


Given


M = 3 kg


r = 40 Cm = .4 m


I = 3 × (.4)2


= .48 kg m2


τ = I α = 12 Nm


I = .48 kg m2


Angular acceleration, α = τ /I


= 12/.48


= 25 rad s-2


Linear acceleration on point P,


a = r α


Where


α is angular acceleration


r is perpendicular distance of point P from axis of rotation


We need to find linear acceleration of rope, which is at distance of .4 m from axis of rotation,


Thus, r = .4 m


α = 25 s-2


Linear acceleration, a = r α


= .4 × 25


= 10m s-2


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