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 = Mr^{2}

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 m^{2}

τ = I α = 12 Nm

I = .48 kg m^{2}

∴ 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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