A rope of negligi

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