A rope of negligible mass is wound round 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? What is the linear acceleration of the rope? Assume that there is no slipping.
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