Question

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.​​

Answer 1

Answer:

Explanation:

Here,

Mass of hollow cylinder (m) = 3kg

Radius ( r) = 40 cm

Force (F) = 30 N

Moment of inertia of hollow sphere about its axis of symmetry ( I) = mr²

= 3 × (0.4)² = 0.48 Kg.m²

Torque acting on cylinder by applied force = r× F

= 30 × 0.4 = 12N.m

Angular acceleration produced in cylinder ,

angular acceleration = torque /M.O.I

= 12/0.48 = 25 rad/s²

Linear acceleration = angular acceleration × radius

= 0.4 × 25 = 10 m/s²

Answer 2

Explanation:

Mass of the hollow cylinder, m = 3 kg

Radius of the hollow cylinder, r = 40 cm = 0.4 m

Applied force, F = 30 N

The moment of inertia of the hollow cylinder about its geometric axis:

I = mr2

= 3 × (0.4)2 = 0.48 kg m2

Torque, τ = F × r = 30 × 0.4 = 12 Nm

For angular acceleration α, torque is also given by the relation:

τ = Iα

α = τ / I = 12 / 0.48 = 25 rad s-2

Linear acceleration = τα = 0.4 × 25 = 10 m s–2