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.
Answers
Answer:
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
Answer:
Explanation:
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