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Asked by Apollo Last Modified
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
Related Questions
The moment of inertia of a body about a given axis is 1.2kgm^2 . Initially the body is at rest. In order to produce a rotational kinetic energy of 1500J, an angular acceleration of 25rad/second ^2 must be applied about that axis for a duration of
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