A hoop of radius 2 m weighs 100 kg. It rolls along a horizontal floor so that its centre of mass has a speed of 20 cm/s. How much work has to be done to stop it?
Radius of the hoop, r = 2 m
Mass of the hoop, m = 100 kg
Velocity of the hoop, v = 20 cm/s = 0.2 m/s
Total energy of the hoop = Translational KE + Rotational KE
Er = 1/2 mv2 + 1/2 I ω2
Moment of inertia of the hoop about its centre, I = mr2
Er = 1/2 mv2 + 1/2 (mr2) ω2
But we have the relation, v = rω
∴ Er = 1/2 mv2 + 1/2 mr2 ω2
∴ Er = 1/2 mv2 + 1/2 mv2
∴ Er = mv2
The work required to be done for stopping the hoop is equal to the total energy of the hoop.
∴Required work to be done, W = mv2 = 100 × (0.2)2 = 4 J
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(b) The instantaneous speed of the point of contact during rolling is zero.
(c) The instantaneous acceleration of the point of contact during rolling is zero.
(d) For perfect rolling motion, work done against friction is zero.
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(b) What is the force of friction after perfect rolling begins?
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(b) work done by gravitational force in the above case,
(c) work done by friction on a body sliding down an inclined plane,
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Very nice answer