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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Welcome to the NCERT Solutions for Class 11 Physics - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 19: A hoop of radius 2 m weighs 100 kg. It rolls along a horizontal floor so that its centre of mass has....
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Very nice answer