A hoop of radius 2 m weighs 100 kg. It r | Class 11 Physics Chapter System of Particles and Rotational Motion, System of Particles and Rotational Motion NCERT Solutions

Question:

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?

Answer:

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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Comments

  • Jaegajal ihaehan
  • 2019-10-16 23:34:45

Very nice answer


Comment(s) on this Question

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....