System of particles and Rotational motion NCERT Solutions Class 11 Physics - Solved Exercise Question 7.19

Question 7.19:
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?
Solution:
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
ET = (1/2)mv2 + (1/2) I ω2
Moment of inertia of the hoop about its centre, I = mr2
ET = (1/2)mv2 + (1/2) (mr2)ω2
But we have the relation, v = rω
∴ ET = (1/2)mv2 + (1/2)mr2ω2
= (1/2)mv2 + (1/2)mv2 = 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

No comments:

Post a Comment