A disc is undergoing pure rolling motion on a flat surface. The radius of the disc is 5cm & velocity of centre of mass is 10m/s. What is the ratio of kinetic energy as seen from the ground frame to a frame moving along the disc with a speed of 10m/s?

Category: QuestionsA disc is undergoing pure rolling motion on a flat surface. The radius of the disc is 5cm & velocity of centre of mass is 10m/s. What is the ratio of kinetic energy as seen from the ground frame to a frame moving along the disc with a speed of 10m/s?
Editor">Editor Staff asked 1 year ago

A disc is undergoing pure rolling motion on a flat surface. The radius of the disc is 5cm & velocity of centre of mass is 10m/s. What is the ratio of kinetic energy as seen from the ground frame to a frame moving along the disc with a speed of 10m/s?
 
(a) 1:1, kinetic energy is frame independant
 
(b) 3:1
 
(c) 2:1
 
(d) 1:3
 
I have been asked this question during an online interview.
 
I’d like to ask this question from Rolling Motion topic in chapter System of Particles and Rotational Motion of Physics – Class 11
Select the correct answer from above options 
Interview Questions and Answers, Database Interview Questions and Answers for Freshers and Experience

1 Answers
Editor">Editor Staff answered 1 year ago

The correct choice is (b) 3:1
 
Easiest explanation: The kinetic energy of a pure rolling disc as seen from the ground frame is given by:                                   1/2MV^2 + 1/2Iw^2.
 
I = MR^2/2                                                                                                                                                                     w = V/R
 
∴ 1/2MV^2 + 1/2Iw^2 = 1/2MV^2 + 1/4MV^2 = 3/4MV^2.
 
Kinetic energy as seen from the moving frame will only be the energy due to rotation about the disc’s axis.
 
∴ Kinetic energy from moving frame = 1/2Iw^2 = 1/4MV^2.
 
∴ Ratio = (3/4MV^2) / (1/4MV^2)
 
= 3:1.


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