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

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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