Question

A uniform disc of radius R is put over another uniform disc of radius 2R of same thickness and density. The edges of the two discs touch each other. What is the position of their centre of mass?

(a) at 5R/3 from the centre of the larger disc

(b) at 2R/3 from the centre of the larger disc

(c) at 3R/5 from the centre of the larger disc

(d) at 2R/5 from the centre of the larger disc

This question was posed to me in an interview.

Asked question is from System of Particles in section System of Particles and Rotational Motion of Physics – Class 11

Answer 1

Right choice is (c) at 3R/5 from the centre of the larger disc

The best explanation: Since both discs have the same thickness and density;

M/(pi x R^2 x t) = M’/((pi x 4 x R^2 x t)

M = M’/4

M = Mass of a smaller disc

M’ = 4M = Mass of larger disc

Sum of masses = M + 4M = 5M

The distance of centres of discs = 3R

The distance of the centre of mass from larger disc = (3R x M)/5M

 = 3R/5.