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?
Answer & explanation
Correct answer: option 1
Since both discs have the same thickness and density :
M/(pi x R2 x t) = M’/((pi x 4 x R2 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.