A solid sphere of radius `3R`, a solid disc of radius `2R` and a ring of radius `R` (all are of mass `m`) roll down a rough inclined plane. Their accelerations are `a,b` and `c` respectively. Find the ratio of `a//b` and `b//c`.

Answer

`(a)/(b) =(15)/(14)` and `(b)/( c) =(4)/(3)`

Explanation

When the rigid body rolls down
For translation motion
`mg sin theta - f = ma` ….(1)
For rotational motion
The torque is provided by the friction
:. `f xx R = I alpha`
`f xx R = I (mK^(2)) (a)/( R)`
`K` is radius of gyration
`f = (mK^(2))/(R^(2)) alpha` ...(2)
Sub. (2) in (1)
`mg sin theta -(mK^(2))/(R^(2)) a = ma rArr a = (g sin theta)/(1 + K^(2)/(R^(2)))`
For sphere (solid) `a = (g sin theta)/(1+(2)/(5)) = (5 g sin theta)/(7)`
For disc `b = (g sin theta)/(1 + 1//2) = (2 g sin theta)/(3)`
For ring `c = (g sin theta)/(1+1) =(g sin theta)/(2)`
Ratio `(a)/(b) =(5 g sin theta)/(7) xx (3)/(2 g sin theta) = (15)/(14)`
& `(b)/( c) = (2g sin theta)/(3) xx (2)/(g sin theta) = (4)/(3)`.
.