A uniform thin circular ring of mass `m (m = 0.4 kg)` and radius `R` has a small particle of the same mass m fixed on it as shown in Fig. The line joining the particle to centre is initially horizontal. The ground is frictionless. Find the contact force (magnitude) exerted by the ground on the ring, when the system is released from rest.

Answer

`3/2mg`

Explanation

Applying torque equation about centre of mas
`tau_(CM)=I_(CM)alpha`………i
`implies N R/2=(I_(CM))alpha`…ii
`I_(CM)=mR^(2)+m(R/2)^(2)+m(R/2)^(2)=3/2mR^(2)`
The `CM` of system (ring+particle) will be at a distance `R//2` from the centre of the ring.

`2mg-N=2m(a_(CM))_(y)` .....iii
using constraint acceleration of centre of circle is zero
`(a_c)_(y)=0`
`implies R/2alpha=(a_(CM)_(y))`.......iv
Form ii, iii and iv we get
`N=3/2mg`