A uniform ring of mass M and radius R=1m is placed horizontally on a frictionless horizontal surface. A particle of mass `M/2` is placed in contact with the inner side of ring as shown in the figure. What is the acceleration of the ring (in SI unit) just after the particle is given velocity `v_(0)=9m//s`
tangentially to the ring?

Answer

`27.00`

Explanation

Acceleration of particle in centre of ring frame is `(v^(2))/r` acceleration of centre of ring is `N/m`.

`N+M/2 a_(CM)=(Mv^(2))/(2r)`
`(3M)/2a_(cm)=(Mv^(2))/(2r)`
`a_(CM)=((v^(2))/(3r))`
`a_(CM)=((9^(2))/(3xx1))=27ms^(-2)`