Question

A uniform ring of mass $M$ and radius $R=1m$ is placed horizontally on a frictionless horizontal surface. A particle of mass $\frac{M}{2}$ 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 1

Answer: 27

Acceleration of particle in centre of ring frame is $\frac{v^2}{r}$ , acceleration of centre of ring is $\frac{N}{m}$ .

$N+\frac{M}{2}{a}_{CM}=\frac{M{v}^2}{2r}$
$\frac{3M}{2}{a}_{cm}=\frac{M{v}^2}{2r}$
$a_{CM}=\left(\frac{v^2}{3r}\right)$
$a_{CM}=\left(\frac{9^2}{3\times 1}\right)=27m{s}^{-2}$