Question

A circular ring of mass m and radius a, lies on a smooth ... Insect of mass m, resting on it starts and walks round it with uniform velocity v relative to the ring. Show that the centre of the ring describes a circle m v with angular velocity

Answer 1

Answer:

The mass of the ring and mass of insect is: {eq}m {/eq}

The radius of the ring is: {eq}r {/eq}

The uniform speed of the insect is: {eq}v {/eq}

The angular velocity of the insect is given as:

{eq}{\omega _i} = \dfrac{v}{r} {/eq}

The moment of inertia of the ring about the insect is given as:

{eq}\begin{align*} {I_r} &= m{r^2} + m{r^2}\\ {I_r} &= 2m{r^2} \end{align*} {/eq}

The moment of inertia of the insect is given by:

{eq}{I_i} = m{r^2} {/eq}

Apply the conservation of angular momentum to find the angular velocity of ring.

{eq}{I_r}\omega = {I_i}{\omega _i} {/eq}

Substitute all the values in the above equation.

{eq}\begin{align*} \left( {2m{r^2}} \right)\omega &= \left( {m{r^2}} \right)\left( {\dfrac{v}{r}} \right)\\ \omega &= \dfrac{v}{{2r}} \end{align*} {/eq}

Thus, the expression for angular velocity of ring is {eq}\omega = \dfrac{v}{{2r}} {/eq}.