- #1

rohanprabhu

- 414

- 2

## Homework Statement

Q] A particle of mass m is moving in a circle of radius r. The centripetal acceleration([itex]a_c = Kt^2[/itex]), where K is a positive contant and t is time. The magnitude of the time rate of change of angular momentum of the particle about the centre of the circle is:

[itex]\textrm{A]}~mKr[/itex]

[itex]\textrm{B]}~\sqrt{m^2 Kr^3}[/itex]

[itex]\textrm{C]}~\sqrt{mKr}[/itex]

[itex]\textrm{D]}~mKr^2[/itex]

## Homework Equations

all concerned with rotational mechanics.

## The Attempt at a Solution

The [itex]\textrm{Key}~\textrm{Idea}[/itex] here is that despite the variable centripetal force, the radius of rotation remains constant. Which means, the net torque on the particle needs to vary with time. At any time, 't' the centripetal force is related to the angular velocity as:

[tex]

\textrm{F}_\textrm{c} = \textrm{Kt}^2 = \textrm{mr}\omega^2

[/tex]

Hence,

[tex]

\omega = \sqrt{\frac{\textrm{Kt}^2}{\textrm{mr}}} = \textrm{t}\sqrt{\frac{\textrm{k}}{\textrm{mr}}}

[/tex]

Now, the moment of inertia of the particle is given by: [itex]\textrm{I} = \textrm{mr}^2[/itex]

Hence, the angular momentum at time 't' is given by:

[tex]

\textrm{l} = \textrm{I}\omega = \textrm{mr}^2 \textrm{t}\sqrt{\frac{\textrm{k}}{\textrm{mr}}} = \textrm{t}\sqrt{mKr^3}

[/tex]

To find the rate of change w.r.t time, we have:

[tex]

\frac{\textrm{dl}}{\textrm{dt}} = \sqrt{mKr^3}

[/tex]

But it ain't one of the options.. i know I'm making a silly mistake somewhere.. just can't catch it..