# Angular momentum question - Strange silly mistake

• rohanprabhu
In summary: The correct answer is Option A] mKr.In summary, the magnitude of the time rate of change of angular momentum of a particle moving in a circle of radius 'r' is given by the formula mKr, where 'm' is the mass of the particle, 'K' is a positive constant, and 't' is the time. This is derived from the relationship between centripetal force and angular velocity, and the moment of inertia of the particle. The options given do not include this formula, but the correct answer is A] mKr.
rohanprabhu

## Homework Statement

Q] A particle of mass m is moving in a circle of radius r. The centripetal acceleration($a_c = Kt^2$), 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:

$\textrm{A]}~mKr$
$\textrm{B]}~\sqrt{m^2 Kr^3}$
$\textrm{C]}~\sqrt{mKr}$
$\textrm{D]}~mKr^2$

## Homework Equations

all concerned with rotational mechanics.

## The Attempt at a Solution

The $\textrm{Key}~\textrm{Idea}$ 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:

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

Hence,

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

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

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

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

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

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

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

alpha=a_c / r = (Kt^2) /r

alpha=d(omega)/dt

put alpha from first eqn. into second one and then integrate.

alpha=a_c / r
put alpha from first eqn. into second one and then integrate.

I don't think that's right. $\alpha = \frac{a}{r}$, where for a particle in rotation, $\alpha$ is the angular acceleration, 'r' is the radius of rotation and 'a' is the linear acceleration (i.e. the tangential acceleration) of the paritcle and not the centripetal acceleration.

## 1. What is angular momentum?

Angular momentum is a physical quantity that describes the rotational motion of an object. It is defined as the product of an object's moment of inertia and its angular velocity.

## 2. How is angular momentum calculated?

Angular momentum is calculated by multiplying an object's moment of inertia (a measure of its resistance to rotational motion) by its angular velocity (the rate at which it rotates).

## 3. What is the principle of conservation of angular momentum?

The principle of conservation of angular momentum states that the total angular momentum of a system remains constant unless acted upon by an external torque. This means that in a closed system, angular momentum is conserved and cannot be created or destroyed.

## 4. How does angular momentum relate to rotational inertia?

Angular momentum is directly proportional to an object's rotational inertia. This means that the larger an object's moment of inertia, the greater its angular momentum will be at a given angular velocity.

## 5. What are some real-life examples of angular momentum?

Some examples of angular momentum in everyday life include spinning tops, figure skaters performing spins, and the rotation of the Earth around its axis. It is also important in the design and functioning of machines such as turbines and gyroscopes.

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