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## Homework Statement

A small ball of mass m suspended from a ceiling at a point O by a thread of length l moves along a horizontal circle with constant angular velocity ##\omega##. Find the magnitude of increment of the vector of the ball's angular momentum relative to point O picked up during half of revolution.

## Homework Equations

## The Attempt at a Solution

I[/B]nitial velocity of ball ##V_{i}=v\hat { j }##

Initial distance of the ball from O is (R)=##lsin\alpha\hat{i}-lcos\alpha\hat{k}##

Final velocity ##V_{f}=-V\hat{j}##

Final distance of the ball from O is ##-lsin\alpha\hat{i}-lcos\alpha\hat{k}##

Initial momentum is ##R## cross ##P##.

##L_{i}=mvLcos\alpha\hat{i}-mvlsin\alpha\hat{k}##

##L_{f}=-mvLcos\alpha\hat{i}+mvlsin\alpha\hat{k}##

##\delta L=-2mvLcos\alpha\hat{i}+2mvlsin\alpha\hat{k}##

So its magnitude is ##2mvL##

I got ##cos\alpha=\frac{g}{\omega^2l}## by writing the force equation.

Now ##v=lsin\alpha\omega##

Using this I got the answer

##2ml^{2}\omega\sqrt{1-\frac{g^{2}}{(\omega^{2}l)^2}}##

But the answer is incorrect.