Angular Momentum of a conical pendulum.

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Satvik Pandey
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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



pun.png

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.
 
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Can you guys tell me some applications like Daum Equation Editor for writing equations in LaTex. I don't why is it not working in my PC.:mad::confused:
 
ehild said:
Check your coordinate system and the components of your vectors, they do not match o0)

I have considered the circle to be in X-Y plane.
pun.png


I have shown the top view also.

For finding angular momentum of anybody about any point(say O) we drop any line from O to the line of the velocity of that body. Then we calculate ##\overrightarrow { R } \times \overrightarrow { p } ##. right?

So ##R_{i}=lsin(\alpha)\hat { i }- lcos(\alpha)\hat { k }##
 
Satvik Pandey said:
I have considered the circle to be in X-Y plane.View attachment 77622

I have shown the top view also.

For finding angular momentum of anybody about any point(say O) we drop any line from O to the line of the velocity of that body. Then we calculate ##\overrightarrow { R } \times \overrightarrow { p } ##. right?

So ##R_{i}=lsin(\alpha)\hat { i }- lcos(\alpha)\hat { k }##

OK, but then the change the top right figure with the unit vectors. And check the signs in Li .
I got

##L_{i}=mvLcos\alpha\hat{i}+mvlsin\alpha\hat{k}##
 
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ehild said:
OK, but then the change the top right figure with the unit vectors. And check the signs in Li .
I got

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

Oh! I made a sign mistake.:mad: Now I got answer

##\frac { 2mgl }{ \omega } \sqrt { 1-{ \left( \frac { g }{ { \omega }^{ 2 }l } \right) }^{ 2 } } ##.

Thank you for the help.:)
 
How is angular momentum conserved from the center of the circle