Angular Momentum of a conical pendulum.

In summary, the conversation discusses a problem involving a small ball suspended from a ceiling and moving in a horizontal circle with constant angular velocity. The goal is to find the magnitude of the increment of the ball's angular momentum relative to a point on the circle. The conversation goes through the steps of finding the initial and final velocity and distance of the ball from the point, calculating the initial and final momentum, and finally arriving at the correct answer after correcting a sign mistake. The concept of angular momentum conservation is also briefly mentioned.
  • #1
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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



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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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  • #2
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:
 
  • #3
Check your coordinate system and the components of your vectors, they do not match o0)
 
  • #4
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.
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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 }##
 
  • #5
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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Likes Satvik Pandey
  • #6
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.:)
 
  • #7
How is angular momentum conserved from the center of the circle
 
  • #8
Poojan said:
How is angular momentum conserved from the center of the circle
You mean, about the centre of the circle?
What is the net force on the mass? Where does it point?
 

What is angular momentum?

Angular momentum is a measure of an object's rotational motion. It is a vector quantity that depends on an object's mass, speed, and distance from the rotation axis.

How is angular momentum calculated for a conical pendulum?

The angular momentum of a conical pendulum is calculated by multiplying the mass of the object attached to the string by its tangential speed, and then multiplying that product by the distance between the object and the rotation axis.

What is the relationship between angular momentum and the length of the string in a conical pendulum?

In a conical pendulum, the length of the string is directly proportional to the angular momentum. This means that an increase in the length of the string will result in an increase in the object's angular momentum, and vice versa.

How does the angle of the string affect the angular momentum of a conical pendulum?

The angle of the string has no effect on the angular momentum of a conical pendulum. This is because the tangential speed and distance from the rotation axis remain constant regardless of the angle of the string.

What happens to the angular momentum of a conical pendulum if the mass of the object is increased?

If the mass of the object attached to the string is increased, the angular momentum of the conical pendulum will also increase. This is because the tangential speed and distance from the rotation axis remain constant, but the mass increases in the calculation of angular momentum.

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