# Bead slides on a spinning hoop

1. Aug 10, 2017

### Kelly Lin

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
I know we can solve it by letting
$$x=acos(\omega t)+acos(\theta+\omega t)\\ y=asin(\omega t)+asin(\theta +\omega t)$$
and put their derivatives in to the Lagrangian.

But, I want to check the other points of view whether it is wrong or something else is missing.
$$a\dot{\theta} = v_{\text{bead}}-v_{\text{hoop}}\\ v_{\text{bead}} = a\dot{\theta}+ a\omega$$
Thus, the Lagrangian will be
$$L=\frac{1}{2}m(a\dot{\theta}+ a\omega)^{2}$$
But, there is a huge difference between two results. Can someone enlighten me??? THANKS!!!

2. Aug 10, 2017

### TSny

You can see that something's wrong here by looking at special cases. For example, suppose the bead remains located at the end of the diameter that passes through the origin of the coordinate system. Then $\theta = 0 = \text{constant}$. Your result wold then say that $v_{\text{bead}} = a\omega$. But the speed of the bead in this case would actually be $2a\omega$.

The relative velocity equation can be written as a vector equation: $$\vec v_{\text{bead/XYZ}} = \vec v_{\text{bead/C}} + \vec v_{C/\text{XYZ}}$$ where

$\vec v_{\text{bead/XYZ}}$ = the velocity of the bead relative to the fixed XYZ coordinate system

$\vec v_{\text{bead/C}}$ = the velocity of the bead relative to a non-rotating reference frame moving with the center of the circular hoop

$\vec v_{C/XYZ}$ = the velocity of the center of the hoop with respect to the fixed XYZ coordinate system.

3. Aug 11, 2017

### Kelly Lin

Is it because the direction of the tangential velocity is always changing? Thus, my equations will only be satisfied when the tangential velocities from both of them are parallel.

4. Aug 11, 2017

### TSny

Even when the tangential velocities due to $\omega$ and $\dot \theta$ are parallel, your formula would not give the correct speed for the bead. When the bead is instantaneously at $\theta = 0$, the speed of the bead would be $a\dot \theta + 2a\omega$; whereas, your formula gives $a\dot \theta + a\omega$.

Note that if the bead is at rest with respect to the hoop and is located at the end of the diameter that passes through the origin (i.e., $\theta = 0$) the bead would have a speed of $2a\omega$ since the bead would be moving in a circle of radius $2a$ with angular speed $\omega$.

5. Aug 11, 2017

### Kelly Lin

How about " the velocity of the bead relative to a non-rotating reference frame moving with the center of the circular hoop". The center of the circular hoop also rotates, so how can I find a non-rotating reference that is moving as the center of the circular hoop?
Sorry!! I still don't get it!!! I still don't know how to use the relative velocity to solve this question.

6. Aug 11, 2017

### TSny

In the figure above, the origin of the primed axes (blue) moves with the center of the hoop. But these axes do not rotate relative to the earth. The primed axes remain parallel to the unprimed axes as the hoop rotates.

The velocity of the bead relative to the earth can be written as $$\vec v_{\text{bead/xyz}} = \vec v_{\text{bead/x'y'z'}} + \vec v_{C/\text{xyz}}$$ The first velocity on the right is the velocity of the bead relative to the primed axes and the second velocity vector on the right is the velocity of the center of the hoop relative to the unprimed axes. These velocity vectors are shown below in green and orange. The velocity of the bead relative to the earth is the sum of these two vectors.

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7. Aug 12, 2017

### Kelly Lin

Thank you very much for your diagrams!!!!!
I got it!!! The relative velocity of the bead is $$a(\dot{\theta}+\omega)$$!!!
Thanks again!!!!!!

8. Aug 12, 2017

### TSny

Yes. That's the magnitude.