Bead and Hoop in 2D: When Does the Particle Fall Off?

[jfy4]

Homework Statement

A particle of mass $m$ is placed on top of a vertical hoop of radius $R$ and mass $M$. The particle is free slide on the outside of the hoop without friction while the hoop is free to roll in a vertical place without slipping. Use the method of Lagrange multipliers to determine the height at which the particle falls off of the hoop.

Homework Equations

$$\frac{d}{dt}\frac{\partial L}{\partial \dot{q}_j}-\frac{\partial L}{\partial q_j}=\lambda_i \frac{\partial f_i}{\partial q_j}$$

The Attempt at a Solution

Let me first venture the Lagrangian, since that is a signifigant portion of the problem, then EL equations can move things along quickly once I have it. The way I read it, the hoop is free to move in the 2D vertical plane, and the bead just rests on top and the whole of dynamics is just between the two, all in a gravitational field.

I then have for the kinetic energy of the hoop
$$T_h=\frac{1}{2}M\dot{X}^2 + \frac{1}{2}I\dot{\phi}^2$$
with the additional constraint that $f_1 = X-R\phi=0$ with $X$ the distance in the x direction the hoop has traveled, and $\phi$ the angle subtended by the hoop. For the bead I have
$$T_b=\frac{1}{2}m(\dot{x}^2+\dot{y}^2) + \frac{1}{2}I_0 \dot{\theta}^2$$
with the constraint $f_2 = r_2-R=0$ with $r_2$ the radius of the hoop. Then I have
$$x = r_1 + r_2 \cos\theta \quad y=r_2 \sin\theta$$
with $x,y$ the location of the bead. By placing the center of the hoop on the x-axis, $r_1$ points to the center of the hoop, and $r_2$ from the center to the bead, $\theta$ is the angle drawn from the x-axis at the center of the hoop to the bead. The potential for the hoop is zero, and for the bead
$$V=-mgy=-mg r_2 \sin\theta$$
The Lagrangian is $L=T_b+T_h -V$. Also $I = Mr_{2}^{2}$ and $I_0 = m r_{2}^{2}$
Does this seem like a good set-up for the problem's lagrangian? Thanks.

 

[gabbagabbahey]

For the bead I have
$$T_b=\frac{1}{2}m(\dot{x}^2+\dot{y}^2) + \frac{1}{2}I_0 \dot{\theta}^2$$
with the constraint $f_2 = r_2-R=0$ with $r_2$ the radius of the hoop. Then I have
$$x = r_1 + r_2 \cos\theta \quad y=r_2 \sin\theta$$
with $x,y$ the location of the bead. By placing the center of the hoop on the x-axis, $r_1$ points to the center of the hoop, and $r_2$ from the center to the bead, $\theta$ is the angle drawn from the x-axis at the center of the hoop to the bead

Why is the $\frac{1}{2}I_0 \dot{\theta}^2$ term in your kinetic energy for the bead/particle? Does the bead/particle roll/spin?

Doesn't your constraint $f_2$ constrain the bead/particle to move on the hoop? Wouldn't a less stringent constraint that allows the bead/particle to lose contact with the hoop make more sense?

 

[jfy4]

In my mind's eye I was thinking that the bead rolls down the side of the hoop until it looses contact with the hoop. That rolling down the side motion I tried to capture mathematically with $\frac{1}{2} I_0 \dot{\theta}^2$. Then I was thinking I could solve the EOM and then let the constraint force associated with the bead on the hoop, $f_2$, equal zero, ie, when the bead leaves the hoop... do you think that method is cumbersome or incorrect?

 

[gabbagabbahey]

In my mind's eye I was thinking that the bead rolls down the side of the hoop until it looses contact with the hoop. That rolling down the side motion I tried to capture mathematically with $\frac{1}{2} I_0 \dot{\theta}^2$.

Does the bead/particle have a finite size? If not, how can you tell it is rolling and not just translating?

Then I was thinking I could solve the EOM and then let the constraint force associated with the bead on the hoop, $f_2$, equal zero, ie, when the bead leaves the hoop... do you think that method is cumbersome or incorrect?

If $f_2=0$, is there anything in your equation that says the bead/particle doesn't fall through the hoop to its interior?

I think a better constraint would be an inequality that says the distance of the bead from the center of the hoop is always ≥ the radius of the hoop.

 

[jfy4]

geez sorry, i shouldn't have said "rolling". the $\frac{1}{2}I_0 \dot{\theta}^2$ isn't for the bead's rolling, its just the rotational energy for a point mass moving in a circle, that circle being the hoop. I feel confident that the bead stays on the hoop until it falls off, and does not move through the hoop.

 

[gabbagabbahey]

geez sorry, i shouldn't have said "rolling". the $\frac{1}{2}I_0 \dot{\theta}^2$ isn't for the bead's rolling, its just the rotational energy for a point mass moving in a circle, that circle being the hoop.

But isn't that energy already accounted for by the translational term $\frac{1}{2}m(\dot{x}^2+\dot{y}^2)$?

When the point mass is moving along the hoop, $x$ and $y$ will have some relationship, but they still completely describe the degrees of freedom for the bead/particle.

I feel confident that the bead stays on the hoop until it falls off, and does not move through the hoop.

So do I, but having no constraining force won't produce that. Without something mathematically constraining the particle's motion so that it never lies inside the hoop, your equations will allow it to do exactly that.

 

[jfy4]

But isn't that energy already accounted for by the translational term $\frac{1}{2}m(\dot{x}^2+\dot{y}^2)$?

I don't know... I suppose so.

So do I, but having no constraining force won't produce that. Without something mathematically constraining the particle's motion so that it never lies inside the hoop, your equations will allow it to do exactly that.

why doesn't $f_2 = r_2 - R$ take care of that? You think that since this constraint doesn't allow for the bead to leave the hoop, just it setting equal to zero later doesn't trap the physics? I kind of was thinking about it like when we do centripital force problems and we set the normal force equal to zero when we want to find when the cart leaves the track-sort-of-thing. You don't think that works here?

 

[gabbagabbahey]

I don't know... I suppose so.

How many degrees of freedom does a classical point particle moving freely in two dimensions have?

why doesn't $f_2 = r_2 - R$ take care of that? You think that since this constraint doesn't allow for the bead to leave the hoop, just it setting equal to zero later doesn't trap the physics? I kind of was thinking about it like when we do centripital force problems and we set the normal force equal to zero when we want to find when the cart leaves the track-sort-of-thing. You don't think that works here?

When you say "setting it equal to zero later" I assume mean setting the forces of constraint $F_i=\lambda_2 \frac{\partial f_2}{dq_i}$ equal to zero? That may work, but my concern is that having zero forces of constraint ignores the fact that whenever the particle is in contact with the hoop, there will be a normal force which keeps it from entering the hoop, so you may get solutions where the trajectory goes inside the hoop at some point. I'll have to think about it a little more.

My approach would be to use the inequality constraint that the distance of the particle from the centre of the hoop is always ≥ the radius of the hoop, which would guarantee you wouldn't get bogus solutions that pass through the hoop at some point on their trajectory.

 

[jfy4]

Yes, I mean setting $\lambda_2=0$.

 

[gabbagabbahey]

Yes, I mean setting $\lambda_2=0$.

I guess that as long as you check that the velocity of the particle when it falls of the hoop is not directed into the hoop, your solution will be valid.