# Lagrangian problem (rigid body+particle)

by fluidistic
Tags: body, lagrangian, particle, rigid
 PF Gold P: 3,173 1. The problem statement, all variables and given/known data A hollow semi-cylinder of negligible thickness, radius a and mass M can rotate without slipping over the horizontal plane z=0, with its axis parallel to the x-axis. On its inside a particle of mass m slides without friction, constrained to move in the y-z plane. This system is set under a uniform gravitational field $-g \hat z$. 1)How many degrees of liberty does the system have? Write down the Lagrangian. 2)Find the configuration of stable equilibrium and write down the corresponding Lagrangian for small oscillations. 3)Find the frequencies of oscillations and their corresponding normal modes and describe these qualitatively. 4)If initially the system is at rest in its position of stable equilibrium and we apply a small impulse to the particle of mass m, write down the corresponding solution of the motion equations. 2. Relevant equations $L=L_1+L_2$. 3. The attempt at a solution 1)1 degree of liberty, using intuition (i.e. you give me the position of either the hollow cylinder or the particle and I can tell you the position of the other body. Their motion isn't independent.). Lagrangian $L_2$ of the hollow cylinder: $L_2=T_2=\frac{I \omega ^2}{2}=\frac{I \dot \phi ^2}{2}$. I'll come back to this angle phi later. Lagrangian of the particle of mass m: $L_1=T_1-V_1$. I choose to use polar coordinates. The origin of my system of reference lies in the y-z plane at the height of the axis of rotation of the hollow cylinder. The kinetic energy of the particle is worth $T_1=\frac{ma^2\dot \theta }{2}$. The potential energy is worth $mga[1- \cos (\theta ) ]$. However $\phi$ is related to $\theta$, they are dependent of each other. By intuition, $\phi I=m \theta$ so that $\dot \phi ^2 =\frac{m^2 \dot \theta ^2 }{I^2}$, I do not know how to prove this though and I know it should be possible. I'd like to know if this relation is correct. All in all this makes the total Lagrangian worth $L=\frac{ma^2 \dot \theta }{2}+mga [\cos (\theta )-1 ]+\frac{m^2 \dot \theta ^2 }{I}$. I know I need to calculate $I$. I tried very hard to do this but I'm really stuck on that (sigh). I considered the hollow semi-cylinder with a length L. I actually derived that the x component of the center of mass lies in the middle of the cylinder (L/2) which was obvious. The much less obvious position to find is the height over the ground of the center of mass (z component). I've drew a sketch. If I'm at a height dz over the ground, I need to find the arc length of a vertical transversal cut of the hollow semi cylinder. I still didn't find it. With intuition it should be proportional to the radius, a. $z_{\text {center of mass}}=\frac{\int z \sigma dM}{\int dM}$ where $dM=SL$ and S is the arc length I'm looking for. I must express it in terms of $dz$. Then integrate from $z_0=0$ and $z_1=a$. I'd appreciate if someone could tell me if what I've done so far is correct and how to get S in terms of dz. Thanks in advance!
 HW Helper P: 3,337 just making sure I am imagining the right thing: There is a hollow semi-cylinder (like a half-pipe), Whose axis is in the x-direction. And it is rotating about an axis which is through its centre of mass, and perpendicular to the x-axis. And we're calling this axis it rotates around as the z-axis. And then the final axis is the y-axis. (And of course the gravitational field is in the negative z-direction). And there is a particle on the inside that must move only on the y-z plane (i.e. it can't move in the direction of the axis of the cylinder). Did I understand everything correctly? In this case, why is the kinetic energy of the particle equal to: $$T_1 = \frac{ma^2 \dot{\theta}}{2}$$ Shouldn't the $\dot{\theta}$ be squared as well? And also, the particle will have some KE due to the fact that the axis of the cylinder is rotating, so there should be another term in the KE of the particle.
 PF Gold P: 3,173 First of all thanks for your concern. Yes there should be theta squared, a mistake I realized later and I forgot to edit my post here. Hmm, I don't think we are seeing the same situation. I've made a sketch to show you what I understand from the problem. So I don't think there should be any term for the kinetic energy of the particle that involves a rotation other than the one in the yz plane. Attached Thumbnails
HW Helper
P: 3,337

## Lagrangian problem (rigid body+particle)

That does look like what I'm imagining. And the cylinder rotates around the z axis. And the particle is going to be on the yz plane, where x=0.

I think there should be another kinetic energy than just the KE from motion in the yz plane. If we imagine the particle is held at constant z, then the cylinder will be turning, so the particle will still be moving, even though it has no motion in the yz plane. (The one exception is if the particle was held at the lowest part of the pipe, then it would have zero motion).
PF Gold
P: 3,173
 Quote by BruceW That does look like what I'm imagining. And the cylinder rotates around the z axis. And the particle is going to be on the yz plane, where x=0. I think there should be another kinetic energy than just the KE from motion in the yz plane. If we imagine the particle is held at constant z, then the cylinder will be turning, so the particle will still be moving, even though it has no motion in the yz plane. (The one exception is if the particle was held at the lowest part of the pipe, then it would have zero motion).
What I understand from the (ambiguous) problem statement is what I sketched. This picture could be at time t=0. At a further time, the mass will slide inside the cylinder, making it swing around an axis parallel to the x-axis. (in fact the axis that passes in its center of mass and at a constant z).
I do not know why the semi-cylinder would rotate around the z-axis. What did you make think this?
You wrote "And it is rotating about an axis which is through its centre of mass, and perpendicular to the x-axis. " but the problem statement is "rotate without slipping over the horizontal plane z=0, with its axis parallel to the x-axis. " Maybe they should have specified "axis of rotation parallel to the x-axis".
For the sake of simplicity I consider this case only.
What do you think? Do you understand the situation as I understand it?
 HW Helper P: 3,337 oooh, I see. the cylinder is rotating around the x-axis. For some reason, I thought it was meant to be rotating around the z-axis. But of course that isn't allowed because the question states that there is no slipping. Thanks for helping me out on that. I'm not sure why you say $\phi I = m \theta$, maybe you should just do the Euler-Lagrange equations with theta and phi as different variables. Also, In the Lagrangian, there should also be a term due to the GPE of the cylinder, since this will change as it rotates around its axis.
PF Gold
P: 3,173
 Quote by BruceW oooh, I see. the cylinder is rotating around the x-axis. For some reason, I thought it was meant to be rotating around the z-axis. But of course that isn't allowed because the question states that there is no slipping. Thanks for helping me out on that.
Glad we are on the same wavelength now. :)
 I'm not sure why you say $\phi I = m \theta$, maybe you should just do the Euler-Lagrange equations with theta and phi as different variables.
Hmm ok but eventually I'll have to express one in term of the other as the system has only 1 degree of freedom hence only 1 generalized coordinate.
 Also, In the Lagrangian, there should also be a term due to the GPE of the cylinder, since this will change as it rotates around its axis.
To me it just look like this gravitational potential energy is constant for the half-cylinder. I think that the Lagrangian is defined up to a constant, I can safely ignore it. For the center of mass of the half cylinder to move vertically, I think the cylinder must have to either fly or get inside the ground.
 HW Helper P: 3,337 hmm. In fact, since the particle has no friction with the cylinder, and the position of the cylinder doesn't affect the height or speed of the particle, then it looks like the problem is two separate motions (that of the cylinder and that of the particle). But that can't be right, because that would be too simple. And the questions seem to imply that the motion of the particle and cylinder are linked. Maybe no friction between the particle and the cylinder means that the particle is like a ball bearing, which rolls on the surface of the cylinder, not causing dissipative friction..? But if that were the case, we would need to know the inertia of the ball bearing. I guess we could assume the ball bearing was a uniform density sphere, but this all seems like too much guesswork, so I'd say this is not on the right track... So we've either got two uncoupled motions (which is not what the question seems to imply), or we need some relation between the phi and theta variables. You've written down $\phi I = m \theta$ but I don't see why this should be true intuitively.
HW Helper
P: 3,337
 Quote by fluidistic To me it just look like this gravitational potential energy is constant for the half-cylinder. I think that the Lagrangian is defined up to a constant, I can safely ignore it. For the center of mass of the half cylinder to move vertically, I think the cylinder must have to either fly or get inside the ground.
I don't think so. think about when the half-cylinder is almost on its side, the centre of mass is equal to the radius of the cylinder. And when the half-cylinder is at its lowest, the centre of mass is definitely below the radius. (these distances are relative to the ground).
PF Gold
P: 3,173
 Quote by BruceW I don't think so. think about when the half-cylinder is almost on its side, the centre of mass is equal to the radius of the cylinder. And when the half-cylinder is at its lowest, the centre of mass is definitely below the radius. (these distances are relative to the ground).
Sigh, this is blowing my mind. I don't understand how a particle of mass m which can only apply a downward force (due to gravity) can make the center of mass of the half cylinder goes high.
I'm at a loss.
P.S.:But I do see what you mean and I agree with you.
 HW Helper P: 3,337 I thought about it a bit more, and I think I understand about how theta and phi are linked. (No ball bearing assumptions required). Its because the speed of the particle also depends on the rotation of the cylinder. For example, if the particle stays at the lowest part of the cylinder, then when the cylinder rotates, the cylinder moves horizontally (since no slip with ground). So even though the particle seems stationary, it also has moved horizontally. You'll need to do some maths to flesh out this principle though.
 PF Gold P: 3,173 I admit I'm very confused.
 HW Helper P: 3,337 How are you doing with the problem now?
 PF Gold P: 3,173 Stuck.
 HW Helper P: 3,337 which part are you working on now?
PF Gold
P: 3,173
I had stopped that problem, gave up. :(
But now I could retry it. When you say
 when the cylinder rotates, the cylinder moves horizontally (since no slip with ground). So even though the particle seems stationary, it also has moved horizontally.
it makes me think I misused the word "slipping" in English (not my native tongue) but I found on the Internet
 The wheel rolls without slipping only if there is no horizontal movement of the wheel at the contact point P (with respect to the surface/ground). Thus, the contact point P must also have zero horizontal movement (with respect to the surface/ground).
So I do not understand how can the cylinder move horizontally if there's no slipping. I need some explanation.
 HW Helper P: 3,337 yes, its saying that the relative movement of the ground and cylinder at the point of contact is zero. If you imagine the centre of mass of the cylinder as the origin, and that the ground is moving at some horizontal velocity v relative to the centre of the mass, then what must be the angular velocity of the cylinder? (Using the fact that there is no relative movement at the point of contact).

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