Physical pendulum with no fixed pole

[bznm]

1. Homework Statement
The violet sleeve has mass M and is free to move horizontally without friction. The green rod has mass 2M and length L and can rotate around a pivot on the sleeve without friction. At t=0, the rod is in vertical position and is rotating with angular velocity w. What are the physical quantities that are constant? What is the maximum angular amplitude of oscillations and maximum sleeve speed?

2. The attempt at a solution
K_{tot} + U_{tot} = constant (there aren't dissipative forces such as friction)
Linear momentum: there are external forces (weight force and vincular reaction) so it isn't constant.
Angular momentum: whatever pole I choose there is some external force that have nonzero torque, so it isn't constant.

I tried to solve the problem calculating the torque with respect to the pivot, but the problem is: that pole is in motion, so I don't have

, but I have

(pole velocity isn't parallel to mass center velocity)... and so I can't go on to calculate angular acceleration and I'm stuck.

Could you help me?

 

[Orodruin]

Are you familiar with Lagrange mechanics?

 

[TSny]

What can you say about the horizontal component of the net external force on the system?

 

[bznm]

@Orodruin: no, I don't know anything about it.
@TSny: There's no net horizontal force, hence the horizontal linear momentum is constant.. but how can I proceed then?

 

[TSny]

@TSny: There's no net horizontal force, hence the horizontal linear momentum is constant.

Yes.

.. but how can I proceed then?

Consider the system at the instant θ is a maximum.

 

[bznm]

Yes.

Consider the system at the instant θ is a maximum.

I think I've got the idea. But there's a doubt. At theta maximum, the rod has zero angular velocity, but with respect to the pole (so the system is all going with velocity V)... So i can extract from there the maximum sleeve velocity... Am I wrong?

 

[TSny]

I think I've got the idea. But there's a doubt. At theta maximum, the rod has zero angular velocity, but with respect to the pole (so the system is all going with velocity V)

Yes, at max θ the rod has zero angular speed. There is no need to say "with respect to the pole".
And, yes, all points of the rod and sleeve have the same velocity at this instant.

... So i can extract from there the maximum sleeve velocity...

Does the sleeve have its maximum speed at the instant that θ is a maximum?

 

[bznm]

Yes, at max θ the rod has zero angular speed. There is no need to say "with respect to the pole".
And, yes, all points of the rod and sleeve have the same velocity at this instant.

I don't understand "There is no need to say "with respect to the pole". ". Well, I think it's important that the rod points speed is zero with respect to the pole, so that sleeve and rod have the same velocity. (Now I'm thinking you were pointing out that I said angular velocity respect to the pole, but I was thinking at corresponding linear velocities of the points).

Does the sleeve have its maximum speed at the instant that θ is a maximum?

I think so, but I can't show why.

Thanks a lot for your support!

(I got a bit confused about my assumptions after seeing this animation https://upload.wikimedia.org/wikipedia/commons/4/4d/Pendulum_movable_support.gif but maybe it's just a different problem)

 

[Orodruin]

I think so, but I can't show why.

Consider the case where the total linear momentum of the entire contraption is zero ...

 

[TSny]

(Now I'm thinking you were pointing out that I said angular velocity respect to the pole, but I was thinking at corresponding linear velocities of the points).

Yes. I was pointing out that angular velocity does not need to be specified relative to some point.

The sleeve does not have its max speed when θ is max. The first question is to find θ_max. You know that all points of the system have the same velocity when θ = θ_max. See if you can use this fact along with the conservation laws to determine θ_max.

(I got a bit confused about my assumptions after seeing this animation https://upload.wikimedia.org/wikipedia/commons/4/4d/Pendulum_movable_support.gif but maybe it's just a different problem)

This animation shows the same system but with different initial conditions. Or, you can think of the animation as showing your system from a particular moving frame of reference. Going to this frame of reference might be helpful for finding the max speed of the sleeve.

[EDIT: After watching the animation more closely, it appears that there's frictional damping in the system. So, it's not quite the same system. Nevertheless, you might want to consider a frame of reference where the total linear momentum of your system is zero. But, it's not necessary to do this.]

 

[bznm]

@TSny ok I followed your hints and I found cos (thetamax) from the conservation of energy (at t=0 the energy is all rotational at thetamax the energy is gravitational and translational (i can find the speed from the conservation of horizontal linear momentum). Now I can't go on with Max sleeve velocity, at first I supposed it was at thetamax, you pointed out i was wrong.. and I can't guess in which position it is!

@Orodruin sorry but I can't get the point :(

 

[Orodruin]

sorry but I can't get the point :(

It is a simple argument for why your statement is false. In that situation, the velocity of the sleeve is zero when $\theta$ is maximal.

 

[bznm]

It is a simple argument for why your statement is false. In that situation, the velocity of the sleeve is zero when $\theta$ is maximal.

Why is that? When theta is max the pendulum is quiet with respect to the pole so it's moving at the sleeve velocity.. Why is the total linear momentum zero?

P.s. could you explain me what is the configuration in which the sleeve velocity is maximum? I just can't see it!

 

[Orodruin]

Why is that? When theta is max the pendulum is quiet with respect to the pole so it's moving at the sleeve velocity.. Why is the total linear momentum zero?

If the pendulum is at relative rest to the sleeve and the total momentum is zero, what is the velocity of the sleeve (and hence also the pendulum)?

 

[bznm]

If the pendulum is at relative rest to the sleeve and the total momentum is zero, what is the velocity of the sleeve (and hence also the pendulum)?

I'm tempted of replying "sleeve-pendulum velocity is zero" but I'm still not sure why at theta max the total linear momentum is 0 .. :(
Moreover, i still can't see how this argument could help me in finding sleeve max velocity..

Thanks for your support!
P.s. what about the reasoning to find theta max that I summerized in a previous message? Is it correct?

 

[TSny]

@TSny ok I followed your hints and I found cos (thetamax) from the conservation of energy (at t=0 the energy is all rotational at thetamax the energy is gravitational and translational (i can find the speed from the conservation of horizontal linear momentum).

Sounds good.

Now I can't go on with Max sleeve velocity, at first I supposed it was at thetamax, you pointed out i was wrong.. and I can't guess in which position it is!

Consider what happens after the rod reaches max angle to the right. As the rod begins to swing back down toward a vertical position, what happens to the horizontal momentum of the rod? So what happens to the horizontal momentum of the sleeve as the rod swings back down?

 

[TSny]

but I'm still not sure why at theta max the total linear momentum is 0

In the frame of reference of the picture given in the problem, the total linear momentum is not zero. However, you can imagine going to a moving inertial frame in which the total momentum is zero. You can work out the max speed of the sleeve in this frame and then transform back to the original frame.

 

[bznm]

But

In the frame of reference of the picture given in the problem, the total linear momentum is not zero. However, you can imagine going to a moving inertial frame in which the total momentum is zero. You can work out the max speed of the sleeve in this frame and then transform back to the original frame.

But why the sleeve velocity is maximum at theta max?

 

[Orodruin]

But

But why the sleeve velocity is maximum at theta max?

It isn't.

 

[bznm]

It isn't.

So I have misinterpreted @TSny message. Let's start again. I choose the frame of reference in which total linear momentum is zero when theta is max. And then? I can't see how to "work out the max speed of the sleeve in this frame" !

@TSny "Consider what happens after the rod reaches max angle to the right. As the rod begins to swing back down toward a vertical position,what happens to the horizontal momentum of the rod?" Well, I think that the rod velocity with respect to the pole has decreased up to zero (in theta max) and then it reverse sign and starts to increase again... But this is only with respect to the pole and this fact makes me impossible to get how things are going. Could you please be more explicit? I'm stuck with it and cannot go on.

 

[Orodruin]

I choose the frame of reference in which total linear momentum is zero when theta is max. And then? I can't see how to "work out the max speed of the sleeve in this frame" !

There is no need to specify "when theta is max". As you have already concluded, the linear momentum is conserved. Think about how you can use this fact to your advantage. What does it mean that the linear momentum is zero?

Edit: Note that both the sleeve and the rod will move in this reference frame!

 

[bznm]

There is no need to specify "when theta is max". As you have already concluded, the linear momentum is conserved. Think about how you can use this fact to your advantage. What does it mean that the linear momentum is zero?

Edit: Note that both the sleeve and the rod will move in this reference frame!

I think you are saying that it means that the two moments are opposite. So I get the maximum (in absolute value) momentum for the sleeve when the rod momentum is maximum (in absolute value). Is it correct? But in that frame, the rod momentum is maximum when?

 

[Orodruin]

Have you tried writing down expressions for the momenta and energy?

 

[bznm]

well, I am not used to write these expressions in particular frame:
Mv (current sleeve velocity)= 2Mw_cm, where w_cm is w(the current angular velocity)*L/3 (the distance between the pole and the center of mass) .. I equate them because the sum should be 0
E=1/2 M v^2+1/2 I(the moment of inertia)*w^2+2M*g*(L-L*cos theta)

I'm pretty sure that's wrong, but please understand me: it's a very unusual exercise for me!

 

[TSny]

@TSny "Consider what happens after the rod reaches max angle to the right. As the rod begins to swing back down toward a vertical position,what happens to the horizontal momentum of the rod?" Well, I think that the rod velocity with respect to the pole has decreased up to zero (in theta max) and then it reverse sign and starts to increase again... But this is only with respect to the pole and this fact makes me impossible to get how things are going. Could you please be more explicit? I'm stuck with it and cannot go on.

In the original frame of reference, when the rod reaches maximum angle, the rod and sleeve are moving together with some speed V toward the right. As the rod swings downward, what happens to the horizontal component of the velocity, v_x, of the center of mass of the rod relative to this frame of reference? Does v_x increase, decrease, or stay at the value V?

If it's difficult to see the answer to this question, it is helpful to think about it from the point of view of the reference frame that moves with constant speed to the right such that the total linear momentum of the system is zero. In this frame, think about what the system is doing at max angle of the rod. Then think about what must happen as the rod swings back down.

 

[bznm]

In the original frame of reference, when the rod reaches maximum angle, the rod and sleeve are moving together with some speed V toward the right. As the rod swings downward, what happens to the horizontal component of the velocity, v_x, of the center of mass of the rod relative to this frame of reference? Does v_x increase, decrease, or stay at the value V?

It should increase because when it swings down some potential energy becomes kinetic... Is it wrong?

 

[TSny]

Does the swinging down contribute velocity to the right or to the left for the CM of the rod?

 

[bznm]

Does the swinging down contribute velocity to the right or to the left for the CM of the rod?

Well, I think left, because it goes towards the vertical position... so vx decreases..

 

[TSny]

Yes. At max height, the rod has some momentum, p_x, to the right (relative to the original reference frame). As it swings down, the velocity of the CM picks up some contribution to the left due to the rotational motion. That is, there is a change in velocity of the CM of the rod toward the left. So, taking x to be positive toward the right, does p_x for the rod increase or decrease as it swings down?

 

[bznm]

Well, I think left, because it goes towards the vertical position...

Yes. At max height, the rod has some momentum, p_x, to the right (relative to the original reference frame). As it swings down, the velocity of the CM picks up some contribution to the left due to the rotational motion. That is, there is a change in velocity of the CM of the rod toward the left. So, taking x to be positive toward the right, does p_x for the rod increase or decrease as it swings down?

I think that the rod momentum decreases while the sleeve momentum increases, because the horizontal momentum must be constant...

 

[TSny]

You're on a skateboard holding a ball and moving toward the right. You and the ball are moving together with a speed V. You then throw the ball toward the left so that the ball has a change in velocity toward the left. Taking x as positive toward the right, does the change in velocity of the ball cause p_x for the ball to increase or decrease? What about your momentum, P_x? Does it increase or decrease?.

 

[bznm]

px for the ball decreases while px for me increases.. I think that you missed my previous post edit, I'm sorry..

 

[TSny]

px for the ball decreases while px for me increases.. I think that you missed my previous post edit, I'm sorry..

Oops, I did miss it. OK, good. So, the momentum of the sleeve increases toward the right as the rod swings down. So, does the sleeve have maximum speed when the rod has its maximum angle?

 

[bznm]

Oops, I did miss it. OK, good. So, the momentum of the sleeve increases toward the right as the rod swings down. So, does the sleeve have maximum speed when the rod has its maximum angle?

Well, It doesn't seem so to me now... I think that while the rod is going down,the sleeve gains horizontal momentum... Is it that the maximum sleeve velocity is when the rod has come back in the vertical position?

 

[TSny]

Well, It doesn't seem so to me now... I think that while the rod is going down,the sleeve gains horizontal momentum...

Yes

Is it that the maximum sleeve velocity is when the rod has come back in the vertical position?

Here's where the zero-momentum reference frame is useful. The animation you posted earlier is similar to being in the zero-momentum frame. What is the orientation of the rod when the sleeve has maximum speed?

 

[bznm]

Yes

Here's where using the zero-momentum reference frame is useful. The animation you posted earlier is similar to being in the zero-momentum frame. What is the orientation of the rod when the sleeve has maximum speed?

Uhmm. it isn't very easy to see unless you already know the answer :P
To me it seems it is with the rod in vertical position.. But I'm not sure.. Please, help me!

 

[TSny]

Uhmm. it isn't very easy to see unless you already know the answer :P
To me it seems it is with the rod in vertical position.. But I'm not sure..

OK, fair enough. It seems "intuitive" that the maximum speed of the sleeve in the zero-momentum frame will occur when the rod is vertical. But "intuitive" is not a good reason. I tried to come up with a good argument without any math, but so far with no luck. But with conservation of momentum and a couple of kinematic relations you can prove the result.

 

[bznm]

OK, fair enough. It seems "intuitive" that the maximum speed of the sleeve in the zero-momentum frame will occur when the rod is vertical. But "intuitive" is not a good reason. I tried to come up with a good argument without any math, but so far with no luck. But with conservation of momentum and a couple of kinematic relations you can prove the result.

Could you please show me how to extract max sleeve velocity and how to transformer it with respect to the inertial frame of the figure? This is the first exercise of this kind that I was given.. It isn't that easy to get the results! I would be very grateful!

 

[bznm]

I still didn't get to the result. Can anybody help me with finding the max sleeve speed?

 

[TSny]

We'll try to show that the maximum speed of the sleeve in the zero-momentum frame of reference occurs when the rod is vertical. It will then be easy to show that the maximum speed of the sleeve in the original frame also occurs when the rod is vertical.

Notation: [edited]

CM stands for center of mass.

U is the instantaneous velocity of the sleeve. Positive toward the right.

v_x is the x-component of the velocity center of mass of the rod. Positive x is to the right.

v_y is the y-component of velocity the center of mass of the rod. Positive y is upward.

θ is the angular displacement of rod from horizontal. Counterclockwise is positive.

ω is the angular velocity of the rod. Counterclockwise is positive.

Goal: show that U takes on its max value when θ = 0.

Step 1. Use conservation of momentum to derive a relation between U and v_x

 

[bznm]

We'll try to show that the maximum speed of the sleeve in the zero-momentum frame of reference occurs when the rod is vertical. It will then be easy to show that the maximum speed of the sleeve in the original frame also occurs when the rod is vertical.

Notation:

CM stands for center of mass.

U is the instantaneous velocity of the sleeve. Positive toward the right.

v_x is the x-component of the center of mass of the rod. Positive x is to the right.

v_y is the y-component of the center of mass of the rod. Positive y is upward.

θ is the angular displacement of rod from horizontal. Counterclockwise is positive.

ω is the angular velocity of the rod. Counterclockwise is positive.

Goal: show that U takes on its max value when θ = 0.

Step 1. Use conservation of momentum to derive a relation between U and v_x

Uhm OK, thanks @TSny for the help. I assume that you mean "v_x is the x-component of the *(velocity of)* center of mass of the rod".

v_x=U+ωRcosθ where R is the distance between the pole and CM.
due to the conservation of horizontal momentum I have:
MU+2Mv_x= const (in the zero-momentum frame is =0).
If I substitute I get
M(U+w)+2M(U+ωRcosθ+w)=0 (where w is the zero momentum frame velocity, i.e. the transformation constant between the frames)
U = -2/3 ω R cosθ - w. This is in the zero momentum frame. In the original frame U= -2/3 ω R cosθ - 2w.
w is a constant, so I don't care about it.
U is maximum in absolute value when theta is 0 and ω is maximum (this also occurs when theta is zero because potential energy is 0).
Is it correct?

 

[bznm]

Furthermore, when θ max, rod-and-sleeve are moving with the same velocity.

This means that if θ max
- in the zero momentum frame the horizontal momentum is 3M*(vx + w) = 0 i.e. w=-v_x
- In the original frame I have that p_x at start = p_x at θ max.
That is 2M ω L = 3M v_x, so v_x=2/3 ω L for θ max.
I have found that the value of w is = -2/3 ω L.

is it correct?
What do I have to do now?

 

[TSny]

Uhm OK, thanks @TSny for the help. I assume that you mean "v_x is the x-component of the *(velocity of)* center of mass of the rod".

Yes, thanks for catching that.

v_x=U+ωRcosθ where R is the distance between the pole and CM.

Yes, good. (You've already worked out my "Step 2" that was coming next!)

due to the conservation of horizontal momentum I have:
MU+2Mv_x= const (in the zero-momentum frame is =0).

Yes. I should have made it clearer that we are going to first work out the maximum speed of the sleeve for the zero-momentum frame. All the symbols in the notation list that I posted should be considered as measured in the zero-momentum frame. So, the right hand side of the above equation is zero.

If I substitute I get
M(U+w)+2M(U+ωRcosθ+w)=0 (where w is the zero momentum frame velocity, i.e. the transformation constant between the frames)
U = -2/3 ω R cosθ - w. This is in the zero momentum frame.

Since all symbols are considered measured relative to the zero-momentum frame, there is no need to introduce w. So, let w = 0. We will get back to the original frame later.

In the original frame U= -2/3 ω R cosθ - 2w.
w is a constant, so I don't care about it.

OK. w = 0 in the zero-momentum frame.

U is maximum in absolute value when theta is 0 and ω is maximum (this also occurs when theta is zero because potential energy is 0).
Is it correct?

Well, this is where it's tricky. The expression U= -2/3 ω R cosθ says that U is maximum when the product ωcosθ is a maximum. Even though cosθ takes on a maximum at θ = 0, that doesn't necessarily imply that the product ωcosθ takes on a maximum at θ = 0. We know that the system as a whole has maximum total KE when θ = 0 because the potential energy of the system is minimum at θ = 0. The total KE of the system is shared between the sleeve and the rod. Just because θ = 0 is the place where the total KE is maximum, it doesn't logically follow from this that the sleeve has its maximum KE at this point. For example, as the rod swings down, v_y increases and then decreases. So, v_y reaches a maximum before the rod becomes vertical. Could U also reach a maximum before the rod becomes vertical? I found it helpful to also have an expression for v_y in terms of ω and θ.

 

[TSny]

Furthermore, when θ max, rod-and-sleeve are moving with the same velocity.

This means that if θ max
- in the zero momentum frame the horizontal momentum is 3M*(vx + w) = 0 i.e. w=-v_x

w is zero in the zero-momentum frame.

- In the original frame I have that p_x at start = p_x at θ max.
That is 2M ω L = 3M v_x

The left hand side is not quite correct. The center of mass of the rod is not located a distance L from the pivot.

 

[bznm]

Yes, thanks for catching that.
The expression U= -2/3 ω R cosθ says that U is maximum when the product ωcosθ is a maximum. Even though cosθ takes on a maximum at θ = 0, that doesn't necessarily imply that the product ωcosθ takes on a maximum at θ = 0. We know that the system as a whole has maximum total KE when θ = 0 because the potential energy of the system is minimum at θ = 0. The total KE of the system is shared between the sleeve and the rod. Just because θ = 0 is the place where the total KE is maximum, it doesn't logically follow from this that the sleeve has its maximum KE at this point.

At first I thought about this problem, but I simply forgot while I was writing the message :D Is the assumption "ω increases while swinging down" is totally wrong?

For example, as the rod swings down, v_y increases and the then decreases. So, v_y reaches a maximum before the rod becomes vertical. Could U also reach a maximum before the rod becomes vertical? I found it helpful to also have an expression for v_y in terms of ω and θ.

Uhmmm..
Isn't v_y=- ω R sinθ? Let's assume 0 <= θ_max < pi/2.. But when θ goes from θ_max to 0, ω increases, while θ (and so sin θ) decreases... But I can't get θ such as v_y is max in abs value.. :/

 

[bznm]

w is zero in the zero-momentum frame.

The left hand side is not quite correct. The center of mass of the rod is not located a distance L from the pivot.

There was a typo.. L should be replaced by R.

 

[TSny]

At first I thought about this problem, but I simply forgot while I was writing the message :D Is the assumption "ω increases while swinging down" is totally wrong?

"ω increases while swinging down" is something we would like to prove (even though intuitively it seems true).

Isn't v_y=- ω R sinθ?

Yes, except if y is positive upward and counterclockwise is positive for ω and θ, then I don't believe there should be a minus sign on the right side.

Let's assume 0 <= θ_max < pi/2.. But when θ goes from θ_max to 0, ω increases, while θ (and so sin θ) decreases... But I can't get θ such as v_y is max in abs value.. :/

At this point, I think you have enough equations to prove that U is max at θ = 0. Let's recap the equations:

v_x = -U/2 (from conservation of momentum) ... Eq. 1

U = -(2ωRcosθ)/3 ..........Eq. 2

v_y = ωRsinθ ...........Eq. 3With these and the fact that the total KE of the system must increase as θ decreases, I think you can prove that U takes on a maximum at θ = 0.

 

[bznm]

"ω increases while swinging down" is something we would like to prove (even though intuitively it seems true).Yes, except if y is positive upward and counterclockwise is positive for ω and θ, then I don't believe there should be a minus sign on the right side.

At this point, I think you have enough equations to prove that U is max at θ = 0. Let's recap the equations:

v_x = -U/2 (from conservation of momentum) ... Eq. 1

U = -(2ωRcosθ)/3 ..........Eq. 2

v_y = ωRsinθ ...........Eq. 3With these and the fact that the total KE of the system must increase as θ decreases, I think you can prove that U takes on a maximum at θ = 0.

Ok @TSny , I know that K_tot= 1/2 M U^2 + 1/2 I ω^2
with a couple of substitions and calculus of moment of inertia, I get
U^2 = 1/M * 2K_tot * (R^2)/(R^2+3L^2) * cos^2 θ
let's group constants:
U^2 = c*K_tot * cos^2 θ
with c positive const.

when I go from θmax to 0,
K_tot increases (due to loss of potential energy) and cos^2 θ increases, hence U^2 increases.
In the domain [0,θmax] 0 is the maximum.
Is it correct?

 

[TSny]

I know that K_tot= 1/2 M U^2 + 1/2 I ω^2

The KE of the rod is not expressed correctly. For example, suppose ω = 0 at some instant when the sleeve is moving. Would the KE of the rod be zero?

But you have the right idea.

 

[bznm]

The KE of the rod is not expressed correctly. For example, suppose ω = 0 at some instant when the sleeve is moving. Would the KE of the rod be zero?

But you have the right idea.

haa sorry! what about Ktot= 1/2 (M+2M) U^2 + 1/2 I ω^2?