# Two falling rods connected by a hinge

• DanielA
In summary: So the total force on the system at the contact points would be...The total force on the system at the contact points would be...The total force on the system at the contact points would be...In summary, the student is struggling with the homework problem and is looking for help. He is convinced that it is much easier than his classmates make it out to be, but does not know where he is wrong. He is looking for a discussion of the starting point so he can fix his intuition and hopefully get better. He has found that the coordinate of the center of mass is y and that the speed of the hinge will be twice the speed of the center of mass.
DanielA

## Homework Statement

I uploaded the homework question. This is #1.

## Homework Equations

None directly given

## The Attempt at a Solution

My main difficulty with the problem is that I am convinced it is much easier than my classmates make it out to be. This is graduate mechanics so I'm pretty sure that the solution can not be this easy, but I don't know where I am wrong. All I am after is a discussion of the starting point so I can fix my intuition and hopefully get better. As a note, one classmate somehow calculated an inertia tensor in his solution which seems unnecessary. He argued symmetry made the solution calculable by examining only one rod with the CoM in the middle of the rod so there would be rotation and translation of the CoM.

The two rods connected to the hinge are perfectly symmetric. Thus the CoM of the system will be on the y-axis at ##y=\frac{l}{4}## (each rod's CoM is half way down its length so the two combined would be on the y-axis). The only force acting on the system is gravity which is acting on the CoM of the system. Thus the problem is essentially a particle (the hinge) being pulled straight down by gravity a distance ##\frac{l}{2}## where ##U=mgy, U(min) = mg\frac{-l}{4} ##and ##U(max)=mg\frac{l}{4}##. ##T = \frac{1}{2}m\dot y^2##. And the rods are relegated to being part of initial conditions to solve the equation of motion. So the Lagrangian is $$\frac{1}{2}m\dot y^2 - mgy$$, the Lagrange equation is trivially $$\ddot y = -g$$ and the solution goes from there to get $$t=\sqrt\frac{3l}{2g}$$ and $$\dot y = -g\sqrt\frac{l}{g} = -\sqrt\frac{3lg}{2}$$

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curious_mind
DanielA said:
The two rods connected to the hinge are perfectly symmetric. Thus the CoM of the system will be on the y-axis at ##y=\frac{l}{4}## (each rod's CoM is half way down its length so the two combined would be on the y-axis). The only force acting on the system is gravity which is acting on the CoM of the system.
What about the normal force from the surface?

Thus the problem is essentially a particle (the hinge) being pulled straight down by gravity a distance ##\frac{l}{2}## where ##U=mgy, U(min) = mg\frac{-l}{4} ##and ##U(max)=mg\frac{l}{4}##. ##T = \frac{1}{2}m\dot y^2##. And the rods are relegated to being part of initial conditions to solve the equation of motion. So the Lagrangian is $$\frac{1}{2}m\dot y^2 - mgy$$, the Lagrange equation is trivially $$\ddot y = -g$$ and the solution goes from there to get $$t=\sqrt\frac{3l}{2g}$$ and $$\dot y = -g\sqrt\frac{l}{g} = -\sqrt\frac{3lg}{2}$$
Could you fill in a little more detail here? Is ##y## the coordinate of the hinge or the center of mass? Is ##m## the mass of the rod or of the system? If you have ##\ddot y=-g##, you get ##y=y_0 - \frac 12 gt^2## where ##y_0## is ##l/2## (hinge) or ##l/4## (center of mass), so where did the factor of 3 come from?

vela said:
What about the normal force from the surface?Could you fill in a little more detail here? Is ##y## the coordinate of the hinge or the center of mass? Is ##m## the mass of the rod or of the system? If you have ##\ddot y=-g##, you get ##y=y_0 - \frac 12 gt^2## where ##y_0## is ##l/2## (hinge) or ##l/4## (center of mass), so where did the factor of 3 come from?
You're right about the normal force. I didn't consider that. Since the system isn't stable in that position without a string to hold the hinges (it was cut in the problem statement), then does that mean that the force on the point of contact will be split in x and y directions? The gravitational force would be split in half and handled at the contact points because of symmetry. We'd only really be concerned about the variable x direction movement of the rod as that would translate to fall, right?

The y coordinate is of center of mass as that is where the force of gravity is pulling on. The 0 on potential is at the ground though. With a different approach, I might need to change that as the CoM can't go "in the ground " like I was kinda treating it with the range. The factor of 3 came from the time. I typoed on the ##\dot y## term. It meant to be -gt and I plugged in t incorrectly in the first term and fixed it on the second

I recognized the typo. I don't see where the factor of 3 came in your calculation for the time.

If ##y## is the coordinate of the center of mass, then the speed of the hinge will be twice the speed of the center of mass, right?

By "points of contact", do you mean where the rods touch the surface? If so, the force has only a y-component since there's no friction. There is an x-component of force on each rod at the hinge, though. The rods push against each other, which accelerates the center of mass of the rods away from the center.

curious_mind
As @vela points out, the translational kinetic energy of each rod will have a contribution from the horizontal motion as well as the vertical motion of the center of the rod. In addition, each rod will have rotational kinetic energy.

If you are going to use a Lagrangian approach, how many degrees of freedom of motion does the system have? If it has only one degree of freedom, then decide what would be a good choice for the generalized coordinate.

TSny said:
As @vela points out, the translational kinetic energy of each rod will have a contribution from the horizontal motion as well as the vertical motion of the center of the rod. In addition, each rod will have rotational kinetic energy.

If you are going to use a Lagrangian approach, how many degrees of freedom of motion does the system have? If it has only one degree of freedom, then decide what would be a good choice for the generalized coordinate.
It seems for this system, the absolute max possible for two 2-d rods would be 6. 1 rotational and 2 translational for each rod. However, of course, there are constraints on the system that reduces that. The symmetric nature of the system will reduce the 2 rotational DoF (one ##\theta## between rod and ground for each side) to 1 (just ##\theta##).

Additionally, the translational DoF are reduced from 4 (x distance of bottom of rod with origin on ground and y height of rod) to 2 by constraint of the hinge. So now, we have 3 x, y, ##\theta##. ##\theta## between a rod and x-axis, the height of the hinge, and the distance of the bottom of the rod from the origin which I placed on the ground under the hinge.

However, We can reduce this further. Because the hinge will not move from the y-axis, (the Center of mass of the system is on the y-axis. The sum of forces in the x-direction on the CoM would cancel due to the symmetry of the rods) we can relate x, y, and theta by right triangles. Since we only actually need one side and an angle to solve a right triangle, we should be able to reduce the 3 remaining DoF to 1 with the GC as ##\theta##.

Now, we need to form a lagrangian. $$U(\theta) = U(\theta_1) + U(\theta_2) = mgl\sin(\theta)$$ ie, the total potential is the sum of the contributions from both rods. Turns out to be the height of the hinge. Next, KE is much the same. We'll just sum up the two rods. Both rods have KE ##\frac{1}{2}m(\dot x^2 + \dot y^2)## where ##\dot x = -\dot \theta \ell \sin(\theta)## and ## \dot y = -\dot \theta \ell \cos(\theta)## where the original relation between x, y, and ##\theta## just came from basic trig on our triangle.

Lagrangian:
$$L = m\ell^2\dot \theta^2 - mg\ell\sin\theta$$

I am pretty sure I still messed this up because the resultant differential equation is ##\ddot \theta + \frac{g}{2\ell}\cos\theta = 0## which I'm pretty sure I don't have a way to solve because of the cosine. Though I will admit, I don't remember every method so I'll look into it.

Interestingly, if the lagrangian is true, the Potential and KE are equivalent to if we had only one rod and the hinge was rotating about it. The only exception is that the KE is missing the factor ##\frac{1}{2}## that would be there with ##T = \frac{1}{2}I\omega^2##

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Yes, there is just one degree of freedom and ##\theta## is a good choice for the generalized coordinate. The value of ##\theta## completely defines the spatial configuration of the entire system.
DanielA said:
$$U(\theta) = U(\theta_1) + U(\theta_2) = mgl\sin(\theta)$$
OK, ##U(\theta)## looks good.
Both rods have KE ##\frac{1}{2}m(\dot x^2 + \dot y^2)## where ##\dot x = -\dot \theta \ell \sin(\theta)## and ## \dot y = -\dot \theta \ell \cos(\theta)##
Your expressions for ##\dot x## and ##\dot y## don't seem to correspond to the center of mass of a rod. As you mentioned in your first post, the center of mass of a rod is located halfway along the rod. It might be helpful to state where you are choosing the origin of your x-y coordinate system. (Maybe you did and I overlooked it.)

Also, your expression for KE does not include the rotational KE of each rod about its center of mass.

I am pretty sure I still messed this up because the resultant differential equation is ##\ddot \theta + \frac{g}{2\ell}\cos\theta = 0## which I'm pretty sure I don't have a way to solve because of the cosine. Though I will admit, I don't remember every method so I'll look into it.
After making the corrections to the KE of the system, you will end up with a differential equation that agrees with your differential equation except that the constant factor will not be ##\frac{g}{2\ell}##.

I don't think the differential equation has a solution that can be expressed in terms of elementary functions. However, you can do an integration that converts it from second order to first order. To do this, multiply the differential equation through by ##\dot \theta## and note that each term is now the time derivative of a simple expression. So, it is easy to integrate to get a first order differential equation involving ##\dot \theta##. As a check, you should be able to get this same equation using just conservation of energy. (So, conservation of energy is all you really need to work this problem.)

TSny said:
Yes, there is just one degree of freedom and ##\theta## is a good choice for the generalized coordinate. The value of ##\theta## completely defines the spatial configuration of the entire system.
OK, ##U(\theta)## looks good.
Your expressions for ##\dot x## and ##\dot y## don't seem to correspond to the center of mass of a rod. As you mentioned in your first post, the center of mass of a rod is located halfway along the rod. It might be helpful to state where you are choosing the origin of your x-y coordinate system. (Maybe you did and I overlooked it.)

Also, your expression for KE does not include the rotational KE of each rod about its center of mass.After making the corrections to the KE of the system, you will end up with a differential equation that agrees with your differential equation except that the constant factor will not be ##\frac{g}{2\ell}##.

I don't think the differential equation has a solution that can be expressed in terms of elementary functions. However, you can do an integration that converts it from second order to first order. To do this, multiply the differential equation through by ##\dot \theta## and note that each term is now the time derivative of a simple expression. So, it is easy to integrate to get a first order differential equation involving ##\dot \theta##. As a check, you should be able to get this same equation using just conservation of energy. (So, conservation of energy is all you really need to work this problem.)

Alright, so I think I fixed it. ##T(\theta) = \frac{1}{12}m\ell^2\dot \theta^2 + \frac{1}{2}m((\frac{\dot x}{2})^2 + (\frac{\dot y}{2})^2)## for each rod. There is a similar triangle between the "big" right triangle of the whole rod and the "small" one from where the rod touches the ground to the CoM of the rod. Turns out the relationship is ##\frac{1}{2}##. So my total would be this multiplied by 2. Substitute in ##\dot x ## and ##\dot y## and we end up with ##T(\theta) = \frac{5}{12}m\ell^2\dot \theta^2## My EoM is ##\ddot \theta - \frac{12g}{5\ell}\cos\theta = 0##. Multiply both sides by ##\dot \theta##, recognize left side is simply a time derivative of ##\dot \theta - \frac{12g}{5\ell}\sin\theta## and do a definite integral of both sides from 0->t and get $$\dot \theta - \frac{12g}{5\ell}\sin\theta = 0$$.

It feels like there should be a t on one of those two terms because my second term doesn't have the right units for an angular velocity since the trig function is unitless. I bet I'm just messing up the simple integration because I only dimly remember the method. After all, there should be a constant of integration because we need to apply initial conditions to the resultant equation. My result says that at ##\theta =30\deg, \dot \theta## isn't 0.

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DanielA said:
Alright, so I think I fixed it. ##T(\theta) = \frac{1}{12}m\ell^2\dot \theta^2 + \frac{1}{2}m(\dot x^2 + \dot y^2)## for each rod.
OK
So my total would be this multiplied by 2. Substitute in ##\dot x ## and ##\dot y## and we end up with ##T(\theta) = \frac{7}{6}m\ell^2\dot \theta^2##
I get a different numerical factor compared to your ##\frac{7}{6}##. Can you show more steps here?

My EoM is ##\ddot \theta - \frac{3g}{7\ell}\cos\theta = 0##. Multiply both sides by ##\dot \theta##, recognize left side is simply a time derivative of ##\dot - \frac{3g}{7\ell}\sin\theta## and do a definite integral of both sides from 0->t and get $$\dot \theta - \frac{3g}{7\ell}\sin\theta = 0$$.
The first term ##\dot \theta## is not correct. Can you show how you got it? Also, you left out the contribution from the lower limit ##t = 0## for the second term.

It feels like there should be a t on one of those two terms because my second term doesn't have the right units for an angular velocity since the trig function is unitless. I bet I'm just messing up the simple integration because I only dimly remember the method. After all, there should be a constant of integration because we need to apply initial conditions to the resultant equation. My result says that at ##\theta =30\deg, \dot \theta## isn't 0.
The units will work out once you correct the integration of the first term. You performed "definite" integration between the limits of ##t = 0## and ##t##. So, there will not be an arbitrary constant of integration. You get an arbitrary constant of integration when you do an "indefinite" integral. That would be another way of doing it: Use indefinite integration and then evaluate the constant of integration using the initial condition at ##t = 0##.

TSny said:
OK
I get a different numerical factor compared to your ##\frac{7}{6}##. Can you show more steps here?

The first term ##\dot \theta## is not correct. Can you show how you got it? Also, you left out the contribution from the lower limit ##t = 0## for the second term.

The units will work out once you correct the integration of the first term. You performed "definite" integration between the limits of ##t = 0## and ##t##. So, there will not be an arbitrary constant of integration. You get an arbitrary constant of integration when you do an "indefinite" integral. That would be another way of doing it: Use indefinite integration and then evaluate the constant of integration using the initial condition at ##t = 0##.
I did a ninja edit on the factors and added an explanation because I realized my ##\dot x## and ##\dot y## were still from origin to ends of the rod and not CoM of the rod. Yeah, I get why the definite integral won't give a constant of integration. I just don't know how to correct my integration.

As far as I know, the indefinite integral of a derivative would just be what is inside the derivative ie ##\dot \theta - \frac{12g}{5\ell}\sin\theta = C## I'm trying to google the integration method, but can't find a good example to review.

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DanielA said:
##\dot \theta - \frac{12g}{5\ell}\sin\theta = C##
I don't get ##\frac{12}{5}##.

In the EOM you have a term ##\ddot \theta##. Recall that the trick was to multiply the EOM through by ##\dot \theta##. So, you now get a term ##\dot \theta \ddot \theta##. Can you express this as the time derivative of a simple expression?

TSny said:
I don't get ##\frac{12}{5}##.

In the EOM you have a term ##\ddot \theta##. Recall that the trick was to multiply the EOM through by ##\dot \theta##. So, you now get a term ##\dot \theta \ddot \theta##. Can you express this as the time derivative of a simple expression?
Wow I jumped the gun on that. The time derivative is of ##\frac{1}{2}\dot \theta^2##. And I messed up the first derivative on the equation of motion. The factor should be ##\frac{6}{5}##. I didn't multiple the ##\frac{5}{12}## by 2 when I took the ##\dot \theta## derivative. Though with how many dumb mistakes I'm making so far, there's probably at least one more left. Still, I have a non simply integrable ##\dot \theta## function. I can get part a on the assignment, but a factor of time hasn't entered any equation yet. And of course, ##\dot \theta## isn't constant or anything so I can't try to derive an equation for time, I think.

DanielA said:
The time derivative is of ##\frac{1}{2}\dot \theta^2##.
Yes
The factor should be ##\frac{6}{5}##.
I get a different factor. I think the mistake is with the evaluation of ##\dot x^2## and ##\dot y^2##.

Still, I have a non simply integrable ##\dot \theta## function. I can get part a on the assignment, but a factor of time hasn't entered any equation yet. And of course, ##\dot \theta## isn't constant or anything so I can't try to derive an equation for time, I think.
If you solve the equation for ##\dot \theta## it will have the form ##\dot \theta = f(\theta)## with some function ##f(\theta)##. Separate variables as ##\frac{d\theta}{f(\theta)} = dt##. Integrate both sides. However, I don't believe the integration of the left side can be done in terms of elementary functions. But, you could do it numerically.

TSny said:
Yes
I get a different factor. I think the mistake is with evaluation of ##\dot x^2## and ##\dot y^2##.

If you solve the equation for ##\dot \theta## it will have the form ##\dot \theta = f(\theta)## with some function ##f(\theta)##. Separate variables as ##\frac{d\theta}{f(\theta)} = dt##. Integrate both sides. However, I don't believe the integration of the left side can be done in terms of elementary functions. But, you could do it numerically.
I honestly don't know where the difference in factors is. And the class isn't one that would expect us to numerically integrate a function to get an answer. However, I clearly would need to for this one. So I don't know what to do since I'm pretty sure something more fundamental than some constant factors in this solution is wrong. I can't do small angle approximations or taylor expansions or anything to simplify things because we clearly aren't using small angles and taylor expands wouldn't be useful for finding cumulative time. I'll just try to talk to my professor during office hours. Maybe I can identify more of my fundamental problems with him. I appreciate all the help, though.

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I think we can clear up the numerical factor if you would give us your expressions for the cartesian coordinates, ##x## and ##y##, of the center of mass of one rod in terms of ##l## and ##\theta##. Also, please state your choice of the location of the origin of your xy coordinate system.

## 1. What is a "Two falling rods connected by a hinge" experiment?

The "Two falling rods connected by a hinge" experiment is a physics experiment where two rods of different lengths and masses are connected by a hinge and then released to fall under the influence of gravity. It is used to demonstrate the principles of rotational motion and conservation of energy.

## 2. How does the length of the rods affect the experiment?

The length of the rods affects the experiment by changing the moment of inertia of the system. A longer rod will have a larger moment of inertia, making it more resistant to changes in rotational motion compared to a shorter rod. This can lead to differences in the final velocity and acceleration of the rods.

## 3. What is the purpose of connecting the rods with a hinge?

The hinge serves as a point of rotation for the rods. Without it, the rods would not be able to rotate freely and the experiment would not be able to demonstrate principles of rotational motion. The hinge also ensures that the rods remain connected throughout the experiment.

## 4. How does the mass of the rods affect the experiment?

The mass of the rods affects the experiment by changing the total energy of the system. A heavier rod will have more potential energy compared to a lighter rod, resulting in a higher final velocity and acceleration. However, the mass of the rods does not affect the principles of rotational motion demonstrated in the experiment.

## 5. What are some real-world applications of this experiment?

The "Two falling rods connected by a hinge" experiment has real-world applications in understanding the physics of pendulums, metronomes, and other oscillating systems. It can also be used to study the behavior of objects in free fall and to demonstrate conservation of energy principles in rotational motion.

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