Challenging problem about an impact with a smooth frictionless surface

  • #1
pepos04
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Homework Statement
Two material points of identical mass are in free fall, connected by a rigid rod of negligible mass and length ##L##. At time ##t=0## both masses have zero initial velocity and are positioned at different heights above the ground, the rod forming an angle ##\theta## with the horizontal. Determine how the system moves immediately after the first contact with the ground, imagining that the ground is perfectly smooth and assuming a fully elastic collision model for the impact.
Relevant Equations
/
The system of two material points of identical mass connected by a rigid rod of negligible mass and length ##L## is an example of a conservation of energy problem. The initial energy of the system is the sum of the kinetic energy of the two points and the potential energy of the rod, which is zero since it is horizontal. The final energy of the system is also the sum of the kinetic energy of the two points and the potential energy of the rod, which is zero since it is vertical. Therefore, we can write: ##E_i=E_f## $$1/2mv_i^2+mgh_i+mgh_f =1/2mv_f^2+mgh_f$$ where ##m## is the mass of each point, ##v_i## and ##v_f## are their final velocities, ##g## is the gravitational acceleration, ##h_i## and ##h_f## are their initial and final heights above the ground, and ##h_f=L\theta##, where ##\theta## is the angle between the rod and the horizontal. Since we assume a fully elastic collision model for the impact, we can use conservation of momentum to relate ##v_i## and ##v_f##. The initial momentum of the system is: $$p_i=mv_i+mgh_i=mL\cos\theta + mgh_i$$ The final momentum of the system is: $$p_f=mv_f+mgh_f=mL\cos\theta+mL\sin\thetag+mgh_f =mL(\cos\theta+\sin\theta)g+mgh_f =mL(\cos\theta+\sin\theta)g−mL(\cos\theta+\sin\theta)g=mL(\cos 2\theta-\sin 2\theta)g=mL(0)g=0$$ Since momentum is conserved, we have: $$p_i=p_f \Rightarrow \ mvi +mghi=0 \Rightarrow \ v_i=−mg/L \Rightarrow v_ f=mg/L \Rightarrow v_f=v_i$$ Therefore, we can substitute this into our energy equation and get: $$E_i=E_f \Rightarrow 1/2mv_i^2 +mgh_i+mgh_f=1/2mv_f^2+mgh_f \Rightarrow h_f-h_i−L \theta g=0 \Rightarrow h_f - h_i - L \theta g=0$$ This means that after the first contact with the ground, both points will have zero height above it. The rod will remain vertical at an angle ##L \theta##. Where do I go wrong? Could you help me understand the physical situation? Thank you.
 
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  • #2
Welcome, @pepos04 !

The first mass to hit the floor will bounce as normal but it will be simultaneously pushed horizontally by the other mass and bar, in such a way that the CM will continue moving vertically down while both masses start rotating around it (both increasing horizontal distance from it).
 
  • #3
The way I read this problem the masses start at different heights ##\Delta h =\pm\frac{L}{2}\sin\theta## above and below the horizontal. Yes, energy is conserved but note that the mass closer to the ground will hit the ground first and immediately bounce back while the other mass is still moving down. This means that the rod will rotate about its middle until the other mass hits the ground. You are asked to describe the motion after the first contact.
 
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  • #4
pepos04 said:
Homework Statement: Two material points of identical mass are in free fall, connected by a rigid rod of negligible mass and length ##L##. At time ##t=0## both masses have zero initial velocity and are positioned at different heights above the ground, the rod forming an angle ##\theta## with the horizontal. Determine how the system moves immediately after the first contact with the ground, imagining that the ground is perfectly smooth and assuming a fully elastic collision model for the impact.

pepos04 said:
Relevant Equations: /
No relevant equations ?

pepos04 said:
The system of two material points of identical mass connected by a rigid rod of negligible mass and length ##L## is an example of a conservation of energy problem. The initial energy of the system is the sum of the kinetic energy of the two points and the potential energy of the rod, which is zero since it is horizontal.
Doesn't the assembly acquire kinetic energy during the descent ?
And the problem statement says
  • 'At time ##t=0## both masses have zero initial velocity' ...
  • 'rigid rod of negligible mass'

I like to sketch situations and at ##t=0## I get something like this

1699986496089.png


with the annotation that the two masses are considered point masses.
If the center of mass at ##t=0## is at height ##h_0##, then one mass point is at ##h_0 - \frac{L}{2}\sin\theta## and the other is at ##h_0 + \frac{L}{2}\sin\theta##.

With the relevant equation for free fall we can calculate that the lower mass will hit the floor at ##t_{1}= \sqrt{ {2\over g} \left (h_0 - \frac{L}{2}\sin\theta\right )}##. Both masses move downwards with a speed ##g\,t_{1}##.
The elastic collision with the floor will instantaneously change the direction of the velocity of the lower mass. So the situation at ##t_{1}## before and after the collision is like

1699987546686.png


And the center of mass is at a standstill at that instant . From then on the contraption is rotating while also dropping further. The next point in time that is of interest is when the other mass hits the floor; call that ##t_{2-}##.
I tried to draw it

1699988374056.png


without calculating. Features are:
  • The center of mass is a bit lower than at ##t_{1}##, so ##\theta' < \theta##
  • mass on the left vertical speed is smaller, right one higher
Now the right mass velocity changes sign, and the center of mass will go up again.
Since there is no loss of mechanical energy, I suspect we will come back to the situation at ##t=0## but I have hard time believing that. So if someone could shed some light on this, that would be great.

I do wonder what exactly the exercise composer has in mind: the scenario up to the second mass hitting the floor ? Qualitatively only ?
pepos04 said:
The final energy of the system is also the sum of the kinetic energy of the two points and the potential energy of the rod, which is zero since it is vertical. Therefore, we can write: ##E_i=E_f## $$1/2mv_i^2+mgh_i+mgh_f =1/2mv_f^2+mgh_f$$ where ##m## is the mass of each point, ##v_i## and ##v_f## are their final velocities, ##g## is the gravitational acceleration, ##h_i## and ##h_f## are their initial and final heights above the ground, and ##h_f=L\theta##, where ##\theta## is the angle between the rod and the horizontal. Since we assume a fully elastic collision model for the impact, we can use conservation of momentum to relate ##v_i## and ##v_f##. The initial momentum of the system is: $$p_i=mv_i+mgh_i=mL\cos\theta + mgh_i$$ The final momentum of the system is: $$p_f=mv_f+mgh_f=mL\cos\theta+mL\sin\thetag+mgh_f =mL(\cos\theta+\sin\theta)g+mgh_f =mL(\cos\theta+\sin\theta)g−mL(\cos\theta+\sin\theta)g=mL(\cos 2\theta-\sin 2\theta)g=mL(0)g=0$$ Since momentum is conserved, we have: $$p_i=p_f \Rightarrow \ mvi +mghi=0 \Rightarrow \ v_i=−mg/L \Rightarrow v_ f=mg/L \Rightarrow v_f=v_i$$ Therefore, we can substitute this into our energy equation and get: $$E_i=E_f \Rightarrow 1/2mv_i^2 +mgh_i+mgh_f=1/2mv_f^2+mgh_f \Rightarrow h_f-h_i−L \theta g=0 \Rightarrow h_f - h_i - L \theta g=0$$ This means that after the first contact with the ground, both points will have zero height above it. The rod will remain vertical at an angle ##L \theta##. Where do I go wrong? Could you help me understand the physical situation? Thank you.
 
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  • #5
BvU said:
The elastic collision with the floor will instantaneously change the direction of the velocity of the lower mass. So the situation at ##t_{1}## before and after the collision is like

View attachment 335348

And the center of mass is at a standstill at that instant .
You can't treat the two masses independently during the bounce. There will be a simultaneous impulse along the rod, changing both velocities.
BvU said:
what exactly the exercise composer has in mind: the scenario up to the second mass hitting the floor ?
It says "immediately after the first contact".

The system is like a simple rigid rod, except that moment of inertia is greater.
It can be solved using conservation of energy and conservation of angular momentum about a well-chosen axis.
kuruman said:
the ground first and immediately bounce back while the other mass is still moving down.
Even if the angle is near vertical?
Lnewqban said:
the CM will continue moving vertically down
Not down.
 
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  • #6
haruspex said:
You can't treat the two masses independently during the bounce. There will be a simultaneous impulse along the rod, changing both velocities.
Fully agree, but trying to think through what the result is gave me a headache :smile:

haruspex said:
The system is like a simple rigid rod, except that moment of inertia is greater.
It can be solved using conservation of energy and conservation of angular momentum about a well-chosen axis.
Would my ##t=t_{1+}## picture be a good starting point ?

##\ ##
 
  • #7
BvU said:
Would my t=t1+ picture be a good starting point ?
No, because it shows both velocities as purely vertical. I considered the vertical velocity of the mass centre and the angular velocity.
 
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  • #8
haruspex said:
Not down.
Why not?
 
  • #9
Lnewqban said:
Why not?
If the objects were unconnected, the velocity of the mass centre would become instantaneously zero. Since they are connected, there is impulsive compression in the rod joining them. That means the impulse from the ground is greater than in the disconnected case. Hence the velocity of the mass centre must now be upward.
Or, if that is unconvincing, do the algebra.
 
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  • #10
haruspex said:
If the objects were unconnected, the velocity of the mass centre would become instantaneously zero. Since they are connected, there is impulsive compression in the rod joining them. That means the impulse from the ground is greater than in the disconnected case. Hence the velocity of the mass centre must now be upward.
Or, if that is unconvincing, do the algebra.
Thank you, @haruspex
I trust your understanding, math and experience; nevertheless, I still can't see it happening like that.
Perhaps you can clarify further and correct any flaw in my view:

1) I understand that a portion of the kinetic energy of the falling system is converted to rotation against the resistance of angular inertia.
2) I also see the vector velocity of the first bouncing mass pointing tangentially to that rotation.
3) As the magnitude of that velocity must be the same before and after the elastic collision, the magnitude of the vector vertical component should be less.
4) If the above is true, the height of the system should be less and less after each alternate collision of both masses.
 
  • #11
Lnewqban said:
2) I also see the vector velocity of the first bouncing mass pointing tangentially to that rotation.
Its velocity relative to the common mass centre will, of course, be tangential to the rotation, but that's not its whole velocity.
Lnewqban said:
3) As the magnitude of that velocity must be the same before and after the elastic collision,
You cannot treat the bounce of the first impacting body as independent of the other.

As I wrote, do the algebra. Then you can check each of your statements in turn.
 
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  • #12
pepos04 said:
Homework Statement: Two material points of identical mass are in free fall, connected by a rigid rod of negligible mass and length ##L##. At time ##t=0## both masses have zero initial velocity and are positioned at different heights above the ground, the rod forming an angle ##\theta## with the horizontal. Determine how the system moves immediately after the first contact with the ground, imagining that the ground is perfectly smooth and assuming a fully elastic collision model for the impact.
Relevant Equations: /

The system of two material points of identical mass connected by a rigid rod of negligible mass and length ##L## is an example of a conservation of energy problem. The initial energy of the system is the sum of the kinetic energy of the two points and the potential energy of the rod, which is zero since it is horizontal. The final energy of the system is also the sum of the kinetic energy of the two points and the potential energy of the rod, which is zero since it is vertical. Therefore, we can write: ##E_i=E_f## $$1/2mv_i^2+mgh_i+mgh_f =1/2mv_f^2+mgh_f$$ where ##m## is the mass of each point, ##v_i## and ##v_f## are their final velocities, ##g## is the gravitational acceleration, ##h_i## and ##h_f## are their initial and final heights above the ground, and ##h_f=L\theta##, where ##\theta## is the angle between the rod and the horizontal. Since we assume a fully elastic collision model for the impact, we can use conservation of momentum to relate ##v_i## and ##v_f##. The initial momentum of the system is: $$p_i=mv_i+mgh_i=mL\cos\theta + mgh_i$$ The final momentum of the system is: $$p_f=mv_f+mgh_f=mL\cos\theta+mL\sin\thetag+mgh_f =mL(\cos\theta+\sin\theta)g+mgh_f =mL(\cos\theta+\sin\theta)g−mL(\cos\theta+\sin\theta)g=mL(\cos 2\theta-\sin 2\theta)g=mL(0)g=0$$ Since momentum is conserved, we have: $$p_i=p_f \Rightarrow \ mvi +mghi=0 \Rightarrow \ v_i=−mg/L \Rightarrow v_ f=mg/L \Rightarrow v_f=v_i$$ Therefore, we can substitute this into our energy equation and get: $$E_i=E_f \Rightarrow 1/2mv_i^2 +mgh_i+mgh_f=1/2mv_f^2+mgh_f \Rightarrow h_f-h_i−L \theta g=0 \Rightarrow h_f - h_i - L \theta g=0$$ This means that after the first contact with the ground, both points will have zero height above it. The rod will remain vertical at an angle ##L \theta##. Where do I go wrong? Could you help me understand the physical situation? Thank you.
I have some questions: 1) I didn't understand one key issue: what size does the text require? What should I calculate? 2) It is not clear to me how I should proceed. Do you have any suggestions or advice on this?
 
  • #13
pepos04 said:
I have some questions: 1) I didn't understand one key issue: what size does the text require? What should I calculate? 2) It is not clear to me how I should proceed. Do you have any suggestions or advice on this?
I joined the thread to reply to post #4, so did not study post #1. I'll address that now:
The initial energy of the system is the sum of the kinetic energy of the two points
We were told it is released from rest.
and the potential energy of the rod, which is zero since it is horizontal
Its PE is zero because we are told its mass is negligible; the rod does not start horizontal, and orientation around the mass centre does not affect PE anyway.
The initial energy of this system is just the sum of the potential energies of the masses.
The final energy of the system is also the sum of the kinetic energy of the two points and the potential energy of the rod, which is zero since it is vertical.
Just after contact there is the changed PE of the points and their KEs.
we can write: ##E_i=E_f## $$1/2mv_i^2+mgh_i+mgh_f =1/2mv_f^2+mgh_f$$ where ##m## is the mass of each point, ##v_i## and ##v_f## are their final velocities, ##g## is the gravitational acceleration, ##h_i## and ##h_f## are their initial and final heights above the ground, and ##h_f=L\theta##, where ##\theta## is the angle between the rod and the horizontal.
This makes no sense.
You have ##h_f## meaning two different things and appearing on both sides of the equation.
The angle of the rod does not change during the time considered, so does not need to be considered in the energy balance. Besides, there is no distance of interest given by ##L\theta##.
There are two masses m; your equation only considers one.
Since we assume a fully elastic collision model for the impact, we can use conservation of momentum
No, fully elastic means mechanical energy is conserved. Momentum is conserved if there are no external forces. Hitting the ground means there is an external force, so momentum of the two mass system is not conserved.
There is also angular momentum. To consider that you have to choose an axis. Angular momentum is conserved about that axis if there are no external torques about the axis. Again, impact with the ground could be such an external torque.
There are two ways to deal with this:
  1. Let the impulse from the ground be ##J##. You can now write equations for the change in momentum and the change in angular momentum. Since ##J## appears in both, you can eliminate it.
  2. The smarter way is to choose your axis such that ##J## has no torque about it. The obvious axis is the point of impact. Now you can write the conservation of angular momentum equation without introducing ##J##.
Here is how to proceed. Follow this as far as you can, posting what you get.
Treat the two mass system as a single rigid body of mass 2m.
Skip the falling phase. Only consider how the state just before impact compares with the state just after impact. Since everything is still in the same position, instantaneously, there is no change to PE.

Start with the instant before contact.
Both masses are falling at speed u, say.
What is the KE of the system?

For the next bit, you need to understand something about angular momentum of rigid bodies. A body mass M moving at velocity V has linear momentum MV. If its path would miss point P by distance R then on the basis of that alone it has angular momentum RMV about P. If the body is also rotating at rate ##\omega## and has moment of inertia ##I## about its mass centre then there is additional angular momentum ##I\omega##, bringing the total to ##RMV+I\omega## (these may have opposite signs).

Setting P to be the point of impact, what is the angular momentum of the system just before impact?

Next, we need some variables for the state just after impact. Since the ground is smooth, the system as a whole does not acquire any horizontal velocity. So I would choose ##v## as the vertical velocity of the mass centre (positive down, to be consistent with ##u##) and ##\omega## as the angular velocity. Define clockwise as positive here.

What is the KE after impact?
Write the conservation equation.

What is the angular momentum about P after impact?
Write that conservation equation.
 
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  • #14
What is the general consensus about what the task "Determine how the system moves immediately after the first contact with the ground ##\dots##" means? Specifically, how quantitative does one have to be? Are we looking for the positions of the masses as a function of time after the collision? If not, what constitutes a satisfactory answer?

What about gravity? Are we to assume that its effects are negligible until the system is done colliding with the ground? Even if the angle is near vertical?
 
  • #15
kuruman said:
What is the general consensus about what the task "Determine how the system moves immediately after the first contact with the ground ##\dots##" means? Specifically, how quantitative does one have to be? Are we looking for the positions of the masses as a function of time after the collision? If not, what constitutes a satisfactory answer?

What about gravity? Are we to assume that its effects are negligible until the system is done colliding with the ground? Even if the angle is near vertical?
Since it specifies immediately, the velocity of the mass centre and the angular velocity must be enough. Since that's not hard, I would have thought it should be quantitative. The OP is clearly familiar with some of the equations and principles involved, even if not applied correctly, though maybe not angular momentum?
 
  • #16
So you're saying that the problem is asking for the initial conditions appropriate to whatever equation of motion is applicable immediately after the first collision. If so, that is not a description of "how the system moves" but how it starts to move at the given point in time. I am not picking nits here, just trying to understand what the parameters of the required task are.

Yes, immediately after the collision is t = 0 for the description of "how the system moves". When must we logically stop the description? In this particular case, since energy is conserved, an idealized system will return to its starting configuration after two bounces off the ground. The end of a complete oscillation would be a logical place to stop the description. Symmetry under time reversal shortens the verbiage and/or math by a factor of 2.
 
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  • #17
kuruman said:
that is not a description of "how the system moves" but how it starts to move at the given point in time
Fair enough… or "how it is moving".
Of course, there is a missing parameter. To write the actual expressions for motion immediately after contact we need either the drop height or the impact velocity. As a matter of convenience I assumed the latter (u below).
kuruman said:
an idealized system will return to its starting configuration after two bounces off the ground
Why? Could it not be chaotic?
I looked for repeating patterns of three bounces and got ##\theta+n\pi=\frac{2u^2}{Lg}\frac{\sin^2(\theta)\cos(\theta)}{(1+\cos^2(\theta))^2}##, where n is related to the number of complete half turns it takes in the air between 1st and second bounce, then in reverse between 2nd and 3rd bounce.
 
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  • #18
haruspex said:
If the objects were unconnected, the velocity of the mass centre would become instantaneously zero
I am sure you don't mean that for the unconnected objects.

Just before first ball contact, V1 = V2.
Just after first ball contact and rebound, V1= V2.

After the rebound, V2 is increasing, V1 decreasing in magnitude.
V2>V1 in magnitude
The COM of the unbound objects has to be decreasing, up to the point of second ball contact.
 
  • #19
256bits said:
I am sure you don't mean that for the unconnected objects.

Just before first ball contact, V1 = V2.
Just after first ball contact and rebound, V1= V2.
If unconnected, just after first ball contact it has velocity -V1, while the upper ball still has velocity V2=V1.
-V1+V2=-V1+V1=0.
 
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  • #20
haruspex said:
I looked for repeating patterns of three bounces and got ##\theta+n\pi=\frac{2u^2}{Lg}\frac{\sin^2(\theta)\cos(\theta)}{(1+\cos^2(\theta))^2}##, where n is related to the number of complete half turns it takes in the air between 1st and second bounce, then in reverse between 2nd and 3rd bounce.
Can I ask you how you got this result? Anyway, @haruspex, I thank you for the advice on the correct process to get to the result. I will try to follow them and post what I get, hopefully it is right. However, vefore your advice (which I have not yet heeded) I had thought that the contact mass, which for an instant must stop at the inversion of its velocity, became home to the instantaneous axis of rotation perpendicular to the page under the action of the torque ##\tau=- mg L \cos \theta## due to the other mass. If Newton's second equation were used, given ##I= mL^2## the moment of inertia of the mass ##m## and ##alpha## its angular acceleration, there would come ##mL^2\alpha=- mgL \cos \theta##, thus an acceleration ##a=-g \cos \alpha## that would be added to ##-g## already possessed. When the latter touches the ground does what is now described occur on the other? The conclusion should be that the initial situation is restored, mechanical energy is conserved, and the system returns to the altitude from which it started. These are just my considerations. Please tell me if they currently convince you. I will proceed with following your advice late
 
  • #21
pepos04 said:
I had thought that the contact mass, which for an instant must stop at the inversion of its velocity, became home to the instantaneous axis of rotation
The ground is smooth, so there are no horizontal forces on the two mass system. Consequently, its mass centre only moves along a vertical line.
 
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  • #22
haruspex said:
I joined the thread to reply to post #4, so did not study post #1. I'll address that now: We were told it is released from rest. Its PE is zero because we are told its mass is negligible; the rod does not start horizontal, and orientation around the mass centre does not affect PE anyway. The initial energy of this system is just the sum of the potential energies of the masses. Just after contact there is the changed PE of the points and their KEs. This makes no sense. You have ##h_f## meaning two different things and appearing on both sides of the equation. The angle of the rod does not change during the time considered, so does not need to be considered in the energy balance. Besides, there is no distance of interest given by ##L\theta##. There are two masses m; your equation only considers one. No, fully elastic means mechanical energy is conserved. Momentum is conserved if there are no external forces. Hitting the ground means there is an external force, so momentum of the two mass system is not conserved. There is also angular momentum. To consider that you have to choose an axis. Angular momentum is conserved about that axis if there are no external torques about the axis. Again, impact with the ground could be such an external torque. There are two ways to deal with this:
  1. Let the impulse from the ground be ##J##. You can now write equations for the change in momentum and the change in angular momentum. Since ##J## appears in both, you can eliminate it.
  2. The smarter way is to choose your axis such that ##J## has no torque about it. The obvious axis is the point of impact. Now you can write the conservation of angular momentum equation without introducing ##J##.
Here is how to proceed. Follow this as far as you can, posting what you get. Treat the two mass system as a single rigid body of mass 2m. Skip the falling phase. Only consider how the state just before impact compares with the state just after impact. Since everything is still in the same position, instantaneously, there is no change to PE. Start with the instant before contact. Both masses are falling at speed u, say. What is the KE of the system? For the next bit, you need to understand something about angular momentum of rigid bodies. A body mass M moving at velocity V has linear momentum MV. If its path would miss point P by distance R then on the basis of that alone it has angular momentum RMV about P. If the body is also rotating at rate ##\omega## and has moment of inertia ##I## about its mass centre then there is additional angular momentum ##I\omega##, bringing the total to ##RMV+I\omega## (these may have opposite signs). Setting P to be the point of impact, what is the angular momentum of the system just before impact? Next, we need some variables for the state just after impact. Since the ground is smooth, the system as a whole does not acquire any horizontal velocity. So I would choose ##v## as the vertical velocity of the mass centre (positive down, to be consistent with ##u##) and ##\omega## as the angular velocity. Define clockwise as positive here. What is the KE after impact? Write the conservation equation. What is the angular momentum about P after impact? Write that conservation equation.
First, I inquired about the text of the problem. Assuming the mass closest to the ground started from height ##h##, the masses will have velocities ##v_0 = \sqrt{2 gh}## immediately before the collision. To post a solution is to explain how the system moves immediately after the collision, qualitatively and quantitatively (i.e., there must be algebraic answers in terms of ##h##, or ##v_0##). This is all I produced from your advice, although I am afraid I made a mess. Let us start with the conservation of energy. The total energy of the system is the sum of its potential energy and kinetic energy. Since the potential energy depends on the height of the masses, we can write: $$PE = mg(h_A +h_B)$$ where ##h_A## and ##h_B## are the heights of mass A and B, respectively, and ##g## is the gravitational acceleration. The kinetic energy depends on the linear and angular velocities of the masses, so we can write: $$KE = \frac{1}{2} m (v_A^2 + v_B^2) + \frac{1}{2} I \omega^2$$ where ##v_A## and ##v_B## are the linear velocities of mass A and B, respectively, ##I## is the moment of inertia of the system about the point of impact, and ##\omega## is the angular velocity of the system. Since the energy is conserved, we can equate the initial and final energies of the system. We can write: $$m g (h_A + h_B) = \frac{1}{2} m (v_A^2 + v_B^2) + \frac{1}{2} I \omega^2$$ This is the first equation that we need to solve. Next, let us use the conservation of angular momentum. The angular momentum of the system is the product of its moment of inertia and angular velocity. Since there is no external torque about the point of impact, we can write: $$I \omega = I’ \omega’$$ where ##I’## and ##\omega’## are the moment of inertia and angular velocity of the system after the collision, respectively. This is the second equation that we need to solve. Now, we have two equations and four unknowns: ##v_A##, ##v_B##, ##I##, and ##\omega##. To solve this system, we need to find expressions for these unknowns in terms of the given variables: ##m##, ##L##, ##\theta##, and ##\Delta h##. Let us start with ##v_A## and ##v_B##. These are the linear velocities of the masses along the y-axis, which is perpendicular to the rod. We can use the geometry of the problem to find the relation between these velocities and the angular velocity ##\omega##. We can write: $$v_A = \omega L \sin \theta$$$$v_B =-\omega L \sin \theta$$ where the negative sign indicates that mass B is moving in the opposite direction of mass A. Next, let us find ##I## and ##I’##. These are the moments of inertia of the system about the point of impact, before and after the collision, respectively. We can use the parallel axis theorem to find the relation between these moments and the moment of inertia of the system about its center of mass. We can write: $$I = I_{cm} + 2 m d^2$$ $$I’ = I_{cm} + 2 m d’^2$$ where ##I_{cm}## is the moment of inertia of the system about its center of mass, ##d## and ##d’## are the distances of the center of mass from the point of impact, before and after the collision, respectively. To find ##I_{cm}##, we can use the formula for the moment of inertia of a rigid body.. Wecan write: $$I_{cm} = m (\frac{L}{2})^2 = \frac{1}{4} m L^2 To find ##d## and ##d’##, we can use the geometry of the problem and the conservation of mass. We can write: $$d = \frac{m h_A- m h_B}{2 m} = \frac{h_A- h_B}{2} = \frac{\Delta h}{2}$$ $$d’ = \frac{m h’_A- m h’_B}{2 m} = \frac{h’_A- h’_B}{2}$$ where ##h’_A## and ##h’_B## are the heights of mass A and B after the collision, respectively. To find ##h’_A## and ##h’_B##, we can use the conservation of energy and the fact that the potential energy of the system is zero at the point of impact. We can write: $$m g h’_A + mgh’_B = 0$$ $$h’_A + h’_B = 0$$ $$h’_A =-h’_B$$ Substituting these expressions into the equation for ##d’##, we get: $$d’ = \frac{h’_A- (-h’_A)}{2} = h’_A$$ Wecan substitute them into the equations for conservation of energy and angular momentum. We get: $$m g \Delta h = m \omega^2 L^2 \sin^2 \theta + \frac{1}{4} m L^2 \omega^2 + 2 m (\frac{\Delta h}{2})^2 \omega^2$$$$\frac{1}{4} m L^2 \omega + 2 m (\frac{\Delta h}{2})^2 \omega = \frac{1}{4} m L^2 \omega’ + 2 m h’_A^2 \omega’$$ Simplifying these equations, we get: $$g \Delta h = \omega^2 L^2 \sin^2 \theta + \frac{1}{4} L^2 \omega^2 + \frac{1}{2} \Delta h^2 \omega^2$$ $$L^2 \omega + 2 \Delta h^2 \omega = L^2 \omega’ + 8 h’_A^2 \omega’$$ These are two quadratic equations in ##\omega## and ##\omega’##. We can solve them by using the quadratic formula. The solutions are: $$\omega = \frac{-\frac{1}{4} L^2 \pm \sqrt{\frac{1}{16} L^4- 4 (\frac{1}{2} \Delta h^2 + L^2 \sin^2 \theta) g \Delta h}}{2 (\frac{1}{2} \Delta h^2 + L^2 \sin^2 \theta)}$$ $$\omega’ = \frac{-L^2 \pm \sqrt{L^4- 32 \Delta h^2 (g \Delta h- \frac{1}{4} L^2 \omega^2)}}{16 h’_A^2}$$ These are the angular velocities of the system before and after the collision, respectively. We can use them to find the linear velocities of the masses by using the equations: $$v_A = \omega L \sin \theta$$ $$v_B =-\omega L \sin \theta$$ $$v’_A = \omega’ L \sin \theta$$ $$v’_B =-\omega’ L \sin \theta$$ These are the linear velocities of the masses along the y-axis, before and after the collision, respectively. Finally, we can use the conservation of momentum along the x-axis to find the linear velocities of the masses along the x-axis, before and after the collision. Since there is no net external force along the x-axis, we have: $$m v_{Ax} + m v_{Bx} = 0$$ $$m v’{Ax} + m v’{Bx} = 0$$ where ##v_{Ax}##, ##v_{Bx}##, ##v’{Ax}##, and ##v’{Bx}## are the linear velocities of the masses along the x-axis, before and after the collision, respectively. Since the masses are initially at rest, we have: $$v_{Ax} = v_{Bx} = 0$$ Therefore, we have: $$v’{Ax} =-v’{Bx}$$To find ##v’{Ax}## and ##v’{Bx}##, we can use the geometry of the problem and the fact that the rod remains rigid. We can write: $$v’_{Ax} = \omega’ L \cos \theta$$ $$v’_{Bx} =-\omega’ L \cos \theta$$ I'm not convinced at all. Could you please help me?
 
  • #23
pepos04 said:
the masses will have velocities ##v_0 = \sqrt{2 gh}## immediately before the collision.
Right.
pepos04 said:
To post a solution is to explain how the system moves immediately after the collision, ….
Since the potential energy depends on the height of the masses, we can write: $$PE = mg(h_A +h_B)$$
There is no change in PE from immediately before impact to immediately afterwards, so there is no need to consider this.
pepos04 said:
Since the energy is conserved, we can equate the initial and final energies of the system. We can write: $$m g (h_A + h_B) = \frac{1}{2} m (v_A^2 + v_B^2) + \frac{1}{2} I \omega^2$$
The initial energy is the KE just before impact, ##mv_0^2##. Equate this to the KE after impact.
Also, your KE after impact expression is wrong. If you want to express the motion after impact in terms of the two velocities, then that is the whole KE, ##\frac{1}{2} m (v_A^2 + v_B^2)##. There is no additional energy from rotation.
But you will find it easier to work in terms of the system considered as a rigid body, so express the motion in terms of the velocity of the mass centre, ##v_1##, and rotation about that centre.

There were a lot of errors in your LaTeX. Before posting a reply with LaTeX in it, check the LaTeX by clicking the "magnifying glass on paper" icon at the right hand end of the toolbar above the text input area. Sometimes it doesn’t work and you have to do a screen refresh and try again. It is a toggle, so click it again to get back into edit mode.
 
  • #24
OK, I will try to write a procedure following your advice. However, is it correct to say that the mass hitting the ground cannot receive a reaction with a horizontal component because the ground is smooth and its impact is perpendicular to it? Then, it should be repelled along the vertical with velocity ##v_0 =\sqrt{2gh}##. This means that the other suspended mass should move perpendicular to the rod by deviating from the vertical fall, making an instantaneous rotation around the other and forming an angle ##\theta## with the vertical. Since its angular momentum before the collision was ##Lm\sqrt{2gh}\cos \theta##, it should also be ##Lm v \cos\theta## immediately after the collision, i.e. the modulus of ##v## should also remain ##v_0##, and thus only change direction. Please I would like to know if I am on the right track or is this nonsense? Thank you.
 
  • #25
pepos04 said:
the mass hitting the ground cannot receive a reaction with a horizontal component because the ground is smooth and its impact is perpendicular to it?
Yes, there cannot be a horizontal reaction from the ground.
pepos04 said:
Then, it should be repelled along the vertical with velocity
No, because at the same time as the reaction from the ground there is a reaction along the rod. This exerts a horizontal impulse equally and oppositely on the two masses, giving them equal and opposite horizontal velocities.
Likewise, it exerts a vertical impulse equally and oppositely on the two masses. That increases the vertical impulse from the ground.
It's this messiness that means it is much simpler to work in terms of conservation laws than worry about all the forces.
 
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  • #26
haruspex said:
Yes, there cannot be a horizontal reaction from the ground. No, because at the same time as the reaction from the ground there is a reaction along the rod. This exerts a horizontal impulse equally and oppositely on the two masses, giving them equal and opposite horizontal velocities. Likewise, it exerts a vertical impulse equally and oppositely on the two masses. That increases the vertical impulse from the ground. It's this messiness that means it is much simpler to work in terms of conservation laws than worry about all the forces.
OK, let me see if I understand. The rod holding the two masses together exerts a force to force them to stay ##L## apart. This force has a horizontal component. In the motion I described, there was a horizontal force acting on the second mass, because now its velocity is no longer vertical. But by the principle of action and reaction, then a horizontal force must also act on the first mass. Right? Would that be equivalent to saying that there are no horizontal forces external to the mass-rod-mass system, but in the description of the motion, the horizontal momentum has changed?
 
  • #27
pepos04 said:
there are no horizontal forces external to the mass-rod-mass system
Right
pepos04 said:
but in the description of the motion, the horizontal momentum has changed?
The horizontal momentum of the mass-rod-mass system has not changed. It is still zero.
 
  • #28
haruspex said:
Right The horizontal momentum of the mass-rod-mass system has not changed. It is still zero.
For “horizontal momentum” I mean “linear momentum”. Could you please explain this point more? However, I tried to work out a procedure under your previous advice, but some points I could not apply (for example, the assertion: "there is no additional energy from rotation," connected with "express the motion in terms of the velocity of the mass center ##v## and rotation about that center") Let me know if the process convinces you, what to fix, what to change altogether, etc. Let’s start ● ##m##isthemassofeachpoint ● ##L##isthelength of the rod ● ##\theta## is the angle of the rod with the horizontal before impact ● ##u##isthevelocity of each point before impact ● ##v##isthevertical velocity of the center of mass after impact ● ##\omega##istheangular velocity of the system after impact ● ##J##istheimpulse from the ground ● ##P##isthepoint of impact ● ##R##isthedistance from the center of mass to the point of impact ● ##I##isthemomentofinertia of the system about the center of mass ● ##g##istheacceleration due to gravity ● ##h##istheinitial height of the lower point above the ground As wesaid, we assume that the system is a rigid body of mass ##2m## and that the collision is fully elastic. Since the system is in free fall, we can use the equation of motion for constant acceleration to find the velocity ##u## before impact: ##u =\sqrt{2 g h}## I applied the conservation of energy. The system has only kinetic energy before and after impact, so we can write: $$\frac{1}{2} (2m) u^2 = \frac{1}{2} (2m) v^2 + \frac{1}{2} I \omega^2$$. Simplifying and rearranging, we get: $$v^2 + \frac{I}{2m} \omega^2 = u^2$$. I also applied the conservation of angular momentum. We choose the point of impact ##P## as the axis of rotation, since there is no external torque about this point. The angular momentum of the system before impact is:
 
  • #29
$$L_p = R (2m) u$$. The angular momentum of the system after impact is: $$L_p = I \omega$$. Equating these two expressions, we get: $$\omega = \frac{R (2m) u}{I}$$ I should find the values of ##R## and ##I##. I used the geometry of the system to f ind ##R##: ##R =L\sin \theta##. I used the parallel axis theorem to find ##I##: $$I = 2 (mL^2/12+mR^2)$$. Substituting these values into the previous equations, we get: $$v^2 + \frac{m L^2}{6} \omega^2 = u^2$$, and $$\omega = \frac{2 u \sin \theta}{L}$$. Wecansolve for ##v## and ##\omega##. We can eliminate ##\omega## from the f irst equation by using the second equation: $$v^2 + \frac{2 m u^2 \sin^2 \theta}{3} = u^2$$. Solving for ##v##, we get: $$v = \sqrt{u^2- \frac{2 m u^2 \sin^2 \theta}{3}}$$. Substituting this value into the second equation, we get: $$\omega = \frac{2 u \sin \theta}{L} \sqrt{1- \frac{2 m \sin^2 \theta}{3}}$$. To summarize, I obtained: ● Thecenter of massofthe system moves vertically downward with a velocity $$v = \sqrt{u^2- \frac{2 m u^2 \sin^2 \theta}{3}}$$. ● Thesystemrotates clockwise about the point of impact with an angular velocity $$\omega = \frac{2 u \sin \theta}{L} \sqrt{1- \frac{2 m \sin^2 \theta}{3}}$$.
 
  • #30
pepos04 said:
the assertion: "there is no additional energy from rotation,"
You can either write the KE treating the masses as independent, ##\frac 12mv_a^2+\frac 12mv_b^2##, or treating them as a single rigid body, ##\frac 12(2m)v^2+\frac 12I\omega^2##. For simplicity, l will use ##r=L/2##.
Since ##v_a^2=(v+r\omega\sin(\theta))^2+(r\omega\cos(\theta))^2##, ##v_b^2=(v-r\omega\sin(\theta))^2+(r\omega\cos(\theta))^2##, and ##I=2mr^2## these two forms are equivalent.
What you had done previously is mix them as ##\frac 12mv_a^2+\frac 12mv_b^2+\frac 12I\omega^2##, thereby counting rotational energy twice.
pepos04 said:
I used the geometry of the system to f ind ##R##: ##R =L\sin \theta##.
Two problems there: ##\theta## is the angle to the horizontal, so you need cos for horizontal displacement, and it is only L/2.
pepos04 said:
I used the parallel axis theorem to find ##I##: $$I = 2 (mL^2/12+mR^2)$$.
I is the moment of inertia about the mass centre of the system, so no need for the parallel axis theorem.
And these are two point masses at the ends of a light rod, not a rod of uniform mass. Each mass has MoI of ##m(L/2)^2## about their common centre, for a total ##\frac 12mL^2##.

Fix up those points and try again.
 
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  • #31
OK@haruspex, thanks. I'll try again. From conservation of angular momentum: $$2m \left(\frac{L}{2}\right) \cos \theta u = I \omega$$, with ##I = \frac{1}{2}mL^2##. So: $$2m u L \cos \theta = m \omega L^2 \Rightarrow \ \omega = 2 \frac{u}{L} \cos \theta = 2 \frac{\sqrt{2gh}}{L} \cos \theta$$. From conservation of mechanical energy: $$\frac{1}{2} (2m) u^2 = \frac{1}{2} (2m) v^2 + \frac{1}{2} I \omega^2 \Rightarrow \ 2 m u^2 = 2 mv^2+\frac{1}{2}mL^2 \Rightarrow \ 4 u^2 = 4 v^2 + L^2 \omega^2$$. Plugging in ##\omega = 2 \frac{u}{L} \cos \theta##, we get: $$4 u^2 = 4 v^2 + L^2 \left(4 \frac{u^2}{L^2} \cos^2 \theta \right) \Rightarrow \ v^2 = u^2- u^2 \cos^2 \theta \Rightarrow \ v = u \sin \theta = \sqrt{2gh} \sin \theta$$. Correct? Or are there fallacies in it?

I am also undecided about the sign, and I am not very clear about the graphic representation. Could you tell me how to do it? I have another doubt, though. The text asks to determine how the system moves immediately after the left mass is impacted. Then, the right mass assumes a direction perpendicular to ##L## with magnitude ##\sqrt{2gh}## unchanged with horizontal component ##\sqrt{2gh} \sin \theta## and vertical component ##-\sqrt{2gh} \cos\theta## while the other mass also assumes the same components with reversed sign. Overall after the collision, the total momentum of the system is therefore zero.

The rigid rod rotates around the CM tending to the horizontal position, and by the conservation of mechanical energy the left mass should reach position ##h## when the right one hits the ground. Probably, however, this solution does not work for ##\theta=\pi/2## since the horizontal component of ##v## cannot be ##v## itself and the vertical one null. The opposite should be true. Can this solution hold if supplemented with the condition ##\theta \neq \pi/2##, while for ##\theta =\pi/2##, ##v## and ##L## are vertical and aligned, the angular momentum is conserved, equal to zero before and after the impact of a mass that remains unchanged, and by conservation of energy the CM starts and returns to the same altitude h? Does that description fit, or are there fallacies?

Perhaps on second thought, this solution conserves energy and horizontal momentum, and meets the condition that the two masses remain at distance L. But I think there are infinite solutions that satisfy these 3 conditions. We have 4 unknowns (the two components of the velocity of each of the 2 masses), and we are using only 3 equations. What is the fourth equation?
haruspex said:
I looked for repeating patterns of three bounces and got ##\theta+n\pi=\frac{2u^2}{Lg}\frac{\sin^2(\theta)\cos(\theta)}{(1+\cos^2(\theta))^2}##, where n is related to the number of complete half turns it takes in the air between 1st and second bounce, then in reverse between 2nd and 3rd bounce.
Moreover, this result caught my attention. Could you tell me how you arrived at it, through what physical principles or maths?
 
Last edited by a moderator:
  • #32
pepos04 said:
OK@haruspex, thanks. I'll try again. From conservation of angular momentum: $$2m \left(\frac{L}{2}\right) \cos \theta u = I \omega$$, with ##I = \frac{1}{2}mL^2##.
Just as we had to consider the angular momentum about P due to the linear velocity u, we have to do the same with v. Need to be careful with signs. For consistency, I will take v as positive down, so it will likely turn out negative. Correspondingly, take ##\omega ## as positive clockwise.
So the angular momentum conservation equation is $$mLu \cos \theta = I \omega+mLv \cos \theta $$.
pepos04 said:
So: $$2m u L \cos \theta = m \omega L^2 \Rightarrow \ \omega = 2 \frac{u}{L} \cos \theta = 2 \frac{\sqrt{2gh}}{L} \cos \theta$$.
No point in substituting for u in terms of g and h yet. That can be done at the end. So ##mLu \cos \theta = \frac 12mL^2\omega+mLv \cos \theta##, or ##u \cos \theta = \frac 12L\omega+v \cos \theta##.

You should end up with ##v=-u\frac{\sin^2(\theta)}{1+\cos^2(\theta)}##.
 
  • #33
@haruspex I now understand. I get two solutions: one is ##v = u## (to be discarded), the other is exactly the one you presented. I can also see the similarity between the equation of ##v## and the equation of ##\theta + n \pi## that you wrote, but I still cannot understand how you got the latter. Could you please help me? Also, what is wrong with my final description (the one about the 4 unknowns and the three equations)? Could you pointly object to this description, which is probably wrong but which I think might work?
 
  • #34
pepos04 said:
We have 4 unknowns (the two components of the velocity of each of the 2 masses), and we are using only 3 equations.
Conservation of energy, conservation of horizontal linear momentum, conservation of angular momentum, constancy of separation.
pepos04 said:
Could you tell me how you arrived at it
One way in which it will be sure to go through a finite repeating sequence is if the descent after the bounce is the mirror image of the ascent. For that to be the case, at the highest point between first and second bounce the rod must be either horizontal or vertical.
If it takes time t to reach the highest point we have ##v=gt## and ##\omega t=\theta+n\pi/4##.
Plugging in the values found for ##v, \omega## leads to a relationship between ##h## and ##\theta##.
But the expression I quoted before wasn't quite right: I only considered the rod being horizontal at the top, so I had ##n \pi/2## instead of ##n\pi/4##. And it is a four bounce sequence, not a three bounce sequence.

There will be many other finite sequences, but they're more complex to analyse.
 
  • #35
Forgive me @haruspex, but I still do not fully understand. In my solution (the one without accounts) did I not consider the four equations (angular momentum, horizontal momentum, separation, energy) correctly? I cannot understand whether this solution is fallacious (most likely yes) and, if so, why it would be. I would be appreciative if you could help me to clear up these perplexities. In addition, I get a result ##\theta+n \frac{\pi}{4}=-\frac{4u^2}{Lg}\frac{\sin^2(\theta)\cos(\theta)}{(1+\cos^2(\theta))^2}##, which therefore differs from yours by a minus sign and a ##4## instead of a ##2##. Even considering a ##2## at the denominator in the LHS, the expression would be different. How is this possible?Forgive me @haruspex, but I still do not fully understand. In my solution (the one without accounts) did I not consider the four equations (angular momentum, horizontal momentum, separation, energy) correctly? I cannot understand whether this solution is fallacious (most likely yes) and, if so, why it would be. I would be appreciative if you could help me to clear up these perplexities. In addition, I get a result ##\theta+n \frac{\pi}{4}=-\frac{4u^2}{Lg}\frac{\sin^2(\theta)\cos(\theta)}{(1+\cos^2(\theta))^2}##, which therefore differs from yours by a minus sign and a ##4## instead of a ##2##. Even considering a ##2## at the denominator in the LHS, the expression would be different. How is this possible?
 

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