Challenging problem about an impact with a smooth frictionless surface

[haruspex]

In my solution

Which post?

differs from yours by a minus sign

In your version, the LHS is positive and the RHS negative for $0<\theta <\pi/2$, implying there is no solution in that range, which seems unreasonable. I cannot tell why you get a different sign without seeing your working.

 

[pepos04]

Which post? In your version, the LHS is positive and the RHS negative for $0<\theta <\pi/2$, implying there is no solution in that range, which seems unreasonable. I cannot tell why you get a different sign without seeing your working.

Post #31, the qualitative description. I get a minus sign because I have substituted the values of $#omega = \frac{4 u \cos \theta}{L (1+ \cos^2 \theta)}$ and $v =- u \frac{\sin^2 \theta}{(1+ \cos^2 \theta)}$, so I get a negative solution because the latter is negative. But you considered $v = gt$, not $v =-gt$, so I should consider $v$ positive. The problem is that I keep getting 4 in the numerator of the RHS, not 2, and a 4 in the denominator of the LHS. How come? Because in your previous equation (which applied to $\theta = n \pi/2$), you have a 2 in the numerator of the RHSandnothing in the denominator of the LHS (where there should be a 2, right?)?

 

[haruspex]

Post #31, the qualitative description.

Maybe you mean this:

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.

I don't know how you arrive at any of that. Seems like guesswork/intuition. Intuition is a good start, but it has to be backed up with equations based on solid principles, the conservation laws.

But you considered $v = gt$, not $v =-gt$, so I should consider $v$ positive.

Yes, I should have written v=-gt.

I keep getting 4 in the numerator of the RHS, not 2, and a 4 in the denominator of the LHS. How come?

I explained that when I wrote post #17 I had only thought of the cases where the rod is horizontal at its highest point. Later, I realised it also works if it is vertical there. That turned a 2 into a 4.
You seem to be saying there is another discrepancy. Maybe, but it was just an interesting aside to the main thread and I see no merit in worrying about the details. The principle was the fun part.

 

[SammyS]

Below is a quote of Post #37, after a little editing. You had a triple # character at the beginning of your first LaTeX expression. It was also missing the \ character for the omega.
I also changed the \frac to \dfrac in both fractions, to make them more readable.

Post #31, the qualitative description. I get a minus sign because I have substituted the values of $\omega = \dfrac{4 u \cos \theta}{L (1+ \cos^2 \theta)}$ and $v =- u \dfrac{\sin^2 \theta}{(1+ \cos^2 \theta)}$, so I get a negative solution because the latter is negative. But you considered $v = gt$, not $v =-gt$, so I should consider $v$ positive. The problem is that I keep getting 4 in the numerator of the RHS, not 2, and a 4 in the denominator of the LHS. How come? Because in your previous equation (which applied to $\theta = n \pi/2$), you have a 2 in the numerator of the RHS and nothing in the denominator of the LHS (where there should be a 2, right?)?

 

[pepos04]

Forgive me, @haruspex and anyone else who would like to answer me (@kuruman, @BvU, @Lnewqban, etc.), but I have some qualitative questions that have arisen from looking at this problem. Why does the form of $v$ not hold $u$? What happens in the case $\theta = \pi/2$? Is it correct that before the collision the linear momentums of both have no horizontal components and that the momentums are equal to $m u$? After the collision due to the effect of $L$ would there then be two opposite horizontal components of the linear momentums that need to be taken into account in the angular momentum? Is the change in linear momentum along y? Why? Why is the linear momentum conserved along x? Can it be said that with respect to the point-mass on the ground before the impact there is only the momentum of the right point-mass that is worth $mL u \cos \theta$, and the momentum of the left one would be zero? The only momentum then would be that of the right one, whose momentum has changed direction, it is no longer vertical. But this is sharply contrasting … If the angular momentum with respect to the ground mass point before the impact is equal to $mL u \cos\theta$, after the impact it should be $M=\vec L \times m\vec v =(L \cos \theta \vec i + L \sin \theta \vec j) \times m(v_x \vec i+ v_y\vec j)$, these being the horizontal and vertical components of $\vec v$. As components one would find $v_x=0, v_y =u$. In other words I would find that they continue in the vertical direction. How to reconcile this?

 

[kuruman]

I have not followed this closely, however since you quoted me, I will answer al least some of your many questions, the ones where I understand what you are asking and can answer without causing confusion.

What happens in the case θ=π/2?

Ideally, the rod will fall straight down vertically and bounce right back up to where it came from. Non-ideally, the leading mass will hot some small angle relative to the vertical which makes the angle $\theta$ as defined here close to $\frac{\pi}{2}.$

Is it correct that before the collision the linear momentums of both have no horizontal components and that the momentums are equal to mu?

If the system is released from rest, yes. Gravity acts vertically. Neither mass has a horizontal component before the collision because (a) there is no horizontal external force acting on either mass and (b) since the masses are equal, the external torque of gravity does not cause a rotation about the center of mass that would provide a horizontal component to the velocities of the individual masses when the masses are unequal.

Is the change in linear momentum along y? Why? Why is the linear momentum conserved along x?

The only external force other than gravity that can act on this system is the frictionless horizontal ground. "Frictionless" means that it can only exert a force perpendicular to its surface. "Horizontal" means that the perpendicular to the surface is along the vertical. Thus, both external forces acting on the system are vertical. There is no horizontal external force acting on the system at any time to change its momentum in the horizontal direction. This means that the linear momentum along x is conserved. The change in linear momentum is along y only because, as I just explained, (a) there are two external forces, gravity and contact force, along y to change it and (b) there is no horizontal external force.

 

[haruspex]

Why does the form of $v$ not hold $u$? What happens in the case $\theta = \pi/2$?

I assume those two questions are connected, but I am not sure what the first one means. If we plug $\theta=\pi/2$ into $v =- u \frac{\sin^2 \theta}{(1+ \cos^2 \theta)}$ we get $v=-u$ as expected.

Can it be said that with respect to the point-mass on the ground before the impact there is only the momentum of the right point-mass that is worth $mL u \cos \theta$, and the momentum of the left one would be zero?

Be careful with the terminology. What you are describing now is angular momentum. If you just write "momentum" then linear momentum is implied.
If a mass m has velocity v and, at its closest, will be distance r from point P then its angular momentum about P is mvr. (The sign depends on your sign convention.)
As a general principle, it is a good idea to choose P as a fixed point. Complications can arise otherwise when discuss torques and angular accelerations.

So with respect to the point on the ground where the first mass will land, the first mass has no angular momentum because r=0. The angular momentum of the other mass is $muL\cos(\theta)$.
Alternatively we can treat the rod and two masses as a single rigid body, so it has mass 2m and $r=L/2\cos(\theta)$, again giving $muL\cos(\theta)$.

If the angular momentum with respect to the ground mass point before the impact is equal to $mL u \cos\theta$, after the impact it should be $M=\vec L \times m\vec v$.

Here you seem to be defining $\vec v$ as the velocity of the second mass just after impact. Are you confusing that with v, which I defined as the (vertical) velocity of the mass centre just after impact?
To clarify, I'll use subscripts c for mass centre and 2 for the second mass. The mass centre has velocity $\vec v_c$ and the second mass has velocity $\vec v_2=\vec v_c+\vec L\times\vec\omega/2$. (I may have the sign wrong.)
The vertical component of that is $v-L\cos(\theta)\omega/2$ and the horizontal component is $L\sin(\theta)\omega/2$.

 

[pepos04]

Thanks @haruspex and @kuruman. I had also thought about this other possible solution. I denote by $v_{1x}, v_{1y}, v_{2x}, v_{2y}$ the components of the velocities of the two particles after the collision. Particle 1 is the one that impacts the ground. The x-axis is horizontal and the y-axis is vertical. The conditions before the collision are $v_{1x}=0, v_{1y}=-u, v_{2x}=0, v_{2y}=-u$ where I took the positive upward direction. Equation 1, horizontal linear momentum conservation: $$0 = mv_{1x} + mv_{2x}$$. Equation 2, energy conservation: $$2 \times \frac{1}{2}m u^2 = \frac{1}{2} m (v_{1x}^2 + v_{1y}^2 + v_{2y}^2)$$. Condition 3 should be the constancy of separation (i.e., the geometric constraint, right?), the one that dictates that the distance between the particles remains $L$. Putting the origin at the point of the collision, the direct verse toward particle 2 is: $\hat l = \cos(\theta) \hat x + \sin(\theta) \hat y$. The condition to be imposed is that the relative velocity between the masses is orthogonal to $\hat l$, so we want to impose that $(\vec v_2- \vec v_1) \cdot \hat l = 0$, and this gives us equation 3: $$\displaystyle (v_{2x}- v_{1x})\cos \theta + (v_{2y}- v_{1y})\sin \theta = 0$$. The angular momentum with respect to the ground mass point before the collision is equal to $\vec M =- m L u \cos \theta \hat z$, after the collision should be $M=\vec L\times m\vec v =( L cos\theta \vec i + L sen\theta \vec j)\times m(v_{2x} \vec i+ v_{2y} \vec j)$. What is wrong with this reasoning? What can be added to it? Should it lead to the same result found or is it just fantasy?

 

[haruspex]

Thanks @haruspex and @kuruman. I had also thought about this other possible solution. I denote by $v_{1x}, v_{1y}, v_{2x}, v_{2y}$ the components of the velocities of the two particles after the collision. Particle 1 is the one that impacts the ground. The x-axis is horizontal and the y-axis is vertical. The conditions before the collision are $v_{1x}=0, v_{1y}=-u, v_{2x}=0, v_{2y}=-u$ where I took the positive upward direction. Equation 1, horizontal linear momentum conservation: $$0 = mv_{1x} + mv_{2x}$$. Equation 2, energy conservation: $$2 \times \frac{1}{2}m u^2 = \frac{1}{2} m (v_{1x}^2 + v_{1y}^2 + v_{2y}^2)$$. Condition 3 should be the constancy of separation (i.e., the geometric constraint, right?), the one that dictates that the distance between the particles remains $L$. Putting the origin at the point of the collision, the direct verse toward particle 2 is: $\hat l = \cos(\theta) \hat x + \sin(\theta) \hat y$. The condition to be imposed is that the relative velocity between the masses is orthogonal to $\hat l$, so we want to impose that $(\vec v_2- \vec v_1) \cdot \hat l = 0$, and this gives us equation 3: $$\displaystyle (v_{2x}- v_{1x})\cos \theta + (v_{2y}- v_{1y})\sin \theta = 0$$. The angular momentum with respect to the ground mass point before the collision is equal to $\vec M =- m L u \cos \theta \hat z$, after the collision should be $M=\vec L\times m\vec v =( L cos\theta \vec i + L sen\theta \vec j)\times m(v_{2x} \vec i+ v_{2y} \vec j)$. What is wrong with this reasoning? What can be added to it? Should it lead to the same result found or is it just fantasy?

I'm not inclined to spend effort checking a method that is unnecessarily complicated.

 

[pepos04]

@haruspex I understand. But I presented this procedure because, with both this method and yours, I arrive at $v_{2y} =- u$ (or, alternately, $v_{2y} = u$, depending on how you position the reference axis, upward or downward). It will be directed upward and have the same magnitude as $u$. But does this solution make physical sense?

 

[haruspex]

@haruspex I understand. But I presented this procedure because, with both this method and yours, I arrive at $v_{2y} =- u$ (or, alternately, $v_{2y} = u$, depending on how you position the reference axis, upward or downward). It will be directed upward and have the same magnitude as $u$. But does this solution make physical sense?

I think I got myself a bit confused over signs.
I took v as positive down to be consistent with u. If I do the same with $v_{2y}$ and take $\omega$ as positive clockwise (as I did) then $v_{2y}=v+L\cos(\theta)\omega/2=-u\frac{\sin^2(\theta)}{1+\cos^2(\theta)}+u\frac{\cos^2(\theta)}{1+\cos^2(\theta)}=u\frac{\cos(2\theta)}{1+\cos^2(\theta)}$.

 

[pepos04]

@haruspex The signs now come back to me, but in my opinion your expression is wrong. The angular velocity should be $\omega = \frac{4 u \cos \theta}{L (1+ \cos^2 \theta)}$, not $\frac{2 u \cos \theta}{L (1+ \cos^2 \theta)}$, so you would be missing a 2 in your expression. We would have $v_{2y}=v+L\cos(\theta)\omega/2=u\frac{\sin^2(\theta)}{1+\cos^2(\theta)}+2u\frac{\cos^2(\theta)}{1+\cos^2(\theta)}=u\frac{\cos( 2\theta)}{1+\cos^2(\theta)} = u\frac{3 \cos^2(\theta)-1}{1+\cos^2(\theta)}$. That way it would all add up for me. Actually, I was referring to $v_{1y}$, which should be $v_{1y} = v_0$. Check to be sure, but I am almost sure because I would have confirmation from both procedures.

 

[haruspex]

@haruspex The signs now come back to me, but in my opinion your expression is wrong. The angular velocity should be $\omega = \frac{4 u \cos \theta}{L (1+ \cos^2 \theta)}$, not $\frac{2 u \cos \theta}{L (1+ \cos^2 \theta)}$, so you would be missing a 2 in your expression. We would have $v_{2y}=v+L\cos(\theta)\omega/2=u\frac{\sin^2(\theta)}{1+\cos^2(\theta)}+2u\frac{\cos^2(\theta)}{1+\cos^2(\theta)}=u\frac{\cos( 2\theta)}{1+\cos^2(\theta)} = u\frac{3 \cos^2(\theta)-1}{1+\cos^2(\theta)}$. That way it would all add up for me. Actually, I was referring to $v_{1y}$, which should be $v_{1y} = v_0$. Check to be sure, but I am almost sure because I would have confirmation from both procedures.

Yes, I think you must be right. I realised just now that the expression I got at the end of post #46 cannot be right since it gives the wrong answer at $\theta=\pi/2$. Yours gives the right answer.
However, I think you meant $v_{2y}=-u\frac{\sin^2(\theta)}{1+\cos^2(\theta)}+2u\frac{\cos^2(\theta)}{1+\cos^2(\theta)}= u\frac{3 \cos^2(\theta)-1}{1+\cos^2(\theta)}$

 

[pepos04]

@haruspex Yes. But do you know an intuitive explanation for why $v_{1y}$ is equal in modulus to $u$ for each angle? This seems surprising to me...

 

[haruspex]

@haruspex Yes. But do you know an intuitive explanation for why $v_{1y}$ is equal in modulus to $u$ for each angle? This seems surprising to me...

Yes, it is interesting. It says that the downward impulse from the rod is matched by the increase in the upward impulse from the ground, but it is not obvious why that should be. It would have to be a consequence of the conservation of mechanical energy.

 

[pepos04]

It would have to be a consequence of the conservation of mechanical energy.

Excuse me, how is mechanical energy conserved? What equations should be set up to verify what you say?

 

[haruspex]

Excuse me, how is mechanical energy conserved? What equations should be set up to verify what you say?

You stated at the beginning that it is, and we used that in the analysis.

 

[pepos04]

You stated at the beginning that it is, and we used that in the analysis.

Of course, you are absolutely right. But I meant to say, actually: how is the fact that $v_{1y} = |u$| a direct consequence of the conservation of mechanical energy? From what equations should it be deduced? I do not grasp the obvious connection between $v_{1y} = |u|$ and conservation of mechanical energy: could you explain it to me? Imagine two masses and a connecting rod are suspended from a ceiling by a string a bit away from CG and string burnt at t=0: can you intuitively explain $v_{1y} = |u|$ by observing this small experiment?

 

[haruspex]

Of course, you are absolutely right. But I meant to say, actually: how is the fact that $v_{1y} = |u$| a direct consequence of the conservation of mechanical energy? From what equations should it be deduced? I do not grasp the obvious connection between $v_{1y} = |u|$ and conservation of mechanical energy: could you explain it to me? Imagine two masses and a connecting rod are suspended from a ceiling by a string a bit away from CG and string burnt at t=0: can you intuitively explain $v_{1y} = |u|$ by observing this small experiment?

I did not say there was an obvious connection. If ME were not conserved the velocity would have been different, so there is not going to be a connection that does not involve it.

 

[pepos04]

Conservation of energy, conservation of horizontal linear momentum, conservation of angular momentum, constancy of separation. 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.

I appreciated the elucidation. But another observation grips me. I went looking for these other finished sequences, but I couldn't figure out how to find them? Why are they more complex to analyze? What would they be?

 

[haruspex]

I appreciated the elucidation. But another observation grips me. I went looking for these other finished sequences, but I couldn't figure out how to find them? Why are they more complex to analyze? What would they be?

If you want to study that further, the first step would be to get more general equations. We only considered the case in which there is no initial rotation.
From there, you may be able to find two bounce sequences using the same approach of setting the orientation to be either horizontal or vertical at the highest point.
Beyond that, it may be possible to prove the existence of more complicated finite sequences without necessarily finding any.