Calculating Theoretical Reaction Force of Ball Traveling in Semi-Circle

[jbriggs444]

sorry 5m/s(sqr)

They why is 2.5 m/s half of 5 m/s? And what happened to the other 7.5 m/s?

 

[haruspex]

Lets say the ball returns to the player at 2.5m/s, having lost half it's momentum to the y direction.

You supposed that the ball was accelerated at 5m/s², not that it reached 5m/s. That would require that the acceleration persisted for 1s.
Ok, suppose it did.

If energy and momentum are both conserved then we can calculate the later velocities from the masses. No need to guess.

 

[TonyCross]

They why is 2.5 m/s half of 5 m/s? And what happened to the other 7.5 m/s?

I think you are talking about Classical mechanics - Kinetic energy which is a fair value to consider.
KE=1/2 mv² (just figured how to superscript) In which case the 1kg mass I describe could be considered to have energy due to it's motion, of 12.5 Joules. so half the velocity would then be 3.125 Joules. So maybe I should have said it lost 6.25 joules of energy and it's velocity was then 3.535m/s.
Is this what you mean?

 

[TSny]

I attempted to use Mathematica to numerically solve the equations of motion and generate some animations.

The tube is represented by a semicircle. I varied the mass ratio M/m, where M is the mass of the tube and m is the mass of the ball. The animations start at the instant the particle enters the tube on the left.

Here is the choice M/m = 2

The ball makes it all the way through the tube. The final angular velocity of the tube is zero. I found this to be the case whenever the ball makes it all the way through the tube. The red dot is the center of mass of the system. The small black dot is the center of mass of the tube.The next case is for M/m = 1/3.

Now the ball does not make it all the way through the tube. It exits the same end of the tube in which it entered. The final angular velocity of the tube is not zero.There is a critical value of the ratio M/m equal to $\large \frac{4\pi-\pi^2}{\pi^2-4} \normalsize\approx 0.46$ where the ball gets stuck at the apex of the tube.

Of course, all of the above depends on the correctness of the equations of motion and the programming.

 

[haruspex]

The final angular velocity of the tube is zero. I found this to be the case whenever the ball makes it all the way through the tube.

Suggests that could be proved without having to get a full analytic solution.

 

[TonyCross]

I attempted to use Mathematica to numerically solve the equations of motion and generate some animations.

The tube is represented by a semicircle. I varied the mass ratio M/m, where M is the mass of the tube and m is the mass of the ball. The animations start at the instant the particle enters the tube on the left.

Here is the choice M/m = 2
View attachment 272194
The ball makes it all the way through the tube. The final angular velocity of the tube is zero. I found this to be the case whenever the ball makes it all the way through the tube. The red dot is the center of mass of the system. The small black dot is the center of mass of the tube.The next case is for M/m = 1/3.
View attachment 272195
Now the ball does not make it all the way through the tube. It exits the same end of the tube in which it entered. The final angular velocity of the tube is not zero.There is a critical value of the ratio M/m equal to $\large \frac{4\pi-\pi^2}{\pi^2-4} \normalsize\approx 0.46$ where the ball gets stuck at the apex of the tube.
View attachment 272196

Of course, all of the above depends on the correctness of the equations of motion and the programming.

I never imagined when I asked this question it would be so complex. Thanks for putting this effort, it must have taken you ages..

 

[TSny]

Suggests that could be proved without having to get a full analytic solution.

Perhaps an argument involving time reversal. At the instant that the ball reaches the apex of the tube, consider the time-reversed motion that takes the ball back from the apex to the entrance of the tube. I think, from symmetry, this reversed motion should be the mirror reflection of the actual forward-time motion beyond the apex (if that makes any sense). These two motions have essentially the same initial conditions.

 

[haruspex]

Perhaps an argument involving time reversal. At the instant that the ball reaches the apex of the tube, consider the time-reversed motion that takes the ball back from the apex to the entrance of the tube. I think, from symmetry, this reversed motion should be the mirror reflection of the actual forward-time motion beyond the apex (if that makes any sense). These two motions have essentially the same initial conditions.

Something is not adding up for me. Consider a point on the final trajectory of the mass centre of the hoop. The hoop starts and finishes with zero angular momentum about that point, so shouldn't the ball have the same angular momentum about that point that it started with?

 

[TSny]

Something is not adding up for me. Consider a point on the final trajectory of the mass centre of the hoop. The hoop starts and finishes with zero angular momentum about that point, so shouldn't the ball have the same angular momentum about that point that it started with?

Yes, I think that's right. But, I don't necessarily see a problem. Suppose you consider the point of intersection of the final trajectories of the ball and cm of hoop. There is certainly zero total final angular momentum about that point if the final angular velocity of the hoop is zero. But, if that point lies on the initial trajectory of the ball, then there would also be zero total initial angular momentum about that point.

 

[TSny]

Here are the trajectory paths for the case where M/m = 2 using the output of Mathematica. The initial and final trajectory lines all intersect at one point.

 

[haruspex]

Yes, I think that's right. But, I don't necessarily see a problem. Suppose you consider the point of intersection of the final trajectories of the ball and cm of hoop. There is certainly zero total final angular momentum about that point if the final angular velocity of the hoop is zero. But, if that point lies on the initial trajectory of the ball, then there would also be zero total initial angular momentum about that point.

Yes, you are right. From studying the animation, I thought I was able to pick a point on the final hoop trajectory for which the ball's angular momentum was not conserved, but I must not have been careful enough.

 

[TSny]

I never imagined when I asked this question it would be so complex. Thanks for putting this effort, it must have taken you ages..

It's an interesting problem. The time spent was worth it. It's fun.

 

[haruspex]

Perhaps an argument involving time reversal. At the instant that the ball reaches the apex of the tube, consider the time-reversed motion that takes the ball back from the apex to the entrance of the tube. I think, from symmetry, this reversed motion should be the mirror reflection of the actual forward-time motion beyond the apex (if that makes any sense). These two motions have essentially the same initial conditions.

Yes, I believe that argument works, very neat.

In this 'pass through' scenario, let the ball start in the +Y direction at speed u, mass m; the hoop finish at angle α to +Y on the +X side, speed v, mass M; the ball finish at β to +Y on the -X side, speed u'.
Angular momentum conservation, using your argument, gives $u\tan(\alpha)=u'\tan(\pi-\beta-\alpha)$.
Linear momentum conservation gives:
X: $mu'\sin(\beta)=Mv\sin(\alpha)$
Y: $mu'\cos(\beta)+Mv\cos(\alpha)=mu$
KE: $mu^2=mu'^2+Mv^2$

Four equations, four unknowns. I'll have a go at solving later.

 

[guv]

I stumbled upon this problem and is fascinated by the result. The lack of spin for the tube is counter intuitive. I like the symmetry argument. It still feels odd that the amount of angular impulse applied for each half the motion with respect to the midpoint of the ring could be exactly equal.