Calculating Theoretical Reaction Force of Ball Traveling in Semi-Circle

In summary, the question asks about the theoretical reaction force on a pipe when a 1kg metal ball is introduced at a constant velocity of 5m/s into a 180-degree bend with a 22mm inner diameter and a radius of 2m. The ball is assumed to roll around the pipe without any friction and the total reactive force acting on the pipe is being calculated. The direction of this force is also questioned, with the assumption that the y force cancels out. The formula for centripetal force (Fc = m v² / r) is used to calculate the force, but there is confusion about the apparent increase in force. The tube is expected to remain in the x direction without any rotation.
  • #1
TonyCross
66
12
Summary:: Theoretical reaction force due to ball traveling in a semi-circle circum

Hi,
Can anyone please help me with the following question:

Setup...
1 x length of pipe bent 180 degrees with an inner diameter of 22mm, it's radius is 2mtrs, the mass of this object is 100kg. Total length of pipe (4*3.14/2) = 6.28 mtr.
1 x Metal ball (sphere) with a Diameter of 20mm and a mass of 1kg.

Question:
Assuming there is no gravity and the tube and the ball are in the vacuum of space, no other forces come to bear such as friction from the pipe sitting on the ground, or tangental friction forces from the rolling ball.
The ball rolls around the pipe causing friction, however I would like to ignore friction for the sake of this problem.

I introduce the 1kg metal ball into the tube at a constant velocity of 5m/s.

What is the total reactive force acting on the pipe?

A followup question is what direction (x/y) is this force acting on the pipe. (Assuming the pipe lays x/y direction with the ball introduced into the x direction and exits -x) What is the linear displacement vector for the tube. I would think that the y force on the tube cancels out.

My confusion lays in the centripetal force equation (Fc = m v² / r ) which gives the figure 25/2 = 12.5N. If the 100kgm pipe is at rest there is inertia to also consider.

My question is what formula would I use to calculate the reactive force in the x direction shown in my diagram, given the time the ball is traveling which is just over 1 second as the pipe is just over 6m long and the ball is traveling at 5m/s. I am trying to workout the acceleration and the displacement of the 100kg pipe, as in Newtons 3rd law.

I hope this makes sense as I am very much trying to find my feet with this problem.

Thanks Tony
 

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  • #2
I'm 65 so I guess it's back to school..
 
  • #3
Hello @TonyCross , :welcome: !

Your profile doesn't show where you are in your curriculum.
My guess is that you are not in an advanced course on Lagrangian mechanics and that you make up this exercise yourself ?
If so, can you forget about the motion of the pipe and assume the pipe is fixed in position ?
If not, you have to ask yourself what kind of encounter this is and what is conserved in such encounters
 
  • #4
And: what makes you so sure the ball comes out in the -x direction ? If the pipe can move, it can rotate, so you'll have to argue that's a reaonable 'outcome' :smile:
 
  • #5
Working in terms of forces and accelerations looks rather complicated. It would be better if we could take an overall view, working in terms of conservation laws.
How many degrees of freedom are there in the state when the ball emerges? How many equations are available?
 
  • #6
Thanks for your reply,
There are no terms of freedom when the ball emerges it must travel in a straight line in-x, the tube is simply providing angular velocity change with a force directed towards the center of the curve. Gravitational force is then applied to the tube in the opposite direction, like spinning a bucket of water on a piece of string, the force is a pulling force from the center, gravity stops the water from coming out the bucket, by reactive force, or in my case the tube.
While in free space before it enters the tube has a force of 0.
The ball has been subjected to 5N initially to propel a 1kg ball to 5m/s. (F=ma)

I know that the formula for centripetal force is (Fc = m v² / r ) which provides a force of 12.5N, my confusion comes from the apparent increase in force.

I know the accelerations look complicated that's the reason for my question.

Cheers Tony
 
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  • #7
BvU said:
And: what makes you so sure the ball comes out in the -x direction ? If the pipe can move, it can rotate, so you'll have to argue that's a reaonable 'outcome' :smile:
Hi,
Thanks for your reply.
My guess is that ignoring friction it will not rotate, this is based on the angular velocity vectors, when the ball travels through the tube the tube causes a constant change in the acceleration vector to the ball (red lines on the image).
My reason for thinking it will travel in the x direction is given in the attached image, if am I wrong if you believe the pipe will rotate, ignoring friction what force will cause this rotation?

Cheers Tony
 

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  • #8
BvU said:
Hello @TonyCross , :welcome: !

Your profile doesn't show where you are in your curriculum.
My guess is that you are not in an advanced course on Lagrangian mechanics and that you make up this exercise yourself ?
If so, can you forget about the motion of the pipe and assume the pipe is fixed in position ?
If not, you have to ask yourself what kind of encounter this is and what is conserved in such encounters
Thanks for the reply,
I don't want to assume the pipe is in a fixed position, that's the whole point of my question.
maybe rather than be sarcastic you could offer, I don't know ... a physics reply!
 
  • #9
TonyCross said:
There are no terms of freedom when the ball emerges
TonyCross said:
My guess is that ignoring friction it will not rotate,
No, it will certainly rotate for part of the time. Having gone a small distance into the tube, the ball will exert a force normal to the contact with the tube. This will point out from the centre if curvature of the tube. The centre of mass of the tube does not lie in that line, so the force will exert a torque that starts the tube rotating.

Will it still be rotating when the ball leaves? Consider the angular momentum of the system about an axis at the point of entry of the ball. On entry, the system has zero angular momentum about that point. But on exit, if the tube is in its original orientation, the ball will have angular momentum, so the tube must do too.

If you drop your assumptions, there are four degrees of freedom for the final state: rotation of the tube, velocity of the tube (two degrees) and speed of the ball relative to the tube.
Can you find four equations?

Edit.. there are more unknowns... see below
 
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  • #10
haruspex said:
No, it will certainly rotate for part of the time. Having gone a small distance into the tube, the ball will exert a force normal to the contact with the tube. This will point out from the centre if curvature of the tube. The centre of mass of the tube does not lie in that line, so the force will exert a torque that starts the tube rotating.

Will it still be rotating when the ball leaves? Consider the angular momentum of the system about an axis at the point of entry of the ball. On entry, the system has zero angular momentum about that point. But on exit, if the tube is in its original orientation, the ball will have angular momentum, so the tube must do too.

If you drop your assumptions, there are four degrees of freedom for the final state: rotation of the tube, velocity of the tube (two degrees) and speed of the ball relative to the tube.
Can you find four equations?
haruspex said:
No, it will certainly rotate for part of the time. Having gone a small distance into the tube, the ball will exert a force normal to the contact with the tube. This will point out from the centre if curvature of the tube. The centre of mass of the tube does not lie in that line, so the force will exert a torque that starts the tube rotating.

Will it still be rotating when the ball leaves? Consider the angular momentum of the system about an axis at the point of entry of the ball. On entry, the system has zero angular momentum about that point. But on exit, if the tube is in its original orientation, the ball will have angular momentum, so the tube must do too.

If you drop your assumptions, there are four degrees of freedom for the final state: rotation of the tube, velocity of the tube (two degrees) and speed of the ball relative to the tube.
Can you find four equations?
Thanks,
I agree that the force may cause the pipe to move about the center of mass, however the mass is large and will resist this change due to inertia. You are also correct about the force pointing out from the center of the curvature. So I would still expect the ball to exit in -x as the same forces exist along the whole curve, any pipe spin induced from the start would be canceled at the exit. However I stand to be corrected.

Once the ball leaves the tube it's angular velocity becomes simply velocity which is constant in the same direction, it can only change when a force is applied to it. If the gravity from the Earth were to simply vanish the moon would instantly continue in a straight line, unless influenced by other forces.

I am really interested in the displacement of the tube in the +x direction, there must be an acceleration in the x direction as the vectored forces always point positive x.

Interestingly I believe the y direction forces cancel out, however I think there would be a displacement in the -y direction, as the total force in the first 90 degrees will give a resultant movement to the -y because the opposing force will simply stop the movement and not correct the -y displacement.

You say"the ball will have angular momentum, so the tube must do too." Correct it must due to Newtons 3rd law. The question is what is the total of this angular momentum in terms of direction.

I have tried to simulate this scenario on Solidworks Motion Analysis software however, I found the results were confusing, it did predict a +x movement but I cannot reproduce using maths. it seems to contravene the conservation of energy which is obviously incorrect.

This is not an exam question simply my curiosity, my expertise is in Electronics so Newtonian Physics is not my normal thing. The reason for the question comes from a project I am currently working on, which might one day be used in a spacecraft .

I have asked several of my friends who are perhaps better placed than me to give the answer, all have declined the challenge.

Cheers Tony
 
  • #11
By the way I was moved to the homework forum by the moderation staff, saying that this is a homework question.
Then giving me a warning about Misplaced Homework thead.
I haven't been in School/University for nearly 50 years ...

So I would just like an answer to the question not probing to help encourage me to solve the problem..
 
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  • #12
TonyCross said:
Thanks for the reply,
I don't want to assume the pipe is in a fixed position, that's the whole point of my question.
maybe rather than be sarcastic you could offer, I don't know ... a physics reply!
No sarcasm intended -- on re-reading I don't see any either. Just a careful probing what direction this thread is taking, and what we have available to work on it. The problem itself, as it takes shape now, is interesting (*) and -- to me -- extremely challenging.
BvU said:
If so, can you forget about the motion of the pipe and assume the pipe is fixed in position ?
So: no. It will move and, since it undergoes off-center force, rotate too. Whether the rotation one way will be compensated the other way later on remains to be seen. @haruspex gives an argument why not.
If not, you have to ask yourself what kind of encounter this is and what is conserved in such encounters
Here is the physics reply you sought but overlooked: I expected you to come up with: conservation of linear and angular momentum and conservation of energy. As in a collision. Only more complicated (not instantaneous and off-center of mass and off-center of gyration)

No friction means the ball slides and does not spin.
Conservation of linear momentum means the tube picks up linear momentum in the x- and y-direction, so the ball must lose some. And conservation of
the ball does lose speed while traveling through the tube. So coming out with -5m/s in minus x-direction is excluded.

TonyCross said:
So I would still expect the ball to exit in -x as the same forces exist along the whole curve, any pipe spin induced from the start would be canceled at the exit. However I stand to be corrected.
@haruspex, in the post #9 just before, explained why pipe spin does not cancel.

- - - - - - - - - - -

(*) I had lots of fun experimenting with my grandchildrens's hot wheels (car 22 g, arc 110 g, so 1:5 instead of your 1:20). The arc clearly rotates clockwise - no cancelling.
 
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  • #13
TonyCross said:
By the way I was moved to the homework forum by the moderation staff, saying that this is a homework question.
Then giving me a warning about Misplaced Homework thead.
I haven't been in School/University for nearly 50 years ...

So I would just like an answer to the question not probing to help encourage me to solve the problem..
Welcome, Tony! :cool:
As I see it:
The ball can’t leave with the same speed because some of its kinetic energy is consumed by the work of rotating and moving the semicircular pipe around.
Otherwise, it would be like a perfectly elastic collision between the ball and a huge mass, in which the ball would bounce back keeping all its energy.

The centroid or center of mass of the pipe shape is off the origin of the radius, reason for which some progressive moments and subsequent rotation should happen.

hqdefault.jpg
 
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  • #14
TonyCross said:
By the way I was moved to the homework forum by the moderation staff, saying that this is a homework question.
Then giving me a warning about Misplaced Homework thead.
Moderators look at the kind of problem, they don't dig into the complexity so much.
I haven't been in School/University for nearly 50 years ...
I can trump your age, but not the 50 years. More like 35.
So I would just like an answer to the question not probing to help encourage me to solve the problem..
Sure. You can ask for the moon if you want. Not so easily answered.
PF is for helping, not for contracting work.
 
  • #15
BvU said:
No sarcasm intended -- on re-reading I don't see any either. Just a careful probing what direction this thread is taking, and what we have available to work on it. The problem itself, as it takes shape now, is interesting (*) and -- to me -- extremely challenging.
So: no. It will move and, since it undergoes off-center force, rotate too. Whether the rotation one way will be compensated the other way later on remains to be seen. @haruspex gives an argument why not.
Here is the physics reply you sought but overlooked: I expected you to come up with: conservation of linear and angular momentum and conservation of energy. As in a collision. Only more complicated (not instantaneous and off-center of mass and off-center of gyration)

No friction means the ball slides and does not spin.
Conservation of linear momentum means the tube picks up linear momentum in the x- and y-direction, so the ball must lose some. And conservation of
the ball does lose speed while traveling through the tube. So coming out with -5m/s in minus x-direction is excluded.

@haruspex, in the post #9 just before, explained why pipe spin does not cancel.

- - - - - - - - - - -

(*) I had lots of fun experimenting with my grandchildrens's hot wheels (car 22 g, arc 110 g, so 1:5 instead of your 1:20). The arc clearly rotates clockwise - no cancelling.
Good point about the ball sliding without friction the ball does not spin, I hadn't considered that.
I concede the ball must loose momentum and must leave the tube at <5m/s.
So could it be that if the ball leaves at let's say 4m/s it has lost 1N of energy to centripetal forces to the tube?
 
  • #16
Lnewqban said:
Welcome, Tony! :cool:
As I see it:
The ball can’t leave with the same speed because some of its kinetic energy is consumed by the work of rotating and moving the semicircular pipe around.
Otherwise, it would be like a perfectly elastic collision between the ball and a huge mass, in which the ball would bounce back keeping all its energy.

The centroid or center of mass of the pipe shape is off the origin of the radius, reason for which some progressive moments and subsequent rotation should happen.

View attachment 272039
I agree that the ball must loose momentum
otherwise where is the energy to move the pipe.
 
  • #17
TonyCross said:
look at the overall project rather than just one aspect
In the thread, sure (sheer curiosity :smile: ) !

TonyCross said:
the tube and the ball are in the vacuum of space
Rocket science :wink: !
 
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  • #18
BvU said:
In the thread, sure (sheer curiosity :smile: ) !

Rocket science :wink: !
Rocket man that's me!
 
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  • #19
haruspex said:
there are four degrees of freedom for the final state: rotation of the tube, velocity of the tube (two degrees) and speed of the ball relative to the tube.
Unfortunately, there are more relevant unknowns. The rotational displacement affects the velocity direction of the exiting ball, and the linear displacement of the tube affects the calculation of the angular momenta. So I cannot see a way to make a simple before-and-after approach work.

At least, maybe we can answer the question of whether, at the finish, the tube will be back to its original orientation, or will once again not be rotating, or both.
In the early stages, the ball will have a high speed relative to the tube, so exert a large torque about the tube's mass centre. Later on, some of the KE of the ball has passed to the tube and the relative speed will be lower, so the torque exerted in the second quadrant should be less. So it appears that at the finish the tube will have a change in orientation and will continue to rotate.
 
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  • #20
haruspex said:
Unfortunately, there are more relevant unknowns. The rotational displacement affects the velocity direction of the exiting ball, and the linear displacement of the tube affects the calculation of the angular momenta. So I cannot see a way to make a simple before-and-after approach work.

At least, maybe we can answer the question of whether, at the finish, the tube will be back to its original orientation, or will once again not be rotating, or both.
In the early stages, the ball will have a high speed relative to the tube, so exert a large torque about the tube's mass centre. Later on, some of the KE of the ball has passed to the tube and the relative speed will be lower, so the torque exerted in the second quadrant should be less. So it appears that at the finish the tube will have a change in orientation and will continue to rotate.
Many thanks for taking a look at the problem.
 
  • #21
Still worthwhile to write down the equations of motion. An integrator should yield quantitative results.

So, rocketman: what's the whole project about ?
 
  • #22
TonyCross said:
Many thanks for taking a look at the problem.
Which part(s) of my argument are you sceptical about?
 
  • #23
BvU said:
Still worthwhile to write down the equations of motion. An integrator should yield quantitative results.

So, rocketman: what's the whole project about ?
Hi,
It involves a mass moving inside a spacecraft .

I am trying to simulate the effects this movement force from my design which may have on the stability of the craft.

Avenue 5 a new comedy series with Hugh Laurie, the basis of the story line is that a fault sends their spaceship on the wrong trajectory while on their way back to earth, to fix the problem they eject their belongings from an air lock which changes the trajectory. This seems to be the correct physics, based on Newtons 3rd.

My expertise is in Electromechanical design (not Newtonian Physics), my project involves the acceleration and deceleration of mass using electromechanics to change the velocity/direction of a spacecraft .

Hint...
Here is something to consider what would be the net reactive force if you took a pinball to space, minus all the lights and bumpers?
You pull the plunger which fires the ball to the top of the playfield, the ball runs around the top and then heads back towards you, then hits the playfield in the opposite direction to which it was fired.

What is the net reactive force experienced by the playfield is it =0N?

Cheers Tony
 
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  • #24
TonyCross said:
What is the net reactive force experienced by the playfield is it =0N?
The units for this quantity are wrong. You likely want net momentum transfer, not net force.

If one makes the simplifying assumption that the mass of the playfield is large compared to the mass of the ball, the momentum transferred during the ball's journey (not counting the launch event) is approximately equal to twice the ball's launch momentum.

If the ball is brought to a stop by some mechanism, it is equally obvious the the net momentum transferred (counting both launch and capture events) is zero. This follows regardless of the mass ratio and is exact.
 
  • #25
jbriggs444 said:
The units for this quantity are wrong. You likely want net momentum transfer, not net force.

If one makes the simplifying assumption that the mass of the playfield is large compared to the mass of the ball, the momentum transferred during the ball's journey (not counting the launch event) is approximately equal to twice the ball's launch momentum.

If the ball is brought to a stop by some mechanism, it is equally obvious the the net momentum transferred (counting both launch and capture events) is zero. This follows regardless of the mass ratio and is exact.
 
  • #26
Agreed.. Net momentum transfer, is the quantity.

When the ball first leaves the ball travels away, the reactive force (In Newtons) is towards the player
When the ball rounds the top of the playfield it transfers some of this force in the opposite direction.
The ball then returns to the player where it impacts the in the direction of the player again transferring the remaining energy towards the player.

The total momentum transfer towards the player is greater than that in the opposite direction.

For this not to be true the force as the ball travels around at the top of the table must be equal to the combined force of the Acceleration (Start) and the Deceleration (end) of the balls travel.
 
  • #27
TonyCross said:
For this not to be true the force as the ball travels around at the top of the table must be equal to the combined force of the Acceleration (Start) and the Deceleration (end) of the balls travel.
[ strike-through is mine ]

I want to nit-pick again that it is not "force" we are talking about. Instead, it is momentum -- force integrated over time.

I think you are agreeing that momentum is conserved, but am not sure.
 
  • #28
jbriggs444 said:
[ strike-through is mine ]

I want to nit-pick again that it is not "force" we are talking about. Instead, it is momentum -- force integrated over time.

I think you are agreeing that momentum is conserved, but am not sure.

Yes momentum must be conserved it's the law 🎓 p=mv ,force integrated over time.
The total momentum lost by object 1 (ball) is equal to the momentum gained by object 2 the (playfield)
example: (y up the table and x across) 1kg ball. 100kg table.
The ball is accelerated with respect to the table at say 5N ball acceleration is a=F/m 5/1=5m/s in y.
conservation of energy (Newton) says that the table must experience the same opposite force 5/100 .05m/s in -y

The ball loses some momentum on the turn transferring this momentum generally in the y direction.

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

This is a deceleration opposing force which is again towards the player in the -y direction 2.5/100 .025m/s in -y
This does not contravene the conservation of energy law as far as I can see.

Maybe I am missing something.
 
  • #29
TonyCross said:
a=F/m 5/1=5m/s in y.
Check your units here. The acceleration is not 5 meters per second.
 
  • #30
sorry 5m/s(sqr)
 
  • #31
TonyCross said:
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?
 
  • #32
TonyCross said:
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.
 
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  • #33
jbriggs444 said:
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?
 
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  • #34
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
Balltube_mr2.gif

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.
Balltube_p333.gif

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.
Balltubecrit.gif


Of course, all of the above depends on the correctness of the equations of motion and the programming.
 
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  • #35
TSny said:
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.
 
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