# I An exception to the principle of equivalence?

1. Nov 18, 2017

### e2m2a

Not sure of the dynamics of the following situation. Suppose there is a mass that can slide on a straight track with one degree of freedom. Assume no friction in this scenario. On top of this mass is a track that forms a quarter circle. On this track is a sphere that can slide(again assume no friction) on this track.

The sliding mass moves along the y-axis. At some point in time the sphere is given an initial velocity, and then enters one end of the quarter-circle curved track. As the sphere moves in a counter-clockwise direction along this track, it eventually reaches the end of this quarter-circle, moving in the negative x-direction, where it makes an inelastic collision at the end of the track.

During the time the sphere rounds the curve, the centrifugal reactive force acting on the sliding mass causes the sliding mass to accelerate at a non-constant rate in the positive y-direction. After the sphere collides at the end of the track, the whole system continues to move in the positive y-direction by Newton’s law of inertia.

Suppose there is an observer on the accelerating mass as the sphere rounds the curve. He will observe a strange gravitational effect that increases as the sphere rounds the curve. He can verify this easily by noting the reading on an accelerometer.

The answer to this question that is not clear to me is will the sphere maintain a constant speed as it rounds the curve, or will it slow down? According to the principle of equivalence the observer in the frame of the moving mass will experience a gravitational effect, even though it is not constant, and would expect the speed of the sphere to slow down due to this gravitational field the observer experiences in his frame. Or would it retain a constant speed?

I ask this question because an observer in an inertial laboratory frame would not experience this gravitational effect, so wouldn’t he correctly expect the sphere to maintain a constant speed as it rounds the curve? The only Newtonian force he would accept as real would be the centripetal force acting on the sphere, which would always be perpendicular to the instantaneous tangential velocity of the sphere.

2. Nov 19, 2017

### Staff: Mentor

No, he wouldn't, because the "gravitational effect" is perpendicular to the velocity of the moving mass. So it will just make the mass do a circular orbit.

3. Nov 19, 2017

### A.T.

How is the quarter circle aligned relative to the straight track. Better post a diagram.

4. Nov 19, 2017

### Staff: Mentor

And some of his own effort.

@e2m2a when posting homework-like questions (even outside of a formal course) it is important to show your own work. PF is not about feeding you answers, but about helping you learn. Please post your work and show where you get stuck, we can help you learn.

5. Nov 20, 2017

### e2m2a

This video should clarify it on youtube: The x and y axis are switched.

6. Nov 20, 2017

### e2m2a

I am not really seeking a quantitative answer, but more of a conceptual understanding. By posting it on the forum and generating a discussion, I am hoping to gain some insights.
Here is what I am struggling with. As the ball rounds the curve in the video you can see that the carriage accelerates momentarily to the right. From what I understand from general relativity, an observer in the frame of the carriage would experience a momentary graviational field. Thus, he should expect the speed of the ball to slow down by the time it reaches the end of the curved track. I know friction in the real world scenario would slow the ball down, but I am asking what if we could make this a frictionless experiment, would the ball still slow down due to this "gravitational effect"?
The conceptual idea I don't grasp is if this is true, then relative to an inertal lab frame, because this slow down in the speed would be Galilean frame invariant, how would the observer in the laboratory frame explain the decrease in the speed? He doesn't observe any graviational effects in his frame.

7. Nov 20, 2017

### Staff: Mentor

Yes. In which direction does this field point?

Not speed; velocity. Velocity in which direction?

8. Nov 20, 2017

### A.T.

The ball transfers kinetic energy to the block, so it's speed must decrease.

9. Nov 20, 2017

### Staff: Mentor

Would there even be a decrease in speed in the lab frame? How much? Again, show your work.

I know you probably feel quite negatively about my repeated request for you to show your work, but you are making complicated inferences and without you showing your work we have no way to know which of the many possible mistakes you are making. This is why it is required in the HW sections.

10. Nov 20, 2017

### e2m2a

Ohj....
The direction of the "gravitational field" would be parallet to the motion of the carriage in the opposite direction. If we did a vector analysis of this gravitational force on the sphere, we can break the force into two components. If we define theta as the angle formed by the line from the center of mass of the ball to the center of curvature of the track with respect to the y-axis, then there is a force component acting on the sphere acting opposite to the tangential velocity of the ball equal to cos(theta) g m, where g is the instantenaeou acceleration of the carriage and m is the mass of the sphere. (In the video the y-axis is the axis perpendicular to the motion of the carriage.) This component would decrease the magnitude of the velocity of the sphere. I get all of this and its not my question.

11. Nov 20, 2017

### e2m2a

I don't know how this question got into the HW section. I never posted it there initially. This thread was transferred from general physics to sr/gr.

12. Nov 20, 2017

### e2m2a

I agree with this conservation of energy approach. But this implies the kinetic energy of the ball is transferred to the carriage somehow. But how?

We might argue that the x-component of the centrifugal reactive force does work on the carriage, imparting kinetic energy on it by the work-energy theorem. But again, the centripetal force acting on the sphere is always perpendicular to the tangential velocity, so it can only do zero work on the sphere. So how could the carriage being gaining a joule of kinetic energy due to this centrifugal reactive force if there is zero work being done on the sphere with respect to the lab frame.

Now I suspect my assumption that there is no real work being done on the sphere with respect to the lab frame is wrong.

13. Nov 20, 2017

### Staff: Mentor

Thanks for describing your analysis. Here is where you go wrong (in the internal frame). The force is normal to the surface, but the velocity is not tangent to the surface.

14. Nov 20, 2017

### e2m2a

I think you mean in the inertial frame not the internal fame. I will have to think about this. Thanks.

15. Nov 20, 2017

### Staff: Mentor

Oops, yes, I must have made a spelling error which got autocorrected wrong.

16. Nov 20, 2017

### Staff: Mentor

Yes. And this field would therefore cause a deceleration of the ball in which direction?

17. Nov 21, 2017

### e2m2a

The cosine component (see previous answers) of the "gravitational" vector would always be opposite the tangential velocity of the ball.

18. Nov 21, 2017

### Staff: Mentor

You're thinking of it backwards. The gravitational vector acts in a particular direction (the minus y direction). So the ball should decelerate in the minus y direction--i.e,. its velocity in the plus y direction should decrease. Does it? (Hint: yes.) The fact that the ball also has an x component of velocity is irrelevant to this particular question.

One thing that is misleading about this setup is that unbalanced forces in the x direction (perpendicular to the motion of the carriage) are ignored, because they are assumed to be cancelled by the track. Therefore, when the ball starts around the quarter circle and begins to move in the x direction, its motion in that direction, as far as the problem is concerned, comes out of nowhere. If this experiment were done floating out in deep space, with no huge Earth to soak up momentum through the track, the carriage would have to shift in the x direction to cancel out the motion of the ball in that direction--the shift would be stopped by the ball colliding at the end of the quarter circle.

19. Nov 21, 2017

### A.T.

As Dale noted, the velocity in the lab frame is different than the velocity in the block frame, while the force is the same. So they aren't perpendicular in the lab frame and negative work is done by the block on the ball.