Tracks

[Gareth56]

AmI correct in thinking that both balls reach the ends of the tracks at the same time due to them having the same initial PE? The intervening route doesn't matter.

 

[CaptainPanic]

At the end-point, they will have the same kinetic energy (assuming zero friction). But they will not arrive at the same time. That's easily visualized by this:

 

If your statement were true, then the balls at A, B and C would all arrive at the same time. But I don't think C is going anywhere soon.

 

The acceleration is different for the different tracks.

[Gareth56]

I'm not sure I follow the acceleration argument. I thought the last "uphill" part of the track in B would have cancelled out the "downhill" part behind it. Also I'm not too sure what track C is supposed to signify.

[insane_alien]

yes, the uphil part returns it to its origional velocity but while it was in the dip it was moving faster than the ball on track A.

 

track C signifies a ball that will not move very fast if at all until it gets the dip at the end. it could take hours for the ball to complete the track while A and B would likely finish in seconds.

[CaptainPanic]

Things similar in tracks A, B and C:

 

All 3 balls will start with velocity = 0 m/s

All 3 balls have equal potential energy

All 3 balls will arrive at the end (right side) with equal velocity, and therefore equal kinetic energy.

 

Things different:

The acceleration is different: balls A and B will start the same (while ball C doesn't move much at all). At the 2nd slope on track B, ball B will accelerate even more. Ball B will leave ball A behind. When ball B reaches the "up" slope (on the other end), it's well ahead of ball A. the "up" slope might slow ball B down, but only to a velocity equal to ball A. Therefore, ball A cannot catch up.

Meanwhile on track C, ball C is still having a siesta.

[D H]

AmI correct in thinking that both balls reach the ends of the tracks at the same time due to them having the same initial PE? The intervening route doesn't matter.

You are incorrect. Suppose the lengths of sloped segments of track B and the segments at the final elevation are short compared to the length of the track at the lower elevation. Almost all of the time on track B will be spend at this lower elevation. The ball is rolling faster here than is the ball on track A. If the length of the lower elevation track is sufficiently long, the ball on track B will reach the destination first.

[swansont]

This falls under the category of the brachistochrone problem. One can show that the fastest path is a cycloid.

[Gareth56]

Thanks to all. I've been trying to find a video clip of this as a demo but to no avail

[CaptainPanic]

Thanks to all. I've been trying to find a video clip of this as a demo but to no avail

 

That's because you ignored the new word you learned from swansont: Brachistochrone

 

Here's a youtube video of our problem:

Youtube has several videos showing the problem... I linked to the shortest of them all.

[Gareth56]

That's because you ignored the new word you learned from swansont: Brachistochrone

 

Here's a youtube video of our problem:

Youtube has several videos showing the problem... I linked to the shortest of them all.

 

 

For interest I didn't ignore the "new word" that had been introduced to me by swanshot to my vocabulary. I also did check out the You Tube website wherupon I observed the videos that you allude to above, however I was hoping that there would be a video of a demo using the tracks as in the diagram above...that's all