Troubleshooting Circular Motion: Solving the Toy Car Loop Puzzle

[Witdouck K]

Homework Statement

A toy car is released from a height of 2R with an initial velocity of 0 on a track with a loop of radius R. At what height does the car lose contact with the track in the loop?

Relevant Equations

Conservation of energy, F = mv^2 / R

I tried using conservation of energy, and using the equations for circular motion, but I can't seem to find a solution. Any help?

 

[scottdave]

Try watching this video for some insight.


They use a little differentiation. If you haven't had Calculus yet, don't be scared away, I think it still will help you.

 

[Delta2]

The key to this problem is to "translate" the statement "It loses contact with the track" to a fairly simple equation. In order to be of more accurate help I will ask you "Which of the forces on the toy becomes zero when it loses contact with the track". Set this force to zero so you have one additional equation together with the equations of conservation of energy and the equations of centripetal force.

Also I find it easier instead of the height to calculate the angle that the radius to the point of lost of contact makes with the horizontal.

 

[Witdouck K]

The key to this problem is to "translate" the statement "It loses contact with the track" to a fairly simple equation. In order to be of more accurate help I will ask you "Which of the forces on the toy becomes zero when it loses contact with the track". Set this force to zero so you have one additional equation together with the equations of conservation of energy and the equations of centripetal force.

Also I find it easier instead of the height to calculate the angle that the radius to the point of lost of contact makes with the horizontal.

I think the normal force becomes zero when the toy loses contact. But when I use this and calculate the angle you specified, and then calculate the height h coresponding to this angle, I get the non sensical anwser of h = -1 / 3 R, which seems really strange. Maybe I made a mistake that I didn't notice, but I still don't know why I can't get the right answer.

 

[scottdave]

You may want to take a look at this thread @Witdouck K

https://www.physicsforums.com/threa...rollercoaster-car-on-a-loop-the-loop.1058011/

Reading through other people's thought process on a similar problem may give you some ideas.

 

[scottdave]

I think the normal force becomes zero when the toy loses contact. But when I use this and calculate the angle you specified, and then calculate the height h coresponding to this angle, I get the non sensical anwser of h = -1 / 3 R, which seems really strange. Maybe I made a mistake that I didn't notice, but I still don't know why I can't get the right answer.

Try drawing a free body diagram for the quadrant where we have potential to lose contact.

 

[Delta2]

I think the normal force becomes zero when the toy loses contact. But when I use this and calculate the angle you specified, and then calculate the height h coresponding to this angle, I get the non sensical anwser of h = -1 / 3 R, which seems really strange. Maybe I made a mistake that I didn't notice, but I still don't know why I can't get the right answer.

Yes that's right the normal force becomes zero. You got to write a detailed post giving all the details of what equations you made and how you worked with them and also a diagram with the track and the forces at the point of lost contact will be nice so I can pinpoint where you might have gone wrong.

If you cant type in Latex and make a diagram with some drawing software, I guess as a last resort you can post screenshots of your handwritten work of the problem.

By the way the answer I get for the angle is $\sin\theta=\frac{2}{3}$ . So the height is $\frac{2}{3}R$. Come to think of it your height of -1/3R means that you might took as zero level of potential energy the level at the starting point. I took as zero level of potential energy the horizontal diameter of the loop.

P.S Actually no, its not where you put the zeroth level of potential energy but where you put your coordinate system. I put it centered at the center of the loop, it seems to me you put it centered at the starting point of the toy.

 

[Witdouck K]

Yes that's right the normal force becomes zero. You got to write a detailed post giving all the details of what equations you made and how you worked with them and also a diagram with the track and the forces at the point of lost contact will be nice so I can pinpoint where you might have gone wrong.

If you cant type in Latex and make a diagram with some drawing software, I guess as a last resort you can post screenshots of your handwritten work of the problem.

By the way the answer I get for the angle is $\sin\theta=\frac{2}{3}$ . So the height is $\frac{2}{3}R$. Come to think of it your height of -1/3R means that you might took as zero level of potential energy the level at the starting point. I took as zero level of potential energy the horizontal diameter of the loop.

P.S Actually no, its not where you put the zeroth level of potential energy but where you put your coordinate system. I put it centered at the center of the loop, it seems to me you put it centered at the starting point of the toy.

On closer inspection, I indeed put the zeroth level of the potential energy at the starting point, I just didn't really realise it untill you pointed it out. When putting it at the center of the loop, I als got $\frac{2}{3}R$, but I know this is just the same answer as $\frac{-1}{3} R$ because of the translation of the origin. Thanks a lot!

 

[Delta2]

On closer inspection, I indeed put the zeroth level of the potential energy at the starting point, I just didn't really realise it untill you pointed it out. When putting it at the center of the loop, I als got $\frac{2}{3}R$, but I know this is just the same answer as $\frac{-1}{3} R$ because of the translation of the origin. Thanks a lot!

I think its not where you put the zeroth of PE but where you put the origin of coordinate system. Usually these two coincide, i.e the zeroth level of PE is the x-axis of the coordinate system but its not always the case.