# Gravity and uniform circular motion involving a loop?

1. Dec 5, 2013

### Ascendant78

1. The problem statement, all variables and given/known data

A 0.27 kg toy car is held at rest against a 1796 N/m spring compressed a distance of 8cm. When released, the car travels a distance of 112cm along a flat surface before reaching a 0.30m high loop. The friction coefficient of the flat surface and the toy car is (0.4, 0.18). Assume friction is so small in the loop that it can be ignored.

Question 1: What is the velocity of the toy car the moment it reaches base of the loop?

Question 2: What is the velocity of the toy car when it reaches the top of the loop?

2. Relevant equations

Solving question 1
Evaluation of toy car from release to bottom of loop using work-energy:

EEi + W = 1/2 (0.27) vf2

1/2 (1796)(0.08)2 + (0.47628)(0.08 + 1.12) cos(180)= 0.135 vf2

5.7472 – 0.571536 = 0.135 vf2

vf = 6.19m/s when toy car reaches the base of the loop

3. The attempt at a solution

As seen in #2 above, I know how to solve for the first question. However, I am at a complete loss as to how we approach the second part? We have dealt with uniform circular motion with no gravity and uniform circular motion with movement perpendicular to gravity, but never one where the direction of movement is constantly changing with respect to the direction of gravity. I am at a loss as to what equation I could even use that would take everything I need into consideration?

2. Dec 5, 2013

### cepheid

Staff Emeritus
Just a comment here: what you have written above is almost completely meaningless, and almost completely useless to someone else who is trying to figure out what you're doing. The reason is because you haven't included any units on your quantities. They are just meaningless numbers. 1796 whats??? It could be kilowatt-hours, for all I know. You might think I'm nitpicking and being pedantic, but I'm actually not. Without the units, it's not a physically meaningful statement, and excluding units is a great way to fall prey to silly mistakes as well.

I'm assuming EEi is the initial "elastic energy?" If that's true, then it's equal to (1/2)kx2, which I think is the first term you have on the lefthand side of the next line (but I had to guess that because there are no units). That term looks okay. Let's talk about the middle term, which I've put in boldface. This term corresponds to what you called "W", by which I assume you mean the work done by friction. You're saying that initial elastic energy + frictional work = final kinetic energy, so that all makes sense by the conservation of energy. But the middle term (the work) has nothing to indicate where any of the numbers came from. The number in red, 0.47628, in particular, has no discernible origin. It could be correct, or it could not be, but I have no way of telling either way, because you didn't write down any algebraic symbols or any of the steps that you used to arrive at it.

Just use the conservation of energy. Gravity is a conservative force: the great thing about conservative forces is that the work they do in moving an object from point a to point b depends only on the difference between those two points. It does not depend on the path taken between those two points. So it doesn't matter that the object travels along a circular loop. To figure out its change in potential energy, all you need to do is figure out the change in height of the object. Potential energy gained tells you kinetic energy lost.

EDIT: this is why conservation of energy is so powerful a concept. It's incredibly useful.

3. Dec 5, 2013

### Ascendant78

Apologies for not including the units and simplifying it all. I didn't want to make a clutter with all the units and figured the velocity I obtained could be checked relatively quickly. Besides, I wasn't really concerned with question 1 anyway, just question 2.

Thanks for the information about using conservation of energy. The only reason I wasn't sure about using that one is because of the force caused by the centripetal when it goes through the circular motion. I figured if it wasn't somehow included in the equation, it would come out wrong. I guess I'm just looking at it wrong conceptually.

4. Dec 6, 2013

### Ascendant78

Still looking for help on this. Can anyone clarify how you can use conservation of energy on this problem when the centripetal force will affect how high the toy car can travel?

5. Dec 7, 2013

### cepheid

Staff Emeritus
Hi Ascendant78,

I replied to your PM regarding this question. Sorry for not getting back to you sooner. For the benefit of other thread goers, the gist of what I said was:

Only two forces act on the object: gravity and the normal force. The normal force does no work.