# Work done by mass on a loop-the-loop track

1. Oct 1, 2007

### delecticious

1. The problem statement, all variables and given/known data
A small block of mass m = 1.8 kg slides, without friction, along the loop-the-loop track shown. The block starts from the point P a distance h = 56.0 m above the bottom of the loop of radius R = 17.0 m. What is the kinetic energy of the mass at the point A on the loop?

What is the downward acceleration of the mass at the point A of the loop?
What is the minimum height h for which the block will reach point A on the loop without leaving the track?
2. Relevant equations

work = mass x gravity x (initial height - final height)

KE = 1/2mv^2

3. The attempt at a solution
For the downward acceleration should I find via forces? If so when the block is at A is both weight and normal force acting on it or just weight, I kind of don't really understand how to do it if that's wrong. For the second part with the minimum height I'm not really sure either do I do that via forces as well?

Last edited: Oct 1, 2007
2. Oct 1, 2007

3. Oct 1, 2007

### delecticious

can you see it now?

4. Oct 1, 2007

### Mentz114

Yes, that's fine.

Looks like a trick. The downward acceleration is g.

initial potential energy - potential energy at A.

For A the mass to stay on the track, the centripetal force must be equal to or greater than m*g. This gives a minimum velocity, and hence a minimum of the starting potential energy.

5. Oct 2, 2007

### delecticious

I found the minimum velocity, but I don't see how that ties into the potential energy part since to find potential energy you all need mass height and gravity.

6. Oct 2, 2007

### atavistic

Find the energy at the bottom most point of the loop...that will only be KE.Now find the PE at the topmost point of the circle and subtract it with the KE obtained b4.That should be the kinetic energy at that point.

For second part I think the downward acceleration is g+v^2/r

THIRD PART (good part)
For the particle to reach A, mv^2/r >= mg

So find v from there. Find mechanical energy at that point. THis mechanical energy should be equal to PE at the min height from which the object is released.SO find the H from there.

7. Oct 2, 2007

### Mentz114

The minimum velocity gives you the amount of kinetic energy you can add into the equation

mgh = 2mgR + 1/2mv^2

solve for h.

8. Oct 3, 2007

### delecticious

well, I almost have this problem solved, but it seems my acceleration is wrong for the downward accleration at A. "g" being the acceleration made sense since at point A gravity is acting on the mass, and when I worked out having Fnet = Fc I got an acceleration that was pretty close to g (9.79 m/s^2), however when I entered the answer in cam out incorrect, but I have no idea why it's wrong, any suggestions?

9. Oct 3, 2007

### Mentz114

At A, if the mass is in equilibrium vertically,

mg = mw^2r and w=v/r, so v = rw and v^2 = rg ( please check)

Plugging into the equation in my post #7,

h = 2R + 1/2 R = 5/2 R

 Rereading your post I infer that you have done this but stuck with the second bit.
If you've solved for the KE at A, you have the velocity, from which the upward force can be got. Subtract this from g to get the nett downward force.

Last edited: Oct 3, 2007
10. Oct 3, 2007

### delecticious

that didn't help. I don't see how you can get the downward force with the velocity because wouldn't you need to know acceleration to get that? Or are you saying that the downward force is the centripetal force? Anyway if I do it exactly as you say subtracting the velocity from g I get a small negative number (-3.1). Do I plug that into F=ma? Or am I just completely wrong? If I plug into F=ma I get -1.7.

11. Oct 3, 2007

### Sourabh N

You know KE at A, so you can find velocity, using which you will find the centripetal acceleration(exactly downward here). Add up this acc. and g to get the final acc.

Last edited: Oct 3, 2007