# Kinetic/Potential Energy

1. Nov 30, 2007

### dominus96

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

An object slides down a slope (from rest) and then immediately around a loop (like a roller coaster). The object has mass m, and the height of the loop is 2r. What is the minimum height required for the object to make it around the loop? Express answer in terms of m, r, g, and v.

2. Relevant equations

K=mv^2/2
U=mgh

3. The attempt at a solution

Would the potential energy of the slope (U1) be equal to the potential energy of the loop (U2) + the kinetic energy, so h=mv^2/2 + mg2r? That's what I think so far, but not sure if it is right.

Last edited: Nov 30, 2007
2. Nov 30, 2007

### Staff: Mentor

Correct. Assuming that the mass starts from rest and fall a height h to the bottom of the loop, then rises to the top of the loop, the difference is the kinetic energy.

What is the condition at the top of the loop that results in a minimum h?

I presume one is assuming a frictionless interaction between mass and path (slope and loop), and no air resistance.

Last edited: Nov 30, 2007
3. Nov 30, 2007

### Bill Foster

$$\frac{v^2}{r}=g$$

4. Nov 30, 2007

### dominus96

Yes it is frictionless and starts at rest. But what do you mean condition at the loop?

5. Nov 30, 2007

### Staff: Mentor

At the top of the loop, what is the condition or requirement with respect to mass to keep it traveling in the loop, and which satisfies a minimum energy (min height) condition with respect to the initial elevation, h?

6. Nov 30, 2007

### dominus96

I have mgh = .5mv^2 + mg(2r). So do I just solve for h? Because that would be h = (.5mv^2 + mg(2r))/mg

7. Nov 30, 2007

See post #3.