# Conservation of energy of a rollercoaster car

1. Nov 26, 2008

### nns91

1. The problem statement, all variables and given/known data
1. A roller coaster car of mass 1500kg starts a distance H =23m above the bottom of a lopp 15m in diameter. If friction is negligible, the downward force of the rails on the car when it is upside down at the top of the loop is ???

2. Lou is trying to kill mice by swinging a clock of mass m attached to one end of a massless stick 1.4m in length on a nail in the wall. The clock end of the stick is free to rotate around its other end in a vertical circle. Lou raises the clock until the stick is horizontal, and when mice peek their heads out from the hole to their den, he gives it an initial downward velocity v. The clock misses a mouse and continues on its circular path with just enough energy to complete the circle and bonk Lou on the back of his head. (a) What was the value of v ? (b) What was the clock's speed at the bottom of it's swing ?

3. A pendulum consists of a string of length L and a bob of mass m. The string is brought to a horizontal position and the bob is given the minium initial speed enabling the pendulum to make a full turn in the verticle plane. (a) What is the maximum kinetic energy K of the bob ? (b) What is the tension in the string when the kinetic energy maximum ??
2. Relevant equations

E=K + U

3. The attempt at a solution

1. I guess at the beginning there will only be U=1500*g*23. At the top of the loop E= U ( I am not sure). Thus, we have the E=U=-W= integral of (Fdx). How can I move on from there ??

2. I am clueless in the way solving this problem.

3. I am kinda confused by :"...the bob is given the minimum initial speed enabling the pendulum to make a full turn in the vertical plane."

2. Nov 26, 2008

### HallsofIvy

Staff Emeritus
Unless you say what "E", "U", and "W" mean, I don't see how anyone can help you with this.

"Just enough energy to complete the circle" is the crucial statement. At the top of the circle, the speed of the weight is just barely above 0. What is the potential energy there? What was the potential energy at the start? What must have been the kinetic energy at the start?

Exactly the same as problem 2. As the pendulum "goes over the top", its speed must be just more than 0. What is the Potential energy there? What is its potential energy at the start? What must have been the kinetic energy at the start?

3. Nov 26, 2008

### nns91

E is Energy, U is potential energy, W is work.

4. Nov 26, 2008

### ak1948

1) The equation you used E= K+U is right. Let's not bring W into this.
Remember E doesn't change in this frictionless problem.
You showed you know how to find initial U at with respect to the bottom of the loop; how about with respect to the top of the loop.
Obviously K is 0 initially since it starts at rest (although the problem doesn't actually say that). So what is K at the top of the loop?
Now you have K, whats v and finally whats F.

5. Nov 26, 2008

### nns91

Respecting to the loop, U=1500*g*15

At the top of the loop, is not K=0 also ??

6. Nov 26, 2008

### ak1948

No at the top of the loop it's moving. You just figured out how much potential energy you gave up getting there. So how much kinetic energy did you pick up if the total energy is unchanged?

7. Nov 26, 2008

### nns91

Right. From there I will probably get the speed. From speed, how can I get work ??

How about the next 2 problems. I am kinda confused in the way to solve it.

Last edited: Nov 26, 2008
8. Nov 26, 2008

### ak1948

2) Hallsofivy gave you the clue. K=0 (or nearly) at the top of the circle. You know how to find the difference in potential energy between the top and the bottom and the top and the starting point. Therefore you can find the kinetic energies and speeds at those points.

9. Nov 26, 2008

### nns91

1. Right. From there I will probably get the speed. From speed, how can I get work ??

10. Nov 26, 2008

### ak1948

You don't want work. You are looking for force, specifically the difference between the centrifugal force and the gravitational force.

11. Nov 26, 2008

### nns91

F=mv^2/2 right ?

12. Nov 26, 2008

### ak1948

no that's kinetic energy. centrifugal force is mv^2/r.

13. Nov 29, 2008

### nns91

For number 2, I set it up like this:

Start: E=(1/2) mv^2

Bottom: E=(1/2)mv^2-mgh

Top: E= mgh.

Do I set Start=top to solve for v ??

14. Nov 29, 2008

### ak1948

I guess the easiest way to say it is
mgh + (1/2)mv^2 = E
at every point. Where E is the same for every point. Since m and g are also the same for every point, you can calculate the change in v between any two points by the change in h. (Or if you needed to, the change in h by the change in v, but in this problem you know the difference in h of all the points of interest.)

15. Nov 29, 2008

### nns91

so will I get v= sqroot(2gh) ? for part b, at the bottom E=(1/2)mv^2-mgh ?

16. Nov 30, 2008

### nns91

Do you guys think I got that right ?

I am still kinda clueless about question 3. I don't get the approach.

17. Dec 1, 2008

### ak1948

Problem 2:

I think you have this right. But I am not exactly clear as to what you have done.

Define E as any convenient value. Let's call it zero; this is completely arbitrary. Since the velocity is 0 at the top, the energy is mgh at the top, so we also have to define h=0 at the top.

At the bottom, then, the height is -2R where R is the length of the stick. Since the total energy is zero,
mg(-2R)+(1/2)mv^2 = 0 , therefore:
v=sqrt(gR)

At the start, 0 = mg(-R) + 1/2(mv^2),
v= sqrt(2gR)

This is what you got, right?
Problem 3 is almost the same problem.