# Roller coaster: kinetic energy

1. Sep 25, 2007

A roller coaster car may be approximated by a block of mass m. The car, which starts from rest, is released at a height h above the ground and slides along a frictionless track. The car encounters a loop of radius R, as shown. Assume that the initial height h is great enough so that the car never loses contact with the track.

http://session.masteringphysics.com/problemAsset/1011023/13/MPE_ug_2.jpg

so, i know kinetic energy = .5m(v^2) and centripetal force in circle is m(v^2)/R so that if i play around with the equations i can get KE = 0.5m(gR) ... but how do i take into account the height if i have to give my answer in terms of m, g, h, and R?

Anyone out there who can solve this thing?

2. Sep 25, 2007

Find an expression for the kinetic energy of the car at the top of the loop.
Express the kinetic energy in terms of m, g, h, and R.

forgot to add that what's above is the problem!

3. Sep 25, 2007

### learningphysics

Just use conservation of energy... I don't think you need to deal with centripetal motion or anything.

4. Sep 25, 2007

i can't work just on that. what do you mean by 'just use conservation of energy'. i need to express my answer in terms of those variables i listed above.

5. Sep 25, 2007

### learningphysics

What is the energy of the coaster at the beginning when it's at rest? Take the gravitational potential energy on the ground to be 0.

6. Sep 25, 2007

The energy of the coaster at the beginning is potential energy which is just mgh.

7. Sep 25, 2007

### learningphysics

Yes, now at the top of the loop, it has potential energy mg(2R), and kinetic energy. can you use conservation of energy to solve for kinetic energy?

8. Sep 25, 2007

thanks! i got it ... now what if i want to find the minimum initial height h at which the car can be released that still allows the car to stay in contact with the track at the top of the loop? how would i approach this one?

9. Sep 25, 2007

### learningphysics

Now you'd use the centripetal acceleration... what does the velocity need to be at the top of the loop for the car to maintain contact?

10. Sep 25, 2007

well acceleration is V^2/R . and so min velocity would be v = gR ...

11. Sep 25, 2007

sorry, v = (gr)^(1/2)

12. Sep 25, 2007

### learningphysics

yup. so what is the total energy of the coaster when it is at the top of the loop? It has that same energy when it is released from rest.

13. Sep 25, 2007

mgh... sorry i'm not getting it

14. Sep 25, 2007

i set mgh-mg2R = 0.5(mv^2) and plugged in what i got for v ... so i ended up with h=3R but i think i'm off by some multiplicative factor ???

the ans should be in terms of R by the way...

15. Sep 25, 2007

### learningphysics

can you show how you get h = 3R, I'm getting h = 2.5R

16. Sep 25, 2007

ah, stupid calc. error. thanks lots learningphysics. you made my day haha

17. Sep 25, 2007

### learningphysics

no prob.

18. Apr 3, 2010

### jakubu

Thats very simple and straight forward. use the conservation of energy law, and solve for the unkwon K=kinetic energy at the top of the loop. NB: th energy at initial position is equal to the energy at final position = top of the loop. initial kinetic energy = 0, So u will left only with final unknown kinetic energy K. Min Velocity, v=gR^0.5

19. Apr 3, 2010

### jakubu

any probs rply