# Velocity w/ kinetic energy

1. Nov 7, 2005

Riding a Loop-the-loop. A car in an amusement park ride rolls without friction around the track shown in the figure. It starts from rest at point A at a height h above the bottom of the loop. Treat the car as a particle.
pic
If the car starts at height h= 95.0 m and the radius is r= 19.0 m, compute the speed of the passengers when the car is at point c, which is at the end of a horizontal diameter.
Take the free fall acceleration to be g= 9.80 m/s^2.
This is a poorly worded problem in my opinion. For one thing, I don't know where C is by their description.

2. Nov 7, 2005

Any help would be appreciated

3. Nov 7, 2005

### whozum

If you treat the loop as a circle, C is at x = pi. You want to use energy methods to solve this problem.

4. Nov 7, 2005

How would I set that up if I don't know the mass or weight of the coaster. The only equation I know is (1/2)*m*v^2 + m*g*y_1 = (1/2)*mv^2 + mg*y_2

5. Nov 7, 2005

### whozum

Notice that any change in kinetic energy would only be caused by a change in potential energy. You know the PE at point A so you know the KE at the bottom of the circle. Now to displace 'r' in the upwards direction you'll need to gain a PE of 'mg(r)' by losing it from your KE at the bottom. Get the idea?

6. Nov 7, 2005

Not Really. The potential energy at point a= mgy = m(9.8)(95) does it not? I still dont know the mass.

7. Nov 7, 2005

### whozum

Okay, the total energy E = PE + KE. Te total energy is constant.
At the top KE = 0 so the total energy is PE.
At the bottom the PE = 0 So the KE = (the old PE).

Get that so far? You don't need to know the mass, it will cancel.

8. Nov 7, 2005

so pe=g*95 ?

9. Nov 7, 2005

### whozum

No the PE = mg(95) like you said. Since you can't get over this hump I"ll give you a push:

The change in kinetic energy will be due to the change in potential energy:

$$\Delta KE = \Delta PE [/itex] [tex] KE_f - KE_i = PE_f - PE_i$$

$$KE_f - 0 = 0 - PE_i$$

$$\frac{1}{2}mv^2 = -mgh$$

10. Nov 7, 2005

Sorry, I just don't understand this at all, but I already said from the beginning that .5mv^2=-mgh, but I still don't know how im suppose to solve for v without m.

11. Nov 7, 2005

### whozum

$$\frac{1}{2}mv^2 = -mgh$$

$$\frac{\frac{1}{2}mv^2}{m} = \frac{-mgh}{m}$$

$$\frac{1}{2}v^2 = -gh$$

Since we are just looking for a change in potential, we don't need a magnitude and can take away the negative.

$$\frac{1}{2}v^2 = gh$$.

Can you find v now?

12. Nov 7, 2005

NO I CANT!!!!

.5v^2=gh
.5v^2=931
v^2=1862

13. Nov 7, 2005

Thank you whozom for completely wasting the past two hours of my life.

14. Nov 8, 2005

### whozum

I showed you how to do your problem. Your incompetance is not my fault.

15. Nov 8, 2005

### gerben

They say "the end of a horizontal diameter", which probably means "at half the circle's diameter above the ground" (i.e. a position on a horizontal line through the middle of the circle). In that case the difference in height between the starting point and point C would be 95-19=76.

Does using this height difference (in the equation that whozum provided) give you the right answer?

16. Nov 8, 2005

### enigma

Staff Emeritus
Try that g*h multiplication again before you get pissy next time...