What is the Velocity of a Car in a Loop-the-Loop Amusement Park Ride?

[ledhead86]

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.
http://community.webshots.com/user/mmaddoxwku"
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.

 

[ledhead86]

Any help would be appreciated

 

[whozum]

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

 

[ledhead86]

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

 

[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?

 

[ledhead86]

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?

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

 

[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.

 

[ledhead86]

so pe=g*95 ?

 

[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:

[tex]\Delta KE = \Delta PE [/itex]<br /> <br /> $$KE_f - KE_i = PE_f - PE_i$$<br /> <br /> $$KE_f - 0 = 0 - PE_i$$<br /> <br /> $$\frac{1}{2}mv^2 = -mgh$$[/tex]

 

[ledhead86]

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 I am suppose to solve for v without m.

 

[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?

 

[ledhead86]

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

NO I CANT!

.5v^2=gh
.5v^2=931
v^2=1862
v=43.15 = INCORRECT ANSWER!

 

[ledhead86]

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

 

[whozum]

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

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

 

[gerben]

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.
http://community.webshots.com/user/mmaddoxwku"
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.

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?

 

[enigma]

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

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