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Conservation of Energy roller coaster car

  • Thread starter kiwikahuna
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  • #1
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Homework Statement


A roller coaster car starts from rest and descends 40 m before rising 20 m to the top of a hill. The passenger of mass 75 kg at the top of the hill feels a normal force of 368 N. The radius of curvature of the hill is...?


Homework Equations



F = m * a
a = v^2/r

The Attempt at a Solution



Because we are given the force and the mass the person, I found the acceleration and set acceleration = v^2/r. Velocity isn't given so I used the kinematics equation: Vf^2 = Vi^2 + 2a(distance)
where Vi =0 and distance = 60. Once I found the velocity I went back and plugged that number into a =v^2/ r to find the radius but I'm not getting the right answer at all. Any help would be great.
 

Answers and Replies

  • #2
The title of your post is "conservation of energy." It doesn't appear that you've applied this concept to your attempt at the solution.

I suggest you find the velocity at the top of the hill by simply considering the change in gravitational potential energy associated with the roller coaster's descent and subsequent ascent.

Please also note the kinematics equation you provided cannot be used because that equation is only valid under constant acceleration.
 
  • #3
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Thank you for the suggestions.
I used the conservation of energy formula:

mg(h1) = 1/2 mv^2 + mg(h2)
where h1 = 40 m and h2 =20 m
I find the velocity to be 19.8 m/s at the top of the 20 m hill.

I plugged the velocity back into a = v^2/r and I get r to be 80 m but it's still not the right answer (which is actually 30m )
 
  • #4
Thank you for the suggestions.
I used the conservation of energy formula:

mg(h1) = 1/2 mv^2 + mg(h2)
where h1 = 40 m and h2 =20 m
I find the velocity to be 19.8 m/s at the top of the 20 m hill.

I plugged the velocity back into a = v^2/r and I get r to be 80 m but it's still not the right answer (which is actually 30m )
Draw a free-body diagram of the car, noting that it as at the TOP of the hill. Also note that the net force must be equal to the quantity (mv^2)/r. How do the normal force and force due to gravity relate? What does this say about the radius of curvature?
 
  • #5
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there's a gravitational force down (with the mass of both the cart and the person)

There's a normal force going up. But this is the normal force that the person feels. To find the net force, you would have to add these two forces together? Any more explanation on this topic would be very helpful. Conceptually I don't understand why you would have to add these two forces together as opposed to subtracting them.
 
  • #6
The normal force and the gravitational force act in opposite directions. Thus, you would in fact subtract them to find the net force. The force due to gravity is stronger than the normal force, so the net force points towards the center of curvature, as it should. However, I am getting 80 m for my answer, just like you. I cannot figure out why 30 m is recorded as the correct answer.

I've done countless problems like this in the past, so I do not know where I'm going wrong.

It'd be great if anyone else could chime in.


there's a gravitational force down (with the mass of both the cart and the person)

There's a normal force going up. But this is the normal force that the person feels. To find the net force, you would have to add these two forces together? Any more explanation on this topic would be very helpful. Conceptually I don't understand why you would have to add these two forces together as opposed to subtracting them.
 

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