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Roller coaster: kinetic energy

by jaded18
Tags: coaster, energy, kinetic, roller
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jaded18
#1
Sep25-07, 07:20 PM
P: 150
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/...3/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?
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jaded18
#2
Sep25-07, 07:53 PM
P: 150
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!
learningphysics
#3
Sep25-07, 07:56 PM
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P: 4,124
Just use conservation of energy... I don't think you need to deal with centripetal motion or anything.

jaded18
#4
Sep25-07, 08:21 PM
P: 150
Roller coaster: kinetic energy

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.
learningphysics
#5
Sep25-07, 08:40 PM
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Quote Quote by jaded18 View Post
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.
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.
jaded18
#6
Sep25-07, 08:43 PM
P: 150
The energy of the coaster at the beginning is potential energy which is just mgh.
learningphysics
#7
Sep25-07, 08:52 PM
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P: 4,124
Quote Quote by jaded18 View Post
The energy of the coaster at the beginning is potential energy which is just mgh.
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?
jaded18
#8
Sep25-07, 09:20 PM
P: 150
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?
learningphysics
#9
Sep25-07, 09:48 PM
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Quote Quote by jaded18 View Post
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?
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?
jaded18
#10
Sep25-07, 09:56 PM
P: 150
well acceleration is V^2/R . and so min velocity would be v = gR ...
jaded18
#11
Sep25-07, 09:57 PM
P: 150
sorry, v = (gr)^(1/2)
learningphysics
#12
Sep25-07, 10:01 PM
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Quote Quote by jaded18 View Post
sorry, v = (gr)^(1/2)
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.
jaded18
#13
Sep25-07, 10:10 PM
P: 150
mgh... sorry i'm not getting it
jaded18
#14
Sep25-07, 10:17 PM
P: 150
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...
learningphysics
#15
Sep25-07, 10:22 PM
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Quote Quote by jaded18 View Post
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...
can you show how you get h = 3R, I'm getting h = 2.5R
jaded18
#16
Sep25-07, 10:25 PM
P: 150
ah, stupid calc. error. thanks lots learningphysics. you made my day haha
learningphysics
#17
Sep25-07, 10:27 PM
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Quote Quote by jaded18 View Post
ah, stupid calc. error. thanks lots learningphysics. you made my day haha
no prob.
jakubu
#18
Apr3-10, 05:30 PM
P: 2
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


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