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Homework Help: Centripetal Force Question

  1. Feb 15, 2010 #1
    1. A roller-coaster car has a mass of 1290 kg when fully loaded with passengers. As the car passes over the top of a circular hill of radius 20 m, its speed is not changing. (a) At the top of the hill, what is the normal force (using the negative sign for the downward direction) FN on the car from the track if the car's speed is v = 9.3 m/s? (b) What is FN if v = 18 m/s?

    2. F=mv2/r

    3. I thought this was going to be a really simple question but apparently I can't figure out the answer. Here's what I did

    a) F=1290(9.3)2/20 = 5580N but for the answer I put -5580N cause of the part that says using the negative sign for the downward direction. I thought that was right

    b) F=1290(18)2/20 = 20900N but I changed it to -20900N for the same reason above.
     
  2. jcsd
  3. Feb 15, 2010 #2

    jhae2.718

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    The normal force is not equal to the centripetal force. The forces acting on car are the weight and Normal force, yielding the centripetal force, which is the net force.
     
  4. Feb 15, 2010 #3
    Oh so would I find the weight now and subtract them?
     
  5. Feb 15, 2010 #4

    jhae2.718

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    No. Create a free body diagram and find the sum of the vertical forces, noting that the weight is in the negative direction and the normal force is in the positive direction. Then set that expression equal to the centripetal force. Then solve for the normal force.

    The normal force is in the positive direction, so you should have a non-negative answer.
     
  6. Feb 15, 2010 #5
    So it should look like this?
    (1290(9.32)/20)+(1290*9.8)=18221N
    (1290(182)/20)+(1290*9.8)=33540N

    Sorry if I'm being annoying I'm on my last attempt for my online homework so I want to make sure it's correct before I submit it.
     
  7. Feb 15, 2010 #6

    jhae2.718

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    I'm looking over the problem again. Is the car traveling over the circle, or is it at the top of a vertical circle on the track, but underneath the track. (See attachment for clarification)

    I assumed the bottom case, but if it's the top case, then the normal force and the weight are both negative, in which case -Fc=-W-N, which can be simplified to Fc=W+N. (Just know that your answer will be negative, because it's in a downward direction.)

    The wording is kind of ambiguous, but "over the top of a circular hill" makes it look like the bottom case as we originally worked out the problem. If that is the case, then we both got the same results.
     

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    Last edited: Feb 15, 2010
  8. Feb 15, 2010 #7
    So now were back to this as the answer?
    (1290(9.32)/20)-(1290*9.8)= -7063N
    (1290(182)/20)-(1290*9.8)= 8256N
     
  9. Feb 15, 2010 #8

    jhae2.718

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    It's dependent on what the problem actually is...I think the answer is your 6:02 post, but since you have to turn the problem in, I don't want to cause you to give an incorrect answer based on my reading of the problem. Does it give any more information?

    I think that the car is on top of the track, in which case the weight would be downwards, and the normal force would be upwards.

    Hopefully, either a recognized Homework Helper or a PF Mentor can weigh in before it's due.

    Sorry...
     
  10. Feb 15, 2010 #9
    Well here's a hint that I get from Wiley Plus

    Write Newton's second law for forces along a vertical y axis. Is the (centripetal) acceleration up or down? What is the gravitational force? (For safety, the car has wheels both above and below the track.)
     
  11. Feb 15, 2010 #10

    jhae2.718

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    What is the exact wording of the problem?

    I'm 90% sure that the car is traveling over, so that N=W+Fc, in which case the answer would be your: (1290(9.32)/20)+(1290*9.8)=18221N
    (1290(182)/20)+(1290*9.8)=33540N.
     
    Last edited: Feb 15, 2010
  12. Feb 15, 2010 #11
    The exact wording is what is above for the question. Alright if youre sure I'll try it out.
     
  13. Feb 15, 2010 #12

    jhae2.718

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    I'm pretty sure it's right. I copied it incorrectly, so remember the speed is squared...
     
  14. Feb 15, 2010 #13
    Well you copied my answers and thats all that matters you know
     
  15. Feb 15, 2010 #14

    jhae2.718

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    Well, let me know how it turns out.
     
  16. Feb 15, 2010 #15

    PhanthomJay

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    i think you're mixing up your plus and minus signs. At the top of the circle, the centripetal acceleration acts down, and therefore, the net force acts down. Calculate the coasters weight, then determine the normal force such that the net force acts down. What's the direction of the normal force in each case? Note the safety feature.
     
  17. Feb 15, 2010 #16
    Nope it was wrong :(
     
  18. Feb 15, 2010 #17

    jhae2.718

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    Hmm, the original post with the problem didn't mention any safety features, so I discounted that...

    That said, you are much more well-qualified than I am, so perhaps you can see what I left out for MFlood7356...
     
  19. Feb 15, 2010 #18

    PhanthomJay

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    Yeah, without that safety feature for part 2 with the coaster speeding at 18m/s, it would leave the tracks and head for parts unknown.

    The net force is the centripetal force, and was calculated correctly for part 1. Let's round it off to 5,000. The net force must always act inward to the center of the circle, Since the coasters weight is about 12,000 down , then the normal force must be about 7,000 up. For part 2, the net force is say 20,000 down. Since the coaster weight is still 12,000, the normal force must be about 8000 down. Safety feature required in order for the track to exert a downward force on the coaster to keep it on track.. I'm using round numbers, not actual numbers.
     
  20. Feb 15, 2010 #19

    jhae2.718

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    I know; I just realized the mistake in the advice I gave him...

    I think I've done too much today; I'm making mistakes I normally don't.
     
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