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Loop-de-loop where the normal force equals the magnitude of the gravitational force

  1. Oct 17, 2011 #1
    1. The problem statement, all variables and given/known data
    The normal force equals the magnitude of the gravitational force as a roller coaster car crosses the top of a 46m--diameter loop-the-loop.


    2. Relevant equations
    What is the car's speed at the top?



    3. The attempt at a solution
    the answer is in m/s
    I have several equations but I really don't know what to do with them...
    the sum of Fr=nr+(Fr)r=n+mg=m(Vtop)^2/r
    and n=m(Vtop)^2/r-mg
    Vc=SQRTrmg/m=SQRTrg
    wc=SQRTg/r

    I'm really lost on this one.
     
  2. jcsd
  3. Oct 17, 2011 #2
    Re: loop-de-loop where the normal force equals the magnitude of the gravitational for

    For the normal force to equal the gravitational force, what forces must balance?
     
  4. Oct 17, 2011 #3

    Mute

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    Re: loop-de-loop where the normal force equals the magnitude of the gravitational for

    It looks like you have everything you need to solve the problem. You just need to figure out what pieces fit where. You should break the problem down a bit: What do you want to find? What information are you given/do you already know? And: What information is necessary to find what you want to find?

    Write down the answers to these questions, and then look at your equations. Which equations contain the thing you want to find, and in those equations, in which ones can you plug in some information given to you in the problem statement? One of the equations is the one you want to start from. Another is a rearrangement of that, but the rearrangement isn't useful because you've solved for something that you know, not something that you don't know.

    Sorry if this seems a bit vague. I'm merely trying to encourage you think about these sorts of problems in a more systematic way without outright giving away the answer, while still trying to help you avoid getting tripped up. Once you take another look at your stuff, if you're still stuck someone will help you out further.
     
  5. Oct 18, 2011 #4
    Re: loop-de-loop where the normal force equals the magnitude of the gravitational for

    I was able to figure this one out.
    Vc=Sqrt 46m (9.8 m/s^2)=
    sqrt 450.8=
    Vc=21.23 m/s
    it was correct
    Thank you!
     
  6. Oct 18, 2011 #5

    PeterO

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    Re: loop-de-loop where the normal force equals the magnitude of the gravitational for

    You have the right answer, but I would have prefered to see 23m (19.6 m/s2).

    That would indicate you realised that the RADIUS of the loop was only 23 m, and the acceleration at the top of the loop was twice that of gravity.

    Please respond if you didn't realise that and I will explain.

    Peter
     
    Last edited: Oct 18, 2011
  7. Oct 18, 2011 #6

    PeterO

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    Re: loop-de-loop where the normal force equals the magnitude of the gravitational for

    There are no forces balancing!! This roller coaster is on the inside of the loop!!! Gravity and the Normal force are in the same direction - they reinforce or supplement each other rather than balancing.
     
  8. Oct 18, 2011 #7
    Re: loop-de-loop where the normal force equals the magnitude of the gravitational for

    Yes, I did not realize this.
     
  9. Oct 18, 2011 #8

    PeterO

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    Re: loop-de-loop where the normal force equals the magnitude of the gravitational for

    The most common "loop-the loop questions" have the cart going through the top of the loop at just the right speed so that there is no need for the track to push at all.
    In those cases, The centripetal force required equals the weight force provided - and the Normal force is zero.

    In this case we are told the Normal force = mg, so the net downward force is twice that of gravity alone.
    Gravity pulls down with mg, the track pushes down with an extra mg, so the total force is 2mg. That means the centripetal acceleration is 19.6, not just 9.8

    Since the track has diameter 46 m the radius is 23 - so you should have used 23 and 19.6 to get your answer, though coincidentally using 46 and 9.8 gave the same result.
     
  10. Oct 20, 2011 #9
    Re: loop-de-loop where the normal force equals the magnitude of the gravitational for

    Oh Ty it makes sense now
     
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