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Circular Motion problem

  1. Feb 23, 2012 #1
    1. The problem statement, all variables and given/known data
    A 'loop a loop' cart runs down an incline and then inside a circular track of radius 3.0 meter and describes a complete circle. Find the minimum height above the top of a circular track from which the cart must be released.

    2. Relevant equations
    So given data is:
    r=3 m
    To find: Height between the top of the circle and the point from which the cart is thrown.
    We have to use law of conservation of energy.

    3. The attempt at a solution
    Okay I'm a little confused here. The final answer is 1.5 m. My solution book has done it like this:

    Then it goes like :

    ##mgh=\frac{1}{2}mv^2##, then ##mgh=\frac{1}{2}mrg## since velocity at top is ##\sqrt{rg}##

    Then ##h=\frac{r}{2}=1.5 m##

    But how can the P.E at the top become zero?

    The way I've done it is, I've taken the reference point at the bottom of the circle. So it's like:

    From the image, the total energy at the top of the circle must equal the energy at the starting point. SO,
    ##mg(2r) + \frac{1}{2}m(v_1)^2=mg(2r+h)+0##
    Evaluating that, we get ##rg=2gh## and therefore ##h=\frac{r}{2}=1.5m##.
    So, who's right ? My book or me? Is the energy at the top of the circle really zero? Answer is the same. Sorry my question is noobish. Please help. I like mechanics but these kind of sums always confuse me.
    Last edited: Feb 23, 2012
  2. jcsd
  3. Feb 23, 2012 #2


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    Both approaches lead to the same result....PE is measured with respect to a reference axis...any arbitrary axis...It's the change in PE that is the same....you assume PE = 0 at the bottom, while the book assumes PE =0 at the top of the loop...it doesn't matter. Suppose the bottom of the loop was elevated 2 m above ground and you chose PE =0 at ground level....you still get the same value for h...
  4. Feb 23, 2012 #3
    So, in reality is the potential energy at the top zero or not?
  5. Feb 24, 2012 #4


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    We can measure potential energy difference only: the work done by the force when an object moves from place A to B: this is equal to the potential difference U(A)-U(B). You can add an arbitrary constant to the potential function, the difference is the same.

    So the gravitational potential energy is U= mgh + const. You can choose the constant that U = 0 at h=0, tehn U=mgh. But you measure the height h also relative to something. If h1=h is the height from the top of the circle and h2 is the height measured from the bottom of the circle, h2=h1+2r. In the first case, U1=mgh and it is U2=mgh2=mg(h+2r)=mgh+mg(2r). But mg(2r) is constant in the problem. U1 and U2 differ only in an additive constant. The potential difference between the starting point (A) and at the top of the circle (B) is U(A)-U(B)=mgh - mg0 =mgh in the first case, and mg(h+2r)-mg(2r) in the second case: both are mgh.

  6. Feb 24, 2012 #5
    Ah...okay now I get it, it's the difference, not the actual P.E. Thanks a lot ehild your explanation was quite helpful!
  7. Feb 24, 2012 #6


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    You are welcome.

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