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Energy conservation of a skier

  1. Feb 18, 2007 #1
    Hello, I'm new here so please let me know if I do something wrong.

    1. Problem 1
    1. A skier of mass M starts from rest at the top of a solid sphere of radius R and slides down its frictionless surface. At what angle will the skier leave the sphere?

    http://img413.imageshack.us/img413/4362/spherele9.jpg [Broken]

    2. Relevant equations
    E = K + U = (.5)mv^2 + mgy

    3. The attempt at a solution
    I started off saying that the E at the top of the sphere will equal the E at the point when the skier leaves the sphere, however, although I can set v = 0 for the initial position, on the other side of the equation, I have two unknowns - the v when the skier leaves the sphere and y. This also doesn't really give me an angle to solve for. I know that the point at which the skier leaves the sphere, her normal force will be 0, but I don't know if/how I can work that into this.

    I think I'm on the wrong track :(
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Feb 19, 2007 #2
    Naah, youre on the right track. Assume, for a moment, that at the point where the skier leaves the sphere, he's moving with a speed v. Let the angle that point makes with the center of the sphere be theta.

    Now, you can calculate the drop in height of the skier from the top using trigonometry, and that potential is converted into kinetic energy. The forces acting are the gravitational force and centripital force. Draw your f.b.d, resolve the forces along the x and y directions, and see how things work out.
  4. Feb 19, 2007 #3
    I don't mean to be a nag, but from what I understand, isn't the centripital force not a real force? I mean, we're taught in basic physics that it is, but isn't the correct term a normal acceleration that FEELS like a force? Just a thought...
  5. Feb 19, 2007 #4
    Centrifugal force is not a real force, its a pseudo force. Centripital force is a real force caused by another real force like gravity/electric force etc.

    The difference is, that centripital force is the same in all frames of motion, but centrifugal force is dependent on the frame from which the object is observed (undergoing rotational motion).

    For example, if you have a merry go round, which is spinning with an angular velocity [tex]\omega[/tex] from the ground frame, a centripital force [tex]m\omega ^2r[/tex] is present.

    If, however, there is another platform above the merry go round, rotating with an angular velocity [tex]\omega _0[/tex], then the centripital force is [tex]m\omega ^2r[/tex] but the centrifugal force is [tex]m(\omega-\omega_0)^2 r[/tex].

    I hope that makes it clear. If you have a problem, just ask. Ill help you if I can.
  6. Feb 19, 2007 #5
    edit: not relevant anymore :)
    Last edited: Feb 19, 2007
  7. Feb 19, 2007 #6
    Alright so I've done a bit of work and I think I'm almost there, let me explain what I've done and where I'm now stuck.

    First of all:
    http://img509.imageshack.us/img509/5949/sphere2lm8.png [Broken]
    I note that the height at the top of the sphere is 2r

    I use this figure to find that the height when the skier leaves the sphere is r+rcos(theta)

    I use conservation of energy to find that the velocity when the skier leaves the sphere is sqrt(2gr(1 - cos(theta))).

    I then move to the Force Body Diagram ( not sure if this is right ):
    http://img412.imageshack.us/img412/6047/fbd1jv4.png [Broken]
    With mg being gravitational force, CP being centripetal force, and N being normal force.

    I observe that mgcos(theta) + CP = N

    I set N equal to zero (for when the skier leaves the sphere)

    I plug in the equation for CP as mv^2 / r and then plug in my value of v from earlier

    After some cancelling, I get mgcos(theta) + 2mg(1 - cos(theta)) = 0

    This then gives me cos(theta) = 2 which can't be right since cos never equals two.
    Last edited by a moderator: May 2, 2017
  8. Feb 19, 2007 #7

    Doc Al

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    Staff: Mentor

    I was just responding to your earlier post--but you deleted it! That earlier post was almost there.

    Your new force diagram seems a bit off (but that depends on your choice of axis, I suppose). Gravity acts down, for one!

    Your signs are off. Find the net force*--set that equal to the formula for "centripetal force". (There are only two forces acting on the skier: gravity and normal force. Centripetal force is just a name given to the net force that acts towards the center.)

    *Edit: Net force in the radial direction. :smile:
    Last edited: Feb 19, 2007
  9. Feb 19, 2007 #8
    Ah so I was right (or at least close earlier) sorry for deleting the post - I thought I was way off before!

    So, (and correct me if I'm wrong here, I've never had a physics course before and I'm afraid I'm dreadfully bad at it), the net force would be the sum of the two forces, gravity and the normal force, so:

    mg + N = mv^2 / r

    (2mgr(1-cos(theta)) / r) - mg = N = 0

    2mg(1 - cos(theta) - mg = 0

    mg = 2mgcos(theta)

    1/2 = cos(theta) -> theta = 60 degrees?

    This looks much better than before, I hope I've done it right this time :/
    Last edited: Feb 19, 2007
  10. Feb 19, 2007 #9

    Doc Al

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    Staff: Mentor

    You need the net force in the radial direction. In what direction is the radial acceleration? What's the radial component of gravity? What are the directions of gravity and the normal force?

    One more time and you've got it.
  11. Feb 19, 2007 #10
    I don't think I quite understand what you mean by radial direction. I'm sorry, I don't mean to be difficult. :frown:

    edit: If you mean the direction that the radius points in at this point (by which I mean using the radius as the y-axis on my FBD), then I get mgcos(theta) + N = mv^2 / r ... would that be the correct way of going about this?
    Last edited: Feb 19, 2007
  12. Feb 19, 2007 #11

    Doc Al

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    Staff: Mentor

    The net force on the skier (and his resulting acceleration) has both radial (parallel to the radius of the circle) and tangential components. For this problem, we only care about the radial components.

    (Radial means the same thing as centripetal.)

    Yes--almost! You still have the directions (signs) wrong. Hint: The radial acceleration is towards the center (as opposed to away from the center). What about the normal force?
    Last edited: Feb 19, 2007
  13. Feb 19, 2007 #12
    I get what you're saying here, but I'm not quite so sure I'm going to get this right, my thinking on this is becoming more and more confusing by the second, but if it's acting towards the center, then it would be (I hope): mgcos(theta) - N = mv^2 / r

    This would then give me an answer of cos(theta) = 2/3, theta = 48.2 degrees approx.
    Last edited: Feb 19, 2007
  14. Feb 19, 2007 #13

    Doc Al

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    Staff: Mentor

    There you go! :approve:

    Just to bang it in, here's how I think of it: First, I choose a sign convention. For example, let's choose away from the center to be positive (it doesn't matter). So, using that convention, the radial components of force are:
    weight= -mgcos(theta)
    normal force= +N​

    And since the radial acceleration is towards the center, the centripetal force is: -mv^2/r

    Put that together: -mgcos(theta) +N = -mv^2/r
    Or: mgcos(theta) - N = mv^2/r (just as you wrote it)
  15. Feb 19, 2007 #14
    Haha, I'm so glad to have finally reached the answer, but I'm even happier that I (now) understand it, as well. Seems so much simpler written out like that, if only my book would do the same!

    Thanks a lot to everyone (especially you Doc Al) for all the help!
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