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Conservation of Energy and Centripital Acceleration

  1. Oct 2, 2007 #1
    I have found similar problems to this elsewhere on the forum, but I still can't seem to understand it:

    1. The problem statement, all variables and given/known data
    A mass rests on the top of a frictionless fixed sphere with radius R, and begins with an infinitesimally small velocity to slide off. At what angle theta (measured from the horizontal) will it loose contact with the sphere?

    3. The attempt at a solution

    I used the conservation of mechanical energy E = 1/2mv^2 - mgR(sin(theta)+1), plugged in -2Rmg for E (since initially the mass is at rest and at the top of the sphere) and derived v^2 = 2gR(sin(theta)-3).
    Assuming that the centripetal force is mg(sin(theta)) - Fn , and equal to v^2/R, I wrote:
    2mgR(sin(theta)-3) = mgR(sin(theta)) - FnR.

    Now, at the point where the mass looses contact, Fn = 0, so
    2mgR(sin(theta)-3) = mgR(sin(theta))

    When I solve for theta I get arcsin(6)???

    I fear I have made some grievous conceptual error, and I would greatly appreciate any help you can give.
    Last edited: Oct 2, 2007
  2. jcsd
  3. Oct 2, 2007 #2


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    Why the minus signs? potential energy = mgh... so why do you have -2Rmg and -mgR(sin(theta) + 1). Where are you setting potential energy = 0?

    Other than this, everything you did seems right.
  4. Oct 2, 2007 #3
    I think the height u took is wrong..for PE.
    Last edited: Oct 2, 2007
  5. Oct 2, 2007 #4
    Thanks for the suggestions.

    I have set the PE = 0 at the bottom of the sphere, so the height at any angle theta is Rsin(theta) + R, the first term for the height above the centre of the circle, and the second for the height of the centre off the ground. In my equations I have simplified this to R(sin(theta)+1). When the mass is at the top of the sphere, this is equal to 2R, which is why I used that for the total energy.
    I tried the problem again without the minus signs (which were clearly wrong, I don't know what I was thinking!) and I ended up with:

    2gR(1-sin(theta)) = gRsin(theta)

    This time the answer is theta = arcsin(2/3) = 0.7297 rad, which at least makes sense, but does anyone know if this is indeed the correct answer?
  6. Oct 2, 2007 #5
    Correct answer is cos(2/3).

    Energy during beginning = 0.Take PE level at the initial pos.

    Energy where the object loses control = mv^2/2 - mg(r-rcos@) = 1/2mv^2 -mgr(1-cos@)

    IMplies mgr(1-cos@) = mv^2/2
    IMplies mv^2 = 2mgr(1-cos@)

    Now forces acting on the body are mg,mv^2/r and N.Breaking components you see mgcos@-N = mv^2/r. Object loses contact when N = 0. So at the pos mgcos@ = mv^2/r

    Use the relation of mv^2 we dervived previously here.

  7. Oct 2, 2007 #6
    Atavistic, I don't quite understand why your value for height is r-rcos(theta), where are you setting theta = 0? I have it set as the horizontal, so that the initial angle is pi/2 radians, which is why I get R(sin(theta) + 1). I guessed that maybe you set the initial angle to 0 radians, but then wouldn't the height by R(cos(theta)+1) rather than R(1-cos(theta))? I apologize if I am missing something, but I would appreciate if you could clear this up.

    Thanks in advance.
  8. Oct 2, 2007 #7


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    You both got the same answer, but Sirkus I believe your answer is the right one since you are using the angle as measured from the horizontal (which is what the question asks for).

    Atavistic is measuring the angle from the vertical... in other words the two angles are complementary.

    Your answer looks right to me Sirkus.
  9. Oct 2, 2007 #8
    I see now why I was confused with Atavistic's post, because he was using the convention that the initial angle was zero and increased as the mass fell, in which case the height is R(1-cos(theta)), which threw me off because in my scenario the angle decreases from pi/2, giving R(sin(theta)+1).

    Thanks for all the help everybody, even though is seems the real problem was just a simple sign error! I'll have to be more careful next time.
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