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Particle mass atop frictionless hemisphere w/ extra force

  1. Sep 29, 2015 #1
    1. The problem statement, all variables and given/known data
    A particle of mass m rests atop a frictionless hemisphere of radius R. A force F = -mkyα acts in the y direction. After an initial small displacement, the particle slides down the sphere under the action of the force.

    Find a) the height above the equator and b) the speed of the particle, both at the moment that it leaves the surface of the hemisphere.

    2. Relevant equations
    I would assume relevant equations include a summation of forces:

    ΣF = FN - mgcosθ - mkyαcosθ

    3. The attempt at a solution
    I feel that the key is that the normal force is equal to 0 when the particle leaves the surface of the hemisphere as there is no longer contact between the two. Thus you get:

    ΣF = - mgcosθ - mkyαcosθ = mac

    at the point when the particle leaves the surface of the hemisphere.

    I would also assume that energy is a part of the equation:

    E(y=R) = mgR
    E(y) = mgy + 1/2mv2

    mgR = mgy + 1/2mv2

    I figured y= Rcosθ and simplified the energy equation to:

    v2 = 2gR(1-cosθ)

    Which is helpful when I know that ac = v2/R

    Now I assume somewhere in that logic I went wrong because when I try using the energy equation and force equation to set up something where I might find cosθ but it keeps coming out to be a very, very ugly equation which I don't see as plausible. My professor hinted that I could also set the problem up in the non-inertial frame of the particle as it slides down but I wouldn't be sure where to start there. Any help is hugely appreciated.
     
  2. jcsd
  3. Sep 29, 2015 #2

    TSny

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    Gold Member

    Hello. Welcome to PF!

    Are you sure you are supposed to include the force of gravity as well as the force F? The problem gives the expression for the force F and then states that the "particle slides down under the action of the force". To me, this wording suggests that maybe gravity is not acting. Without gravity, the algebra is not too bad.

    Your energy equation does not take into account the work (or potential energy) associated with the force F.
     
  4. Sep 29, 2015 #3
    Without the force of gravity though, what will the potential energy equation look like? Will there even be a potential energy equation or would it just be the change in energy, W = Fd?
     
  5. Sep 29, 2015 #4

    TSny

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    The force F is not constant, so you cannot use W = Fd. You need to use calculus to calculate the work. You can consider the work as the change in a potential energy function. But getting the expression for the potential energy is essentially the same as calculating the work. So, you can just concentrate on finding the work.
     
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