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Simple Energy conservation, PE = KE + PE

  1. Mar 25, 2008 #1
    1. The problem statement, all variables and given/known data

    A block mass M slides down the side of a frictionless circle Radius R. At an angle Theta the mass M flies off the circle, what is the angle?

    2. Relevant equations

    PE(top) = KE(point it flies off) + PE(at that point)

    Arad = V^2 / R

    Sum Of Forces = Mass * Acceleration

    3. The attempt at a solution

    Okay I actually did this one before and I was trying to do it again but somehow I don't seem to be able to get it. The answer was Inverse Cosine of 1/1.5 or 48 degrees.

    The problem was done with energy equations

    PE(top) = KE(point) + PE(point)

    I set 0 PE to be the middle of the circle so

    mgR = .5 mv^2 + mg(Rcos(theta))

    mass cancels

    gR = .5v^2 + gRcos(theta)

    I think I'm going wrong here but I said the only force acting on the block is weight or mg, because at the point is leaves the normal force goes to 0.

    so sum forces = mass * acceleration

    mg = ma

    mass cancels

    g = a

    then v^2 / R = a , so gR = V^2

    then plugging that in

    gR = .5v^2 + gRcos(theta)

    gR = .5gR + gR cos(theta)

    gR cancels

    1 = .5 + cos(theta)

    and I end up with 60 degrees so I think I missed out a number somewhere but I don't know where.
  2. jcsd
  3. Mar 25, 2008 #2

    Shooting Star

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    Homework Helper

    It'll leave the surface of the sphere when the normal reaction N becomes zero.

    If [itex]\theta[/itex] is the angle made with the upward vertical with the radius at the position of the particle, then the centripetal force is given by,

    mv^2/r = mgcos[itex]\theta[/itex] - N, which will give you mv^2 when N=0.

    Also, KE = change in PE from top position, which gives,

    mv^2/2 = mg(...) [you find it, in terms of r and [itex]\theta[/itex], using geometry].

    From this, you'll get [itex]\theta[/itex].
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