1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    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].
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?