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Ice on an inverted bowl: Radial force?

  1. Oct 16, 2016 #1
    Hello! I am just stuck on one part of this question and would be grateful for any help.

    Question

    A small block of ice slides from rest from the top of an inverted frictionless bowl of radius
    R (above right). How far below the top x does the ice lose contact with the bowl?

    Equations
    • mgx = mg*sin(θ)
    • mgy = mg*cos(θ)
    • ∑Fr = (mα) + (mg*cos(θ)) - (Fn)
    • ∑Fθ = mg*sin(θ)
    • x(t) = A*cos(ωt + φ) ... maybe needed?

    Attempt
    I found a solution here (http://www.feynmanlectures.info/solutions/particle_on_sphere_sol_1.pdf), and they assert that the ice-cube falls off the bowl when the normal force is zero. That makes perfect sense.

    But they also assert that Fn = mg*cos(θ) - mα. The problem is easy to solve from there. But I just can't figure out why the force of centripetal acceleration (mα) is being subtracted from the radial gravitational force (mg*cos(θ)). It seems like the answer is right in front of me -- I just cannot see it.
     
    Last edited: Oct 16, 2016
  2. jcsd
  3. Oct 16, 2016 #2

    haruspex

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    It's better to think of it as ΣF=ma, where a is the resulting acceleration in the outward radial direction, i.e. the negative of the centripetal acceleration. What is the sum of forces in the outward radial direction?
     
  4. Oct 16, 2016 #3
    Thanks for that pointer. I think this is where my confusion lies.

    I figure that, in the radial direction, you have three different forces: Fn in the outward (negative) direction, and mα + mg*cos(θ) in the centripetal (positive) direction. But something about that feels problematic...

    Maybe my question really is, what is mα?
     
  5. Oct 16, 2016 #4

    haruspex

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    This is a common misunderstanding. The normal force and the force from gravity are both real, applied forces. Centripetal force is not - it is the radial force needed to provide a certain radial acceleration. I.e. it is the radial component of the resultant of the applied forces. Some authors spurn use of "centripetal force", referring only to centripetal acceleration.
     
  6. Oct 16, 2016 #5
    Aah, thank you again.

    So, what I understand is that the centripetal force is just a result of the real, applied forces (namely Fn and mg*cos(θ)). If it's a result of the two...then is it true that mα = mg*cos(θ) - Fn ? Then, said another way, mα is used to represent the sum of all forces in the radial direction?
     
  7. Oct 17, 2016 #6

    haruspex

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    Yes.
     
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