1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: Puck on a sphere, energy & Newton's 2nd

  1. Sep 4, 2013 #1
    1. The problem statement, all variables and given/known data

    Consider a small frictionless puck perched at the top of a fixed sphere of radius R. If the puck is given a tiny nudge so that it begins to slide down, through what vertical height will it descend before it leaves the surface of the sphere? [Hint: Use conservation of energy to find the puck's speed as a function of its height, then use Newton's second law to find the normal force of the sphere on the puck. At what value of this normal force does the puck leave the sphere?]

    2. Relevant equations


    3. The attempt at a solution

    I use y=y(t) as the height of the puck at time=t.

    I managed the first part of the hint, getting down to [itex]v=\sqrt{2g(R-y)}[/itex]

    The next part is what's throwing me. Intuitively, I would think that the puck leaves the sphere when the normal force is zero. This would happen when the normal force vector is perpendicular to the weight vector, making it so that the puck falls of once it gets to the equator of the sphere. That would then be my answer, but if it were correct, I wouldn't have to find the velocity of the puck. It also stands to reason that the velocity of the puck at any given time affects whether it's about to fly off the surface.

    Analytically, I tried: (N for normal and W for weight)
    N=[itex]\dot{p}[/itex]-W , setting N=0 gives

    I considered integrating both sides with respect to dt so I could sub in my velocity equation on the left, but the right hand side would give me a function of t, so I would get something like


    If I solve that for y I don't have y explicitly, I have function y(t). Also, I didn't put arrows above all the vectors, but consider that W points down, Wt would also point down. This would mean that p also has to point down, which shouldn't happen even after the puck leaves the sphere, which is nonsense.

    The only other thing I can think of is the "centrifugal force" lessening the magnitude of N, but since it's not a real force, I don't really know how to do math with it.
  2. jcsd
  3. Sep 5, 2013 #2
    I am not sure if you can integrate this equation, because you set [itex]\displaystyle{N=0}[/itex], so it's valid only for a specific moment.

    But you can solve the exercise easier without integrating. Just take the Newton's 2nd law for the normal components of the forces. Imagine that every moment you take the tangent of the sphere at the point where the puck is and you take the components that are perpendicular to this line.

    [itex]\displaystyle{\vec{N}}[/itex] is always perpendicular so you don't have to do something for it. But as you said [itex]\displaystyle{\vec{W}}[/itex] points down so you have to take only the perpendicular component. Take an angle [itex]\displaystyle{\theta }[/itex] and use it to analyze [itex]\displaystyle{\vec{W}}[/itex].

    The resultant of these components is the centripetal force! You know how to find this force in terms of [itex]\displaystyle{v}[/itex] (known equation). So you get an equation and ofcourse set [itex]\displaystyle{N=0}[/itex]. Now you just have to do some geometry - trigonometry to find [itex]\displaystyle{\theta }[/itex] in terms of [itex]\displaystyle{y}[/itex] and the problem is solved.

    There's also one solution with a little bit more mathematics, using polar coordinates, but I think that if you know the centripetal force the previous one is easier.

    Centrifugal is a non-inertial force. So if you want to use it you have to take a non-inertial reference frame. Here you can take a reference frame in which the puck is not moving. Then Newton's first law says: [itex]\displaystyle{\Sigma \vec{F}=\vec{0}}[/itex]. This can be done only using an (imaginary) non-inertial force. The normal component of this is centrifugal force which turns out to have the same length and opposite direction of centripetal force.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted