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Sphere-particle problem

  1. Oct 24, 2008 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    A friend of mine is stuck on a problem (that I saw about 2 years ago on the internet) so I thought about how to solve it but was totally unable.
    Here it comes : Consider a particle on the top of a sphere of radius r. If there's no friction between the particle and the particle is falling, at what point will it leave out the sphere?


    2. Relevant equations



    3. The attempt at a solution
    I was tempted to use the formula [tex]\sum \vec{F}=m\vec{a}[/tex] until I realized that there was some complications. For example I got that the acceleration in the x-axis is worth [tex]g\sin \left( \theta(t) \right)[/tex] and I have no clue about how to express [tex]\theta[/tex] in function of the time. It's not like the particle is following a circular path. More precisely, there's no centripetal force... the only force that influences the movement is the weight of the particle (assume it has a mass m).
    I've also thought about conservation of energy... but since I don't know at what height it will leave the sphere, I can't get the velocity which could have helped me maybe.
    Can you help me a very bit? Thanks in advance.
     
  2. jcsd
  3. Oct 24, 2008 #2

    alphysicist

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    Hi fluidistic,

    It is following a circular path up until the time it leaves the surface. So you know a formula for the radial acceleration right up until that time.

    That's not true while it is on the sphere. After it leaves the sphere, the only force is the force of gravity, but until then the sphere is putting a normal force on it.


    Yes, use conservation of energy, which will give you a relationship between height and velocity. Also, use a force diagram right at the moment it loses contact, which will give you another relationship with velocity. You will need to do some trig, and think about what it means when something loses contact, but that should be enough to solve the problem.
     
  4. Oct 24, 2008 #3

    fluidistic

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    Thanks a lot alphysicist!


    You're right. I did the free body diagram at a point "x" on the sphere and put the weight and normal force. Maybe what I wanted to mean was that the body is not accelerating in the "y axis" which constantly changes.
    It's almost 3am now, so I'll use what you told me tomorrow.
    Good night!
     
  5. Oct 24, 2008 #4

    alphysicist

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    Yeah, 3am or so is usually my limit for work, too. I'm a couple of time zones behind you, though, so with the help of a lot of caffeine I plan on still going a while. Let me know how the problem turns out.
     
  6. Oct 25, 2008 #5

    fluidistic

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    Hi,
    I didn't do that well. I realized that the normal force is not equal to [tex]mg\cos \theta(t)[/tex] because if it was so, then the particle would lose contact when [tex]\theta (t)=\frac{\pi}{2}[/tex] but this is not true.
    Nevertheless I tried a few things... I thought that the centripetal force was worth [tex]mg\cos \theta[/tex] (same magnitude as the normal force), thus [tex]a_c=g\cos \theta=\frac{v^2}{r} \Rightarrow v=\sqrt{rg\cos \theta}[/tex].
    With a little trigonometry, [tex]h=r\cos \theta[/tex] where h is the height at which the particle will leave the sphere. So it remained to find [tex]\theta[/tex]... Then I tried to figure out a good free body diagram when the particle leaves the sphere, but I'm not sure. When it leaves the sphere, the normal force is worth 0N but there's still the weight. I can't find anything related to velocity in such a position.
    WAIT!! I just got it! Using the conservation of energy at this point I found out that [tex]\cos \theta[/tex] is worth [tex]\frac{2}{3}[/tex]! So [tex]h=\frac{2r}{3}[/tex] which is the answer if I remember well!
    Thanks so much alphysicist! I can't say I've all the credit, but at least I've done something.
     
  7. Oct 25, 2008 #6

    alphysicist

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    That looks perfect, fluidistic!
     
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