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Angular momentum

  1. Apr 4, 2008 #1
    Hello!
    [​IMG]

    1. The problem statement, all variables and given/known data
    A particle with mass m is thrown in lateral speed [tex]V_0[/tex] inside a hollow half-ball with radius [tex]R[/tex]. At the beginning of it's motion the ball has an angle of [tex]\theta_0[/tex] from the perpendicular.
    The gravitational force will pull the particle toward the center of the ball, while the centrifugal force will push it outwards.
    Calculate the speed [tex]V_0[/tex], as a function of [tex]\theta_0[/tex], needed for the particle to reach the top of the half-ball in the peek of its motion.
    Important! there's no string attached to the ball. The line on the image just indicates the radius.
    2. Relevant equations
    [tex]\overline J=m\overline r \times \overline v[/tex]
    [tex]\overline \omega=\overline{ \omega_0} + \overline\alpha t[/tex]

    3. The attempt at a solution

    Well, the problem is I don't understand the forces involved.
    I know there some sort of [tex]J_0[/tex] here, because there's an [tex]\overline r[/tex] and a [tex]\overline v[/tex]. I can also draw a forces equation. Then there's the Normal force against mg and centrifugal force (btw - can I use the centripetal force instead?), but I don't quite know how to combine the two - F and J - together.

    Thank you.

    [edit]
    I thought of something: there are three forces: [tex]N, mg, \frac{mv^2}{R}[/tex].
    also, I can do something like this: [tex]\Delta J = J_{end}-J_{start}[/tex], and [tex]J_{end}=0[/tex], because on the peak of the motions happens when v=0. also, [tex]J_{start}=mv_0R(sin\theta+cos\theta)[/tex].
    and also [tex]\frac{dJ}{dt}=r \times F[/tex]
    so if I only knew how to play the forces right, I would have it.
    Is it correct? if so, how do I know the force equation?
     
    Last edited: Apr 4, 2008
  2. jcsd
  3. Apr 4, 2008 #2
    I'm not completely sure, but the N force is probably the centripetal force because it's always pointed towards the center. It is the component of ball's weight and it's equal to weight at the bottom of the half-ball.

    Maybe you can solve this by using energies.
     
  4. Apr 4, 2008 #3
    Thanks!
    I did it with energies and it worked perfectly (through one line :) ).
     
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