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Angular momentum

  1. Jan 6, 2014 #1
    1. The problem statement, all variables and given/known data
    A finger presses a ball and throws it away with angular velocity ω0 and velocity V0 like in the picture. the coefficient of friction is μ. what is the ratio ω0 to V0 so that they both are zeroed at the same time

    2. Relevant equations
    [tex]I_C=\frac {2}{5}mr^2[/tex]
    [tex]Torque=\dot{L}[/tex]
    From kinematics:
    [tex]V^2=V_0^2-2ax,\quad \omega^2=\omega_0^2-2\alpha\theta[/tex]

    3. The attempt at a solution
    The friction decelerates the ball:
    [tex]mg\mu r=\frac{2}{5}mr^2\cdot\alpha\Rightarrow\alpha=\frac{5\mu g}{2r}[/tex]
    from kinematics:
    [tex]0=\omega_0^2-2\alpha\theta\Rightarrow\theta=\frac{\omega_o^2 r}{5g\mu}[/tex]
    The distance travelled:
    [tex]x=\theta r=\frac{\omega_o^2 r^2}{5g\mu}[/tex]
    I insert this distance into the kinematics formula:
    [tex]0=V_0^2-2g\mu\frac{\omega_o^2 r^2}{5g\mu}\Rightarrow V_0=\sqrt{\frac{2}{5}}\omega_0 r[/tex]
    And the answer should be:
    [tex]V_0=\frac{2}{5}\omega_0 r[/tex]
     

    Attached Files:

  2. jcsd
  3. Jan 6, 2014 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Is this true if the ball slips?

    Try starting with kinematic equations that involve time, since you know that ω and v come to zero at the same time.
     
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