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Angular velocity of a ball after change of direction

  1. Nov 22, 2014 #1

    Vir

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    1. The problem statement, all variables and given/known data
    A ball with radius, R, and moment of inertia, I = 2/5 mR2 is rolling on a horizontal plane with angular velocity ω0. At point A the ball rolls onto an incline with the angle θ wrt the horizontal plane. No air resistance and ω0R = V, where V is the velocity of CM.

    What is the ball's angular velocity just after it has started rolling up the incline?

    2. Relevant equations


    3. The attempt at a solution
    I'm thinking it has something to do with the ball's angular momentum. However I'm not sure if I should be looking at the momentum with respect to a specific point or something else.
     
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  3. Nov 22, 2014 #2

    haruspex

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    Angular momentum is always in respect of some chosen reference point. For conservation of angular momentum to be used, that reference point must be either the mass centre of the object or a point fixed in the inertial frame.
    On contacting the incline, what force (impulse) occurs? Since you do not care what the magnitude of that is, it will be useful to choose a reference point about which it has no moment.
     
  4. Nov 22, 2014 #3

    Vir

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    So when the ball hits the incline it experiences a force through its CM which reduces the x component of the velocity and increases the y component. Reference point at CM seems reasonable. I can also assume that Δt of the impulse is so short that the force of gravity and the frictional force has not yet accelerated the ball downwards?

    If that is the case then the angular momentum about CM is conserved:

    \begin{equation} Iω_0 = Iω_1 \end{equation}

    Where do I go from here? I assume that ω1 < ω0, so I seem to be missing something in my equation.
     
  5. Nov 22, 2014 #4

    haruspex

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    How do you deduce it's through the centre of mass? (It isn't.)
     
  6. Nov 23, 2014 #5

    Vir

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    Alright, I guess then it would make sense to use the point where the ball makes contact with the incline as my reference point. So let's call this point's height over the horizontal plane x. Then I get:

    \begin{equation}
    m(R-x)V_0 + Iω_0 = m(R-x)V_1 + Iω_1
    \end{equation}

    Doesn't feel like I am any closer.
     
  7. Nov 23, 2014 #6

    haruspex

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    The right hand side is wrong. What direction is the velocity after hitting the ramp? How far is the point of contact from that?
    You know the slope of the ramp, so use some geometry to determine x.
    What is the relationship between angular velocity and linear velocity for a rolling ball?
     
  8. Nov 23, 2014 #7

    Vir

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    So, as the ball touches the incline, the angle between the two contact points is θ which gives me \begin{equation} x = R - R cos θ. \end{equation}

    wR = v

    I'm not exactly sure what you mean. The velocity is at an angle θ of the horizontal.
     
  9. Nov 23, 2014 #8

    haruspex

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    Right.
    You are taking moments about the point where it hits the ramp. If the new velocity is v1 parallel to the ramp, what is the moment of it about that point? Remember that you want the perpendicular distance from the reference point to the line of travel of the mass centre.
     
  10. Nov 23, 2014 #9

    Vir

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    Thank you for pointing this out, as it was the part I was missing.

    The new velocity vector is at a distance R from my reference point, which gives me the equation:

    \begin{equation}
    m R \cos θ V_0+ Iω_0 = mRV_1 + Iω_1
    \end{equation}

    which simplifies to

    \begin{equation}
    ω_1 = \frac{5}{7}ω_0(\frac{2}{5}+\cos θ).
    \end{equation}

    Thanks alot for the help! I guess the main point in this kind of problem is to pick a reference point such that dL/dt = 0 and remembering that |RxV| is only dependent on the component of R that is perpendicular to V.
     
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