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Total linear momentum of swinging bat

  1. Apr 29, 2013 #1
    1. The problem statement, all variables and given/known data

    A bat for use in a ball game has total mass M = 1:5 kg and total length ` = 1 m;
    assume it can be approximated by a thin uniform rod of length `. What is its radius
    of gyration k[itex]_{0}[/itex] about an axis perpendicular to the bat and through its centre of
    mass? A player grips the bat and swings it about an axis 0.8m above the top of the
    handle and perpendicular to the bat (see diagram); what is its moment of inertia
    about the rotation axis?

    While swinging the bat, the player strikes a ball with the bat vertical (see diagram).
    If the bottom end of the bat is moving horizontally at 10 ms1 at the moment of
    impact,find at that instant (i) the angular velocity of the bat, (ii) the angular
    momentum of the bat about the rotation axis, (iii) the bat's kinetic energy and (iv)
    its total linear momentum.

    The player has been told by a coach to aim to strike the ball in such a way that it
    needs no impulse from his hands at the moment of collision. Suppose the ball strikes
    the bat a distance d below the centre of mass (see diagram). Assuming that the
    impulse J from the ball is perpendicular to the bat, find in terms of J the changes
    during the collision in both the angular velocity of the bat and the linear velocity
    of its centre of mass. If the player grips the bat a distance d[itex]_{grip}[/itex] above the centre
    of mass, show that the point of grip suffers no sudden change in velocity only if
    d = k[itex]^{2}_{0}[/itex]/d[itex]_{grip}[/itex]
    . If the player grips the bat 0.1m below the top, use this result to find
    the value of d he should aim for.

    2. Relevant equations

    p =mv
    L = Iω
    K =1/2 * Iω[itex]^{2}[/itex]
    v= ωr
    I = mk[itex]^{2}[/itex]
    I = I[itex]_{0}[/itex] + Md[itex]^{2}[/itex]

    3. The attempt at a solution

    I'm stuck on part (iv), I don't know how to find the total linear momentum.

    What I have done so far is tried using p=mv, with v = 10 but it gave a wrong answer. How can a rotating object have linear momentum?

    The answers p = 10.83 kgms[itex]^{-1}[/itex]
     

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    Last edited: Apr 29, 2013
  2. jcsd
  3. Apr 29, 2013 #2

    SammyS

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    attachment.php?attachmentid=58347&d=1367271639.png

    It might help if you would show your answers for the earlier questions.


    Where is the center of mass for the bat?

    What is the velocity of the center of mass of the bat?
     
  4. Apr 29, 2013 #3
    Radius of gyration about axis through centre of mass: 0.29m; moment of inertia about actual rotation axis 2.66 kg m2.
    (a) Angular velocity 5.56 rad s−1;
    (b) Angular momentum 14.78 kg m2 s−1;
    (c) Kinetic energy 41.05 J;
    (d) Total momentum 10.83 kg m s−1.

    The centre of mass of the bat is at the mid point of the bat.

    I don't know its velocity, never have I come across a question where you needed to find the Vcm for a single body.
     
  5. Apr 29, 2013 #4

    SammyS

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    So, how did you find the linear momentum?
     
  6. Apr 29, 2013 #5
    I took a guess and tried using the horizontal velocity of 10m/s but that's incorrect as you can see from the answers.

    So to find the total linear momentum I need the Vcm?
     
  7. Apr 29, 2013 #6

    SammyS

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    How far is the CM from the axis of rotation?
     
  8. Apr 29, 2013 #7
    0.8+0.5 = 1.3m
     
  9. Apr 29, 2013 #8

    SammyS

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    Use that together with the angular velocity to find VCM .
     
  10. Apr 29, 2013 #9
    Yes, got it now. Thanks

    Vcm = 1.3 * 10/1.8

    P = 1.5 * Vcm = 10.83 Ns
     
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