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Bat Speed vs Bat weight

  1. Jun 23, 2011 #1
    Ok So I have been wondering this for awhile.

    Could someone help me find an equation to figure this out.

    I play slowpitch softball and was curious about swing speed and the weight of the bat.
    If we keep the mass of the ball constant (.198 kg), and the pitch speed constant (8.94 m/s), and have 4 different bats (22oz-24oz-26oz-28oz):(.624kg .680kg .737kg .794kg) that correspond with swing speeds (fastest swing speed for the lightest bat) lets call them 60mph-66.67mph-73.34mph-80mph:(26.82m/s-29.80m/s-32.79m/s-35.76m/s). so....
    .624kg@35.76m/s - .680kg@32.79m/s - .737kg@29.80m/s - .794kg@26.82m/s.

    So I guess im trying to find the velocity of the ball after the collision with the bat (v1a).

    At some point with a higher swing speed does the lighter massed bat become infective vs the higher massed bat even though the swing speed is higher??

    Does anyone have an equation for this?? would it be...
    (m1v1b + m2v2b = m1v1a + m2v2a) ????????

    Does anyone know how to find the velocity of the bat after collision.???
    .....is there a force equation for this?

    m1: mass of ball (.198kg)
    v1b: Velocity Of ball before collision (8.94m/s)
    m2: mass of bat
    v2b: velocity of bat before collision
    v1a: velocity ball after collision
    v2a: velocity bat after collision ((??)another equation??)

    please help, thanks
     
  2. jcsd
  3. Jun 23, 2011 #2

    sophiecentaur

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    That's about the basics of it.
    There is another factor, though. The speed that the bat ends up at will depend upon its mass and your 'strength' (i.e. power output) and how much you can accelerate the bat before it makes contact.
    The optimum will depend on the particular batter.
    I remember reading, many years ago, a New Scientist article on the topic. I seem to remember that Babe Ruth had a very heavy bat.
     
  4. Jun 23, 2011 #3
    Welcome to Physics Forums.

    I see that you posted this previously and got no responses. I don't have an answer, but I might have some useful speculation.

    The approach you suggested ignores the fact that the bat is rotating approximately about the point where it is gripped. As a result the point along the bat where the collision occurs is a factor.

    A Google search for the physics of sweet spots might help you frame the problem. http://farside.ph.utexas.edu/teaching/301/lectures/node107.html" [Broken] focused n the sweet spot being where to hit the ball such that the least impulse is conveyed to the batter's hands. It relates (in equation 394) the impulse to the hands and the impulse on the ball. That analysis might give you insight into how to take a wider range of factors into account.

    http://www.physics.usyd.edu.au/~cross/baseball.html" [Broken] presents experimental results and tells you how to find the velocity of after being hit if you have the "apparent coefficient of restitution". That site also includes stuff on the effects of bat weight on swing speed.
     
    Last edited by a moderator: May 5, 2017
  5. Jun 23, 2011 #4

    turbo

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    Another factor is that the bat is not "hinged" to rotate around the center of gravity of the batter. That is a simplification you can make to make the numbers easier to crunch. A good batter can time the swing such that the bat is sweeping through the location of the ball at max speed during the time of contact.

    A similar effect is seen in golfing. You don't have to be the largest or strongest person to hit the longest drive. You have to have a feel for when to "release" and let that club pivot so the head sweeps through the ball's position at max speed at the time of contact.
     
  6. Jun 24, 2011 #5
    hi man pretty interesting problem.. i have derived equation in simpler way

    you have given example of impulsive force that is large force acting on short interval of time
    equation for this force is
    J=F*t
    here you are taken time as 1 second therefore
    J=F
    Now let consider two case
    1st case- force exerting by bat on ball
    J=F1
    F1=m1(v1a-v1b)
    case 2: let ball exerting force on bat
    J=F2
    F2=m2(v2a-v2b)
    according to newton 3rd law
    F1=-F2
    therefore m1(v1a-v1b)=-m2(v2a-v2b)
    m1\m2=-(v2a-v2b)\v1b-v1a
    therefore velocity of bat after collision
    - v2a=m1\m2(v1b-v1a)-v2b
    v2a=-m1\m2(v1b-v1a)+v2b
    it is the equation for velocity of bat after collision
    substitute your numerical value you will get the answer
     
  7. Jun 24, 2011 #6
    ok but here is my problem,
    I am looking for ball velocity after collision to see how hard the ball is hit/how far it goes.

    If I dont have the ball velocity after collision I cant find the bat velocity after collison... right?
     
  8. Jun 24, 2011 #7
    skanda9051,

    You need not make the assumption that t=1 second. The times will drop out of the expression when you set J1=-J2.

    I am assuming you are keeping the convention arsenal1021 established and m1 and m2 are the masses of the ball and bat, respectively. This would work if the bat and the ball were a closed system. You need, though, to take into account the fact that the bat is effectively anchored at one end and is rotating. Experience probably tells you that where along the length of the bat the hit occurs matters, so the equation to predict v2a has to include a variable for that
     
  9. Jun 24, 2011 #8
    hi sorry for not mentioning before..
    Ball velocity can be find easily
    impulsive force that is J=F*t
    here you have taken time 1 second therefore
    J=F
    force acting on the ball is given
    F=m1*a
    here acceleration of ball is given a=35.76
    calculate you will get force then
    you know force is given by change of momentum
    F=m1(v1a-v1b)
    you already know velocity of ball before collision v1b=(8.94m/s)
    and mass of ball m1=(.198kg)
    and above you will find the force substitute on F
    therefore final equation is
    velocity ball after collision that is v1a=F\m1+v1b
    substitute all the known value you will get velocity ball after collision
     
  10. Jun 24, 2011 #9

    sophiecentaur

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    No one seems to have considered how near the collision is to being elastic. This could have a huge effect on the situation and, without having a representative value for the coefficient of restitution, you can't calculate anything with certainty.
     
  11. Jun 24, 2011 #10
    hi we are taking here time as constant so how can time changes.......... and also i never played soft ball so i don't have idea where to hit and other things........
     
  12. Jun 24, 2011 #11
    ya you are right elasticity of bat and ball depends.. but we are considering ideal case but experimentally if you want calculate than it is very difficult...........
     
  13. Jun 24, 2011 #12

    sophiecentaur

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    The time can't seriously be 1 second. Time, here, refers to the time taken for the impact to take place and is the same for the momentum changes of both bat and ball (a few milliseconds, probably - but it's not relevant unless, possibly, you are trying to consider the losses as the ball deforms during contact)
     
  14. Jun 24, 2011 #13

    sophiecentaur

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    I agree - but, if we're ignoring this factor then why bring in the angular momentum / sweet-spot issue? You get a very good idea by reducing the whole thing to a linear elastic collision and straight transfer of momentum. (Which is what's described in the OP)
     
  15. Jun 24, 2011 #14
    so taking integration on that equation can solve problem
     
  16. Jun 24, 2011 #15

    sophiecentaur

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    Why use integration?
    To find the final velocities of both bat and ball, all you need to remember that, in a perfectly elastic collision, the relative velocity of separation is just the negative of the relative velocity of approach. (I don't think that was mentioned yet.) Combine that with the equation in the OP. Work out one and you have the other - two simultaneous equations with two unknowns --- simples. Force is not involved in the solution.
     
  17. Jun 24, 2011 #16
    The situation is one where the ball hitting the bat at the center of percussion causes a pure change in angular momentum of the bat and maximizes the change in momentum of the ball.

    Hitting the bat at ther sweet spot or centre of percussion transfers more momentum to the ball than hitting at the center of gravity. The centre of percussiion of a straight rod has a velocity 30% greater than its centre of gravity. The ball will thus have a greater initial speed from the impact or rebound. For a straight rod the center of gravity lies at the midway point - the center of percussion lies 66.6% from the batters end.

    For a slow pitch bat the two may coincide or be very close together in which case you can use direct translational momentum. If not you have to perform calculations with the ball hitting at the centre of percussion and change in angular momentum of the bat.

    As a very rough estimate of the error involved, for every 10ft/sec difference in change in velocity of the ball and second of flight time is a 10 foot distance in travel. If the ball's flight time is 5 seconds that is 50 feet - the game winning home run or the ninth inning 3 rd out. ( 7 innings in slow pitch I vaugely think ) Very crud error calculation by the way.
     
  18. Jun 24, 2011 #17

    sophiecentaur

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    I should have thought that the crude calculation, involving just translational momentum should be sorted out for a start - and it doesn't seem to have been, yet. Then you'd have to look at the effects of inelasticity and rotation later (which effect is, in fact greater?). It's no good trying to run before you can walk. I am a pragmatist and a 'near enough' result is a significant goal to aim for, initially.
     
  19. Jun 24, 2011 #18
    I agree with you. do the 5 minute calculation and tweak it along the way.

    Your statement that the relative velocity of separation is just the negative of the relative velocity of approach to find a solution to the eqations benefitted greatly.

    My point was the amount of error if the analysis is short changed and what ( maybe) to strive for in the end.

    The discussion is a good one and more ideas are out there I am sure.

    My question is how much force, power does the batter himself put into the velocity of the ball, as he is still pulling the bat during the brief interval when the ball and bat make contact.
     
  20. Jun 24, 2011 #19

    turbo

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    And that will be tricky to model. Like in golf, you can start the swing with your dominant upper-body muscles, but the follow-through is going to fall on the off-side arm. In other words, if you are right-handed, you'll be making the follow-though primarily by pulling through the remainder of the swing with your left.
     
  21. Jun 24, 2011 #20
    ok well this is just way over my head now....

    saying its
    - the same bat just different weights (like how they sell them)
    - same bat length
    - the ball hits at the same sweet spot
    - same ball
    - same pitch speed or even hit it off a tee
    - same angle of swing
    - Only difference is higher swing speed generated from a lighter bat

    Do you think "generally" that using a lighter bat say 22 oz could potentially hit the ball further than the same bat that is 26 or 28 oz because someone like me could generate a higher swing speed with the lighter bat.???
     
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