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Bowling ball problem

  1. Dec 20, 2014 #1
    A bowling ball is released with no rotation (linear KE only) at a given speed, the friction between the ball and the lane creates a torque which tends to rotationaly accelerate the ball, robbing it of linear KE and reducing the linear speed until synchronous speed is reached.
    My question is :
    The rotational torque i got from : m * g * µ * r
    m = mass
    g = local gravity rate
    µ = friction co-efficient
    r = ball radius
    Is all the friction force absorbed in rotating the ball, or does part of it decelerate the ball directly ?
    Comments please
    Thanks
    Dean
     
  2. jcsd
  3. Dec 20, 2014 #2

    Nugatory

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    Try applying conservation of energy to this problem. Before the ball touches the floor it has a certain amount of kinetic energy from its forward motion and a certain amount of kinetic energy (zero, because it hasn't started rotating yet) from its rotation. Once it is rolling across the floor in steady state it has the same total amount of kinetic energy, but now the rotational component is non-zero. What does this mean for its forward motion?
     
  4. Dec 22, 2014 #3
    Its diminished, im aware of the translation from linear to rotational, my problem is that : does all of the the force from sliding friction rotate the ball ?
    Ive attached the word sheet i put together for reference.
    Thanks for your time.
    Dean Barry
     

    Attached Files:

  5. Dec 22, 2014 #4

    jbriggs444

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    Force from the sliding friction acts both as a torque that spins the ball and as a linear force which decelerates the ball. The full force acts in both manners. None of it is "used up" for one purpose and thereby made unavailable for the other.
     
  6. Dec 22, 2014 #5
    ive sent the excel sheet, sorry, heres the word doc.
     

    Attached Files:

  7. Dec 22, 2014 #6
    So, the friction force = m * g * µ = 10.791 N , i attributed all of that to the torque rotating the ball, what do i do now ? half to rotation and half to linear deceleration ?
    Thanks
    Dean
     
  8. Dec 22, 2014 #7

    jbriggs444

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    All to rotation and all to deceleration.
     
  9. Dec 22, 2014 #8
    Dean,
    In real life some minuscule amount of the kinetic energy of a bowling ball is converted into heat energy through friction. Similarly energy is expended on pushing all those troublesome air molecules out of the way like trillions of little bowling pins. In the grand scheme of things, though, these are not worth considering.
    Since we are ignoring the lesser effects:
    All of the energy given to the rotation is taken from the linear motion.
    All of the energy taken from the linear motion is given to rotation.

    As jbriggs444 stated, the force you calculated is applied equally and oppositely to each.
     
  10. Dec 22, 2014 #9

    Nathanael

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    How can you say the kinetic energy is the same?

    I can understand for, say, a ball rolling down an inclined plane. The point of contact (where the force of friction is acting) is stationary, and thus no work is being done by friction (right?).

    But in this case, the point of contact is sliding along the ground until it reaches a steady state, so wouldn't you have to account for energy lost to friction?
     
  11. Dec 22, 2014 #10

    Nugatory

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    You do if you want to calculate the exact final speed, but the correction is small. You don't if you just want to answer OP's question.
     
  12. Dec 22, 2014 #11

    jbriggs444

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    Energy conservation is the wrong way to solve this problem. Kinetic energy is not conserved. Not even approximately.

    An approach based on angular momentum and linear momentum is appropriate. First, pick a reference axis about which to compute angular momentum. Two choices suggest themselves. Either choice should allow you to determine a ratio k between the rate at which linear velocity is reduced and the rate at which tangential velocity is increased.

    That gives you one equation: linear-velocity-reduction = k x tangential-velocity-increase

    When the final stable state has been attained, the ball's linear velocity will match its tangential velocity. That gives you a second equation:

    starting-linear-velocity - linear-velocity-reduction = starting-tangential-velocity + tangential-velocity-increase.

    You know the starting linear velocity and the starting tangential velocity. So that's two equations and two unknowns. Solve them.
     
  13. Dec 23, 2014 #12
    Right gents, im sticking with the conservation of KE applies for now.
    Calculate the friction force (f).
    Then :
    1) calculate linear deceleration using a = f / m
    2) calculate deceleration due to energy transfer
    add the above

    Thanks for all your input
    Dean Barry
     
  14. Dec 23, 2014 #13

    ehild

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    The kinetic energy is not conserved as there is sliding friction.
    Treat translation and rotation of the ball separately.
    Your first steps are right: The force of friction is f=μmg.The translation of the ball decelerates : a = -f/m . How does the velocity change with time?
    The force of friction acts with a torque τ=rf around the centre of the ball and accelerates rotation. The angular acceleration is α=τ/I (I is the moment of inertia)
    How does the angular velocity change with time?
    When pure rolling, the angular velocity and the translational velocity of the CM are related, how? From that you get the time when the ball starts to roll. Knowing the time, you get the velocity and angular velocity.
     
  15. Dec 23, 2014 #14

    sophiecentaur

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    I think this is basically a flawed idea. If any of the energy is transferred from linear to rotational by friction then there needs to be Work Done, turning the ball and slowing its linear velocity. That work will, I think, also involve work against the friction force. I would go so far as to say that the energy dissipated through the friction will be the same as the energy transferred to rotational KE.
    There is an electrical analogue to this when an uncharged Capacitor is connected across a charged one. (an old chestnut that students are often presented with) The total electrostatic energy is always less than the original, irrespective of the actual value of the resistance (of which there must always be some). For two equal value capacitors, the energy dissipated is actually half. So I would say that the actual coefficient of friction would not affect the bowling ball result. Once slipping has stopped, the tangential speed is the same as the linear speed. The simplified, ideal model with no slippage will not actually work as there will be an unacceptable infinite force at first contact.
     
  16. Dec 24, 2014 #15
    im thinking maybe the friction force (and therefore torque) is maximum at the beginning and ramps down to zero at synchronous speed, deceleration not being at a constant rate.
    im going to sit with this for now, thank you all for your help
    dean
     
  17. Dec 24, 2014 #16

    sophiecentaur

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    When dynamic friction is the same as static friction, this is not the case and the force would be constant, giving linear acceleration. But your point is a good indication that the Energy is not conserved (a force times a distance) so that makes doing it your way much harder. As with most interaction problems, it is much easier to stick with conservation of Momentum for your calculations. It simplifies things so much that it almost seems like cheating!.
     
  18. Dec 24, 2014 #17
    Thank you, i will ponder this.
    Dean
     
  19. Dec 24, 2014 #18
    return jan 5th
     
  20. Dec 24, 2014 #19

    sophiecentaur

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    Taking the MI of a solid sphere as 2Mr2/5, it looks like the final linear velocity of the ball is 0.7 of the original. and about 30% of the original energy is lost to friction. I won't show the workings until someone comments / agrees with me. Wotcha think?
     
  21. Dec 24, 2014 #20

    Nathanael

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    I agree. I find the final velocity and the final energy to be 5/7 of their respective initial values (using the MI of a solid sphere).

    Interesting idea! I would have never noticed this without doing the math.
     
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