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Energy from torque in a ball launcher

  1. Aug 16, 2012 #1


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    hi, i was given

    [itex]\frac{1}{2}[/itex]mv[itex]^{2}[/itex] = 2[itex]\tau[/itex]Δ[itex]\theta[/itex]

    for working out kinetic energy transferred to a ball in a tennis ball launcher type scenario, where two wheels are counterrotating and the ball goes between them.

    Is this right? is the speed of the wheels not taken into account?

    If the drive motor was geared down to give the wheels outrageous torque at an incredibly low rpm then surely the ball would just fall out of the end with no kinetic energy at all.
  2. jcsd
  3. Aug 16, 2012 #2

    Simon Bridge

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    It the ball leaves at the same speed as the rim of the wheel, then [itex]E_K = \frac{1}{2}m r \omega[/itex] ... where [itex]\omega[/itex] is the angular speed of the wheels.

    If may not: - if the wheel provides a torque to the ball for an angular distance [itex]\Delta \theta[/itex] then the change in kinetic energy would be [itex]\Delta E_K = \tau \Delta \theta[/itex] (rotational equiv. for W=Fd right?) The speed of the wheel is then already accounted for in the kinematics.
  4. Aug 17, 2012 #3
    The ball will not move any faster than the drive wheel of the launcher; so once the ball reaches the same angular speed of the drive wheel it will not experience any more angular acceleration and the torque on it would reduce to zero. Remember, torque is the rate of change of angular speed.

    A very low speed in a friction style launcher will mean that the ball will reach it's top velocity in a very short angular distance.
  5. Aug 17, 2012 #4


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    I thought torque was supposed to be a force, if its rate of change of angular speed then its an angular acceleration?

    so i can't just say the torque on the ball is the same as the stall torque of the drive motor?

    is the original equation right then, and i just need a different value for torque, or do i need to change it to account for the angular speed of the wheel and therefore maximum speed of the ball?
  6. Aug 17, 2012 #5

    Simon Bridge

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    You don't have to apply a torque to the wheels to keep them moving at a constant angular speed ... (except to overcome friction). Without an applied torque, inserting the ball would just slow the wheels down (kinetic energy to the ball has to come from the wheel). To keep the wheels at a constant speed while the ball is inserted, a torque has to be applied.

    If the torque is applied for the short time T that the ball is between the wheels, then the energy imparted to the ball in that time, by each wheel, is [itex]\tau \omega T = \tau\Delta\theta[/itex] ... now do you see?

    Since the balls are a constant size, then [itex]\Delta\theta[/itex] is likely to be the same no matter what the angular speed is. That's just how far the wheels rotate while the ball is between them.
  7. Aug 17, 2012 #6
    Apologies, I meant to say that torque causes angular acceleration, which is the rate of change of angular speed.
  8. Aug 17, 2012 #7


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    In this case, the torque is that required to maintain or restore the angular speed of the wheels while they accelerate the ball. The problem simplifies things by assuming the torque is constant (I think the work done could be expressed as the integral of 2 τ(θ) dθ ).
    Last edited: Aug 17, 2012
  9. Aug 24, 2012 #8


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    ok, thanks guys now i get what torque im supposed to put in the equation, but i don't know how i would work that out without doing it backwards using the energy given to the ball and then finding the torque.
    I suppose you need to know the the energy transferred to the ball to know the torque required to replace it.

    I cant just use the speed of the wheels as the speed of the ball because obviously the wheels will slow down as the ball is launched because of the extra torque.

    I have the speed of the wheels, i just want to find the speed of the ball.
    Would it help if i figured out the inertia of the wheels?
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