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Tennis ball launchers

  1. May 20, 2005 #1
    Hey not technically a college student but i do have a problem aimed at this sort of level. I'm attempting to build a motorised tennis ball launcher using two counter rotating wheels. they are approx 15cm dia and compress a standard tennis ball two just over 3cm height. unfortunatly the motors are required to be DC fed from a 12V supply and i can't find any specs for what sort of speed or torque I'm gonna need. please somebody out there must have had a similar problem. oh an my only real constraint is budget, limited to about £100 to get both motors. you can e-mail responses to noahj1984@hotmail.com
     
  2. jcsd
  3. May 21, 2005 #2

    Andrew Mason

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    The principle is simple enough. The wheels supply a force to the ball over a small distance (the product of which is the energy imparted). The kinetic energy of the tennis ball is equal to the work done on the ball (Force x distance).

    The force on the ball is a little complicated because it is a combination of the torque of the motors and the compression of the ball (which causes the ball to push off against the motor wheels). Ultimately, though, the energy comes from the motors.

    [tex]\frac{1}{2}mv^2 = 2\tau\Delta\theta[/tex]

    where [itex]\tau[/itex] is the torque provided by each motor and [itex]\theta[/itex] is the angle of rotation of the motor while the motor wheel is in contact with the ball.

    Lets' suppose that [itex]\Delta\theta[/itex] is 90 degrees or [itex]\pi/2[/itex] radians.

    [tex]\frac{1}{2}mv^2 = 2\tau\pi/2[/tex]

    [tex]\tau = \frac{1}{2\pi}mv^2[/tex]

    [tex]v = \sqrt{\frac{2\pi\tau}{m}}[/tex]

    The mass of a tennis ball is .057 kg and if you want a speed of 30 m/sec, you would need a motor that produces a torque of:

    [tex]\tau = \frac{1}{6.28}*.057 * 900 = 8.2 Nm[/tex]

    which is about 6 foot pounds (pounds-force) of torque.

    AM
     
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