I am going to design a bowling machine. In this design, I'm going to use two motors that are being placed vertically and their shaft are attached with two horizontal identical wheels with a space of ball in between these two wheels. enter image description here

The wheen diameter is 360mm and the weight is 3kg each. the ball is 75mm dia and the weight is 160 grams. I need to calculate torque and rpm required to throw the ball at 170km/h velocity. The cricket pitch length is 22 meter = 2200mm

Required Torque:

The required acceleration is a need to throw the ball at 170km/h means 47.22m/s

$v^2=u^2+2fs$ where intial velocity is $u=0$

So, $f = v^2/2s$

accleration $f = 50m/s^2 $

The force required $ F = ma$ where $m =$ total weight of two wheels + wight of ball $= 6+0.160$ $= 6.016$ kg.

So, $F = 308N$ i.e, $154N$ in each wheel.

Assume rubber to leather kinetic friction $u = 0.4$ it's not actually

kinetic Frictional force $Fr = 1.4 * 154 = 215.6$

Required torque $Ta = F * r$ $= 215.6 * (0.36/2) = 36.8 Nm$

Required RPM:

If the linear velocity of the ball is $v=47.22m/s$ The velocity of the ball is an average velocity of two-wheel $v = (v1+v2)/2$

for the maximum velocity of the ball with no spin, the velocity of two-wheel must be the same.

so, $v1 = v = v2$

ao, the angular velocity of each wheel should be $v = d × w × 0.001885$

So, $w = v/(d * 0.001885)$RPM


Please let me know I am right or wrong?

  • $\begingroup$ What happens if you calculate from an energy perspective instead? Do you get a similar answer? This is usually a good method to 'sense check' numbers. $\endgroup$ – Jonathan R Swift Nov 4 '19 at 12:19
  • $\begingroup$ I haven't tried an energy perspective way. This way seems easy to me. $\endgroup$ – Prayuktibid Nov 4 '19 at 12:21
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    $\begingroup$ Your calculations have errors. Can you explain how you got from "So, $f = v^2/2s$" to "accleration $f = 0.5m/s^2$" You have said that $v=47.22\text{ m/s}$, so $v^2=2300$. What value of $s$ have you used, and where has this come from? $\endgroup$ – Jonathan R Swift Nov 4 '19 at 13:03
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    $\begingroup$ "F=ma where m= total weight of two wheels + wight of ball ..." That's true if you are throwing the wheels. In your case you are not and (1) you would use the moment of inertia in your calculations, not the weight and (2) you would most likely have the wheels up to speed before the chucking the ball in. That way you slowly transfer energy into the wheels as it accelerates with a low powered motor and almost instantly from the wheels to the ball. Major rethink required, methinks. $\endgroup$ – Transistor Nov 4 '19 at 20:51
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    $\begingroup$ Right, so edit your question then. Forget about the motor for now and instead figure out how much kinetic energy is to be transferred from the wheels to the ball. You know the exit velocity so therefore you know the wheel speed after the throw. You need to work out the wheel speed decrease during the throw due to energy transfer. Once you have that you can work out the required speed before the throw. $\endgroup$ – Transistor Nov 5 '19 at 10:16

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