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I'm designing a ball launcher that uses 2 wheels to accelerate a ball: enter image description here Each wheel is driven by its own motor.

I selected a desired speed for the balls to be launched at, $v_{\text{ball}}$, and so I know the required angular velocity for the wheels, $$ \omega_{\text{wheel}} ~{\equiv}~\dot{\theta}_{\text{wheel}} ~=~\frac{v_{\text{ball}}}{r_{\text{wheel}}} \,,$$where $r_{\text{wheel}}$ is the radius of a wheel.

I also know work done and time needed so I can calculate power needed, but I have two values of work (and time changes depend on it):

  • $W_{\text{no-ball}}\equiv$ the work when no ball is touching the wheels;

  • $W_{\text{ball}}\equiv$ the work when one ball is touching the wheels.

Question: In order to find how much torque is required, which of the following formulas should I use?

$$ \begin{array}{rccrc} && \hspace{1.5cm} && \\ \left(1\right) & \large{\tau = I \cdot \alpha} && \left(2\right) & \large{t \cdot \tau = I \cdot \omega} \\ \phantom{|} \\ % \left(3\right) & \large{p = \tau \cdot \omega} && \left(4\right) & \large{\tau = \frac{W}{t \cdot \omega}} \end{array} $$ $$ \begin{array}{l} \text{where:} \\ \phantom{\text{whe}} \begin{array}{lll} \tau~\equiv~\text{torque;} & t~\equiv~\text{time;} & p~\equiv~\text{power;} \\ W~\equiv~\text{work;} & I~\equiv~\text{moment of inertia;} & \omega~\equiv~\text{angular velocity;} \\ \alpha~\equiv~\text{angular acceleration.} \end{array} \end{array} $$ I think that I should use $\operatorname{Eq}{\left(4\right)}$ to calculate each torque required for 2 work values, $W_{\text{no-ball}}$ and $W_{\text{ball}}$.

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    $\begingroup$ Do you have a diagram of the ball thrower that you're making? $\endgroup$ – Nat Apr 29 '18 at 1:28
  • $\begingroup$ It is not the torque that throws the balls, but the angular momentum of the two wheels that transfer momentum to the ball. Well in reality, when the ball deforms it might need some torque to squeeze between the wheels and none of the equations above will help you. $\endgroup$ – ja72 Apr 29 '18 at 1:43
  • $\begingroup$ @ja72 thanks for your answer . yes you are right . here there is no load so no torque load . but torque needed to make rollers starts to spin. also i want calculate torque to choose motor . $\endgroup$ – Ahmed elmenshawie Apr 29 '18 at 1:51
  • $\begingroup$ @Nat i have uploaded a photo for it $\endgroup$ – Ahmed elmenshawie Apr 29 '18 at 1:56
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The torque to spin up the wheels depends on the friction and the amount of time you require.

IF you want to reach $\omega$ in $t$ time, then you need average acceleration $\alpha = \frac{\omega}{t}$.

The torque needed for this acceleration is $\tau = I \alpha = \frac{I \omega}{t}$

So it looks like equation (2) would be appropriate.

But you have to consider friction also. The torque needs to be more. For example if free spinning from $\omega$ it takes $t_{f}$ to slow down and stop, the average friction torque is $\frac{I \omega}{t_f}$ which needs to be added to the spin-up torque

$$ \tau = \frac{I \omega}{t} + \frac{I \omega}{t_f} = I \omega \left( \frac{1}{t} + \frac{1}{t_f} \right)$$

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