How do I compute the rapid angular rate of a pin offered a spooled distance and also called for feed rate?

In an effort to design the cassette tape spooling activity, I'm wanting to address this trouble.

Called for feed rate 5cm/second. 2 Spindles with beginning distance of S1 : 0.8 centimeters (vacant spindle) S2 : 1.8 centimeters (complete spindle)

The spooled tape density is unidentified yet can be thought to be around 15micrometer if required. The moment to finish the spooling from one spindle to the various other spindle is called for to be 45 mins.

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2019-12-02 02:53:22
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Answers: 1

Suppose that the feed pin has external distance $r_f(t)$ and also the consumption pin has external distance $r_i(t)$, in a similar way allow the pin is angular rate be offered by $\omega_f(t)$ and also $\omega_i(t)$. Keep in mind that both $r_f$ and also $r_i$ hinge on the interval $[r_0, R_0]$, where $r_0= .8$ centimeters and also $R_0 = 1.8$ centimeters.

The only restraint is that the tape has to pass in between both pins at a constant price of $s = 5$ cm/s. For the tape to continue to be educated under this problem the digressive rate of each pin have to be $s$. Hence,

$$s = r_f(t)\cdot\omega_f(t) = r_i(t)\cdot\omega_i(t).$$

So

$$\omega(t) = \frac{s}{r(t)}$$

for either spindle. So currently we merely require to locate a summary of the external distance of the pin. Take into consideration that the tape has density $a \ll r_0$ and also size $w$, hence the quantity of tape moved each time is

$$ \frac{dV}{dt} = \pm a s w$$

and also given that $V = \pi w r(t)^2$ after that we additionally have

$$ \frac{dV}{dt} = 2\pi r(t) \dot{r}(t) w$$

and afterwards by relating these expressions and also addressing the IVP for the feeding pin returns,

$$r_f(t) = \sqrt{R_0^2 - \frac{as}{\pi} t}.$$

Similarly, the IVP for the consumption pin returns

$$r_i(t) = \sqrt{r_0^2 + \frac{as}{\pi} t}.$$

From those expressions it is very easy to see that the correct rates to drive the pins are about

$$\omega_f(t) = \left(\sqrt{\left(\frac{R_0}{s}\right)^2 - \frac{a}{s\pi} t}\right)^{-1}$$

and also

$$\omega_i(t) = \left(\sqrt{\left(\frac{r_0}{s}\right)^2 + \frac{a}{s\pi} t}\right)^{-1}.$$

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2019-12-03 04:20:28
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