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In an attempt to model the cassette tape spooling action, I'm hoping to solve this problem.

Required feed speed 5cm/second. Two Spindles with starting radius of S1: 0.8cm (empty spool) S2: 1.8cm (full spool)

The spooled tape thickness is unknown but can be assumed to be around 15micrometer if needed. The time to complete the spooling from one spool to the other spool is required to be 45 minutes.

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Are you trying to find the correct angular speed to drive the full spindle such that the tape passes between them with a fixed speed of 5 cm/s? –  Alex Troesch Nov 28 '10 at 4:31
    
Yes. I'm also trying to find the correct angular speed for the feed intake. –  Azeworai Nov 28 '10 at 4:32
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1 Answer

Suppose that the feed spindle has outer radius $r_f(t)$ and the intake spindle has outer radius $r_i(t)$, similarly let the spindle's angular velocity be given by $\omega_f(t)$ and $\omega_i(t)$. Note that both $r_f$ and $r_i$ lie in the interval $[r_0, R_0]$, where $r_0= .8$ cm and $R_0 = 1.8$ cm.

The only constraint is that the tape must pass between the two spindles at a constant rate of $s = 5$ cm/s. For the tape to remain taught under this condition the tangential velocity of each spindle must be $s$. Thus,

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

So

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

for either spool. So now we simply need to find a description of the outer radius of the spindle. Consider that the tape has thickness $a \ll r_0$ and width $w$, thus the volume of tape transfered per unit time is

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

and since $V = \pi w r(t)^2$ then we also have

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

and then by equating these expressions and solving the IVP for the feeding spindle yields,

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

Similarly, the IVP for the intake spindle yields

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

From those expressions it is easy to see that the proper speeds to drive the spindles are approximately

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

and

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

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Thanks Alex that is what I needed. To make sure my understanding is correct. If i take r0 as 1.8, s=5cm/s, a=15x10^-6, t = 0 and pi=3.142. I get a spindle angular speed at 2.778 degrees/second. Is this correct? Or am I working in radians/second? –  Azeworai Nov 28 '10 at 11:00
    
Based on the information in your question $r_0 = .8$ cm and $R_0 = 1.8$ cm. Also, these expressions are in radians per second since $v_t = r\cdot\omega$ only holds in radians. –  Alex Troesch Nov 28 '10 at 18:30
    
Ah i got mixed up on the parameters. Cheers. –  Azeworai Nov 29 '10 at 9:55
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