Hello, turkeyy!

As a cable is pulled from a large spool, it unwinds in layers of constant radius.

If the cable is pulled at a steady 6 ft/sec, use the equation $\displaystyle s\:=\:r\theta$

to find how fast (radians/sec) the spool is turning when the radius is 1.2 ft. Code:

* * *
* *
* *
* *
* O *
* * *
* | * *
| θ *
* 1.2 | * *
* | o Q
* | *
* o * → → → → → →
P 6 ft/s

The cable is pulled horizontally at 6 ft/sec.

Then point $\displaystyle P$ is moving around the circle at the same rate.

We have: .$\displaystyle r = 1.2$

$\displaystyle \text{Let }\,s \:=\:\text{arc(PQ)}\text{, then: }\,\frac{ds}{dt} \,=\,6$

From $\displaystyle s \:=\:r\theta$, we have: .$\displaystyle s \:=\:1.2\,\theta$

Differentiate with respect to time: .$\displaystyle \frac{ds}{dt} \:=\:1.2\,\frac{d\theta}{dt} \quad\Rightarrow\quad \frac{d\theta}{dt} \:=\:\frac{1}{1.2}\,\frac{ds}{dt}$

Since $\displaystyle \frac{ds}{dt} = 6$, we have: .$\displaystyle \frac{d\theta}{dt} \:=\:\frac{1}{1.2}(6) \:=\:5\text{ rad/sec}$