# another question:)

• Nov 7th 2008, 11:17 AM
turkeyy
another question:)
As television cable is pulled from a large spool to be strung from the telephone poles along a street,it unwinds from the spool in layers of constant radius.If the truck pulling the cable moves at a steady 6ft/sec (a touch over 4 mph),use the equation s=rθ to find how fast (radians per second )the spool is turning when the layer of radius 1.2ft is being unwound.
• Nov 7th 2008, 01:10 PM
Soroban
Hello, turkeyy!

Quote:

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