A printing press has a roller 23 inches in diameter. A point on the rollers surface moves at a speed of 90 feet per second. What is the rollers angular speed? Round to the nearest hundredth.
I did c=d*π which got me c=23π
then I did $90/23π =c/2π$ which after cross multiplying got me 77.24038227. However the answer is 93.91 radians per second.
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$\begingroup$The circumference of the roller is $2 \pi r = 23\pi$ inches as you have it.
The linear velocity $v$ of a point on the surface of the roller is $90$ feet per second, or $1080$ inches per second.
So the roller has a period $T$ of $\frac{23 \pi}{1080} \approx 0.06690428791$ seconds per revolution. Since there are $2 \pi$ radians per revolution, this becomes $0.06690428791/(2 \pi) \approx 0.01064814814$ seconds per radian.
The angular velocity $\omega$ is the reciprocal of this: $1/0.01064814814 \approx 93.9130434783 \approx 93.91$ radians per second.
Or, an even easier relationship (now that I've worked it all out the long way!) is this:
$$v = r \omega \implies \omega = v/r = \frac{1080 \text{ in/sec}}{11.5 \text{ in}} \approx 93.91 \text{ rad/sec}.$$
$\endgroup$ 3 $\begingroup$First, the linear speed is feet/sec while the diameter is inches, so you need a factor $12$ that gives you one revolution in $(23 \pi)/(90\cdot 12)$ seconds. Is your angular rate in degrees/second? Another conversion.
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