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Chapter 5: Problem 66

A high-speed dental drill is rotating at $3.14 \times 10^{4} \mathrm{rad} / \mathrm{s}$ Through how many degrees does the drill rotate in 1.00 s?

Short Answer

Expert verified
Answer: The dental drill rotates approximately \(1.80 \times 10^{6}\) degrees in 1.00 second.

Step by step solution

01

Convert radians per second to degrees per second

To convert radians to degrees, we will use the following conversion factor: \(1 \mathrm{rad} = \frac{180}{\pi} \mathrm{degrees}\). Given the angular speed in radians per second, \(3.14 \times 10^{4} \mathrm{rad} / \mathrm{s}\), we will multiply it by the conversion factor. $$ \text{Angular speed in degrees per second} = (3.14 \times 10^{4} \mathrm{rad} / \mathrm{s}) \times \frac{180}{\pi} \mathrm{degrees} $$
02

Calculate the total degrees rotated in 1.00 second

To find the total degrees rotated in 1.00 second, multiply the calculated angular speed in degrees per second by the time (1.00 s): $$ \text{Total degrees rotated} = (\text{Angular speed in degrees per second}) \times (1.00 \mathrm{s}) $$
03

Evaluate the expression

Calculate the angular speed in degrees per second and the total degrees rotated: $$ \text{Angular speed in degrees per second} = (3.14 \times 10^{4} \mathrm{rad}/\mathrm{s}) \times \frac{180}{\pi} \mathrm{degrees} \approx 1.80 \times 10^{6} \mathrm{degrees}/\mathrm{s} $$ $$ \text{Total degrees rotated} = (1.80 \times 10^{6} \mathrm{degrees}/\mathrm{s}) \times (1.00 \mathrm{s}) \approx 1.80 \times 10^{6} \mathrm{degrees} $$ So, the high-speed dental drill rotates approximately \(1.80 \times 10^{6}\) degrees in 1.00 second.

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Most popular questions from this chapter

A curve in a stretch of highway has radius \(R .\) The road is unbanked. The coefficient of static friction between the tires and road is \(\mu_{s} .\) (a) What is the fastest speed that a car can safely travel around the curve? (b) Explain what happens when a car enters the curve at a speed greater than the maximum safe speed. Illustrate with an FBD.
A spy satellite is in circular orbit around Earth. It makes one revolution in \(6.00 \mathrm{h}\). (a) How high above Earth's surface is the satellite? (b) What is the satellite's acceleration?
Grace, playing with her dolls, pretends the turntable of an old phonograph is a merry-go-round. The dolls are \(12.7 \mathrm{cm}\) from the central axis. She changes the setting from 33.3 rpm to 45.0 rpm. (a) For this new setting, what is the linear speed of a point on the turntable at the location of the dolls? (b) If the coefficient of static friction between the dolls and the turntable is \(0.13,\) do the dolls stay on the turntable?
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