Question

equivalent ratios: 2π radians = 360° = 1 revolution = circumference 2πr \units\
example 5: the second - hand of a clock is 10.2 cm long. find the linear speed of the tip of the second hand in cm/s.
10.2·2π = 20.4π cm
\frac{20.4π}{60}=\frac{64.056}{60}=1.0676 cm/s
example 6: a 15 - inch diameter tire on a car makes 9.3 revolutions per second.
a. find the angular speed of the tire in rad/sec
9.3·2π = 18.6π rad/sec
b. find the linear speed of the car in in/sec
\frac{15}{2}·18.6π=\frac{279π}{2}
≈438.25 in/sec
example 7: the circular blade on a saw has a diameter of 7.25 inches and rotates at 4800 revolutions per minute.
a. find the angular speed of the blade in rad/sec
\frac{4800}{60}=80·2π
≈160π rad/sec
b. find the linear speed of the saw teeth (in ft/sec) as they contact the wood being cut.
\frac{7.25}{2}·160π=\frac{1160π}{2}
≈1822.12
example 8: a woman is riding a bicycle whose wheels are 30 inches in diameter. if the wheels rotate at 150 rpm, find the speed at which she is traveling in mi/hr.
150·60 = 9000 rph
9000·30π = 270000π
=848230.0165
≈13.39 mi/hr

Explanation:

Step1: Recall angular - linear speed formulas

The relationship between angular speed $\omega$ (in rad/s), linear speed $v$ (in units/s), and radius $r$ (in units) is $v = r\omega$. Also, 1 revolution = $2\pi$ radians.

Step2: Solve Example 6a

Given the tire makes 9.3 revolutions per second. To find the angular speed $\omega$, we use the conversion factor. Since 1 revolution = $2\pi$ radians, $\omega=9.3\times2\pi = 18.6\pi$ rad/sec.

Step3: Solve Example 6b

The diameter of the tire is 15 inches, so the radius $r=\frac{15}{2}$ inches. Using $v = r\omega$, with $\omega = 18.6\pi$ rad/sec, we have $v=\frac{15}{2}\times18.6\pi=\frac{279\pi}{2}\approx438.25$ in/sec.

Step4: Solve Example 7a

The blade rotates at 4800 revolutions per minute. First, convert to revolutions per second: $\frac{4800}{60}=80$ revolutions per second. Then, find the angular speed $\omega$: $\omega = 80\times2\pi=160\pi$ rad/sec.

Step5: Solve Example 7b

The diameter of the blade is 7.25 inches, so the radius $r = \frac{7.25}{2}$ inches. Using $v=r\omega$ with $\omega = 160\pi$ rad/sec, $v=\frac{7.25}{2}\times160\pi=\frac{1160\pi}{2}= 580\pi$ inches per second. To convert to feet per second, since 1 foot = 12 inches, $v=\frac{580\pi}{12}\approx151.84$ ft/sec.

Step6: Solve Example 8

The wheels rotate at 150 rpm. First, convert to revolutions per hour: $150\times60 = 9000$ rph. The diameter of the wheel is 30 inches, so the radius $r = 15$ inches. The distance traveled in one revolution is the circumference $C = 2\pi r=30\pi$ inches. The distance traveled per hour in inches is $9000\times30\pi$ inches. To convert to miles per hour, since 1 mile = 5280 feet and 1 foot = 12 inches (1 mile = $5280\times12$ inches), the speed $v=\frac{9000\times30\pi}{12\times5280}\approx13.39$ mi/hr.

Answer:

Example 6a: $18.6\pi$ rad/sec
Example 6b: $\frac{279\pi}{2}\approx438.25$ in/sec
Example 7a: $160\pi$ rad/sec
Example 7b: $\frac{580\pi}{12}\approx151.84$ ft/sec
Example 8: $\approx13.39$ mi/hr