QUESTION IMAGE
Question
- the carousel at the county fair makes 3 revolutions per minute.
a) find the linear speed in ft/sec of a person riding a horse that is 22.5 ft from the center.
3 - 45π = 135π = 424.12 = 7.07 ft/sec
b) the linear speed of the person on the inside of the carousel is 3.1 ft/sec. how far is this person from the center?
3.1 · 60 = 186 = 59.21 = 19.74
c) how much faster is the rider on the outside going than the rider on the inside?
Step1: Convert revolutions per minute to radians per second
The carousel makes 3 revolutions per minute. Since 1 revolution = $2\pi$ radians and 1 minute = 60 seconds, the angular - speed $\omega$ is $\omega=\frac{3\times2\pi}{60}=\frac{\pi}{10}$ radians per second.
Step2: Find the linear speed in part (a)
The formula for linear speed $v$ is $v = r\omega$, where $r$ is the radius. Given $r = 22.5$ ft and $\omega=\frac{\pi}{10}$ radians per second, then $v=22.5\times\frac{\pi}{10}=2.25\pi\approx7.07$ ft/sec.
Step3: Find the radius in part (b)
We know $v = r\omega$, and we are given $v = 3.1$ ft/sec and $\omega=\frac{\pi}{10}$ radians per second. Rearranging the formula for $r$ gives $r=\frac{v}{\omega}=\frac{3.1}{\frac{\pi}{10}}=\frac{31}{\pi}\approx9.87$ ft.
Step4: Find the difference in speeds in part (c)
The speed of the outside rider is $v_{1}\approx7.07$ ft/sec and the speed of the inside rider is $v_{2}=3.1$ ft/sec. The difference $\Delta v=v_{1}-v_{2}=7.07 - 3.1 = 3.97$ ft/sec.
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a) $7.07$ ft/sec
b) Approximately $9.87$ ft
c) $3.97$ ft/sec