A radial saw has a blade with a 7-inch radius. Suppose that the blade spins at 1300 rpm. (radians per minute). a.) find the angular speed of the blade in rad/min. b.) find the linear speed of the saw teeth in ft/s (feet per second). USE (3.1416 for pi)
ok no, it did not specify that rpm meant radians per second
in part "a" it says radians per minute so I didn't know if that corresponded to the question
I'll assume 1300 rpm = 1300 revolutions per minute,
then, for part a) : angular speed = \(2 \pi f = 2\pi*1300 = 2600 \pi \) rad/min
for pat b) : linear speed = \(r \omega \) \(r\) = radius \(\omega\) = angular speed
you will need to convert the units .. .
i have never seen these formulas before lol
u will see it more... if you're doing uniform-circular motion problems.... :)
okay is that all i will need to do? ima keep this question open in case something comes up
what does the long f mean, after 2 pi?
f = frequency = 1300 revolutions per minute
okay lol thanks
linear speed = \(\large r \omega = 7 \times 2600 \pi \) inches/ minute
convert it to feet/second
so 7/12 for inches to feet and...8168(2600pi) would be 490080 in seconds?
1/12 for inches 1/60 for seconds
careful, minute is in
careful, minute is in denomunator
\(\large \mathbb{7\times 2600 \pi \frac{inch}{min} }\)
\(\large \mathbb{7\times 2600 \pi \frac{12inch}{12min} }\)
\(\large \mathbb{7\times 2600 \pi \frac{feet}{12min} }\)
\(\large \mathbb{7\times 2600 \pi \frac{feet}{12\times 60 seconds} }\)
\(\large \mathbb{\frac{7\times 2600 \pi}{12\times 60} \frac{feet}{ seconds} }\)
simplify
okay thank you so much!!!
you should get 79.4 feet/second
u wlc :) btw, u may use wolfram for calculator : http://www.wolframalpha.com/input/?i=%287*+2600+%5Cpi%29%2F%2812%5Ctimes+60%29
wow that is an awesome website thanks for all the help!
yes it is.... np =)
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