anonymous
  • anonymous
I REALLY NEED HELP!!!!!!!!!! A truck with 32-in.-diameter wheels is traveling at 60 mi/h. Find the angular speed of the wheels in rad/min. How many revolutions per minute do the wheels make?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
does any1 know how to do this?
anonymous
  • anonymous
?
jdoe0001
  • jdoe0001
heheh

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anonymous
  • anonymous
u know what i mean any way do u know how t do this problem?
jdoe0001
  • jdoe0001
the truck is going 60 miles per hour how many inches would that be? well keep in mind there are 5250 feet in 1 mile, and 12 inches per foot \(\bf 60\ miles \implies (60 \times 5280) ft = 316800\\ \textit{how many inches? well, }316800 \times 12 = 3801600\\ \textit{so there are 3801600 inches in 60 miles}\)
jdoe0001
  • jdoe0001
so the truck is really going at 3801600 miles per hour well, what's that in minutes? well, there are 60 minutes per hour, so 3801600/60 well, how many is that in seconds? well, since there are 60 secs per minute it'd be THAT AMOUNT divided by 60
anonymous
  • anonymous
wait isnt it 3801600 inches per hr?
jdoe0001
  • jdoe0001
ohh shoot, yes, inches per hour indeed
jdoe0001
  • jdoe0001
right, we were converting to inches 60 * 5280 * 12
jdoe0001
  • jdoe0001
and so, how many inches is the truck going per second? well \(\bf \cfrac{3801600}{60(mins)} = 63360 \qquad \qquad \cfrac{63360}{60(secs)} = 1056\)
jdoe0001
  • jdoe0001
so we can say that the truck is really going at 1056 inches per second actually we don't need the secs unit, just the minute
jdoe0001
  • jdoe0001
so the truck is going at 63360 inches per minute
anonymous
  • anonymous
so we want 63360 in/ min?
anonymous
  • anonymous
is the radians the inches?
jdoe0001
  • jdoe0001
so now let's take a peek at the wheel the diameter is 32 inches long that means the radius is half that, 16 inches so we want to know the circumference of the wheel|dw:1377550085206:dw|
jdoe0001
  • jdoe0001
radians is the angular unit
jdoe0001
  • jdoe0001
when asked on "angular speed" all they're really asking is "how many angles per time unit are there"
jdoe0001
  • jdoe0001
in this case is just " how many angles per minute is the truck going"
anonymous
  • anonymous
angles per minute?
jdoe0001
  • jdoe0001
well, heehhe, radians in this case, but yes, I gather angles might sound more something you may know
anonymous
  • anonymous
ohh i understand
jdoe0001
  • jdoe0001
you know what a geometric angle is right? just the opening between two lines
anonymous
  • anonymous
ya
jdoe0001
  • jdoe0001
and to get that, we need to know from the point a spot on the whee touches the road and goes around, till the point it gets back again to the same point that is the circumference |dw:1377550450904:dw|
jdoe0001
  • jdoe0001
that will tell us, how far the truck goes on one "go around" of the wheel
anonymous
  • anonymous
so its 100.531
jdoe0001
  • jdoe0001
yes it s 100.531 so we can say that in one "go around" of the wheel, the truck is only doing 100.531 inches well, how many radians is that? well, one "go around" is \(\large 2\pi\) so we can say that the amount of radians done in 1 minute is => \(\bf \cfrac{100.53}{2 \pi}\)
anonymous
  • anonymous
so 157.914
jdoe0001
  • jdoe0001
well, \(\bf \cfrac{100.53}{2 \pi} = 157.9 \ radians\)
anonymous
  • anonymous
so thats the rad/min?
anonymous
  • anonymous
or the revolutions/min?
jdoe0001
  • jdoe0001
can't be radians, since it's inches over radians, the \(2\pi\) is the radian unit already, thus \(\bf \cfrac{100.53in}{2 \pi}\)
anonymous
  • anonymous
so its revolutions
jdoe0001
  • jdoe0001
hmmm gimme a sec
anonymous
  • anonymous
ok
jdoe0001
  • jdoe0001
\(\bf \textit{is going at 100.53in per revolution}\\ 63360\ in/min\\ \cfrac{63360 \frac{in}{min}}{100.53 in} \implies \cfrac{63360in}{min}\times \cfrac{1}{100.53in} \implies 63360min\)
jdoe0001
  • jdoe0001
hmmm well, so much for much typing
jdoe0001
  • jdoe0001
\(\bf \textit{is going at 100.53in per revolution}\\ 63360\ in/min\\ \cfrac{63360 \frac{in}{min}}{100.53 in} \implies \cfrac{63360in}{min}\times \cfrac{1}{100.53in} \implies 630.25min\) rather
jdoe0001
  • jdoe0001
the angular speed will be angle/time, thus that is 630.25 per each revolution since we were using the circumference value of 100.53 which is 2pi
anonymous
  • anonymous
so whats the rad/min?
jdoe0001
  • jdoe0001
so the speed will end up as \(\bf \cfrac{2\pi}{630.25}\\ \textit{dividing both sides by } 2 \pi\\ \cfrac{1}{\frac{630.25}{2\pi}}\) to get the value for 1 radian per min
jdoe0001
  • jdoe0001
hmm so the value per radian ends up as 100.53 rad/min
jdoe0001
  • jdoe0001
lemme recheck myself
anonymous
  • anonymous
so thats the answer?
anonymous
  • anonymous
kk
jdoe0001
  • jdoe0001
no.... can't be if the circumference is 100.53 inches long how many radians does it do? well, dividing that by 2pi I get 16 so it does 16 inches per radians and we have 63360 inches in one minute so the amount of radians is, just the equivalence of it in inches, and dividing \(\bf \cfrac{63360\frac{in}{min}}{16\frac{in}{rad}} \implies \cfrac{63360 in}{min} \times \cfrac{rad}{16in}\implies 3960 \cfrac{rad}{min}\)
jdoe0001
  • jdoe0001
the revolutions are easy since we already have the amount of inches it does per 2pi and per min so just dividing 63360/100.53 will give 630.25 revolutions per minute
anonymous
  • anonymous
k...and the radians per min are 100.531rad/min
jdoe0001
  • jdoe0001
3960 rad/min you see the circumference is 100.53 which is 2pi to know how many inches per radian, we divide 100.53, and get 16 thus we know there are 16 inches for each radian then as above, we divide the radian = 16 inches by 63360 inches
jdoe0001
  • jdoe0001
so we're converting the circumference from inches to rad then using the radian value to divide the in/min the truck wheel does
anonymous
  • anonymous
thx so much
jdoe0001
  • jdoe0001
yw

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