A truck with 48-in.-diameter wheels is traveling at 45 mi/h.

Question

(a) Find the angular speed of the wheels in rad/min.

(b) How many revolutions per minute do the wheels make?

Answer 1

First of all, this is physics problem.

We know that the diameter of the wheel is d = 48 inches and its speed v = 45 mi / hr.

We need to find the angular speed ω in rad /min and frequency f in rev / min. Usually I recommend my students to convert all units to SI, but in this case they ask the answers in units per minute. So, I will do the problem using different conversion of units. I convert the speed that is in mi / hr to inch / min. We know that 1 mile is 63 360 inches, and 1 hour is 60 minutes. Then if I multiply speed by 63 360 and divide by 60, I will get that the speed is 47 520 inch / min.

From the formula

v = ω·R = ω·d / 2

it is easy to get that angular velocity is

ω = 2v / d .

Plugging in the numbers we have

ω = 2·47 520 inch / min / 48 inch = 1980 rad / min.

Knowing that angular velocity

ω = 2πf,

We can find the frequency from the formula above.

f = ω / 2π.

After substitution of known values, we get

f = 1980 rad / min / 2·3.14 = 315 rev / min.

Keep in mind that because the data were given to us in whole numbers, the answer also we round to the whole number.

Answers: 1980 rad / min, 315 1 / min

Answer 2

Hi again Diea! Hoping to redeem myself - I promise to slow down and double-check my thoughts this time. Even seasoned Precalc teachers can make silly mistakes sometimes; hopefully you can remember to give yourself some grace then too, the next time you make one :)

Ok, so this is basically the reverse of your last problem; this time we have an answer in miles/hour and are hoping to convert back to both revolutions/minute, as well as the more trig-specific radians/min.

We again begin with the initial ratio and try to organize other known relationships to have units we don't want shift into units we do via cancelation:

we start with

45 miles

------------

1 hour

and incorporate

1 hour

----------

60 mins

63360 inches

-------------------

1 mile

to get to a linear speed of 45*63360/60 inches/min - these steps are the same for both (a) and (b); now let's continue to finish each:

a) the relevant relationship with respect to radians is that one full revolution of the circle links the following two pieces:

1. the circumference, 2*pi*24 inches (since in general the circumference is 2*pi*r, and here we're given a diameter - twice the radius - of 48)
2. 2*pi radians

we can again look at the units we have - inches/min - to see we want 2*pi*24 inches to be in our denominator so that inches will cancel with inches and we'll be left with our hoped-for units of radians/min.

So we multiply further by

2*pi radians

-----------------

2*pi*24 inches

to get a final answer of 45*63360/60*2*pi/(2*pi*24) - things work out nicely here, and we're left with an exact value of 1980 radians/min... no worrying about rounding instructions needed!

b) here we instead want a relationship linking inches to revolutions instead of to radians. We can still do this by thinking about what happens with one full rotation of the wheel:

1. 2*pi*24 inches is still the circumference, and we now link it to
2. 1 revolution, which unfortunately does mean we're not gonna get a pi/pi like last time so we should anticipate needing to figure out where the rounding instructions are noted on the site you're using.

Same deal as before: we put inches in the denominator in order to cancel with the inches in the numerator of the work we've done before part (a), and so multiply by

1 revolution

-----------------

2*pi*24 inches

to get 45*63360/60*1/(2*pi*24) which is about 315.127 revolutions/min but we're gonna need clearer insight about what format the site wants its answers in in order to know which level of rounding it's going to accept.

If you're able to find that info, I'd be happy to help you better understand how to apply it both here and more generally.