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The earth spins on its axis once a day and orbits the sun once a year $\left(365_{4}^{1} \text { days). Determine the average angular velocity (in rad/s) of the }\right.$ earth as it $(\text { a) spins on its axis and }(b) \text { orbits the sun. In each case, }$ take the positive direction for the angular displacement to be the direction of the earth's motion.

$+2.0 \times 10^{-7} \mathrm{rad} / \mathrm{s}$

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Cornell University

Rutgers, The State University of New Jersey

Hope College

University of Sheffield

for this problem. We have two parts and they're both asking is basically the same thing. So for part A, we're being asked, determined to determine the average angular velocity of mega as the earth spins on its axis. So to find Omega, we have a displacement in angle over a time. And if those if the angular velocity is constant, we can just take the total displacement and divided by the total time to find that angular velocity. So that's what we're gonna do for these problems. So for spinning on its axis, the earth rotates once per day. So the total angle displaced is going to be two pi radiance. And if we work in radiance and seconds, then we'll get an answer and readings per second, which is what we're looking for. So the total time involved in one day is 24 hours times 3600 seconds per hour. So when we calculate this out, we get seven point 27 times 10 to the minus five radiance per second and the way that I did this calculation here is one day is 24 hours. So if we want to insert something that is equivalent to one day and the denominator there, then we can do. Then we can put in 24 hours times 3600 seconds her, our and then the hours cancel, and we're left with 24 times 3600. So that's part a now for Part B. We need to do the same thing, but with a different amount of time. Because we have one full revolution. So are Delta Theta that we need to put in two this equation. It's still too piratey ins, but in this case we have one year for orbit around the sun. So in order to do this one one year is equivalent to 365 0.25 days. Okay, times a fraction with day on the bottom one day so that we can cancel the days in each day. There's 24 hours, and in each one hour there's 3600 seconds. So if we multiply all this out will cancel our days. Well, cancel our hours and we'll have a number that's in seconds. So if we insert these numbers into a calculator for a year, we get 31 point six 1,000,000 seconds But if we just plug all these numbers directly into the equation above, where we end up getting is our answer for Omega, here is about two times 10 to the minus seven radiance per second.

University of Washington