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The earth rotates once per day about an axis passing through the north and south poles, an axis that is perpendicular to the plane of the equator. Assuming the earth is a sphere with a radius of $6.38 \times 10^{6}$ m, determine the speed and centripetal acceleration of a person situated (a) at the equator and (b) at a latitude of 30.0 north of the equator.

464$m / s$$3.37 \times 10^{-2} m / s^{2}$402$m / s$$2.92 \times 10^{-2} m / s 62$

Physics 101 Mechanics

Chapter 5

Dynamics of Uniform Circular Motion

Newton's Laws of Motion

Applying Newton's Laws

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in this problem, we have to find the velocity and angular acceleration at two points on Earth, one of the earth's equator and 13 degrees north. So the only information that we're given in this problem is Earth Radius, which is equal to 6.3 day times 10 to the sixth meters. Well, we actually do know other information. You know what the earth, uh, unearth a day past 24 hours, Which means that earth must have a period of rotation of 24 hours. Now we have to rewrite this in seconds, since that's the unit that we always work with. You re doing physics. So our tea is gonna be 24 hours. I'm 60 time 60 which equal to 400. Okay, so from here, we can start solving our problem first. Let's look at the scenario where we're on the equator. Now. We know that we can rewrite team as two pi over omega, which could in turn me were written as two pi r overby. Now we can write the in terms of our known quantities. B is equal to two pi r over tea. So when we plug in the two results that we have above. We have this for a radius and this routine. We find that the is equal to 464 meters per second roughly. And if you want the truck that the unit's workout since our period, it is always seconds and our radios adjusted meters. Our units are in fact meters per second so that the first result now from this we can find the angular acceleration of the equator. We know that Alfa is equal to B squared. We are The skirt is applying that we just found an r was getting in the problem so we can plug in a result for earlier. And we find that a result is equal to 3.37 times 10 to the negative too. When we don't have our units, we know that these units of meters per second squared, centimeters squared per second squared ours nudes of meters. So our meters are units by that to end up being meters per second square and that's gonna be the second part over answer. Okay, now let's consider the second scenario. So at this point, as we told the problem, we are now at a latitude of three degrees north of the equator. That just means that all of her answers are going to be multiplied by the quantity CO signed 30 in order to take that into account. All right, so let's start just like we did earlier by solving for our velocity. So you referred earlier. V is equal to two pi r over tea, but in this case are gonna be multiplied by a co sign of 30. Now, once again, this is all just in quantity. Since our absence, the radius of the earth remains the same, no matter where you were on it, of course, and we're going to get that our velocity is equal to ah, 402 meters per second per cent of 30 is just a number. So it doesn't have any units and air units still work out the same as they did earlier. Now for the second part of B, Come on, go ahead once again and just do what we did earlier. Alfa is equal. He's going on my ring that we have to multiply by coast of 30. All we have left to do. Now it's plug in, how we do in fact, find that are Elsa. It's equal to 2.92 times 10 to the night of two, and once again, or, you know, two meters per second squared and the problem itself.

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