 Now, let's work through some exercises. Let's think about the effect that uniform circular motion has on people. So let's imagine somebody standing at the equator on Earth. The Earth is rotating about its axis once every day. What is the acceleration of a person standing at the equator due to the Earth's axial spin? To solve this, we need to approximate the Earth as a circle for this problem, so we can use our uniform circular motion tools. We know that the centripetal acceleration is equal to the tangential velocity squared divided by the radius, and we know that the speed is given by 2 pi r divided by big t, the period. If we substitute in our expression for the speed into our expression for acceleration, we find that the centripetal acceleration is 4 pi r on big t squared. So now we can substitute values in. The radius of the Earth is about equal to 6400 kilometers, and the period of the rotation of the Earth is one day, or 24 hours. But it's generally good to give answers in normal SI units, so we're going to convert kilometers into meters and hours into seconds. One kilometer is 1000 meters, and one hour is 60 minutes, which is itself 60 seconds, so one hour is then equal to 3600 seconds. So in our expression, replacing kilometer with 1000 meters and hour with 3600 seconds, we find that the total centripetal acceleration of a person standing on the surface of the Earth at the equator is 0.034 meters per second squared. This acceleration comes from the gravitational force of the Earth. Now we're going to try a related problem that's a bit more complicated. What is the centripetal acceleration of a person standing at 28 degrees north of the equator? This is similar to the last problem, so a similar approach should work. The centripetal acceleration and speed are still given by the same equations, and the period will still be 24 hours for the person standing at 28 degrees north of the equator. However, the radius of the path that the person follows will be different. If we sketch out the path the person follows alongside the radius of the Earth, we can see that the person will follow a different circular path with a different radius than they did in the problem before. To work out the length of the radius R, we can use trigonometry. The radius of the path and the axis of the Earth's rotation meet at right angles. If we label this angle theta, we can see that the angle theta is 90 minus 28 degrees since the angle between the radius of the equator and the axis of the Earth is 90 degrees in total. So now that we have an angle, the radius of the Earth, and a right angle triangle, we can use trigonometry to work out R. From Sokotoa, we know that sine of theta will equal R. The radius of the uniform circular motion path of a person standing at 28 degrees north of the equator divided by R Earth, the radius of the uniform circular motion of the person standing at the equator, which is equal to the radius of the Earth. Rearranging, we find that R is equal to R Earth times sine theta. So now we have everything we need. As before, we can substitute our equation for velocity into our equation for centripetal acceleration. Big T is still 24 hours of 3600 seconds, and R is the radius of the Earth times sine theta, or 6.4 millimetres times sine theta. Now, we could plug these numbers into a calculator to get our answer, or we could make life a bit easier by recognising that all of these terms are the same as in our previous example, except for our sine theta. And all of these terms will multiply together to give us 0.034 meters per second squared. So we can then see that the centripetal acceleration in this example is 0.034 meters per second squared times sine of 90 minus 28 degrees. Calculating, we find that this comes to 0.030 meters per second squared. Does this look right? Let's check. If our person was back on the equator, theta would be 90 degrees minus 0 degrees, which is equal to 90 degrees, and sine of 90 degrees is 1. So we get the same answer as we got previously. That's a good sign. What about the poles, though? At the poles, the person isn't travelling along a circular route anymore, so I'd expect that the centripetal acceleration should be 0. If the person were up at the north pole, 90 degrees north of the equator, then theta would be equal to 90 minus 90, or 0, and sine of 0 is 0, so this centripetal acceleration would be 0. So our answer at 28 degrees, which is in between these two, seems reasonable.