Oops! I forgot to preface that the radial distance from the earth's center is the same at both the north pole and equator, so that shouldn't be taken into account.
Is someone at the north pole undergoing uniform circular motion? Are they at the equator? How would that affect the weight?
Are you sure the radius is the same? According to the formula, v = (w) x (r), where w is the angular velocity and r is the radius, we can figure out the velocity that someone is moving when they stand at the equator and at the north pole. The angular velocity (w) is the same in both cases since it takes the same amount of time (24 hours) to move around the globe (2pi) regardless of where you stand. The linear velocity (v), therefore, is dependent on the radius of of where your standing that will ultimately determine how fast you move around the globe. If you stand at the north pole,you moving slowly because the circle you move around the pole is small while if you stand at the equator the distance around is much larger. Thus to be able move around the globe in 24 hrs at the equator, wouldn't you have to move at a quicker linear velocity?
In regards to weight, because our radius is smaller, then our weight (force) is less so we feel lighter.
Did I say that correctly?
And more specifically, the net sum of the forces that have a vector that points in + or - direction towards the center. For example, if you have a ball on a string that you're swinging like a ferris wheel, at pi/2 and 3/2pi, only tension = mv^2/r, NOT tension and weight.
Really? Why is that? I thought that gravity would be acting on the object at all points. Looking at a diagram I can see that pi/2 and 3pi/2 are at the sides while pi and 2pi are at at the top and bottom, respectively (assuming we get on the ride at the bottom, lol). Why wouldn't you have a weight at those 2 points?
It would read less on a rotating earth than a non rotating earth at the equator. Gotta be able to view physics conceptually. Centripetal acceleration at the equator is the highest, this question could ask the difference between someone at the equator and someone at a pole. You want to travel in a straight line (newton law), if the earth is spinning (a(c)=v^2) the earth is trying to throw you off the planet but gravity is fighting against that and winning but if there was no centripetal acceleration, gravity doesn't have to fight anything against centripetal acc and has a stronger downward force on you. The scale would read greater on non rotating earth. The key to physics is conceptual intuition, you really don't need to know equations if you can think logically about them and know the important relationships. The MCAT is designed to test your intuition in physics which catches a lot of people by surprise because the math was what was stressed most heavily in most of our schools
See, now I am a bit confused by what centripetal force means. As far as I know, its the net force that points in the direction perpendicular to your tangential motion. The way you make it sound is that it is a force that wants you to keep moving in a straight line. However, I thought that centrifugal force (which is not a force) was the tendency for you to move in a straight line. If anything, it seems as if centripetal force is essentially gravity (i.e. the force keeping attracting us towards the centre of the earth), because it if it wasn't then we would fly off the earth along a tangent.
So the first half of your post makes sense in that the earth, due to its spin, wants to throw us off, however, gravity (or the centripetal force) which pushes us against the centre of the earth is fighting against that tendency to move along the tangent.
So conceptually, on a rotating earth the net force is (mg - Fn = (mv^2)/r), where mg is your weight and Fn is your normal force and all of that is equal to your mass times centripetal acceleration (since we are moving in a circular motion). Thus, how much the scale reads (normal force (Fn)), will be less because from our weight (mg), we have to subtract the mv^2/r so we get a lower value. However, on the a non-rotating earth, we do not have a centripetal acceleration and so we don't have a centripetal force. In this case, if I stood on a earth that didn't rotate, my weight (mg) would be equal to the normal force exerted by the surface (Fn = mg). Therefore, Fn is now larger relative to the equation above.
What I just wrote, did that make sense conceptually and mathematically or am I wrong?
