The title above was a question on an exam that was marked wrong for me. I answered that if the Earth rotated slower (i.e. longer days), my apparent weight would increase. I based this on the observations that weights of objects of equal masses increase the higher in latitude you go. At the poles, the apparent weight of things is slightly higher than at the equator. I tried to argue my case with analogies, but it didn't work and the counter-argument was that $F_C = m * r * ω^2$ and that if the tangential velocity decreased, the centripetal force would also decrease. Since the centripetal force pulls inward, this decrease in centripetal force would lessen our apparent weight.
Who's correct here? If I am, how can I prove that this argument above is flawed and that an increase in the Earth's rotational speed will decrease an object's weight? I also want to circumvent the possible argument that "well, centrifugal force is a fictitious force, so it doesn't count."
Best Answer
The counter argument seems to be based on a flawed understanding of the centripetal force. Let's call $W$ your weight and $N$ the normal force that the scale applies on you. The net force $W$ - $N$ is the centripetal force on you that allows you to rotate along with the Earth. If the tangential speed goes down, the centripetal force goes down, so $N$ must go up, i.e. the weight you read on the scale is higher.
If you are not on the equator, there is a minor complication that the centripetal force doesn't point exactly downward but toward the spin axis. This isn't an essential consideration for the argument above, but technically the relevant (downward) component of the centripetal force is $m\omega^2r\cos\theta$, where $\theta$ is your latitude and $r=R\cos\theta$, $R$ being the Earth's radius.
"The centrifugal force is fictitious so it doesn't count" is flawed in its own right, but maybe that's a discussion for another time.