# About the uniform circular motion

1. Oct 16, 2004

### achulovex

Hi..
I have problem solving this question..
if anyone could help me out, i would really appreciate it.

Q. Taking the Earth to be a sphere of radius 6380 km, rotating about its own axis once every 24 hours, find the speed of rotation at the Equator. What is the apparent weight of a person of mass 65kg standing on the Equator as a result of this rotation? What must be the minimum period of the Earth rotation so that the person would feel weightless?

i got 464m/s for the speed and 634.something for the apparent weight but the answer was 73.8m/s for the speed and 636.9 for the apparent weight. when i tried to find the apparent weight with the velocity 63.8m/s, i got 636.9 N (Fg - Fc = W apparent)

maybe there's something wrong my logic..

i want to know how i have to solve it..

and one more..

Q. A car gies around a circular track 500m in diameter at a constant speed of 20 m/s (a) How long does it take for the car to go halfway around the track? (b) what is its average velocity in that time interval?

ok.. for (a) i just divide 500x 3.14 x 0.5 by 20 m/s

and for (b) i divided 500m by the time i got in (a).

is this correct? i want to make sure.. =)

thanks for the help.. :rofl:

2. Oct 16, 2004

### Skomatth

For #1 I got the same speed you did using v=2(3.14)(6380 X 10^3m)/86400s
Are you sure the supposed answer is right?

I then used that answer and got 637 N for the apparent weight.

Draw a force-body diagram for the person. You have mg acting down and normal force(the apparent weight) acting up. They combine to create a centripetal force of mv^2/r.