About the uniform circular motion

[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:

 

[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.

 

[wais]

Hi there,

For the first question about the Earth's rotation, your calculation for the speed at the equator is correct. The formula for the apparent weight is Wapparent = mg - mv^2/r, where m is the mass, g is the acceleration due to gravity, v is the speed, and r is the radius of the Earth. So, for a person with a mass of 65 kg, the apparent weight would be:

Wapparent = (65 kg)(9.8 m/s^2) - (65 kg)(464 m/s)^2/(6380000 m) = 636.9 N

For the second part of the question, if we want the person to feel weightless, that means the apparent weight should be zero. So, we can set the formula equal to zero and solve for the minimum speed:

0 = (65 kg)(9.8 m/s^2) - (65 kg)(v_min)^2/(6380000 m)
(v_min)^2 = (65 kg)(9.8 m/s^2)(6380000 m)/65 kg
v_min = 73.8 m/s

Therefore, the minimum speed at which the person would feel weightless is 73.8 m/s.

For the second question about the car going around a circular track, your calculations are correct. To find the time it takes for the car to go halfway around the track, we can use the formula d = vt, where d is the distance, v is the velocity, and t is the time. So, we have:

250 m = (20 m/s)(t)
t = 12.5 s

And for the average velocity, we can use the formula v_avg = d/t, where d is the total distance and t is the total time. In this case, the total distance is 500 m and the total time is 12.5 s, so:

v_avg = (500 m)/(12.5 s) = 40 m/s

I hope this helps! Let me know if you have any other questions. Good luck with your studies!