Need help with question regarding circular motion.

Click For Summary

Homework Help Overview

The problem involves calculating the speed and centripetal acceleration of a person at the equator and at a latitude of 30.0° north of the equator, based on the Earth's rotation. The context is circular motion, specifically relating to the effects of rotation on objects at different latitudes.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to calculate speed and centripetal acceleration using formulas related to circular motion, but expresses uncertainty about the correctness of their results, particularly regarding the calculations at 30° latitude.
  • Some participants question the assumptions made about the radius at different latitudes and suggest visual aids to clarify the geometry involved.
  • Others provide insights into the correct radius for calculations at 30° latitude, indicating it should be less than the radius at the equator.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem and providing guidance on the necessary adjustments for the radius at 30° latitude. There is no explicit consensus yet, but some productive direction has been provided regarding the geometry of the situation.

Contextual Notes

Participants note the importance of accurately determining the radius of rotation at different latitudes, which is essential for correct calculations. The original poster's calculations are based on assumptions that may need to be revisited.

Fungamania
Messages
2
Reaction score
0
Thanks for checking out my thread. Okay, first off, the question.

The Earth rotates once per day about an axis passing through the north and south poles, an axis that is perpendicular to the plane of the equator. Assuming the Earth is a sphere with a radius of 6.38 X 10^6 m, determine the speed and centripetal acceleration of a person situated (a) at the equator and (b) at a latitude of 30.0° north of the equator.

I'll show what I've done so far.

a) V=2(pi)r/t
plugged in the appropriate values where r is (6.38 x 10^6) and t is (86400).

my results from that equation was 463.97 m/s.

I then continued to calculate the centripetal acceleration:

a=v^2/t
a=(463.97^2)/6.38 x 10^6)
a=0.03 m/s^2
(I wasn't expecting it to be this small.

b) the troubling thing now is obviously the angle. Do I use the following formula?

v=sq(rgtan(theta))
v=sq((6.38 x 10^6)(9.8)(tan30))
v=6,008 m/s

then to caluculate the centripetal accceleration:
a=v^2/r
and I end up getting 5.6 m/s^2


I'm not sure if the answers are correct (most likely not) Any help would be greatly appreciated. Thanks for your time!
 
Physics news on Phys.org
bump, anyone?
 
Sorry, I don't have anything to help you out with; but I just wanted to tell you your problem helped me solve my problem. Apparently I had incorrect values for the radius of the Earth and the period of the Earth. Thanks, Fungamania.
 
You need to reconsider the radius at 30deg. It should be LESS then the radius at the equator. I recommend drawing a picture, or look at a globe to figure out just what the radius is. (in this case it is the distance from the axis of rotation that you need).
 
You can give someone a sin but you cannot make them read.
 
Find the radius of the smaller circle. for a degree od 30, i got the radius as 5525242.076 m. I am pretty sure it is right, i don't know how to explain it in words. You have to draw it out and find the radius using the coine of 30 times the radius of the earth. Hope it helps. If you still have a question, email me.
 

Similar threads

Replies
15
Views
2K
  • · Replies 2 ·
Replies
2
Views
954
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
Replies
2
Views
1K
Replies
6
Views
4K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
4
Views
2K
Replies
2
Views
1K