Need help with question regarding circular motion.

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SUMMARY

The discussion focuses on calculating the speed and centripetal acceleration of a person at the equator and at a latitude of 30.0° north due to Earth's rotation. For the equator, the speed is determined to be 463.97 m/s with a centripetal acceleration of 0.03 m/s². At 30.0° latitude, the radius of rotation is calculated to be approximately 5,525,242.08 m, leading to a speed of 6,008 m/s and a centripetal acceleration of 5.6 m/s². Participants emphasized the importance of correctly determining the radius at different latitudes for accurate calculations.

PREREQUISITES
  • Understanding of circular motion principles
  • Familiarity with trigonometric functions, specifically sine and tangent
  • Basic knowledge of physics equations for speed and centripetal acceleration
  • Ability to perform calculations involving Earth's radius and rotation
NEXT STEPS
  • Learn how to derive the radius of circular motion at various latitudes
  • Study the effects of Earth's rotation on objects at different latitudes
  • Explore the relationship between angular velocity and linear speed
  • Investigate the implications of centripetal acceleration in real-world scenarios
USEFUL FOR

Students studying physics, educators teaching circular motion concepts, and anyone interested in the dynamics of Earth's rotation and its effects on objects at different latitudes.

Fungamania
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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!
 
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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.
 

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