Physics Word Problem

Physics
Tutor: None Selected Time limit: 1 Day

The radius of the Earth is about
6.37 ×10^6 m.
A) What is the centripetal acceleration of a point on the equator due to the rotation of earth about its axis?
Answer in units of m/s^2

B) What is the centripetal acceleration of a point at the North Pole?
Answer in units of m/s^2.

Jun 23rd, 2015

Thank you for the opportunity to help you with your question!

A) On the surface the centripetal acceleration a = v^2/R 

where v is the velocity of a point on the surface and R is the radius of the Earth at the equator. 

Now v = w*R 

where w is the angular velocity and w = 2*Pi / T where T is the period of rotation of the Earth in seconds ie 24*3600s. 

So a = w^2*R = 4*Pi^2*R / T^2 

a = 4*Pi^2*R / T^2 

a=4*(3.14)^2*6.37*10^6/(24*3600)^2=33.6*10^-3m/s^2

B) At north pole the answer is 0 as R = 0 in the last equation above as this radius is the distance from the axis of rotation and this passes through the North (and South) Pole.

Please let me know if you need any clarification. I'm always happy to answer your questions.
Jun 23rd, 2015

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