How Does Angular Velocity Vary from Top to Bottom of a Tower on the Equator?

AI Thread Summary
A 300m tower at the equator has a different angular velocity at its top compared to its bottom due to the Earth's rotation. The circumference at the Earth's surface is calculated as c=2π*6400000m, while the circumference at the top of the tower is c=2π*6400300m. To determine the speed difference, the period of rotation must be considered, using 24 hours for the Earth's angular velocity. The period for the tower's angular velocity needs to be established to accurately calculate the difference. Understanding these concepts is essential for solving the problem effectively.
StephenDoty
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A 300m tower is built on the equator. How much faster does a point at the top of the tower move than a point at the bottom?

The Earth is 6400km
thus c=2pi * 6400000m
and c of the top of the tower is c= 2pi *6400300m

Now what?

Thanks.

Stephen
 
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Do you use the period of 24hr for both the Earth angular velocity and the tower angular velocity?
I need the period for the tower to solve the problem
 
since
angular velocity= 2pi r/ period

how do I find the period of the tower so I can find the angular velocity of the tower and subtract it from the angular velocity of the earth?
 

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