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

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SUMMARY

The discussion focuses on calculating the difference in angular velocity between the top and bottom of a 300m tower located on the equator. The Earth's radius is approximately 6400km, leading to a circumference of 40,030,000m at the equator. The angular velocity of the Earth is derived using a 24-hour period, while the period for the tower must be determined to accurately compute its angular velocity. The key calculation involves using the formula for angular velocity: angular velocity = 2πr / period.

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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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