- #1
mugzieee
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A 300-m tall 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?
Velocity in uniform circular motion
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SUMMARY
The discussion focuses on calculating the difference in linear velocity between the top and bottom of a 300-meter tall tower located at the equator. The formula used is derived from the period of Earth's rotation, T = 86400 seconds. The bottom speed is calculated as 2πR/T, while the top speed is 2π(R + 300 meters)/T. The resulting difference in speed is approximately 0.02182 m/sec, confirming that the top of the tower moves faster than the bottom due to the additional height.
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Physics students, educators, and anyone interested in understanding the principles of motion related to Earth's rotation and circular dynamics.
- #2
xanthym
Science Advisor
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{Period of Earth's Rotation} = T = (86400 sec)mugzieee said:
{Radius of Earth} = R
{Tower Bottom Speed} = 2*Pi*R/T
{Tower Top Speed} = 2*Pi*(R + 300 meters)/T
{Tower Top Speed} - {Tower Bottom Speed} = {2*Pi*(R + 300 meters)/T} - {2*Pi*R/T} =
= 2*Pi*(300)/T
= 2*Pi*(300)/(86400)
= (0.02182 m/sec)
~~
- #3
mugzieee
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hey bro how did you get the T=86400 sec? thanks for your help, it was the correct answer!
- #4
brewnog
Science Advisor
Gold Member
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Mugzieee, how long is a day?
- #5
mugzieee
- 77
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sorry, that was a thoughtless question which i realized afterwards. thank you 
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