Solving Tangential Speed at Daytona Beach FL

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SUMMARY

The discussion focuses on calculating the latitude at which the tangential speed due to Earth's rotation is 200 m/s less than that of Daytona Beach, FL. The tangential speed at Daytona Beach is derived from the circumference of the Earth at that latitude divided by the number of seconds in a day (86400). The formula used is C_{Daytona Beach}/86400 = C_X/86400 - 200, where C_X represents the circumference at the unknown latitude. The problem also involves understanding the geometry of the situation, specifically the relationship between the radius of the Earth and the cosine of the latitude.

PREREQUISITES
  • Understanding of tangential speed and circular motion
  • Basic knowledge of trigonometry, specifically cosine functions
  • Familiarity with Earth's rotation and time calculations
  • Concept of right triangles and their properties
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  • Calculate the tangential speed at various latitudes using the formula provided
  • Explore the relationship between latitude and tangential speed in a physics context
  • Study the geometry of circles and how it applies to rotational motion
  • Review the principles of physics related to angular velocity and its applications
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Students in physics, particularly those studying mechanics or engineering, as well as educators looking to explain concepts of rotational motion and tangential speed.

ERAUin08
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Hey everybody I have some trouble with this problem.

Homework Statement


At what latitude is the tangential speed of a point due to the Earth's rotation 200 m/s less than is Daytona Beach FL?




The Attempt at a Solution


I honestly don't know where to start. I think I have to do something with measures of triangle angles and lengths of their side. I know I need to give it an honest effort however I don't know where to go. Thank you in advanced.
 
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No, this has nothing at all to do with "measures of triangle angles and the lengths of their sides". It has everything to do with how fast a point moves around a circle. The City of Daytona Beach FL rotates around the axis of the Earth in 24 hours= 24(60)(60)= 86400 seconds. Its "tangential speed" is the circumference of that circle divided by 86400 seconds. you are looking for another point (there are 2 latitudes of points that fit) at which the circumference about which the point moves, also divided by 86400 seconds is that number minus 200:
\frac{C_{Daytona Beach}}{86400}= \frac{C_X}{86400}- 200

No here is where angles and triangles may come into play. If you draw a line from the center of the Earth to Daytona Beach, that line makes an angle a line to the equator equal to the Latitude of Daytona Beach. The line from the center of the Earth to Daytona Beach, the line from Daytona Beach to the axis of the earth, and the axis of the Earth make a right triangle with hypotenuse length equal to the radius of the earth. The radius of the circle Daytona Beach makes around the Earth is radius of the Earth times cos(latitude of Daytona Beach).
 
Ok thank you very much HallsofIvy I really appreciate your help. This class is physics 1 for engineers. I believe I was just lost on what to do because of the wording and experience. Thank you very mcuh
 

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