Calculating Centripetal Acceleration for a Frisbee in Circular Motion

Thread starter GensisLee
Start date Jun 13, 2011
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Circular Circular motion Motion

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The discussion focuses on calculating the centripetal acceleration of a Frisbee with a diameter of 28.0 cm that completes one turn in 0.110 seconds. The formula used for the calculation is ac = 4π²r/T². The calculated centripetal acceleration at the Frisbee's outer edge is 4.57 x 10² m/s². A question arises regarding the conversion of the time from 0.110 seconds to 0.100 seconds, suggesting a potential typo. Clarification on this time adjustment is sought within the context of the calculation.

Jun 13, 2011

GensisLee

You throw a Frisbee to your friend. The Frisbee has a diameter of 28.0cm and makes one turn in 0.110s. What is the centripetal acceleration at its outer edge?
a_c=4π²r/T²
I actually know the answer and how to solve this question...
a_c=4π²r/T²
a_c=π²(0.14m)/(.100s)²
= 4.57 x 10² m/s²
My question is why is how 0.110s is changed to 0.100s ?

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Jun 13, 2011

lewando

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Typo?

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