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    After one rotation of 2pi radians, why isn't the angular displacement zero?

    I ask because the formula for Time period, T is 2pi/omega, this is for the time taken to complete one rotation. I guess it makes more mathematical sense to use 2pi, but the angular displacement is zero, so why isn't it 0/omega?

    Thanks
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    (Original post by Mathsqueen)
    After one rotation of 2pi radians, why isn't the angular displacement zero?

    I ask because the formula for Time period, T is 2pi/omega, this is for the time taken to complete one rotation. I guess it makes more mathematical sense to use 2pi, but the angular displacement is zero, so why isn't it 0/omega?

    Thanks
    Haven't done physics side of circular motion, but if it was 0/omega then the time would be 0, as expected because no rotation has been performed at 0 seconds. At 2\pi you have traveled a full circle so it gives you the time that it took to complete one full circle. As for angular displacement being 0, well it must be because you come back to where you start at with multiples of 2\pi
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    (Original post by RDKGames)
    Haven't done physics side of circular motion, but if it was 0/omega then the time would be 0, as expected because no rotation has been performed at 0 seconds. At 2\pi you have traveled a full circle so it gives you the time that it took to complete one full circle. As for angular displacement being 0, well it must be because you come back to where you start at with multiples of 2\pi
    Thank you so much! So the angular displacement is 0, but 2pi is used in calculations to indicate that a single full rotation has been made, so the computation makes sense.
 
 
 
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