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Velocity in uniform circular motion

  1. Oct 23, 2009 #1
    a Ferris wheel has a radius of 20.0m and it completes 1 revolution in 8.42 seconds

    a) what is the speed of a point on the edge of the wheel?

    The circle is divided into 8 triangles. I have determined from that that the arcs of the circle (8 of them) each equal 20m. Because the triangles are similar and each side of the triangle will equal 20m. Is my thinking here correct?

    so velocity equals 20/8.42=19.00

    b)The x component of acceleration at the top of the circle would equal 0. Correct?

    c)The y component of acceleration at the top of the circle would equal a=v^2/r 19.00^2/20.00=-18.05 since the acceleration vector is pointing toward the center of the circle it is negative.

    Is this correct?
     
  2. jcsd
  3. Oct 23, 2009 #2
    I have a change in the velocity
    v=2pir/t
    v=14.92

    acceleration would equal 11.14
     
  4. Oct 23, 2009 #3
    a)v=2[tex]\pi[/tex]r/T=2*[tex]\pi[/tex]*20/8.42=14.92m/s
    and the length of the arcs will equal circumference divided by 8
    b)do u mean the horizontal component? yes that would equal zero as only acceleration is towards center of Ferris wheel
    c)yes but the v value is wrong. look at part my explanation for part a. and as for the negative sign that depends on how u assign upwards and downwards direction. it would be better to say downwards or to center of the wheel.

    EDIT: dont know why my pis are above. they shouldnt be lol
     
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