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Two children are riding on a rotating merry-go-round. Child A is at a greater distance from the axis of rotation than child B. Which child has the larger tangential speed?

1 Answer

10 votes

Answer:

Child A.

Step-by-step explanation:

  • For all points on the merry-go-round, the angular speed ω is the same, as it's the rate of change of the angle rotated regarding time, and all points along the same radius rotate at the same time.
  • Based on the definition of angular velocity, and the definition of angle, we find that there exists a fixed relationship between the angular speed and the tangential speed, as follows:


v_(t) = \omega * r (1)

  • So, since ω remains constant, the tangential speed is directly proportional to the distance from the axis of rotation r.
  • This means that it will be larger for the child A, who is at a greater distance from the axis of rotation than child B.
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