Problem16
Section 6.2: Centripetal Acceleration

Question

Olympic ice skaters are able to spin at about 5 rev/s. (a) What is their angular velocity in radians per second? (b) What is the centripetal acceleration of the skater's nose if it is 0.120 m from the axis of rotation? (c) An exceptional skater named Dick Button was able to spin much faster in the 1950s than anyone since—at about 9 rev/ s. What was the centripetal acceleration of the tip of his nose, assuming it is at 0.120 m radius?

Question by OpenStax is licensed under CC BY 4.0

Final Answer

  1. 30 rad/s30 \textrm{ rad/s}
  2. 1×102 m/s21\times 10^{2}\textrm{ m/s}^2
  3. 4×102 m/s24\times 10^{2}\textrm{ m/s}^2

Solution video

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Video Transcript

This is College Physics Answers with Shaun Dychko. Olympic skaters can rotate at a speed of about 5 revolutions per second. Expressing that in radians per second requires multiplying by the conversion factor 2π radians for every revolution; this is 30 radians per second. The centripetal acceleration of the skater's nose will be the distance from the axis of rotation to the nose which we take to be about 0.120 meters, multiply it by this angular speed and I have written down the unrounded answer here for the speed and square that and you get about 1 times 10 to the 2 meters per second squared. For Dick Button—very fast spinner— the centripetal acceleration of his nose will be 0.120 meters times 9 revolutions per second converted into radians per second squared which is 4 times 10 to the 2 meters per second squared.

Comments

NoOneReally·
Answer for part a is incorrect -- the angular velocity is 31 rad/sec... as clearly noted in parts b, and the calculator screenshots.
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NoOneReally·
Disregard my last. The answer is restricted to one significant figure; 30 has one, 31 has two.
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ShaunDychko·
Yes, that makes sense. Thanks for posting.
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