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A gyroscope slows from an initial rate of 32.0 rad/s at a rate of .
  • How long does it take to come to rest?
  • How many revolutions does it make before stopping?
Question by OpenStax is licensed under CC BY 4.0.
Final Answer

a) $45.7 \textrm{ s}$

b) $116 \textrm{ rev}$

Solution Video

OpenStax College Physics Solution, Chapter 10, Problem 7 (Problems & Exercises) (1:44)

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Video Transcript

This is College Physics Answers with Shaun Dychko. The gyroscope has an initial angular velocity of 32 radians per second and slows to a final angular velocity of zero. Its angular acceleration we're told has a magnitude of 0.7 radians per second squared and I've put a negative here to indicate that it's in the opposite direction to the initial angular velocity since the thing is slowing down. So we're going to find how long it takes to slow down to stop, and we know the final angular velocity is the initial angular velocity plus the angular acceleration times time. We'll subtract <i>omega i</i> from both sides and then divide both sides by <i>alpha</i> and we get that the time is the difference in angular velocity divided by the angular acceleration. So that's zero minus 32 divided by negative 0.7, and this gives 45.7 seconds to come to a stop. The number of revolutions that it does while it is stopping we can figure out by looking at this formula. <i>Theta</i> is the angular displacement and we'll get an answer in radians first of all then we'll convert that into revolutions by multiplying by one revolution for every two pi radians. We get to this point by starting here and solving for <i>theta</i> and we subtract <i>omega i</i> squared from both sides and then divide both sides by two <i>alpha</i>. So we get <i>theta</i> is <i>omega f</i> squared minus <i>omega</i> initial squared divided by two <i>alpha</i>. So that's zero minus 32 radians per second squared, divided by two times negative 0.7 radians per second squared. This gives 731.43 and after we divide by two pi, this is 116 revolutions.