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PhysicsBowl 2019 Q12

  • Writer: keshprad
    keshprad
  • Mar 28, 2020
  • 1 min read

12) A spring is attached to a horizontal disk. When at rest, we are told that the spring is unstretched at a distance d. And when it is moving with an angular velocity ω, it stretches to a distance 2d.


Let's start by looking at the force diagram when it is moving.

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Where m is mass, d is distance, and F[s] is the spring's restoring force

Seeing that F[s] is the only force acting towards the center of the motion, we can equate F[s] and the centripetal force. We will need to use Hooke's law to find F[s]. Although the mass is at a distance 2d, from the center of the disk, it is only a distance d from the spring's equilibrium position. However, in our centripetal motion formula, the radius is 2d, because this radius is only concerned with the distance from the center of motion.

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Where F[s] is the spring's restoring force, k is the spring constant, x is distance from equilibrium, d is distance, F[c] is centripetal force, m is mass, ω is angular velocity, and r is radius

Answer: B

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