Rotating Mass-Spring System

In the \(xy\)-coordinate system, a \(\SI{1}{\kilo\gram}\) mass is attached to one end of a massless ideal spring. The other end of the spring is fixed at the origin. The spring constant is \(k = \SI[per-mode=symbol]{10}{\newton\per\meter}\), and the spring's unstretched length is \(\SI{1}{\meter}\).

The mass is initially being held in the air horizontally at \((x,y) = (\SI{1}{\meter}, \SI{0}{\meter}),\) and then is released so that gravity pulls it downwards. How far is the mass from the origin right at the moment it first crosses the (vertical) \(y\)-axis?

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Details and Assumptions:

  • Give your answer in meters, to 2 decimal places.
  • There is an ambient downward gravitational acceleration of \(\SI[per-mode=symbol]{10}{\meter\per\second\squared}\).
  • For the sake of this problem, assume that the spring can be stretched to great lengths.
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