A kid of mass 20 kg is playing with a swing with an arm length of 5 m. His siste

Question

A kid of mass 20 kg is playing with a swing with an arm length of 5 m. His sister rst pushes him such that the maximum angular deviation from vertical position is 0.2 radian (11.46deg.). After 10 cycles, the maximum angular deviation is reduced by a factor of e. What is the quality factor Q of the swing? If the sister has to push the kid on every cycle to maintain a constant maximum angular deviation of 11.46deg, how much work should the sister do on the kid on every push? [Hint: recall the variousexpression of the quality factor of a lightly-damped oscillator.]

Explanation / Answer

initial potential energy is,

   PE = mgL[1 - cos@] = 19.94 J

Final potential energy is.

PE = mgL[1 - cos(11.46+e)] = mgL[2sin2(0.2e/2)]

for small angles, we get

PE = mgL[2(0.2e/2)2] = 20e2

energy dissipated in 10 cycles = 19.94 - 20e2

energy dissipated in 1 cycles = [19.94 - 20e2]/10

the quality factor is,

   Q = 2*pi*[energy stored / energy dissipated in 1 cycles] = 1252.2 / [19.94 - 20e2]

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the energy dissipated in each cycle is,

      [19.94 - 20e2] = [1.994 - 2e2] J

so the work done per each cycle:

      [1.994 - 2e2] J

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