anonymous
  • anonymous
Section 22: Concept Questions: how does the period of the uniform rod compare to the period of the simple pendulum for small oscillations?
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schrodinger
  • schrodinger
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anonymous
  • anonymous
here it is: http://ocw.mit.edu/courses/physics/8-01sc-physics-i-classical-mechanics-fall-2010/torque-and-the-simple-harmonic-oscillator/MIT8_01SC_quiz22.pdf
anonymous
  • anonymous
How'd this question get overlooked??? for a simple pendulum of length d and mass m acted on by gravity it is easy to show (by breaking the force mg into radial and tangential components) that \[F _{t}=-mgsin(\theta)\]negative because it is a restoring force. Then\[a _{t} = \frac {F _{t}} {m}=-gsin(\theta)=d \alpha=d \frac {d^{2} \theta} {dt^{2}}=\]or\[\frac {d^{2} \theta} {dt^{2}}+\frac {g} {d}\sin(\theta)=0\]In the small angle approximation\[\sin(\theta)=\theta\]so\[\frac {d^{2} \theta} {dt^{2}}+\omega^{2} \theta=0\]which has the solution \[\theta=\theta _{\max}\cos(\omega t + \phi)\]and since \[\omega T=2 \pi\]\[T=\frac {2 \pi} {\omega}=2 \pi \sqrt {\frac {d} {g}}\] for a physical pendulum like a rod with CM a distance d from the pivot\[\tau _{P} = I _{P} \alpha=-mgdsin(\theta)\]or\[\frac {d^{2} \theta} {dt^{2}}=- \frac {mgd} {I _{P}}\sin(\theta)=-\frac {mgd} {I _{P}}\theta=-\omega^{2} \theta\]in the small angle approximation. Once again the solution is \[\theta=\theta _{\max}\cos(\omega t + \phi)\]and since \[\omega T=2 \pi\]\[T=\frac {2 \pi} {\omega}=2 \pi \sqrt {\frac {I _{P}} {mgd}}\]\[I _{p} = \frac {1} {3} mL^{2}= \frac {1} {3} m(2d) ^{2}=\frac {4} {3}md^{2}\]Substituting you can see that \[T _{rod}=\frac {2 \pi} {\omega}={2 \pi} \sqrt{\frac {4d} {3g}}=\frac {2} {\sqrt{3}}T _{pendulum}=1.15 \times T _{pendulum}\]
anonymous
  • anonymous
Of course in the small angle approximation\[T=\frac {2 \pi} {\omega}=2 \pi \sqrt {\frac {I _{P}} {mgd}}\]is a general solution, so you can just substitute in the I of any object including a simple pendulum \[md^{2}\]

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