## LeoMessi Group Title Section 22: Concept Questions: how does the period of the uniform rod compare to the period of the simple pendulum for small oscillations? 3 years ago 3 years ago

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 $T _{rod}=\frac {2 \pi} {\omega}={2 \pi} \sqrt{\frac {4d} {3g}}=\frac {2} {\sqrt{3}}T _{pendulum}=1.15 \times T _{pendulum}$
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 $md^{2}$