The rotational inertia of a uniform thin rod about its end is ML²/3, where M is the mass and L is the length.Such a rod is hung vertically from one end and set into small amplitude oscillation.If L = 1.0 m this rod will have the same period as a simple pendulum of length: A)33 cm B)50 cm C)67 cm D)100 cm E)150 cm

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Quizplus · Accepted Answer

The period of a physical pendulum, like a rod swinging about one end, is given by $T = 2\pi\sqrt{\frac{I}{mgd}}$, where $I$ is the moment of inertia about the pivot point, $m$ is the mass, $g$ is the acceleration due to gravity, and $d$ is the distance from the pivot to the center of mass. For a uniform rod of length $L$ pivoted at one end, $I = \frac{1}{3}mL^2$ and $d = \frac{L}{2}$. Substituting these into the formula gives $T = 2\pi\sqrt{\frac{\frac{1}{3}mL^2}{mg\frac{L}{2}}} = 2\pi\sqrt{\frac{2L}{3g}}$.For a simple pendulum with a small amplitude, the period is given by $T = 2\pi\sqrt{\frac{l}{g}}$, where $l$ is the length of the pendulum. Setting the periods equal to find when the rod and the simple pendulum have the same period, we get $\sqrt{\frac{2L}{3g}} = \sqrt{\frac{l}{g}}$. Solving for $l$ gives $l = \frac{2L}{3}$.Substituting $L = 1.0 m$ into $l = \frac{2L}{3}$ gives $l = \frac{2(1.0)}{3} = \frac{2}{3} m = 0.67 m = 67 cm$.

The Rotational Inertia of a Uniform Thin Rod About Its