A uniform rod of mass M and length L is free to swing back and forth by pivoting a distance x from its center. It undergoes harmonicoscillations by swinging back and forth under the influence of gravity.Randomized VariablesM = 2.8 kgL = 1.7 mx = 0.29 mPart (a) In terms of M, L, and x, what is the rod's moment of inertia I about the pivot point.I= ((ML²)/12) + Mx² ✓ Correct!Part (b) Calculate the rod's period I' in seconds for small oscillations about its pivot point.Part (c) In terms of L, find an expression for the distance xm for which the period is a minimum.

Given, Mass , M = 2.8 kg Length, L = 1.7 m x =0.29 m

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A uniform rod of mass M and length L is free to swing back and forth by pivoting a distance x from its center. It undergoes harmonic oscillations by swinging back and forth under the influence of gravity.