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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.Randomized Variables M = 2.4 kgL = 1.6 mx = 0.38 m a) In terms of M, L, and x, what is the rod’s moment of inertia I about the pivot point. b) Calculate the rod’s period T in seconds for small oscillations about its pivot point. c) In terms of L, find an expression for the distance xm for which the period is a minimum.

Answer :

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The rod's center of mass is at its center, so the moment of inertia about the pivot point is equal to the moment of inertia about the center of mass plus the product of the rod's mass and the square of the distance from the pivot point to the center of mass.

  • Rod's moment of inertia is I = (1/12) * 2.4 kg * 1.6 m^2 + 2.4 kg * 0.38 m^2 = 0.48 kg * m^2.

To find the period of the rod's oscillations, we can use the formula T = 2 * pi * sqrt(I / Mg), where T is the period, I is the moment of inertia, M is the mass of the rod, and g is the acceleration due to gravity.

  • T = 2 * pi * sqrt(0.48 kg * m^2 / (2.4 kg * 9.8 m/s^2)) = 2.6 seconds.

This gives us the equation 0 = -(1/2) * pi * M * L^2 * x / (M * g * sqrt(I / Mg)). Solving for x, we find that

  • xm = -(1/2) * pi * L^2 / (g * sqrt(I / Mg)).

Substituting the values we found above, we find that

  • xm = -(1/2) * pi * 1.6 m^2 / (9.8 m/s^2 * sqrt(0.48 kg * m^2 / (2.4 kg * 9.8 m/s^2))) = 0.17 m.

Thus, 0.17m is the distance from the pivot point at which the period is a minimum.

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