(11%) Problem 7: A uniform rod of mass M and length L is free to swing back and

Question

(11%) Problem 7: A uniform rod of mass M and length L is free to swing back and forth by pivoting a distance r from its center. It undergoes harmonic oscillations by swinging back and forth under the influence of gravity Randomized Variables M4.8 kg L=1.5m x=021 m 33% Part (a) In terms of M, L, and x, what is the rod's moment of inertia , about the pivot point. 33% Part (b) Calculate the rod's period T in seconds for small oscillations about its pivot point. 33% Part (c) In terms of L, find an expression for the distance xm for which the period is a minimum. Grade Summa Deductions Potential

Explanation / Answer

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part A :

MOment of inertia about pivot axis is obatined from parallel axis theorem

I = Io + Mx^2

Io = ML^2/12

so

Ip = (ML^2/12) + Mx^2
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part B:

time period T = 2pi sqrt(I/Mgx)

T = 2pi sqrt((ML^2/12 + Mx^2))/Mgh

Solving for time T

T = 2pi *sqrt( (L^2+12x^2)/12gx))
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Part C:

use dx/dt = 0

from U/v rule

2pi *[(L^2 + 12x^2)/(12 g x)}-3/2 *-1/2 * 2 {(24x *12 gx) -(12g *(L^2 +12x^2)/2gx)^2 = 0

Solving fro x,

gives out that

x = 0.29 L

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