A uniform rod of mass M and length L is pivoted at distance x from its center an

Question

A uniform rod of mass M and length L is pivoted at distance x from its center and undergoes harmonic oscillations.

Randomized VariablesM = 4.6 kg
L = 1.5 m
x = 0.33 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. A uniform rod of mass M and length L is pivoted at distance x from its center and undergoes harmonic oscillations.

Randomized VariablesM = 4.6 kg
L = 1.5 m
x = 0.33 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.

Explanation / Answer

A) moment of inertia is I = Icm+ (M*d^2)

Icm = M*L^2/12 = (4.6*1.5^2)/12 = 0.8625 kg*m^2

here d = x

then required moment of inertia is (1/12)*M*L^2 + (M*x^2)

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B) T = 2*pi*sqrt(L^2+12x^2/(12*g*x))

T = 2*3.142*sqrt((1.5^2+(12*0.33^2))/(12*9.81*0.33)) = 1.901 S

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C) minimum period is occured at xm = L/sqrt(12)

then xm = 1.5/sqrt(12) = 0.433 m

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