A 114-kg sculpture hangs from a horizontal rod that serves as a pivot about whic

Question

A 114-kg sculpture hangs from a horizontal rod that serves as a pivot about which the sculpture can oscillate. The sculpture's moment of inertia with respect to the pivot is 10.3 kg·m2, and when it is swung at small amplitudes, it is found to oscillate at a frequency of 0.403 Hz. How far is the pivot from the sculpture's center of mass?

Map Sapling Learning macmillan learning 114-kg sculpture hangs from a horizontal rod that serves as a pivot about which the sculpture can oscillate. The sculpture's moment of inertia with respect to the pivot is 10.3 kg m2, and when it is swung at small amplitudes, it is found to oscillate at a frequency of 0.403 Hz. How far is the pivot from the sculpture's center of mass? Number 1.53 Im Incorrect. Previous Give Up & View Solution @Try Again (?) Next el Exit Explanation

Explanation / Answer

let m = 114 kg

I_support = 10.3 kg.m^2

f = 0.403 Hz

we know,

T = 1/f = 1/0.403 = 2.48 s

Time period of physical pendulum,

T = 2*pi*sqrt(I_support/(m*g*Lcm))

here I_support is the moment of inertia of the object about the pivot.

Lcm is the is the distance from pivot to center of mass of the object.

so,

T^2 = 4*pi^2*I_support/(m*g*Lcm)

Lcm = 4*pi^2*I_support/(m*g*T^2)

= 4*pi^2*10.3/(114*9.8*2.48^2)

= 0.0592 m

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