Answer to Question #77753 in Physics for Assertiveguy

Question

A thin rod of mass 0.3 kg and length 0.2 m is suspended horizontally by a metal wire
which passes through its centre of mass and is perpendicular to its length. The rod is set
to torsional oscillation with a period of 2 s. The thin rod is replaced by an equilateral
triangular lamina which is suspended horizontally from its centre of mass. If its period
of oscillations is found to be 10 s, the moment of inertia of the triangular lamina about
the axis of suspension is approximately
(A) 5.0 × 10−3 kg m2
(B) 0.5 × 10−3 kg m2
(C) 0.5 × 10−4 kg m2
(D) 2.5 × 10−2 kg m2

Answer 1

The period of a torsional pendulum is
T=2π√(I/κ)
where I is the moment of inertia and κ is torsion constant.
The rod moment of inertia about its center of mass is for mass M kg and length L
I_rod =1/12 ML^2=1/12 (0.3 kg) (0.2 m)^2=0.001 kg m^2
Calculate the torsion constant using the equation for the period:
κ=I_rod (2π/T_1 )^2

For lamina
I_lamina=(T_2/2π)^2∙κ=(T_2/2π)^2∙I_rod (2π/T_1 )^2=I_rod (T_2/T_1 )^2=(0.001) (10/2)^2=0.025kg m^2

Answer: (D) 2.5 × 10−2 kg m2