5.5 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 x 10-3 kg m²
(B) 0.5 x 10-3 kg m
(C) 0.5 x 10-4 kg m
(D) 2.5 x 10-2 kg mº
Answers
Answer:
2.5 x 10^-2 kg m^2
Explanation:
Given 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
Torsional pendulum period is given by T = 2π√I /K , I is moment of inertia and K is torsion constant
The rod moment of inertia about the centre of mass M kg and length L is given by
I (rod) = 1/12 ML^2
= 1/12 x 0.3 kg x 0.2 m^2
= 0.001 kg m^2
So lamina = I rod (T 2 / T1)^2
= 0.001(10/2)^2
= 0.025 kg m^2
= 2.5 x 10^-2 kg m^2
, the moment of inertia of the triangular lamina about
the axis of suspension is approximately 2.5 x 10^-2 kg m^2