A uniform thin bar of mass 3 kg and length 0.9 m is bent to make an equilateral triangle. The moment of inertia of the triangle through its centre of mass and perpendicular to plane of triangle is :

Question

A uniform thin bar of mass 3 kg and length 0.9 m is bent to make an equilateral triangle. The moment of inertia of the triangle through its centre of mass and perpendicular to plane of triangle is : (A)  0.030 kg-m2  (B)  0.015 kg-m2  (C)  0.090 kg-m2  (D)  0.045 kg-m2

Tardigrade Admin · Accepted Answer

From parallel axis theorem, I=IC M+M a2 where IC M is moment of inertia about COM and a is distance of mass from axis of rotation. Moment of inertia of thin bar is (M l2/12) ∴ I=(M l2/12)+M a2 From Δ O A O, tan 30°=(a/l / 2) ⇒ a=(l/2) tan 30° ∴ I=(M l2/6) ∴ Total M I=(M l2/6)+(M l2/6)+(M l2/6) =0.045 kg - m 2

Q. A uniform thin bar of mass $3 k g$ and length $0.9 m$ is bent to make an equilateral triangle. The moment of inertia of the triangle through its centre of mass and perpendicular to plane of triangle is :

$0.030 k g - m^{2}$

$0.015 k g - m^{2}$

$0.090 k g - m^{2}$

$0.045 k g - m^{2}$

Q. A uniform thin bar of mass 3kg and length 0.9m is bent to make an equilateral triangle. The moment of inertia of the triangle through its centre of mass and perpendicular to plane of triangle is :

0.030kg−m2

0.015kg−m2

0.090kg−m2

0.045kg−m2

Q. A uniform thin bar of mass $3 k g$ and length $0.9 m$ is bent to make an equilateral triangle. The moment of inertia of the triangle through its centre of mass and perpendicular to plane of triangle is :

$0.030 k g - m^{2}$

$0.015 k g - m^{2}$

$0.090 k g - m^{2}$

$0.045 k g - m^{2}$