Question

13.
Point masses 1, 2, 3 and 4 kg are lying at the point (0, 0.0). (2.0.0), (0.3.0) and (-2.-2, 0)
respectively. The moment of inertia of this system about X-axis will be-
(A) 43 kg-mo? (B) 34 kg-m?
(C) 27 kg-m
(D) 72 kg-m?
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Answer 1

Answer:

(A)43 kg - m^2

Explanation:

Given Point masses 1,2,3 and 4 kg are lying at the point (0,0,0)(2,0,0)(0,3,0) and (-2,-2,0) respectively.The moment of inertia of this system about X-axis will be

Sum of moments of inertia of all particles is equal to moment of inertia of whole system about axis of rotation.

So I = I1 + I2 + I3 + I4

I = m1r1^2 + m2r2^2 + m3r3^2 + m4r4^2

We know that point masses 1, 2, 3,4 kg lies at (0,0, 0), (2,0,0), (0,3,0) and (-2,-2,0).

So I = 1 x 0 + 2 x 0 + 3 x 3^2 + 4 x - 2^2

 I = 0 + 0 + 27 + 16

I = 43 kg m^2

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