Point masses 1,2,3 and 4 kg are lying at the points...

Point masses

1,2,3

and 4 kg are lying at the points

(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 Point masses 1,2,3 and 4 kg are lying at the points (0,0,0),(2,0,0),(0,3,0) and (-2,-2,0) respectively. The moment of

Options

A $43 \mathrm{~kg}-\mathrm{m}^2$
B $34 \mathrm{~kg}-\mathrm{m}^2$
C $27 \mathrm{~kg}-\mathrm{m}^2$
D $72 \mathrm{~kg}-\mathrm{m}^2$

Correct answer

A. $43 \mathrm{~kg}-\mathrm{m}^2$

Step-by-step solution

Moment of inertia of the whole system about the axis of rotation will be equal to the sum of the moments of inertia of all the particles. Point masses 1,2,3 and 4 kg are lying at the points (0,0,0),(2,0,0),(0,3,0) and (-2,-2,0) respectively. The moment of

I=I_1+I_2+I_3+I_4

\therefore I=m_1 r_1^2+m_2 r_2^2+m_3 r_3^2+m_4 r_4^2

I=(1 \times 0)+(2 \times 0)+

\left(3 \times 3^2\right)+4(-2)^2 I=0+0+27+16=43 \mathrm{~kg}-\mathrm{m}^2

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