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

Question

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-m2 (B) 34 kg-m2 (C) 27 kg-m2 (D) 72 kg-m2

Tardigrade Admin · Accepted Answer

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,

I = I_{1} + I_{2} + I_{3} + I_{4}

∴ 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) + (3 \times 3^{2}) + 4 (- 2)^{2}

I = 0 + 0 + 27 + 16 = 43 k g - m^{2}

Q. Point masses $1, 2, - 3$ and $4 k g$ 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

$43 k g - m^{2}$

$34 k g - m^{2}$

$27 k g - m^{2}$

$72 k g - m^{2}$

Q. Point masses 1,2,−3 and 4kg 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

43kg−m2

34kg−m2

27kg−m2

72kg−m2

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, I=I1​+I2​+I3​+I4​ ∴I=m1​r12​+m2​r22​+m3​r32​+m4​r42​ I=(1×0)+(2×0)+(3×32)+4(−2)2 I=0+0+27+16=43kg−m2

Q. Point masses $1, 2, - 3$ and $4 k g$ 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

$43 k g - m^{2}$

$34 k g - m^{2}$

$27 k g - m^{2}$

$72 k g - m^{2}$