NEET2011PhysicsRotational MotionActual
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
Options
- A43 ~kg - m ^2
- B34 ~kg - m ^2
- C27 ~kg - m ^2
- D72 ~kg - m ^2
Correct answer
A. 43 ~kg - 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. I=I₁+I₂+I₃+I₄ I=m₁ r₁^2+m₂ r₂^2+m₃ r₃^2+m₄ r₄^2I=(1 0)+(2 0)+ (3 3^2 )+4(-2)^2I=0+0+27+16=43 ~kg - m ^2