Rotational Motion Question 199
Question: 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:
A) $43kgm^{2}$
B) $34kgm^{2}$
C) $27kgm^{2}$
D) $72kgm^{2}$
Show Answer
Answer:
Correct Answer: A
Solution:
[a] 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.
$\therefore I=I _{1}+I _{2}+I _{3}+I _{4}$
$=0+0+27+16=43kgm^{2}$
BETA
