CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

Four point masses, each of value $$m$$, are placed at the corners of a square $$ABCD$$ of side $$l$$. The moment of inertia of this system about an axis passing through $$A$$ and parallel to $$BD$$ is:

20442_042d196113bf4680993440ea4f661a04.jpg


A
3ml2
loader
B
3ml2
loader
C
ml2
loader
D
2ml2
loader

Solution

The correct option is C $$ml^{2}$$
$$I_A=0$$   (on the Axis .)
$$I_B=I_D=m  r^2$$
$$=m\left ( \dfrac{l}{\sqrt2{}} \right )^2$$
$$=\dfrac{ml^2}{2}$$
$$I_C=m(\sqrt{2}l)^2$$
$$I_C=2ml^2$$
$$I=I_A+I_B+I_C+I_D$$
$$I = 0 +\dfrac{ml^2}{2}+\dfrac{ml^2}{2}+2ml^2$$
$$I=3  ml^2$$

Physics

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image