Cheenta
How 9 Cheenta students ranked in top 100 in ISI and CMI Entrances?
Learn More

Moment of Inertia

Four small spheres, each of which you can regard as a point of mass (0.2)Kg are arranged in a square (0.4)m on a side and connected by extremely light rods. Find the moment of inertia of the system about an axis through the centre of the square.
Discussion:

The length of each side of a square is (0.4)m. Thus, the perpendicular bisector is at (0.2)m. The length of the bisector dropped from the point of intersection of the diagonals is (0.2)m. Therefore, the distance of the vertices from the point of intersection of diagonals is $$ r=\sqrt{(0.2)^2+(0.2^2)}=0.2828m$$
The moment of inertia $$ I=MR^2= 4(0.2)(0.2828)=0.0640kg m^2$$

Knowledge Partner

Cheenta is a knowledge partner of Aditya Birla Education Academy
Cheenta

Cheenta Academy

Aditya Birla Education Academy

Aditya Birla Education Academy

Cheenta. Passion for Mathematics

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.
JOIN TRIAL
support@cheenta.com