How Cheenta works to ensure student success?
Explore the Back-Story

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 $$ I=MR^2= 4(0.2)(0.2828)$$=$$0.0640kg m^2$$

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 $$ I=MR^2= 4(0.2)(0.2828)$$=$$0.0640kg m^2$$

Leave a Reply

This site uses Akismet to reduce spam. Learn how your comment data is processed.

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
Menu
Trial
Whatsapp
Physics Program
magic-wandrockethighlight