Moment of Inertia: 4 Spheres Connected by Rods in Square

[amm617]

Homework Statement

Four small spheres, each of which you can regard as a point of mass 0.200 , are arranged in a square 0.400 on a side and connected by light rods.

The picture has 4 spheres connected by rods in the shape of a square. Theres a point O in the middle of the square and a horizontal line through the point O with point A on one end and point B on the other end.

a)Find the moment of inertia of the system about an axis through the center of the square, perpendicular to its plane (an axis through point O in the figure).

b)Find the moment of inertia of the system about an axis bisecting two opposite sides of the square (an axis along the line AB in the figure).

c)Find the moment of inertia of the system about an axis that passes through the centers of the upper left and lower right spheres and through point O.


I=m1(r1^2)+m2(r2^2)+...

 

[Nate810]

Homework Equations I=m1(r1^2)+m2(r2^2)+...The Attempt at a Solution a)I=m1(r1^2)+m2(r2^2)+m3(r3^2)+m4(r4^2)I=(0.200)(0.2^2)+(0.200)(0.2^2)+(0.200)(0.2^2)+(0.200)(0.2^2)=0.0032 kgm^2b)I=(0.200)(0.2^2)+(0.200)(0.2^2)+(0.200)(0.2^2)+(0.200)(0.2^2)+(0.200)(0.4^2)=0.0192 kgm^2c)I=(0.200)(0.2^2)+(0.200)(0.2^2)+(0.200)(0.4^2)=0.0128 kgm^2

 

[Moetasim]

+mn(rn^2) is the formula for moment of inertia, where I is the moment of inertia, m is the mass of each sphere, and r is the distance from the axis of rotation to each sphere.

a) To find the moment of inertia about an axis through the center of the square, we can use the parallel axis theorem. This theorem states that the moment of inertia of a system about an axis that is parallel to the original axis of rotation is equal to the moment of inertia about the original axis plus the product of the mass and the square of the distance between the two axes. In this case, the original axis of rotation is through point O and the new axis is through the center of the square. The distance between these two axes is the half the length of one side of the square, which is 0.200 m. Therefore, the moment of inertia about the new axis is:
I = (0.200)(0.200^2) + (0.200)(0.200^2) + (0.200)(0.200^2) + (0.200)(0.200^2) = 0.016 kg•m^2

b) To find the moment of inertia about an axis bisecting two opposite sides of the square, we can use the perpendicular axis theorem. This theorem states that the moment of inertia of a planar object about an axis perpendicular to the plane is equal to the sum of the moments of inertia about two perpendicular axes within the plane. In this case, the two perpendicular axes are through the centers of the top and bottom spheres, and through the centers of the left and right spheres. The moment of inertia about these two axes can be calculated using the formula above, and then added together. Therefore, the moment of inertia about the axis bisecting two opposite sides of the square is:
I = 2(0.200)(0.200^2) + 2(0.400)(0.400^2) = 0.176 kg•m^2

c) To find the moment of inertia about an axis passing through the centers of the upper left and lower right spheres and through point O, we can again use the parallel axis theorem. The distance between this axis and the original axis through point O is the diagonal of the square, which is 0.400 m. Therefore, the moment of inertia about this axis is:
I =