Q28 E - University Physics with Modern Physics

Question

Four small spheres, each of which you can regard as a point of mass of 0.200 kg, are arranged in a square of 0.400 m on a side and connected by extremely light rods (Fig. E9.28). 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);

 bisecting two opposite sides of the square (an axis along the line AB in the figure);

 that passes through the centers of the upper left and lower right spheres and point O.

Step-by-Step Solution

Answer

 (a) the center of the square, perpendicular to its plane is  0.0640km.m2

 (b) Bisecting two opposite sides of the square is ${0.032kg.m^2}^{}$.

(c) The center of the upper left and lower right sphere and through point o is $0.0320kg.m^2$

Step:-1 explanation

If we want to solve our question. we will be applying an equation that determines the moment of inertia 

 $I=\sum _{m_i{r_i}^2}$

Where, a $I$ moment of the inertia 

$m_i=$ mass

 $r_{i=}$ distance from the axis of the rotation.

Step:-2  Concept

  The center of the square, perpendicular to its plane

Here, we will assume applying the Pythagorean theorem,

 

First, we will calculate the distance from the axis.

 ____________(1)