Four particles, one at each of the four corners of a square with 2.0-m-long edges, are connected by massless rods. The masses are m1=m3=3.0 kg and m2=m4=4.0 kg. Find the moment of inertia of the system about the z axis. (the z axis runs through m2, which is at the origin, m1 is on the y axis, and m3 is on the x axis. Use the parallel-axis theorem and the result for Problem 41 to find them moment of inertia of the four-particle system about an axis that passes through the center of mass and is parallel with the z axis. Check your result by direct computation.

Question

Mayra Bishop

Physics

14 June, 22:30

Four particles, one at each of the four corners of a square with 2.0-m-long edges, are connected by massless rods. The masses are m1=m3=3.0 kg and m2=m4=4.0 kg. Find the moment of inertia of the system about the z axis. (the z axis runs through m2, which is at the origin, m1 is on the y axis, and m3 is on the x axis. Use the parallel-axis theorem and the result for Problem 41 to find them moment of inertia of the four-particle system about an axis that passes through the center of mass and is parallel with the z axis. Check your result by direct computation.

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Konnor · Answer 1

The particles are in x-y plane with coordinates of masses as follows

m₂ at (0,0) m₁ at (0,2), m₄ at (2,2) and m₃ at (2,0)

Moment of inertia about z axis

I_z = 0 + 3 x 2² + 4 x (2√2) ² + 3 x 2²

= 12 + 32 + 12

= 56 kgm²

Now let us find out moment of inertia about axis through CM

According to theorem of parallel axis

I_z = I_g + m x r²

Here m is total mass that is 14 kg and r is distance between two axis which is √2 m

56 = I_g + 14 x (√2) ²

I_g = 56 - 28

= 28 kgm²

We can directly compute I_g as follows

I_g = 4 x (√2) ² + 3 x (√2) ² + 4 x (√2) ²+3 x (√2) ²

= 8 + 6 + 8 + 6

= 28 kgm²

So the result obtained earlier is correct.