# Four particles, one at each of the four corners of a square with 2.5-m-long edges, are connected by mass less rods. The masses of the particles are m1 = m3 = 2.0 kg and m2 = m4 = 7.0 kg. Find the moment of inertia of the system about the z axis.

Question ID
525250

Created
April 4, 2011 3:58pm UTC

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0

2

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1. Equation: I=M*L^2

M is the mass of the object.
I is the moment of inertia.
L is the length squared or radius squared.

I=(2.5^2(m2+m1))+ (2.5^2(m2+m3))
I= 6.25*(9)+6.25*(9)
I=112.5 kg*m^2

525777

Created
April 5, 2011 1:54pm UTC

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4

2. Well, to find the moment of inertia of the system about the z axis, we can think of the square as a spinning circus act. Picture those particles as clowns on unicycles!

Now, let's calculate the moment of inertia. Since the rods are mass-less, we only need to consider the masses of the particles. Let's label the corners of the square as A, B, C, and D, with A being the bottom left corner and going clockwise.

To calculate the moment of inertia, we need to consider the distance from each particle to the z-axis and square it, then multiply it by the mass of that particle:

Iz = (m1 * r1^2) + (m2 * r2^2) + (m3 * r3^2) + (m4 * r4^2)

Since the square is symmetrical, the distances from the particles to the z-axis are the same. Let's represent that distance as "r."

Iz = (m1 + m2 + m3 + m4) * r^2

Now, we just need to plug in the values:

Iz = (2.0 kg + 7.0 kg + 2.0 kg + 7.0 kg) * r^2

Iz = 18.0 kg * r^2

So, the moment of inertia of the system about the z axis is 18.0 kg times r squared. Just don't let those clown particles start juggling chainsaws on those unicycles!