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Four identical particles (mass of each = 0.24 kg) are placed at the vertices of a rectangle (2.0 m × 3.0 m) and held in those positions by four light rods which form the sides of the rectangle. What is the moment of inertia of this rigid body about an axis that passes through the center of mass of the body and is parallel to the shorter sides of the rectangle?

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boffeemadrid

Answer:

2.16 kgm²

Explanation:

Width = 3 m

According to the question distance would be

[tex]r=\dfrac{Width}{2}\\\Rightarrow r=\dfrac{3}{2}\\\Rightarrow r=1.5\ m[/tex]

Moment of inertia is given by

[tex]I=mr^2\\\Rightarrow I=0.24\times 1.5^2\\\Rightarrow I=0.54\ kgm^2[/tex]

Moment of inertia of system from the center of mass

[tex]I_s=4I\\\Rightarrow I_s=4\times 0.54\\\Rightarrow I_s=2.16\ kgm^2[/tex]

The moment of inertia of this rigid body about an axis that passes through the center of mass of the body and is parallel to the shorter sides of the rectangle is 2.16 kgm²

samueladesida43

The moment of inertia of the rigid body about an axis that passes through the center of mass of the body and is parallel to the shorter sides of the rectangle is 2.16kgm².

How to calculate moment of inertia?

The moment of inertia can be calculated by using the following formula:

I = mr²

  • I = moment
  • m = mass of substance
  • r = distance

According to this question, the width of the rectangle is 3m.

r = width/2 = 3/2 = 1.5m

Moment of inertia is calculated as:

I = 0.24 × 1.5²

I = 0.54 kgm²

However, the moment of inertia of system from the center of mass is calculated as follows:

I(s) = 4I

I(s) = 4 × 0.54

I(s) = 2.16kgm².

Therefore, moment of inertia of the rigid body about an axis that passes through the center of mass of the body and is parallel to the shorter sides of the rectangle is 2.16kgm².

Learn more about mass at: https://brainly.com/question/19694949

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