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Four small spheres, each of which you can regard as a point of mass 0.200 kg, are arranged in a square 0.400 m on a side and connected by light rods. Find the moment of inertia of the system about an axis through the center of the square, perpendicular to its plane.

Respuesta :

johanrusli

The moment of inertia of the system about an axis through the center of the square, perpendicular to the plane is 0.064 kg.m²

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Further explanation

Let's recall Moment of Inertia formula as follows:

[tex]\boxed{ I = m R^2 }[/tex]

where:

I = moment of inertia

m = mass of object

R = distance between the object and the axis of rotation.

Given:

mass of sphere = m = 0.200 kg

length of side = x = 0.400 m

Asked:

net moment of inertia = ΣI = ?

Solution:

Let's ilustrate this question as shown in the attachment.

Firstly , let's find distance between center of the square and the sphere:

[tex]R = \frac{1}{2} \sqrt{x^2+x^2}[/tex]

[tex]R = \frac{1}{2} \sqrt{2x^2}[/tex]

[tex]R = \frac{1}{2}\sqrt{2} x[/tex]

[tex]R = \frac{1}{2} \sqrt{2} (0.400)[/tex]

[tex]\boxed{R = 0.200\sqrt{2} \texttt{ m}}[/tex]

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Next , we could find total moment of inertia as follows:

[tex]\Sigma I = mR^2 + mR^2 + mR^2 + mR^2[/tex]

[tex]\Sigma I = 4mR^2[/tex]

[tex]\Sigma I = 4(0.200)(0.200\sqrt{2})^2[/tex]

[tex]\boxed{\Sigma I = 0.064 \texttt{ kgm}^2}[/tex]

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Learn more

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Answer details

Grade: High School

Subject: Physics

Chapter: Rotational Dynamics

Ver imagen johanrusli
snehashish65

The moment of inertia of the system about an axis through the center of the square, perpendicular to its plane is [tex]0.0636 \;\rm kg-m^{2}[/tex].

Given data:

The mass of each sphere is, [tex]m = 0.200 \;\rm kg[/tex].

Length of side of square is, [tex]L = 0.400 \;\rm m[/tex].

The expression for the moment of inertia of the system about an axis through the center of the square, perpendicular to its plane is,

[tex]I = 4 mR^{2}[/tex]

Here,

R is the distance between center of the square and the sphere. And its value is,

[tex]R =\dfrac{1}{2}\sqrt{L^{2}+L^{2}}\\R =\dfrac{1}{2}\sqrt{0.400^{2}+0.400^{2}}\\R = 0.282 \;\rm m[/tex]

Then, moment of inertia is,

[tex]I = 4 mR^{2}\\I = 4 \times 0.200 \times 0.282^{2}\\I = 0.0636 \;\rm kg-m^{2}[/tex]

Thus, the moment of inertia of the system about an axis through the center of the square, perpendicular to its plane is [tex]0.0636 \;\rm kg-m^{2}[/tex].

Learn more about moment of inertia here:

https://brainly.com/question/2176093?referrer=searchResults

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