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.

Answers

Answer 1

Answer:

The moment of inertia of the system about an axis through the center of the square, perpendicular to its plane is $0.0636 \;\rm kg-m^(2)$ .

Given data:

The mass of each sphere is, $m = 0.200 \;\rm kg$ .

Length of side of square is, $L = 0.400 \;\rm m$ .

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

$I = 4 mR^(2)$

Here,

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

$R =(1)/(2)\sqrt{L^(2)+L^(2)}\nR =(1)/(2)\sqrt{0.400^(2)+0.400^(2)}\nR = 0.282 \;\rm m$

Then, moment of inertia is,

$I = 4 mR^(2)\nI = 4 * 0.200 * 0.282^(2)\nI = 0.0636 \;\rm kg-m^(2)$

Thus, the moment of inertia of the system about an axis through the center of the square, perpendicular to its plane is $0.0636 \;\rm kg-m^(2)$ .

Learn more about moment of inertia here:

brainly.com/question/2176093?referrer=searchResults

Answer 2

Answer:

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

$\texttt{ }$

Further explanation

Let's recall Moment of Inertia formula as follows:

$\boxed{ I = m R^2 }$

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:

$R = (1)/(2) √(x^2+x^2)$

$R = (1)/(2) √(2x^2)$

$R = (1)/(2)√(2) x$

$R = (1)/(2) √(2) (0.400)$

$\boxed{R = 0.200√(2) \texttt{ m}}$

$\texttt{ }$

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

$\Sigma I = mR^2 + mR^2 + mR^2 + mR^2$

$\Sigma I = 4mR^2$

$\Sigma I = 4(0.200)(0.200√(2))^2$

$\boxed{\Sigma I = 0.064 \texttt{ kgm}^2}$

$\texttt{ }$

Learn more

Ratio of Moment of Inertia : brainly.com/question/2176655
Net Torque on the objects : brainly.com/question/9612563
Impacts of Gravity : brainly.com/question/5330244
Effect of Earth’s Gravity on Objects : brainly.com/question/8844454
The Acceleration Due To Gravity : brainly.com/question/4189441

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Answer details

Grade: High School

Subject: Physics

Chapter: Rotational Dynamics

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Answers

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6. A 73-kg woman stands on a scale in an elevator. Thescale reads 810 N. What is the magnitude and
direction of the elevator's acceleration?

A. 0.23 m/s2 up
B. 1.3 m/s2 up
C. 6.5 m/s2 down
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Answers

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Learn more :brainly.com/question/18266310

Answer:

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Answers

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