Moment of Inertia for Hexagon w/ Point Masses & Length 7 cm

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Homework Help Overview

The problem involves calculating the moment of inertia for a regular hexagon with point masses located at each vertex. The hexagon has sides of length 7 cm, and the moment of inertia is to be determined about an axis through the center and perpendicular to the plane of the hexagon.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the use of the formula I=mr² and the method of calculating the radius for each mass. There are questions about the specific radius values used for the point masses at the vertices and how these contribute to the total moment of inertia.

Discussion Status

Some participants are exploring different methods for calculating the radius to each mass, with one participant noting a specific calculation for the top and bottom vertices. There is acknowledgment of the method being correct, but uncertainty remains about the overall calculation leading to the expected answer.

Contextual Notes

Participants mention that the sides of the hexagon are made of rods with negligible mass, which may influence the moment of inertia calculation. There is also a reference to the geometric properties of the hexagon being composed of equilateral triangles.

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Homework Statement



a regular hexagon with sides of length 7cm has a point mass of 1kg at each vertex. what is the moment of inertia for rotation about an axis which goes through the center of the hexagon and is perpendicular to the place of the hexagon? note the sides are made of rods with negligible mass.

so i thought.. I=mr^2. so i went on to find the radius of to each mass and would sum them. but i do not get the answer. the answer by the way is .0294

thanks.
 
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hellothere123 said:

Homework Statement



a regular hexagon with sides of length 7cm has a point mass of 1kg at each vertex. what is the moment of inertia for rotation about an axis which goes through the center of the hexagon and is perpendicular to the place of the hexagon? note the sides are made of rods with negligible mass.

so i thought.. I=mr^2. so i went on to find the radius of to each mass and would sum them. but i do not get the answer. the answer by the way is .0294

thanks.

What radius to each mass did you get?

Your method is OK.
 
well... for the top and bottom two.. i used.. sqrt[ (.07/2)^2 + ((.07/2)sqrt(3))^2 ] which is to the top corners.. so i thought it would be the same for the other corners so multiplied that by 4.

then i have the ones on the sides which would be just .07 as the radius.
 
hellothere123 said:
well... for the top and bottom two.. i used.. sqrt[ (.07/2)^2 + ((.07/2)sqrt(3))^2 ] which is to the top corners.. so i thought it would be the same for the other corners so multiplied that by 4.

then i have the ones on the sides which would be just .07 as the radius.

A hexagon is made up of 6 equilateral triangles. That means that they are all at .07 m - all 6 of them.
 
oh, haha, i see now... THANKS A BUNCH!
 

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