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

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The discussion focuses on calculating the moment of inertia for a regular hexagon with point masses at each vertex. The correct formula, I=mr^2, is applied, but confusion arises regarding the radius for each mass. One participant calculates the radius for the top and bottom vertices using the Pythagorean theorem, while others clarify that all vertices are equidistant from the center. Ultimately, the correct moment of inertia is confirmed to be 0.0294 kg·m². The conversation highlights the importance of accurately determining the radius for each mass in such calculations.
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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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