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

[hellothere123]

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.

 

[LowlyPion]

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.

 

[hellothere123]

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.

 

[LowlyPion]

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.

 

[hellothere123]

oh, haha, i see now... THANKS A BUNCH!