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

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In summary, the problem asks for the moment of inertia for rotation about an axis passing through the center of a regular hexagon with sides of length 7cm and point masses of 1kg at each vertex. The sides of the hexagon are made of rods with negligible mass. The formula used is I=mr^2, and the answer is .0294. The method used involves finding the radius for each mass, with the top and bottom two being calculated using the formula sqrt[ (.07/2)^2 + ((.07/2)sqrt(3))^2 ], while the ones on the sides are just .07. It is important to note that a hexagon is made up of 6 equilateral triangles, each
  • #1
hellothere123
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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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  • #2
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.
 
  • #3
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.
 
  • #4
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.
 
  • #5
oh, haha, i see now... THANKS A BUNCH!
 

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

What is the formula for calculating the moment of inertia for a hexagon with point masses and length 7 cm?

The formula for calculating the moment of inertia for a hexagon with point masses and length 7 cm is I = (m1r1^2 + m2r2^2 + m3r3^2 + m4r4^2 + m5r5^2 + m6r6^2), where m is the mass of each point and r is the distance from the center of mass to each point.

How do you find the center of mass for a hexagon with point masses and length 7 cm?

The center of mass for a hexagon with point masses and length 7 cm can be found by dividing the sum of each point's mass multiplied by its distance from an origin point by the total mass of the system. This can be represented by the formula xcm = (m1r1 + m2r2 + m3r3 + m4r4 + m5r5 + m6r6) / (m1 + m2 + m3 + m4 + m5 + m6).

Can the moment of inertia for a hexagon with point masses and length 7 cm be negative?

No, the moment of inertia for a hexagon with point masses and length 7 cm cannot be negative. It is a measure of an object's resistance to changes in rotational motion, so it must be a positive value.

What is the significance of the moment of inertia for a hexagon with point masses and length 7 cm?

The moment of inertia for a hexagon with point masses and length 7 cm is significant because it helps determine an object's angular acceleration when subjected to an external torque. It is also important in designing and analyzing the stability and balance of rotating systems.

How does the distribution of point masses affect the moment of inertia for a hexagon with length 7 cm?

The distribution of point masses affects the moment of inertia for a hexagon with length 7 cm because the distance of each point from the center of mass is a factor in the calculation. If the point masses are evenly distributed, the moment of inertia will be lower compared to a system with more mass concentrated at the edges or corners of the hexagon.

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