A triangle's rotational kinetic energy problem

[Nadialy]

Homework Statement

The three 300 g masses in the figure are connected by massless, rigid rods to form a triangle.

What is the triangle's rotational kinetic energy if it rotates at 8.00 revolutions per s about an axis through the center?

Values given:
Mass of each ball at the corner of the triangle: 0.3 kg
Sides of Equilateral Triangle: 0.4m
It is rotating at 8.00 revolutions per second

Homework Equations

I= (m)(r$$^{2}$$) + (m)(r$$^{2}$$) + (m)(r$$^{2}$$)
K$$_{rot}$$=(1/2)Iw$$^{2}$$

The Attempt at a Solution

First thing i did was try to find the radius of the triangle. I cut it in half down the middle, making a right angle with a side of .2m and a hypotenuse of .4m.

(0.2$$^{2}$$) + (height$$^{2}$$) =(0.4$$^{2}$$)
height = 0.346

I halved the height to get the radius.

Radius: 0.173


Then i used the Equation I= (m)(r$$^{2}$$) + (m)(r$$^{2}$$) + (m)(r$$^{2}$$)

I = (0.3)(0.173$$^{2}$$) + (0.3)(0.173$$^{2}$$) + (0.3)(0.173$$^{2}$$)

I = 0.027

Then i converted 8 rev per sec into 50.24 rad per sec.

Then i used K$$_{rot}$$=(1/2)Iw$$^{2}$$

K$$_{rot}$$=(1/2)(0.027)(50.24)$$^{2}$$

K$$_{rot}$$ = 136

I don't know what I'm doing wrong. Please help.

 

[Dick]

The radius of rotation isn't 0.173m. Think about it. It's the hypotenuse of a triangle, one of whose sides is 0.2m. It should be the distance from the center of the triangle to the rotating masses, not the "height".

 

[Nadialy]

wouldn't that be .4 then? I don't think I understand what you're trying to tell me.

 

[Dick]

The 'r' in your formula should be the distance from the axis to the rotating masses. It isn't 0.173m nor is it 0.4m. Rethink your triangle.

 

[Nadialy]

Thank you. That was a silly place for me to make mistakes.