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.