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**1. 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

**2. Homework Equations**

I= (m)(r[tex]^{2}[/tex]) + (m)(r[tex]^{2}[/tex]) + (m)(r[tex]^{2}[/tex])

K[tex]_{rot}[/tex]=(1/2)Iw[tex]^{2}[/tex]

**3. 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[tex]^{2}[/tex]) + (height[tex]^{2}[/tex]) =(0.4[tex]^{2}[/tex])

height = 0.346

I halved the height to get the radius.

Radius: 0.173

Then i used the Equation I= (m)(r[tex]^{2}[/tex]) + (m)(r[tex]^{2}[/tex]) + (m)(r[tex]^{2}[/tex])

I = (0.3)(0.173[tex]^{2}[/tex]) + (0.3)(0.173[tex]^{2}[/tex]) + (0.3)(0.173[tex]^{2}[/tex])

I = 0.027

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

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

K[tex]_{rot}[/tex]=(1/2)(0.027)(50.24)[tex]^{2}[/tex]

K[tex]_{rot}[/tex] = 136

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