# Homework Help: A triangle's rotational kinetic energy problem

1. Apr 2, 2008

1. The problem statement, all variables and given/known data
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. Relevant equations
I= (m)(r$$^{2}$$) + (m)(r$$^{2}$$) + (m)(r$$^{2}$$)
K$$_{rot}$$=(1/2)Iw$$^{2}$$

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

I halved the height to get the radius.

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

2. Apr 2, 2008

### 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".

Last edited: Apr 2, 2008
3. Apr 2, 2008

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

4. Apr 2, 2008

### 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.

5. Apr 2, 2008