The rigid body shown in Figure 10-66 (it is a triangle with the upper mass = to M and the bottom two corners/masses equal to 2M. The bottom side length equals .70 cm with point P in the middle of the side and the other two sides equal .55 cm)

consists of three particles connected by massless rods. It is to be rotated about an axis perpendicular to its plane through point P. If M = 0.40 kg, a = 35 cm, and b = 55 cm, how much work is required to take the body from rest to an angular speed of 5.0 rad/s?
*My work so far*
I believe I need to use:
W = (1/2)(I)(ang speed)^2
I am not sure how to find I though. Would I use I = I1 + I2 + I3 and calculate I for each of the 3 masses?

1.51

To find the moment of inertia (I) of the rigid body, you can indeed calculate the moment of inertia for each individual mass and then add them together.

The moment of inertia for a point mass is given by:

I = m * r^2

where m is the mass of the point and r is the perpendicular distance from the axis of rotation to the point mass.

For the upper mass (M), the perpendicular distance from point P to the axis of rotation is a/2, so the moment of inertia for this mass is:

I1 = M * (a/2)^2

For the bottom masses (2M), the perpendicular distance from point P to the axis of rotation is b/2, so the moment of inertia for each of these masses is:

I2 = 2M * (b/2)^2

Now, you can sum up the individual moments of inertia to find the total moment of inertia for the entire system:

I = I1 + I2 + I3

where I3 = I2 since the two bottom masses are identical.

Once you have the moment of inertia, you can use the equation you mentioned:

Work (W) = (1/2) * I * (angular speed)^2

Plug in the values of I and the given angular speed to find the work required to take the body from rest to an angular speed of 5.0 rad/s.