Please help with I, rotational inertia, calculation - requires integration.?

Question

See the figure at http://i14.tinypic.com/40dz9eb.gif.

The masses, of mass m, are 'particles'. The rods, of mass M and length L, are 'thin'. I'm trying to help someone answer the question "What is the rotational energy and kinetic energy?" I think given I, the rest I can do OK. Many thanks to anyone who can help.

Answer 1

The moment of intertia about an axis of rotation is given by:

I = Integral of {r^2 dm}

For a point mass M located at a distance X from the axis, this reduces to

I = M*X^2

where r is the distance from the axis of rotation, and dm is the differential mass.

If you define a linear mass density (mass per unit length) for the "thin rod" sections in this problem as:

D = dM/dr

then dM = D dr

Note that if the thin rods have uniform density, then

D = M/L

Substituting this into the integral above, and assuming D is a constant, we have that:

I = D*Integral of {r^2 dr}

In this problem, we need to split up the integral to account for the point masses located at distances L and 2L from the axis of rotation:

I = D*Integral from 0 to L {r^2 dr}
+ m*L^2 + D*Integral from L to 2L {r^2 dr}
+ m*(2L)^2

I = (D*L^3)/3 + mL^2 + (8*D*L^3)/3 - (D*L^3)/3 + 4*m*L^2
I = 5*m*L^2 + (8*D*L^3)/3

But if D = M/L

I = 5*m*L^2 + (8*M*L^2)/3

The rotational kinetic energy is given by:

E = 0.5 * I * w^2

where w is the rotational velocity in rad/sec.