Calculating Inertia

1. Dec 5, 2007

Destrio

Two particles, each with mass m, are fastened to each other and to a rotation axis by two rods, each with length L and mass M. The combination rotates around the rotation axis with angular velocity ω. Obtain an algebraic expression for the rotational inertia of the combination about the axis.

I = m1r1^2 + m2r2^2
I = ML^2 + M(2L)^2
I = 5ML^2

where am I going wrong? or am I not taking some factors into consideration?

thanks

2. Dec 5, 2007

Destrio

anyone?
this seems like it should be an easy problem but it really has me stumped

thanks

3. Dec 5, 2007

Staff: Mentor

You forgot to include the rotational inertia of the rods.

4. Dec 5, 2007

Destrio

so inertia of the particles is found with:
I = m1r1^2 + m2r2^2
I = mL^2 + m(2L)^2
I = 5mL^2

so inertia of the rods:
I = (1/3)(2M)(2L)^2
I = (8/3)ML^2

total inertial = 5mL^2 + (8/3)ML^2

thanks

5. Dec 5, 2007

Staff: Mentor

Looks good to me. (Assuming the rods and masses are arranged in a straight line perpendicular to the axis.)

6. Dec 5, 2007

they are