Calculating Rotational Inertia: Two Masses and Two Rods

[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

 

[Destrio]

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

thanks

 

[Doc Al]

You forgot to include the rotational inertia of the rods.

 

[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

 

[Doc Al]

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

 

[Destrio]

they are

thanks again for your help
much appreciated