Rotational inertia - globes connected by a thin rod

[naji0044]

Homework Statement

Finding the rotational inertia

Relevant Equations

I=∑m*r^2

Problem Statement: Finding the rotational inertia
Relevant Equations: I=∑m*r^2

A rigid body of 2 massive globes with homogenous mass distribution and a thin rod is connecting the 2 globes. The globes has radius R1 = 0.18 m and R2 =0.28 m and masses m1=193 kg and m2=726 kg. The thin rod has the mass ms=10kg and the length L=0.88 m. THe body can rotate frictionless around the rotation axis in the middle of the thin rod.

a) Calculate the the body's overall inertiamoment and show that the inertiamoment is I=476 kgm^2

my calculations are m1*(1/2*L+r1)^2+m2*(1/2*L+r2)^2 and i doubt if this is even correct and can't tell how the inertiamoment of the rod would look. Ignore the F(force)

 

[Doc Al]

Hint: Don't treat the spheres as if they were point masses. (Think parallel axis theorem.)

 

[haruspex]

cant tell how the inertiamoment of the rod would look

You should have a standard formula for the MoI of a uniform stick about its centre. Similarly for a uniform solid sphere.

 

[naji0044]

okay so for a solid sphere i use 2/5*m*r^2 and for the rod it is 1/12*m*L^2?

 

[Doc Al]

okay so for a solid sphere i use 2/5*m*r^2 and for the rod it is 1/12*m*L^2?

Yes, about their respective centers.

 

[naji0044]

Okay so you use the formula I=Icm+Md^2 and therefore:
2/5*m1*r1^2+m1*(r1+1/2*L)^2 + 2/5*m2*r2^2+m2*(-r2-1/2*L)^2 + 1/12*m3*L+m3*1

 

[haruspex]

m3*1

Where does this term come from?
The rest is fine.

 

[naji0044]

Where does this term come from?
The rest is fine.

honestly i am
Where does this term come from?
The rest is fine.

 

[naji0044]

I am not sure how to calculate the rod when the rotation axis is right on the middle of it

 

[haruspex]

I am not sure how to calculate the rod when the rotation axis is right on the middle of it

That's what the $\frac 1{12}mL^2$ formula gives you. There is nothing to add in this case.