Moment of Inertia and two uniform solid spheres

[meganw]

Homework Statement

Two uniform solid spheres of mass M and radius R0 are connected by a thin (massless) rod of length 1.6R0 so that the centers are 3.6R0 apart.

(a) Determine the moment of inertia of this system about an axis perpendicular to the rod at its center.

________________ MR^2

(b) What would be the percentage error if the masses of each sphere were assumed to be concentrated at their centers and a very simple calculation could be made?

Homework Equations

mr^2 = moment of inertia for a point mass

The Attempt at a Solution

I'm confused. I guess it's not a point mass because I keep doing it that way but I keep getting the wrong answer. (2*1.8)=3.6 1.8 is the R and 2 for the second M.

What's the real formula I should be using for this? Thanks! I need help. :)

 

[physixguru]

Calculate the distance between the centre of rotation and the spheres...
You can't treat the spheres as point masses...their MI should be calculated about the axis of rotation as 2MR^2/5 AND added to get the inertia for the system.for this you must use the parallel axis theorem to calculate the MI of spheres about the axis of rotation...

 

[meganw]

Okay I tried it again using the parallel axis theorem but I got it wrong:

I=Icm +Mh^2

Icm=2/5*MRo^2
Mh^2=M*(1.6Ro)^2

Multiply them by two for the two masses:

I=4/5MR0^2+5.12MRo^2
=5.92 * MRo^2

It says that's wrong. What am I doing incorrectly?

 

[Doc Al]

How did you get h = 1.6Ro?

 

[meganw]

h

OH, Dr. Al, you're right! I was using 1.6 instead of 1.8.

Which gives you 11% error because you just do 2(M)(1.8 Ro)^2 which is 6.48 and % error is Theoretical (7.28) -Actual (6.48) / Theoretical = 10.98% error.

Thank you! =) You've really helped me and I really appreciate it!