Moment of Inertia of this combined system?

[jaewonjung]

Homework Statement

A rigid rod of mass M and length L has moment of inertia 1/12 ML^2 about its center of mass. A sphere of mass m and radius R has moment of inertia 2/5 MR^2 about its center of mass. A combined system is formed by centering the sphere at one end of the rod and placing an axis at the other.


What is the moment of inertia of the combined system?

Relevant Equations

SumI=I1+I2+I3...
I=MR^2

Moment of inertia for a rod rotated around one end is 1/3ML^2 using the parallel axis theorem
Moment of inertia for a sphere is 2/5 MR^2

Itotal=Isphere+Irod=2/5MR^2+1/3ML^2

However, the answer is 2/5MR^2+1/3ML^2+ML^2

Why is there an extra ML^2 added to the moment of inertia?

 

[Master1022]

Moment of inertia for a rod rotated around one end is 1/3ML^2 using the parallel axis theorem
Moment of inertia for a sphere is 2/5 MR^2

Itotal=Isphere+Irod=2/5MR^2+1/3ML^2

However, the answer is 2/5MR^2+1/3ML^2+ML^2

Why is there an extra ML^2 added to the moment of inertia?

So you need to use the parallel axis theorem properly here. When you have compound systems, you need to shift both of their axes to the same point. You have done so for the rod, but not for the sphere. That moment of inertia is still through the centre of the sphere.

Also, always best to shift both separately to the same point and then add, rather than vice versa.

Hope that helps.

 

[jaewonjung]

Just for my understanding, which is the direction of the axis? Just to confirm that it is out of the plane of the rod?

Sorry for not including a picture. The axis is perpendicular to the rod, and is placed at the end of the rod that doesn't have the sphere attached to it.

 

[jaewonjung]

So you need to use the parallel axis theorem properly here. When you have compound systems, you need to shift both of their axes to the same point. You have done so for the rod, but not for the sphere. That moment of inertia is still through the centre of the sphere.

Also, always best to shift both separately to the same point and then add, rather than vice versa.

Hope that helps.

I see. shifting the axis from the center of the sphere to the other end of the rod is ML^2.
Thanks!