# Moment of inertia of the system of a sphere attached to a rod rotating in the axis shown In this system moment of inertia of the whole system about the axis shown in the question is given ( sphere is able to freely rotate about its axis)

"i want to know why we are not treating sphere as a point mass? why are we considering its rotation about its own axis and adding its moment of inertia *as all the points of the sphere are moving in the same way around the axis * .i am not able to see any rotational motion of sphere about its axis so why are we considering its moment of inertia about its axis

But according to me Moment of inertia of this system should be like this. treating sphere as point mass located at L+R distance from the axis of rotation • Perhaps you might want to revisit the question. The question might have explicit statement saying that the sphere can freely rotate about an axis, very likely the rod would extent as an axis and the sphere could rotate about this axis. Mar 9, 2021 at 11:22
• Why do you think that the spheres should be point masses? They have a nonzero radius, besides if they were point masses your handwritten expression would still be incorrect, as the last term wouldn't have $(L+R)^2$. Mar 9, 2021 at 11:23
• because all the point of the sphere are moving in the same way around the axis and the last term would be correct beacuse if we treat it as a point mass than all its mass can be assumed at its geometrical center which is at a distance (L+R) from the axis of rotation Mar 9, 2021 at 11:31
• i am not able to see any rotational motion of the sphere about its axis so why are we considering its moment of inertia about its axis Mar 9, 2021 at 11:46
• All points of the sphere are "moving the same way around the axis", but the points do not all have the same radius from the axis. But the sphere is rotating in the same way the Earth's moon is rotating. Mar 9, 2021 at 11:47

In the limit when $$L>>R$$, Then you might ignore the distribution of mass and can consider the sphere as a point mass. That is reflected from the fact $$\lim_{L\rightarrow \infty} I_{\text{on image}}=I_{\text{by OP}}$$