1
$\begingroup$

Consider a particle of mass m that collides and sticks to a rod of mass M hanging from a pivot at some distance d from the center of mass of the rod. The result of such a collision would alter the center of mass of the rod, along with the moment of inertia.

I understand the trivial case of the parallel axis theorem is to derive the moment of inertia about the end of a rod using it's moment of inertia for the center of mass, and the theorem only works when starting with the moment of inertia about the center of mass.

I have the following inquiry:

Given the new center of mass, call it CM', and as a result the distance between the old and new center of mass, can one recover the moment of inertia about the new center of mass through the following formula?

If so, can the following formula then be applied to find the moment of inertia about the pivot point of the rod?

My suspicion is, as long as the first equation is valid, the second equation is valid.

$\endgroup$
2
  • $\begingroup$ The notion of a new center of mass is a bit confusing. The combined center of mass was always there, as well as each individual center of mass. As far as the combined center of mass nothing changes before or after the impact. $\endgroup$ Oct 17, 2021 at 19:04
  • $\begingroup$ From where are you measuring the quantity (CM')? $\endgroup$
    – R.W. Bird
    Oct 18, 2021 at 13:05

2 Answers 2

1
$\begingroup$

You first equation is missing a term. You are summing up MMOI about the combined COM so you need the parallem axis theorem for both the rod and the particle

$$ \begin{aligned} I_c' & = I_c^{\rm rod} + M (\tfrac{L}{2} - c')^2 + m ( d - c')^2 \\ \end{aligned} \tag{1}$$

where $I_c^{\rm rod} = \tfrac{M}{12} L^2$, $d$ is the distance of the particle from the pivot, and $c'$ is the combined center of mass distance from the pivot. Of course the two are related since $c' (M+m) = M \tfrac{L}{2} + m d$.

So if we elimitate $d$ from (1) and express it in terms of the new COM it is

$$ \begin{aligned} I_c' & = I_c^{\rm rod} + (M+m) \tfrac{M}{m} \left( \tfrac{L}{2} - c' \right)^2 \\ \end{aligned} \tag{2}$$

If you go the other way and eliminate $c'$ from (1) then you have

$$ \begin{aligned} I_c' & = I_c^{\rm rod} + \tfrac{M m}{M+m} \left( \tfrac{L}{2} - d \right)^2 \\ \end{aligned} \tag{3}$$

Either way, once $I_c'$ is evaluated, then the parallel axis theorem is indeed used to transfer to the pivot

$$ \begin{aligned} I_R' & = I_c' + (M+m) \left( c' \right)^2 \\ & = I_c' + (M+m) \left( \tfrac{M \tfrac{L}{2} + m d}{M + m} \right)^2 \\ & = I_c' + \tfrac{1}{M+m} \left( M \tfrac{L}{2} + m d \right)^2 \end{aligned} \tag{4}$$

Here $I_R'$ is the MMOI of the combined system (and hence the ') at the pivot.

$\endgroup$
0
$\begingroup$

If the original CM is a distance (d) from the pivot, then the original (I) is (1/12)M$L^2$ + M$d^2$. If (m) strikes the rod at a distance (x) from the pivot, then the new (I) is (1/12)M$L^2$ + M$d^2$ + m$x^2$. I know this does not answer your question, but why complicate the problem?

$\endgroup$
1
  • $\begingroup$ You are right - I discussed this with my professor and realized I had a complete misconception of moments of inertia and when it is appropriate to simply add them - thank you! $\endgroup$
    – Phonon_DOS
    Oct 20, 2021 at 10:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.