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Consider this scenario: There are $n$ tiny balls (tiny means we can fix several balls in one place) with mass $m_1,m_2,...$ fixed in a stick which we may ignore the mass of the stick. Now we rotate the system(stick and balls) around the center of mass. I feel like the greatest rotation inertia will be achieved when half the mass of balls is at one point of the stick while the other half is at the opposite point of the stick. But I don't know how to prove it mathematically?

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  • $\begingroup$ Is it always possible to divide $n$ different masses into two halves? $\endgroup$
    – Righter
    Dec 12, 2021 at 11:15
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    $\begingroup$ Why bother proving it mathematically, if the physical intuition is correct? After all, if we had to wait for rigorous mathematics, we would not have classical mechanics until three centuries after calculus had been invented. Proof comes in various flavours in physics as well as maths. $\endgroup$ Dec 12, 2021 at 12:06
  • $\begingroup$ @Righter Yes, you're right. $\endgroup$
    – Sherlock
    Dec 12, 2021 at 12:52
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    $\begingroup$ @MoziburUllah: But the OP did not say "rigorous mathematics". And your condition "if the physical intuition is correct" massively limits the scope and relevance of your comment. Math can help confirm or disconfirm our physical intuition; it can help elicit implicit conditions/assumptions needed for the conclusion to hold; it can help communicate the result in a convincing way to those who don't (yet) share our physical intuition; and it can help extend the result to situations that we don't (yet) have physical intuition for. $\endgroup$
    – ruakh
    Dec 13, 2021 at 1:37

1 Answer 1

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If there are $n$ balls, the total mass of all the balls is $m$ and the length of the stick is $L$

An arrangement such as $A$ wouldn't be maximum as we need the moment of inertia and it depends on $m_id^2$ and the $d_i$ could be greater by going to arrangement $B$

If we moved a proportion of the mass $km$ from one end of $B$ to make $C$, the COM is now a distance $kL$ from one end and $(1-k)L$ from the other.

enter image description here

The moment of inertia for $B$ is $$\frac{mL^2}{4}\tag1$$

For $C$ it's $(kL)^2(1-k)m+((1-k)L)^2km$ and that simplifies to $$k(1-k)mL^2\tag2$$

Expression 2) is only greater than 1) if $$k(1-k)\gt \frac{1}{4}\tag3$$ $$4k^2-4k+1 \lt0\tag4$$

$$(2k-1)^2\lt0\tag5$$

That's not possible, although we can get $0$ if $k=1/2$.

So changing the proportion of the mass at the ends of $B$ can't increase the moment of inertia. If some of the mass was moved from one end of $B$ to a place not at the other end of the rod, that wouldn't help, so arrangement $B$ gives the maximum moment of inertia.

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    $\begingroup$ The $k-1$ expressions on the diagrams should have been $1-k$ $\endgroup$ Dec 12, 2021 at 11:38
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    $\begingroup$ Thanks for the answer. I'm mainly confused about the arrangement $A$. Let's build arrangement $A$ from arrangement $C$.To be more concrete, suppose there is one tiny mass run from the left of the stick to the right of the center of mass(we may call it $O$ hereafter) of the stick while not reaching the right end. This case is not obvious, and I'm mainly confused with this. $\endgroup$
    – Sherlock
    Dec 12, 2021 at 12:01
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    $\begingroup$ @ Sherlock, yes it's not obvious, but from the main answer moving just one mass from the left to the far right would have a lower moment of inertia than $B$. Also moving one from the far left to a point not all the way to the right would be even lower as it's like moving all the way to the right with a smaller stick, so it would have to end up lower than $B$. $A$ would definitely be lower than $B$ as the $\sum m_i r_i^2$ would be lower as for many of the masses the $r_i$ are lower than $B$ and the COM is in the middle for both cases. $\endgroup$ Dec 12, 2021 at 12:47

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