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It is a rather simple question from my assignment but I have a confusion that needs to be resolved.

Find the mass $m$ that needs to be kept at one end of a thin rod of mass $M$ and length $L$, hinged at a distance $\frac{L}{4}$ from $m$ so that the rod stays horizontal.

I propose two solutions:

Solution 1: Let hinge be called O. Since rod is in equilibrium, balancing the torques about O gives equation: $$mg×\frac{L}{4} = Mg × \frac{L}{4} $$ (Assuming the mass $M$ at centre of mass of the rod, $\frac{L}{2}$ from $m$ and $\frac{L}{4}$ from hinge.) This gives $m= M$

Solution 2: (This is what I did the second time I tried the question) I realized that in taking the mass of rod at its centre of mass, I had neglected the presence of $m$. I rewrote the equation as: $$mg\frac{L}{4} = \left[\frac{ML}{2(m+M)} -\frac{L}{4}\right] (m+M)g$$ The quantity in square brackets is the distance of centre of mass from hinge and the total mass at centre of mass is of both the weight added and the rod's. This gives $m=M/2$

Solution 1 is what my teacher did but I feel that solution 2 is more appropriate and wholesome. I still wonder why he wouldn't count in $m$ for the rod's torque. Is it that the mass is not counted because the effect of (or consequence of) adding the mass is what we're calculating assuming the mass yet to be added? Or is it that I'm just overthinking and solution 2 is right?

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    $\begingroup$ "I realized that in taking the mass of rod at its centre of mass, I had neglected the presence of $m$". I can't deduce what you. mean by "presence" form the right side of your solution 2. Solution 1 is obviously the correct one. $\endgroup$
    – Bob D
    Commented Mar 26, 2021 at 13:15
  • $\begingroup$ @BobD try this way: the "rod plus mass m" has a centre of mass different from the "only rod" system so their distance from the pivot must differ. That's what case 1 and 2 show. $\endgroup$
    – Rew
    Commented Mar 26, 2021 at 13:33
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    $\begingroup$ But you really need to consider the contribution of the torque due to the mass M separately from the contribution of the torque due to m for the purposes of determining equilibrium about the hinge (pin support). $\endgroup$
    – Bob D
    Commented Mar 26, 2021 at 13:43
  • $\begingroup$ @BobD yeah that's what I'm asking about. Could you elaborate on why is it so? $\endgroup$
    – Rew
    Commented Mar 26, 2021 at 13:47
  • $\begingroup$ First, could you tell me exactly what you did with m to get the second equation? It's still not clear to me. $\endgroup$
    – Bob D
    Commented Mar 26, 2021 at 13:53

2 Answers 2

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Taking moments about the hinge O gives you Solution 1 as the correct one.

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Let's go back to why you wrote the first equation. You want to start with the idea that the linear combination of all torques on the rod sum to find the angular acceleration times some constant. And since the rod must stay horizontal, we assume the rotational acceleration is zero.

$$ \sum{\tau_i} = \alpha I = 0$$

We can decompose the torques in different ways, but they need to be accounted for exactly once. In the first equation you have separated the torques due to the mass of the rod and the torque due to the small mass.

$$ \tau_{\text{mass}} = -\tau_{\text{rod}} $$

In the second equation you have incorporated the mass and the rod together (which is fine), but then you set it equal to the small mass by itself. This is incorrect.

$$ \tau_{\text{mass}} = -\tau_{\text{rod+mass}} $$

If you want to do it that way, set it equal to zero, not to the torque of the small mass.

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