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I'm currently working my way through an introductory mechanics course, using University Physics with Modern Physics by Young and Freedman.

I understand that the first condition for equilibrium, i.e. $\sum \pmb{F} = \mathbf{0}$, ensures that the center of mass of the rigid body in question not accelerate.

What I don't understand is how the second condition of equilibrium, that the net torque $\sum \pmb{\tau}$ with respect to a point $O$ be zero, implies that the rigid body is not rotating around about $O$ (by "rotating about $O$", I assume my textbook really means rotating about some axis through $O$ - is this correct?). Perhaps it doesn't, but this is what I intuitively feel like we are trying to accomplish with the second condition of equilibrium.

The textbook goes on to claim that if a rigid body is not rotating about $O$, then its angular momentum with respect to $O$ is zero. Furthermore, since the rate of change of this angular momentum is given by $\sum \pmb{\tau}$, letting the latter be equal to zero ensures that the rigid body continues to not rotate about $O$, the authors argue.

Could somebody please provide me with a proof of why the second condition of equilibrium is equivalent to the body not rotating about any point, if this is so? If not - exaplain what the second condition of equilibrium does accomplish.

I hope my question is clear - if not, tell me, and I will try to clarify.

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  • $\begingroup$ The actual text in the book (equation 11.2) says: "for a nonrotating object to stay nonrotating" as the significance of the net torque being zero. Where did you get this interpretation from? What edition of the text are you using? $\endgroup$
    – nasu
    Jan 7, 2022 at 13:35
  • $\begingroup$ You are right @nasu, my it doesn't say that. I will clarify this in an edit. $\endgroup$
    – Simon SMN
    Jan 7, 2022 at 14:28

1 Answer 1

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First, recheck the textbook. Are you * sure * you read and quoted it correctly? It sounds like perhaps you misread. See comments below.

That said, lets answer the question as asked....

Something's wrong with these statements as asked and quoted....

$\sum \pmb{F} = \mathbf{0}$, ensures that the center of mass of the rigid body in question not accelerate.

Correct. For a rigid body and in a classroom context, no net force, means no net acceleration. It means whatever velocity it has, it keeps ("objects in motion stay in motion unless acted on by a force"). That's worth saying explicitly, given your question....

$\sum \pmb{\tau}$ with respect to a point $O$ [being] zero, implies that the rigid body is not rotating around about $O$

Incorrect. Torque is the rotational "equivalent" of a force, in simple terms. This would be like saying that if there is no net force, an object has no velocity. Whereas what it has, is no net acceleration (again in a simple classroom sense).

What it should say is, net torque $\sum \pmb{\tau}$ with respect to a point $O$ being zero, implies that the rigid body has no net angular acceleration around $O$. (In a simple classroom sense). As with linear forces, it can certainly have an angular velocity.

And yes, that means "around any axis through $O\$", as you surmise.

The textbook goes on to claim that if a rigid body is not rotating about $O$, then its angular momentum with respect to $O$ is zero.

Correct. If it's not rotating around $O$, then it has zero angular velocity (hence also angular momentum) around $O$. Note that this doesn't mean there is no net torque. For example if a flywheel spins clockwise, and you constantly apply a counterclockwise torque, it will instantaneously have zero angular velocity and momentum before starting to spin counterclockwise. That's the same as applying a constant linear force to an object that changes velocity from +ve to -ve.

Since the rate of change of this angular momentum is given by $\sum \pmb{\tau}$, letting the latter be equal to zero ensures that the rigid body continues to not rotate about $O$ [= zero angular velocity], the authors argue.

Incorrect. If the rate of change (of anything!!) is zero, that doesn't mean the thing itself is zero. Basic calculus 101. If net force is zero, it doesn't "ensure" an object has zero velocity. If net torque is zero, it doesn't "ensure" an object has zero angular velocity. End of.

Put another way, using their own words, "since the rate of change of this angular momentum is given by $\sum \pmb{\tau}$, letting the latter be equal to zero" ..... If you set the rate of change of angular momentum to zero, then by definition the angular momentum doesn't change. So it's constant. But nothing at all says it has to be constant and zero. Could be constant and any value.

Bottom line.....

Could somebody please provide me with a proof of why the second condition of equilibrium is equivalent to the body not rotating about any point?

If quoted accurately, your textbook authors sound like they need to get better technical proofreaders before their next textbook. But honestly, it sounds like you just misread it or didn't read it as written. It happens.

What's good is, you had the instinct in the topic, to realise that what you thought you read, sounded wrong. And indeed it was.

Don't sweat it, sounds like you understand it just fine.

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  • $\begingroup$ The text does not say this. The OP should provide the actual statement from thebook and not his "interpretation". $\endgroup$
    – nasu
    Jan 7, 2022 at 13:32
  • $\begingroup$ That's possible. And from your brevity, I assume you know the textbook and don't just have the OP's question to go on. I think this is still valuable because whatever the book says, the OP states this confuses them, and we can clarify. Perhaps "check you read the text carefully" is a second point, I've now added that up top, but the fact they are confused on this is still the basis of their question. $\endgroup$
    – Stilez
    Jan 7, 2022 at 13:36
  • $\begingroup$ Yes, I am using this book for the introductory physics course I am teaching and I checked with the text before I wrote my comments, of course. $\endgroup$
    – nasu
    Jan 7, 2022 at 13:39
  • $\begingroup$ Thought so, thank you for confirming. Hopefully the explanation will be of use anyhow. $\endgroup$
    – Stilez
    Jan 7, 2022 at 13:40
  • $\begingroup$ Sure, as the OP question show, the equilibrium conditions need some details. It is not uncommon to state the conditions of equilibrium without qualification. $\endgroup$
    – nasu
    Jan 7, 2022 at 13:47

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