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A uniform rod of length $4x$ is rotating about the edge $O$ of the table. (The rod does not fall off the table.) The centre of mass $G$ of the rod is distance $x$ away from $O$. The rod is making an angle $\theta$ with the horizontal.

The only forces present are the weight $W$ of the rod, the normal reaction $N$ of the table on the rod and the frictional force $S$ that prevents the rod from slipping off the table as it rotates. Let the Radial direction point from O to G, and the Transverse direction be anticlockwise.

I apologise for not including a diagram but it should be very quick to sketch.

I would like to set up equation(s) of motion for the scenario above.

  1. Would it be appropriate to approach this problem using Newton's 2nd Law and then resolving the equation into radial and transverse components? If so, am I suppposed to be considering the motion of a point on the rod, or the motion of the rod as a whole body?

  2. Would it be appropriate to approach this problem by taking torques (say, about $O$), i.e. using $\tau=I \alpha$, where $\tau$ is the total external torque on the rigid rod?

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Hi Ryan - I think your question is easily fixed, I just removed the extra part that was causing confusion. Now, if you'd like to change what you're asking, feel free to edit it again, but keep in mind that the edit history is saved so you shouldn't include the original text of your question except where it becomes part of the new text. Think of it as rewriting more than editing. – David Zaslavsky♦ Apr 29 '12 at 18:11
Thank you. BTW for future reference, am I correct to understand that the "homework" tag doesn't necessarily have to mean actual homework problems? – Ryan Apr 29 '12 at 18:42
That is exactly correct. Basically, it applies to any question where you are doing a practice problem of some sort in order to learn the method of doing that type of problem. (As opposed to when you are doing a physics problem because you actually need the answer for something else.) Sometimes it's a little tricky to tell when the tag is needed, but mostly people are pretty good about editing it in or reminding you about it when something looks homework-y. See this meta question for more info. – David Zaslavsky♦ Apr 29 '12 at 18:51

1 Answer

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Given that the frictional force prevents the rod from slipping, this is purely a rotational problem. Therefore, it should be possible to solve it using only torque. When doing so, you consider torques on the entire rod, and the equation of motion you write will apply to the rod as a whole, not a single point of the rod. This is because the rod is a rigid body.

The moment the rod begins to slip, this becomes no longer a purely rotational problem, and after that you would need to use Newton's second law as well. The same qualifications apply, though; you consider forces on the entire rod, and the equation of motion you write using Newton's second law will apply to the rod as a whole, because it's a rigid body. Of course, depending on how it's worded, it's quite possible that in the actual problem you don't need to consider this case (after the rod starts slipping).

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So can you confirm that for purely rotational problems such as this, it doesn't make sense to try to use Newton's 2nd Law (for translation, not rotation, i.e. $F=ma$)? Cos I was given a "model answer" to this problem that had $S−Wsinθ=−m(xω^2)$ (using N2L and then resolving in the radial direction)...is this equation nonsense? – Ryan Apr 29 '12 at 19:05
$S-Wsin\theta = -m(x\omega^2)$ looks just like "resultant force in the radial direction = m times centripetal acceleration". So that should be OK shouldn't it? – twistor59 Apr 29 '12 at 19:14
@david If it were a particle rotating (revolving) around an axis, I understand that it has centripetal acceleration. But why is it ok to equate the resultant force on the rod (LHS) to the $ma$ of its centre of mass? Does our equation of motion here refer to the motion of the rod as a whole, or does it refer to the motion of the centre of mass of the rod?? – Ryan Apr 29 '12 at 20:04
@Ryan (3 comments up): using $\sum\vec{F}=m\vec{a}$ is also a possible way to do it. But that requires you to calculate the forces exerted by the corner of the table on the rod. The advantage of using torque is that you can ignore those forces. (1 up) Because the rod is a rigid body. Any translational motion of the rod as a whole is also translational motion of its center of mass. There is a mathematical derivation of this which you can look up in a good textbook (like K&K IIRC). – David Zaslavsky♦ Apr 29 '12 at 20:58
@david Assuming that the eqn above is meaningful (is it??), why is it ok to equate the resultant force on the rod (LHS) to the $ma$ of its centre of mass (RHS)?? This is the main crux of my struggle. In any case, am I right to think that even if both methods (taking torque, using N2L) are correct, they aren't interchangeable (i.e. whichever method to use depends on what info we require)? – Ryan Apr 29 '12 at 21:31
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