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I have been asking and learning about angular momentum. I know that an object traveling straight has ang. momentum about a point of rotation.

Just curious, this example shows the torque acting on a meteor traveling in a straight line. Therefore, wouldn't there be work done by torque? Normally work would be $W=\int \vec{\tau} \cdot \overrightarrow{d \theta}$, but here that doesn't work. https://courses.lumenlearning.com/suny-osuniversityphysics/chapter/11-2-angular-momentum/

I know that it is the same as $W=\vec{F}\cdot\vec{d}$, and can calculate as such, but was trying to see how they are the same.

What angle would you use since it doesn't change like it would due in true circular motion?

Kevin C Speltz
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  • Unclear: Why do you say that $W= \int \vec{\tau}\cdot \vec{d\theta}$ doesn't work? do you know that for a fact, are you guessing, or have you done a derivation? – Bill N Sep 21 '20 at 16:49
  • I can't open your link? – Eli Sep 21 '20 at 17:17
  • I updated with a new link. Sorry the other one doesnt work – Kevin C Speltz Sep 21 '20 at 17:18
  • Bill N, I was thinking of using W=∫τ⃗ ⋅dθ→ , but i am not sure of the angle. As the angle doesn't change much due to the fact the radius is changing. So i was thinking we maybe have to use the distance traveled, and the relationship between DS= R time theta, and working it backwards, which would be the angle as if the torque was applied in a circular motion. But i am not sure. I know in circular motion, the work done by torque is equal to the sumation of the linear work. But in a case where we look at torque about a particle moving straight line, i must be missing something – Kevin C Speltz Sep 21 '20 at 17:24
  • How do you apply a torque on a particle again? – JAlex Sep 21 '20 at 19:51
  • Did you look at the link JAlex? There is an example where they ask one to calculate the torque created on the Meteor about the origin which is where the person is standing. So there is a torque on the meteor about that axis, and in that case, how you calculate teh work done on it via the torque? If you take the force and distance, in terms of linear motion it works, but not in rotational values – Kevin C Speltz Sep 21 '20 at 20:56
  • @KevinCSpeltz The angle is changing along with T as the meteor falls. You have the picture. Do the geometry. You shouldn't assert "that doesn't work here" without actually doing the math. – Bill N Sep 22 '20 at 03:13
  • @KevinCSpeltz For some reason I didnt get a notification. Be sure to use the "@" before the name just to make sure. I read somewhere a long time ago that it wasn't needed, but I guess it is. – Bill N Sep 22 '20 at 03:15
  • @Bill N, they say the torque is constant, as the perpendicular radius relative to the origin doesn't change. So if torque is constant, then it doesn't work when I do the math. That's why I'm asking – Kevin C Speltz Sep 22 '20 at 13:12
  • Does this answer your question? Calculate force between rotating objects – John Alexiou Sep 22 '20 at 23:59
  • Not really. Did you see the meteor example? They explain there is a torque that is constant on the meteor about the origin. My curiosity is if we can view a torque on the meteor versus just a linear force, how does that work via work done. I can calculate the kinetic energy from force time distance. Shouldn't I be able to do same from torque, but radisu changes and theta is not uniformly changjng, so it doesn't – Kevin C Speltz Sep 23 '20 at 00:08

1 Answers1

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Work is defined as $$W=\int\vec{F}\cdot\mathrm{d}\vec{s}.\hspace{1in}(1)$$ The relationship for torque and work which states $$W=\int\vec{\tau}\cdot\mathrm{d}\vec{\theta}\hspace{1in}(2)$$ is only true if the particle experiencing the torque is at a constant distance from the axial point, i.e., rigid body rotation or circular orbits. Derivations which develop equation (2) are assuming circular motion.

That is not the case here. The meteor is getting closer to the origin (the point chosen for torque calculations). To account for the change in distance one must add a term to equation (2), $$W_r=\int\vec{F}\cdot\hat{r}\mathrm{d}r.\hspace{1in}(3)$$

Edit

For this problem the $x$ cooridinate is fixed at $X$, but $y$ is changing. So we can write $$r=\frac{X}{\cos\theta}.\hspace{1in}(4)$$ $\theta$ is the angle with respect to the horizontal ($x$-axis). We also note that $\vec{F}$ makes an angle $\pi/2 - \theta$ with the $\hat{r}$ so $$\vec{F}\cdot\hat{r}=F\sin\theta.\hspace{1in}(5)$$ From (4) we can determine d$r$: $$\mathrm{d}r=\frac{X\sin\theta}{\cos^2\theta}\mathrm{d}\theta.$$

Now we can put all the pieces together for (3) and integrate from $\theta_1$ to $\theta_2$ and see that (2)+(3) gives the same result as (1). I'll leave the details of the integrals to the curious person, but it does work out properly.

End Edit

If one considers elliptical planetary orbits, this makes sense: the kinetic energy of the satellite or planet or moon is increasing and decreasing which means work is being done, but the torque is zero which means equation (2) gives zero. The work being done is done by the radial component of the attractive force as the object moves toward or away from the attractive center.

Bill N
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  • Okay, so I cannot use Work = T x Theta since this isn't true circular motion. Most problems are linear anyways so a rotation about a point is ignored. We just focus on the force about its C of M. But regardless and this was my main wonder, there is a constant torque about the origin on the meteor, and if i use equation #1 or #3 i will get same equal work Correct? I know there has to work done, because as you said, the kinetic energy is increasing. And if i integrate right(calc was long time ago) would it be Work= F x (x^2)/2. – Kevin C Speltz Sep 22 '20 at 16:49
  • @KevinCSpeltz If you use (1) or the sum of (2) and (3) you will get the same values. The integral in (3) would be require you to account for the change in angle between the downward force and the changing $\vec{r}$ direction. – Bill N Sep 22 '20 at 16:55
  • so I have to add #2 and #3? Can you explain how or show me the derivation how #3 makes up the angle difference given constant torque about the origin in this example? – Kevin C Speltz Sep 22 '20 at 18:46
  • @KevinCSpeltz See my edit. – Bill N Sep 22 '20 at 19:58
  • having a hard time following step#5. R(hat) is the radius/lever arm the torque is acting on right? So would F*R(hat) =Frsintheta. Are we missing an R?. Cos(π/2- theta) equals sin theta – Kevin C Speltz Sep 22 '20 at 22:27
  • @KevinCSpeltz $\hat{r}$ is the unit vector pointing from the origin to the meteor. It has length 1. There is an X distance in the dr value. No, there isn't a missing R. Yes, $\cos(\pi/2-\theta)=\sin(\theta).$ – Bill N Sep 22 '20 at 23:10
  • Still not following. Okay F is straight up and down in the problem. If we wanted to find Force along r(hat), wouldn't that be F/(sin theta). Because the triangle I draw has Rhat following origin to meteor, F is up/down,lever arm is parallel to x axis. Maybe I still missing something, sorry – Kevin C Speltz Sep 22 '20 at 23:27
  • @KevinCSpeltz Do you know how to calculate a dot product of two vectors? That's what is in equation (3), and yes it is the component of $\vec{F}$ along the radial, but F is the hypotenuse. – Bill N Sep 23 '20 at 02:11
  • Sorry, I wasn't thinking. I wasn't taking into account F would be hypotenuse give the radial part. Making more sense now. Thank you – Kevin C Speltz Sep 23 '20 at 03:11