# Applying same torque equation but getting different results

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1. Nov 17, 2015

### Better WOrld

1. The problem statement, all variables and given/known data

As shown in the attached figure, 2 forces $F_1=3N$ and $F_2=5N$ are applied on the rod of mass $1$ kg.The separation between the points of application of the 2 forces is 0.2 m. If the rod undergoes only translational motion, find the length of the rod.

Note that both the forces pass above the CM of the rod.Also, the rod is in the vertical plane ie gravity passes through its CM along the rod itself.

2. Relevant equations

3. The attempt at a solution

I first tried the problem by taking torque of $F_1$ and $F_2$ about the CM and equating it with $0$ since the angular acceleration is 0. On solving, I got the correct answer: the length of the rod is $1m$. However, when I tried to take torque about the top of the rod (ie the point of application of $F_1$) I get a result I can't understand.

$$0 \times F_1+ 0.2 \times F_2+\dfrac{l}{2} \times g=\dfrac{ml^2}{3} \alpha$$
$$0+1+0=0$$
$$1=0$$ since $\alpha$ of any point on the rod about the CM is $0$.

I just cannot understand the flaw with my reasoning, and would be truly grateful for any assistance with my second method. Many thanks in advance!

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2. Nov 17, 2015

### TSny

The equation $\tau_{net} = I\alpha$ is not always valid. It is valid for an object rotating about an axis that is fixed in an inertial reference frame. It is also valid for the case where the origin is chosen at the center of mass and moves with the center of mass (even if the center of mass is accelerating). This is why you get the correct answer when taking torques about the center of mass.

If you take the origin at the top of the rod, you need to be careful. First, you need to be clear about what it means to choose the origin at the top of the rod. One meaning would be that the origin is fixed in an inertial frame, but coincides with the top of the moving rod at the instant of analysis. Another meaning would be to let the origin move with the top of the accelerating rod, in which case the origin is accelerating relative to an inertial frame.

If you want to go with the first meaning where the origin is fixed in an inertial frame, then you can use the more general law $\tau_{net} = \frac{dL}{dt}$, where $L$ is the total angular momentum of the rod about the origin.

If you go with the second option where the origin moves with the top of the rod as the rod accelerates, then you can use $\tau_{net} = I\alpha$ or $\tau_{net} = \frac{dL}{dt}$. But you will need to take into account that you are working in an accelerated frame of reference by introducing an appropriate fictitious (pseudo) force.

3. Nov 17, 2015

### Better WOrld

Thanks a lot for clearing my doubt!

4. Nov 18, 2015

### Better WOrld

This is how I had solved the original question: since the CM was accelerating at $2ms^{-1}$, I included a pseudo force to the left. Thus, my torque equation became as shown below:
$$0\times F_1+(-0.2\hat{j})\times F_2\hat{i}+\dfrac{-l}{2}\hat{j}\times F_p\hat{-i}=0$$
$$0+1-l=0$$

However, I can't understand how to solve the problem when the rod isn't accelerating either translationally or rotationally. In this case, the axis at the top of the rod is an inertial frame of reference. Thus, we cannot include pseudo forces on the CM. Taking torque about the axis on top, I get the same result as I had go in my original doubt.

Lastly, I wasn't able to understand how to solve the problem using $\tau=\dfrac{dL}{dt}$. Could you please clear my doubts? Many thanks!

5. Nov 18, 2015

### Better WOrld

Sorry, but I have some fresh confusions in my mind:(

This is how I had solved the original question: since the CM was accelerating at $2ms^{-1}$, I included a pseudo force to the left. Thus, my torque equation became as shown below:
$$0\times F_1+(-0.2\hat{j})\times F_2\hat{i}+\dfrac{-l}{2}\hat{j}\times F_p\hat{-i}=0$$
$$0+1-l=0$$

However, I can't understand how to solve the problem when the rod isn't accelerating either translationally or rotationally. In this case, the axis at the top of the rod is an inertial frame of reference. Thus, we cannot include pseudo forces on the CM. Taking torque about the axis on top, I get the same result as I had go in my original doubt.

Also, in Rotation, when we speak of Inertial/Non-Inertial Frames, do we refer to the axis (say about which we tae torque) as the frame of reference? As in would an accelerating axis be termed as a non inertial frame of reference? What exactly is a frame of reference in Rotation?

Lastly, I wasn't able to understand how to solve the problem using $\tau=\dfrac{dL}{dt}$. Could you please clear my doubts? Many thanks!

6. Nov 18, 2015

### insightful

How do you know the acceleration is 2ms-2 if mass is not known?

7. Nov 18, 2015

### TSny

If you want to us the law $\vec{\tau}=\dfrac{d\vec{L}}{dt}$ for an inertial frame in which you choose the origin to coincide with the top of the rod at the instant that you want to apply the law, you need to think about how to write the angular momentum of the rod relative to the chosen origin.

The angular momentum $\vec{L}$ of a system about a point (origin) of some reference frame is the sum of $\vec{r} \times(dm \, \vec{v})$ for all mass elements $dm$, where $\vec{r}$ is the position of the mass element relative to the origin and $\vec{v}$ is the velocity of the mass element relative to the reference frame.

A theorem of mechanics states that you can express the angular momentum of any system in the form $$\vec{L} = \vec{R}_{cm} \times (M \, \vec{V}_{cm}) + \vec{L}_\text{rel to cm}$$ Here, $\vec{R}_{cm}$ is the position of the center of mass, $M$ is the total mass of the system, $\vec{V}_{cm}$ is the velocity of the center of mass, and $\vec{L}_\text{rel to cm}$ is the angular momentum of the system relative to the center of mass.

So, for your rod, the first term on the right of the above equation says that there is a contribution to the angular momentum that comes from considering the entire mass of the rod as concentrated at the center of mass of the rod and treating the rod as a single particle. The second term would be due to any rotation of the rod about the center of mass.

8. Nov 18, 2015

### Better WOrld

Sir, in the original question, it was mentioned that the mass of the rod is 1kg, and that it translates with $a=2ms^{-2}$

9. Nov 18, 2015

### Better WOrld

Thank you Sir But Sir, please could you explain how to take the time derivative? Would it be by expressing angular momentum first as a function of t by using the standard equations of motion?

Aslo Sir, could you please clear the other 2 doubts I had mentioned in my previous post?

10. Nov 18, 2015

### insightful

Sorry; missed that. It seems to me the answer (1 m) is independent of the mass.

11. Nov 18, 2015

### Better WOrld

It's alright Sir. Sir, could you please explain the fallacy in my method? In the modified question, I took both F1 and F2 as 5N. Thus, the rod does not translate with any acceleration; it continues to move with its initial (unspecified) constant velocity. Sir, I am to calculate the length of the rod.

I took the torque about the top of the rod, but only got an incomprehensible result.
Since the rod has no angular acceleration, torque about the top of the rod is 0.

$$0\times F_1+(-0.2\hat{j}\times F_2\hat{i})=0$$
$$1=0$$

Please could you show me how to solve the question using torque at the top of the rod? Also could you please explain the fallacy in my reasoning?

12. Nov 18, 2015

### TSny

Consider the term $\vec{R}_{cm} \times (M \, \vec{V}_{cm})$. Try taking the time derivative of this term and simplifying. You do not need to know the explicit expressions for $\vec{R}_{cm}$ and $\vec{V}_{cm}$ as functions of time in order to get a simple result. You just need to think about the meaning of the time derivatives of $\vec{R}_{cm}$ and $\vec{V}_{cm}$.

13. Nov 18, 2015

### insightful

In this case, can you draw a picture of the problem, and see what the length must be to have no angular acceleration? (Specifying the 0.2m in this case adds to the confusion.)

14. Nov 18, 2015

### insightful

Going back to Post #1, in l/2 x g, what is g?

15. Nov 18, 2015

### Better WOrld

Sorry Sir, it should be mg ie the weight.

16. Nov 18, 2015

### Better WOrld

Sir I was unable to do so. Sir I was just thinking about the physical possibility of the situation, and it seems that t=such a situation (2 forces in opposite directions with a separation between their points of impact, and yet no angular acceleation) is not possible for the rod. Is that why I get 1=0?

17. Nov 18, 2015

### insightful

Doesn't the weight act downward? How is the weight relevant here?

18. Nov 18, 2015

Yes.