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Why can the total force be taken as if applied on the COM. (rotational mechanics?) 
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#1
Jan713, 05:54 PM

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Why can the total force exerted on an object be taken as if a single force was applied on the center of mass?
I think at most the total force must be the sum of tiny equal forces uniformly distributed. The mass must at most be uniformly distributed too. And this only matters when we start talking about rotational mechanics, right? i.e: Torques. So that the torque can be calculated as [itex]\tau[/itex]=[itex]R \times{F}_{T}[/itex]. But why is this true? R being the vector of the position of the COM. I think it can be explained this way:Using the known equation: [itex]\boldsymbol{L}=\boldsymbol{L}_{CM}+\boldsymbol{L}_{spin}[/itex] [itex]{L}_{CM}=[/itex] is the L of a point particle in the COM with mass M. [itex]{L}_{spin}[/itex] is the L of body relative to it's center of mass. Now by symmetry. Since [itex]{L}_{spin}[/itex] is on the center of mass it must be 0. There's an equal amount of tiny forces on each side of the body which makes the torque 0. Can anyone just give me some insight into this? I just want to understand this basic clearly and there may be a simpler broader explanation. 


#2
Jan813, 01:05 AM

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Why can the total force exerted on an object be taken as if a single force was applied on the center of mass?[/quote]In general, it can't. You express the applied force in components through the com and perpendicular to that.



#3
Jan813, 04:16 PM

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luckily, I have just been going over rotational dynamics, so here's what I got! Firstly, I should say this is all for the motion of a rigid body. These equations won't work if the object is not rigid. First up, for any two points which are both fixed with respect to the rigid body:
[tex]\frac{d( \vec{r}' \vec{r})}{dt}= \vec{\omega} \wedge (\vec{r}'\vec{r})[/tex] Which I think is just another way of saying Euler's rotation theorem. Now, from this, if we calculate the total momentum of a rigid body we get: [tex]\vec{P}=M \frac{d \vec{r}}{dt} + M \ \vec{\omega} \wedge (\vec{R}  \vec{r})[/tex] Where [itex]\vec{P}[/itex] is the total momentum of the rigid body, [itex]M[/itex] is the total mass of the rigid body, [itex]\vec{R}[/itex] is the centre of mass of the rigid body and [itex]\vec{r}[/itex] is any arbitrary point which is fixed with respect to the rigid body. Finally, [itex]\vec{\omega}[/itex] represents the instantaneous rotation of the rigid body. So, now if we choose [itex]\vec{r}=\vec{R}[/itex] (which is always allowed, since the centre of mass is always a fixed point with respect to the rigid body), then the equation simplifies very nicely: [tex]\vec{P}=M \frac{d \vec{R}}{dt} [/tex] So we see the total momentum of a rigid object is due to the velocity of its centre of mass, times by the total mass of the object. Now, using a similar, but longer calculation, to get the total angular momentum of the rigid body, gives the equation: [tex]\vec{L}= \vec{R} \wedge \vec{P} + \underline{I_{(\vec{r})}} \cdot \vec{\omega}  M(\vec{r}  \vec{R}) \wedge \frac{d(\vec{r}  \vec{R})}{dt}[/tex] Where [itex]\underline{I_{(\vec{r})}}[/itex] is the inertia matrix, calculated around the point [itex]\vec{r}[/itex] (which, again, is any point which is fixed with respect to the rigid body). So from the above equation, you can see that the total angular momentum is due to the angular momentum of the centre of mass with respect to the origin (the first term), and the second term is due to the angular momentum of the rigid body around some point which is fixed with respect to the rigid body, and the third term has no easy interpretation, but it is sort of like the negative of the angular momentum of the fixed point with respect to the centre of mass. Now, remember the fixed point can be any point which is fixed with respect to the rigid body. so if you are given the the inertia matrix around some point which is not the centre of mass, then you can use the above formula. But if you are given the inertia matrix around the centre of mass, then you are very lucky, because the third term disappears, and we get: [tex]\vec{L}= \vec{R} \wedge \vec{P} + \underline{I_{(\vec{R})}} \cdot \vec{\omega}[/tex] Which is nicer. (and this is why the inertia matrix around the centre of mass is most often used). 


#4
Jan813, 04:25 PM

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Why can the total force be taken as if applied on the COM. (rotational mechanics?)
So is this nice property of the angular momentum around the centre of mass due to lots of equal and opposite forces? Not necessarily. You don't have to assume that to get these equations. (Although it is probably a sufficient condition).
To get these equations, the only thing you really need to do is assume that it is a rigid body, and then the rest will follow. Edit: I'm not even that sure that equal and opposite forces are a sufficient condition... Obviously, they don't imply a rigid body, so I don't think equal and opposite forces are a useful assumption. 


#5
Jan1113, 03:41 PM

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Sorry if my exposition wasn't so good.
I had done this when I posted but I was doubtful about something, it seems fine though. I just have to take the derivative applying the product rule: [itex]\boldsymbol{L}=\boldsymbol{L}_{CM}+\boldsymbol{L}_{spin}[/itex] [itex]\boldsymbol{\tau}=\boldsymbol{\tau}_{CM}+ \boldsymbol{\tau}_{spin}[/itex], but [itex]\boldsymbol{\tau}_{spin} = 0[/itex] * [itex]\boldsymbol{\tau}=\boldsymbol{\tau}_{CM}[/itex] QED *If a is constant: [itex]\boldsymbol {\tau}_{spin} = \sum {r_{i} \times m_{i}a} =( \sum {m_{i}r_{i}}) \times a=0 \times a=0[/itex] Bruce, I think you demonstrated the first equation correct? I accepted that but I was suspicious of it's direct result through derivation, I can't remember why but oh well. If I misunderstood something I'm sorry. I didn't know about Euler's rotational theorem. Thanks! Edit, clear up: The question was basically this. If a pencil is nailed to wall. If there's no friction and the pencil's horizontal what's the torque on the pencil? We just take the total force, put it on the center of mass of the pencil and torque=r*F, with r=length/2. Why is this legal? It's now answered I think. 


#6
Jan1213, 04:10 PM

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Also, it's important to remember that this is only true when the gravitational field is uniform. (That is why you were able to say that the acceleration due to gravity is the same on each small element of the rigid body). So if the gravitational field varied throughout space, then generally it would cause a torque around the centre of mass. 


#7
Jan1313, 06:42 AM

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#8
Jan1413, 07:27 PM

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mount the pencil on a pivot at the wall and hold it horizontal then let go. At the instant you let go  the total torque is taken as MgL/2 where M is the total mass of the pencil and L it's overall length. This assumes a uniform massdistribution for the pencil  so it is very blunt and has no eraser on the other end ;) This is "legal" because it agrees with the results of experiments.... i.e. the Universe works like this. Physics is an empirical science, so this is the bottom line. We can see that it is mathematically consistent with the rest of our physical models by considering how the total torque is made up ... gravity, after all, acts on the entire length of the pencil. Thus, you can work out the torque about the pivot due to gravity acting on each bit of the pencil and add them up (provided you are comfortable with torques adding up?) I doubt it will be very convincing if I do the math here  so you will have to. Hows your calculus? Try for differently shaped objects. Try it for an arbitrarily shaped mass distribution. Then you will see. Since this is the same result all the time, there is no need to repeat the whole calculation every single time we have to do one of these problems. 


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