Confusion About Rotational Motion

[BrandonInFlorida]

I watched a video that showed how to calculate the center of gravity of a horizontal bar suspended from two wires, one attached to each end. Each wire was then attached to a vertical wall. The angle each wire made with the wall it was attached to was given. They treated it as an a example of a rigid body in equilibrium, meaning that the vector sum of the forces is zero and the vector sum of the torques is zero.

They first used the condition that the horizontal components of the tensions on the two wires add to zero because there is no translational motion.. They then calculated the sum of the torques about a point in the bar and set it to zero. They used the variable "d" to be the distance of the point from the left end of the bar. They were able from these considerations to get a value for d, which they then called the center of gravity. What confuses me is this. Shouldn't the net torque be zero about each and every point in the bar and not just the center of gravity?

Here is the video:

 

[kuruman]

What confuses me is this. Shouldn't the net torque be zero about each and every point in the bar and not just the center of gravity?
Here is the video:

At equilibrium, the net torque is indeed zero about any point in space and not necessarily on the bar. By selecting the center of gravity, one limits the algebra that needs to be done. Try calculating torques about one of the ends of the bar. If you do, the weight of the bar enters the picture which means you have to also write the vertical equilibrium equation to eliminate it. Using the C.G. limits the number of equations needed from 3 to 2.

 

[BrandonInFlorida]

The problem is that they used the requirement that the net torque is zero to calculate d and then announced that they had calculate the center of gravity. If the net torque is zero about every point, then how did it even permit them to get a value for d?

 

[Lnewqban]

For the case of a uniform horizontal bar the CM should be located at $L/2$.

Please, see:
http://hyperphysics.phy-astr.gsu.edu/hbase/cm.html

http://hyperphysics.phy-astr.gsu.edu/hbase/cmms.html

 

[BrandonInFlorida]

Not what I asked. Since the zero net torque condition they applied is true for every single point, how did it permit them to get any specific point, much less the center of gravity? They found the center of gravity from the requirement that the net torque equal zero, but that is true at every point, so how did it permit them to calculate the center of gravity?

 

[Lnewqban]

Not what I asked. Since the zero net torque condition they applied is true for every single point, how did it permit them to get any specific point, much less the center of gravity? They found the center of gravity from the requirement that the net torque equal zero, but that is true at every point, so how did it permit them to calculate the center of gravity?

The weight force applied at the CM will tend to produce a torque around each of the supported ends of the bar.
The bar does not rotate because clockwise torque is cancelling counter-clockwise torque.
Remove one of the wires and the potential torque becomes a reality and a new balance is found (CM relocates itself directly below the hanging point).

What makes the location of the CM unique is that all natural potential torques induced by any portion of the body is canceled by an opposite one.
If you hang your bar from that unique point, you don’t need two wires but one to conserve equilibrium.

 

[BrandonInFlorida]

Sorry, but either I don't understand or you are not answering my question. How does the requirement used in the video that the net torque be zero around the point locate the center of gravity if the condition would be true at every point?

 

[kuruman]

Here is how.
We have already established in the video that $$T_1\sin\theta_1=T_2\sin\theta_2~~\rightarrow ~T_2=\frac{\sin\theta_1}{\sin\theta_2}T_1.$$Now consider the vertical equilibrium equation in which we replace $T_2$ with the expression already found: $$T_1\cos\theta_1+\frac{\sin\theta_1}{\sin\theta_2}T_1\cos\theta_2=W~\rightarrow~W=\left(\cos\theta_1+\frac{\sin\theta_1}{\tan\theta_2}\right)T_1$$Now calculate torques about the right end of the rod using $x$ as the distance from that end to the C.G. and the convention that clockwise torques are negative: $$-T_1\cos\theta_1 L+W x=0 ~\rightarrow~x=\left(\frac{T_1 }{W}\right)L\cos\theta_1.$$ Using the previous equation, $$x=\frac{L\cos\theta_1}{\cos\theta_1+\frac{\sin\theta_1}{\tan\theta_2}}.$$Is this answer correct? Let's see.

We know that $\sin\theta_1=0.6;~\cos\theta_1=0.8;~\tan\theta_2=\frac{4}{3}.$ Then $$x=\frac{2~(m)\times 0.8}{0.8+\frac{0.6}{4/3}}=1.28~\mathrm{m}$$That's from the right end of the rod. From the left end of the rod we have $d=L-x=2.00-1.28=0.72~\mathrm{m}$ which is the answer obtained in the video.

 

[Ibix]

Sorry, but either I don't understand or you are not answering my question. How does the requirement used in the video that the net torque be zero around the point locate the center of gravity if the condition would be true at every point?

If you take moments about anywhere except the center of gravity, the weight force has a moment. If you do a calculation that is not about the center of gravity and forget to include the weight force's torque, then you will find that the net torque is non-zero (because you forgot one of the forces).

If you calculate the net torque excluding the weight, it will be non-zero unless you work around the center of mass. Thus asking "about what point is the net torque excluding the weight zero?" is the same as asking "where is the center of gravity?".

 

[kuruman]

Sorry, but either I don't understand or you are not answering my question. How does the requirement used in the video that the net torque be zero around the point locate the center of gravity if the condition would be true at every point?

The torque equation has a look that depends on the choice of reference point. Nevertheless, the sum of the torques is zero. Compare the torque equations in my solution with the one in the video. It's like saying $3 -5+2 =0$ and $6-6=0$. If you are comfy with this, you should be comfy with the equilibrium torque equations.

 

[BrandonInFlorida]

Here is how.
We have already established in the video that $$T_1\sin\theta_1=T_2\sin\theta_2~~\rightarrow ~T_2=\frac{\sin\theta_1}{\sin\theta_2}T_1.$$Now consider the vertical equilibrium equation in which we replace $T_2$ with the expression already found: $$T_1\cos\theta_1+\frac{\sin\theta_1}{\sin\theta_2}T_1\cos\theta_2=W~\rightarrow~W=\left(\cos\theta_1+\frac{\sin\theta_1}{\tan\theta_2}\right)T_1$$Now calculate torques about the right end of the rod using $x$ as the distance from that end to the C.G. and the convention that clockwise torques are negative: $$-T_1\cos\theta_1 L+W x=0 ~\rightarrow~x=\left(\frac{T_1 }{W}\right)L\cos\theta_1.$$ Using the previous equation, $$x=\frac{L\cos\theta_1}{\cos\theta_1+\frac{\sin\theta_1}{\tan\theta_2}}.$$Is this answer correct? Let's see.

We know that $\sin\theta_1=0.6;~\cos\theta_1=0.8;~\tan\theta_2=\frac{4}{3}.$ Then $$x=\frac{2~(m)\times 0.8}{0.8+\frac{0.6}{4/3}}=1.28~\mathrm{m}$$That's from the right end of the rod. From the left end of the rod we have $d=L-x=2.00-1.28=0.72~\mathrm{m}$ which is the answer obtained in the video.

Strange that I have consistently asked why the point found is the center of gravity and not some other point, and you have consistently not answered.

 

[BrandonInFlorida]

If you take moments about anywhere except the center of gravity, the weight force has a moment. If you do a calculation that is not about the center of gravity and forget to include the weight force's torque, then you will find that the net torque is non-zero (because you forgot one of the forces).

If you calculate the net torque excluding the weight, it will be non-zero unless you work around the center of mass. Thus asking "about what point is the net torque excluding the weight zero?" is the same as asking "where is the center of gravity?".

Finally an answer to what I actually asked. The fact that the weight was not included in the calculation means we are talking about the center of gravity. Thank you.

 

[vanhees71]

I'm a bit puzzled what all this is about. By definition the center of mass is
$$\vec{x}_{\text{cm}}=\frac{1}{M} \int_{\mathbb{R}^3} \mathrm{d}^3 x \vec{x} \rho(\vec{x}),$$
where $\rho$ is the mass density of the body and $M$ its total mass.

This has a priori nothing to do with the question, whether there's a torque wrt. to the center of mass. This depends on the forces acting on the body and is given by
$$\vec{\tau}=\int_{\mathbb{R}^3} \mathrm{d}^3 x (\vec{x}-\vec{x}_{\text{cm}}) \times \vec{f}(\vec{x}),$$
where $\vec{f}$ is the force density acting on the body.

A body is in equilibrium if the total force as well as the total torque is 0.

 

[BrandonInFlorida]

I'm a bit puzzled what all this is about. By definition the center of mass is
$$\vec{x}_{\text{cm}}=\frac{1}{M} \int_{\mathbb{R}^3} \mathrm{d}^3 x \vec{x} \rho(\vec{x}),$$
where $\rho$ is the mass density of the body and $M$ its total mass.

This has a priori nothing to do with the question, whether there's a torque wrt. to the center of mass. This depends on the forces acting on the body and is given by
$$\vec{\tau}=\int_{\mathbb{R}^3} \mathrm{d}^3 x (\vec{x}-\vec{x}_{\text{cm}}) \times \vec{f}(\vec{x}),$$
where $\vec{f}$ is the force density acting on the body.

A body is in equilibrium if the total force as well as the total torque is 0.

I'm puzzled that you're puzzled. As I've said in this thread repeatedly, I didn't understand how applying a condition that is true for every single point would locate the center of gravity. Ibix explained it to my satisfaction.

 

[etotheipi]

I'm puzzled that you're puzzled. As I've said in this thread repeatedly, I didn't understand how applying a condition that is true for every single point would locate the center of gravity. Ibix explained it to my satisfaction.

I'm puzzled that you're puzzled that @vanhees71 is puzzled. A condition for equilibrium of a body is that the total torque (of all forces), which can be taken about any origin in space, is zero$$\vec{\tau}=\int_{\mathbb{R}^3} \vec{x} \times \vec{f}(\vec{x}) \mathrm{d}^3 x= \vec{0}$$Next, the centre of mass is a weighted mean over all mass elements and isn't affected by whatever fields you apply to the rigid body. The centre of gravity is a little different, and is defined as the point about which the gravitational force (only) has zero torque. Specifically,$$\vec{\tau}_g=\int_{\mathbb{R}^3} (\vec{x}-\vec{x}_{\text{cg}}) \times \vec{f}_g(\vec{x}) \mathrm{d}^3 x= \vec{0}$$That is to say you can replace all of the gravitational forces equivalently with a single force acting through the uniquely defined position $\vec{x}_{cg}$. The centre of gravity coincides with the centre of mass if the gravitational field is uniform, since you can eventually take the $\times \vec{g}$ out of the integral.

 

[hutchphd]

I did not plough through the video but here is the deal. Somewhere they use the fact that "the weight can be considered as being concentrated at the center of gravity" when they did the calculation. Then they find where that point must be for equilibrium. Not self-referential. (The existence of such a point is simply assumed)

Strange that I have consistently asked why the point found is the center of gravity and not some other point, and you have consistently not answered.

Strange that you would be so confrontational to people trying to help you.

 

[Lnewqban]

Sorry, but either I don't understand or you are not answering my question. How does the requirement used in the video that the net torque be zero around the point locate the center of gravity if the condition would be true at every point?

Example 2 of the video shows the same technique to horizontally locate the center of gravity of a non-uniform body respect to reactive forces.
You just weight each wheel and perform a geometrical calculation about the horizontal location of that theoretical center.
You do basically the same for the problem of the bar supported by two wires.

Please, note that those calculations are based only on the proportion of the magnitude of those weights or reactive forces respect to the distance between them.
Other than to establish that the body is in balance, you don't really need to invoke torques in balance.

That technique tells you absolutly nothing about the vertical location of the center of gravity, which may be important in some problems or real life situations.
It exists only because the total weight force, theorically located at that point, generates each reactive force via a torque and a lever.
Just imagine the front wheel to be the fulcrum, the weight vector pulling down at the CG of the car and the measured weight at the rear wheel as the reactive vertical force.

3-D location of the center of mass of body of a irregular shape does require consideration of all idividual masses and their distance to that theotical spatial center.
We don't need reactive forces or even gravity for that calculation.

 

[haruspex]

I'm puzzled that you're puzzled that @vanhees71 is puzzled.

Strange that you would be so confrontational to people trying to help you.

I'm with the OP on this. @Ibix picked up the key fact that the equation used in the video omitted a torque due to gravity, thereby implying the point is the c.o.g. Up to then, and even some later posts it seems, nobody had grasped what the OP's difficulty was. The OP's frustration is understandable, and perhaps a little emphasis was in order to make well-intentioned posters pay closer attention.
I do not read it as confrontational.

 

[Ibix]

I'm with the OP on this. @Ibix picked up the key fact that the equation used in the video omitted a torque due to gravity, thereby implying the point is the c.o.g. Up to then, and even some later posts it seems, nobody had grasped what the OP's difficulty was.

To be fair, the answer is implicit in #2 at least. That the OP hadn't accepted it as the answer was what pointed me at his specific difficulty.

 

[hutchphd]

And perhaps I misinterpreted frustration as snarkiness... if so I tender my apologies.