rotational equilbrium Watch

MEPS1996
Badges: 4
Rep:
?
#1
Report Thread starter 5 years ago
#1
I have read that the conditions for rotational equilibrium are that the sum of anticlockwise moments=sum of clockwise moments about any given point. I have also read that a body can be in rotational equilibirum if a single force acts on it and that this force acts through the centre of mass. However, surely if you take any point on the line of action of the force on the object, the wight of the object is the only moment about that point so the condition i stated earlier is not satisfied. I am thinking the original condition may be for both rotational and translational equilibrium. If i havent made myself clear please ask me. Can anybody shed some light on this?
thanks
0
reply
natninja
  • Community Assistant
Badges: 20
Rep:
?
#2
Report 5 years ago
#2
(Original post by MEPS1996)
I have read that the conditions for rotational equilibrium are that the sum of anticlockwise moments=sum of clockwise moments about any given point. I have also read that a body can be in rotational equilibirum if a single force acts on it and that this force acts through the centre of mass. However, surely if you take any point on the line of action of the force on the object, the wight of the object is the only moment about that point so the condition i stated earlier is not satisfied. I am thinking the original condition may be for both rotational and translational equilibrium. If i havent made myself clear please ask me. Can anybody shed some light on this?
thanks
because the mass distribution on either side of the centre of mass is effectively symmetric, the symmetry from this means all the individual moments will cancel and we only get linear acceleration and everyone s happy, if this weren't the case then any force would cause rotation.... I think that answers what you wanted to know but I'm not entirely sure what you were asking....
0
reply
Stonebridge
Badges: 10
#3
Report 5 years ago
#3
(Original post by MEPS1996)
I have read that the conditions for rotational equilibrium are that the sum of anticlockwise moments=sum of clockwise moments about any given point. I have also read that a body can be in rotational equilibirum if a single force acts on it and that this force acts through the centre of mass. However, surely if you take any point on the line of action of the force on the object, the wight of the object is the only moment about that point so the condition i stated earlier is not satisfied. I am thinking the original condition may be for both rotational and translational equilibrium. If i havent made myself clear please ask me. Can anybody shed some light on this?
thanks
The centre of mass of the object is defined as being the point about which all the moments sum to zero. So if you find this point and push there, there will be no rotation. If it did rotate you would have been pushing in the wrong place.
reply
MEPS1996
Badges: 4
Rep:
?
#4
Report Thread starter 5 years ago
#4
(Original post by Stonebridge)
The centre of mass of the object is defined as being the point about which all the moments sum to zero. So if you find this point and push there, there will be no rotation. If it did rotate you would have been pushing in the wrong place.
. I assume when you referred to moments you were referring to the moments of all the atoms in the object about the centre of mass. Imagine a diagram of a body with the lines of action of only its weight from its centre of mass and one other force whose line of action passes through the centre of mass. If you take moments about a point on the line of action of the force, the weight has a non-zero moment about that point. So moments about this point are non-zero, but the object is in rotational equilibrium? I can see their is some problem here in my definitions i am just trying to find it. please help
0
reply
MEPS1996
Badges: 4
Rep:
?
#5
Report Thread starter 5 years ago
#5
(Original post by natninja)
because the mass distribution on either side of the centre of mass is effectively symmetric, the symmetry from this means all the individual moments will cancel and we only get linear acceleration and everyone s happy, if this weren't the case then any force would cause rotation.... I think that answers what you wanted to know but I'm not entirely sure what you were asking....
thanks for your reply. My understanding is that if an object has all forces acting through its centre of mass, then it is in rotational equilbrium? :/ However, if you consider the case of 2 forces acting though the centre of mass, which do not share a common line of action, taking moments about a point on the line of action of one force will mean a moment acts about that point from the other force, which means that the object shouldnt be in rotational equillbrium. Do you see my problem? There is a contradiction, so there is a fault in one of my two definitions of rotational equilbrium. Can you see it?
thanks
0
reply
natninja
  • Community Assistant
Badges: 20
Rep:
?
#6
Report 5 years ago
#6
(Original post by MEPS1996)
thanks for your reply. My understanding is that if an object has all forces acting through its centre of mass, then it is in rotational equilbrium? :/ However, if you consider the case of 2 forces acting though the centre of mass, which do not share a common line of action, taking moments about a point on the line of action of one force will mean a moment acts about that point from the other force, which means that the object shouldnt be in rotational equillbrium. Do you see my problem? There is a contradiction, so there is a fault in one of my two definitions of rotational equilbrium. Can you see it?
thanks
there is no contradiction, add the forces vectorially and if the net force is through the centre of mass then it is in equilibrium.
0
reply
Stonebridge
Badges: 10
#7
Report 5 years ago
#7
(Original post by MEPS1996)
. I assume when you referred to moments you were referring to the moments of all the atoms in the object about the centre of mass. Imagine a diagram of a body with the lines of action of only its weight from its centre of mass and one other force whose line of action passes through the centre of mass. If you take moments about a point on the line of action of the force, the weight has a non-zero moment about that point. So moments about this point are non-zero, but the object is in rotational equilibrium? I can see their is some problem here in my definitions i am just trying to find it. please help
If you have a force acting on the body through the c of m and confine the body to rotate about some other point (this I assume to be what you call "a point on the line of action of the force") where this point is not the centre of mass, then the body is not in rotational equilibrium. This is because in order to do that you need to apply some sort of force at the other point in question in order to create this situation. These two forces create a couple which rotates the object.
If you are saying the object is not confined to rotate about this second point then taking moments about it is pointless,because you can now go on to choose any number of similar points on the object and make the same claim about moments about those points. The sum of all of the moments about all of the possible points will be zero.

If this isn't what you mean, please draw a diagram to explain.
reply
MEPS1996
Badges: 4
Rep:
?
#8
Report Thread starter 5 years ago
#8
(Original post by natninja)
there is no contradiction, add the forces vectorially and if the net force is through the centre of mass then it is in equilibrium.
I see what you mean, but it is my understanding that moments have to be zero about any point for rotational equilbirium. If you combine all the forces into one by vector addition, then by taking moments about any point which is not on the line of action of this combined force has a non-zero moment, contradicting my definition that moments have to be zero about any point for the rotational equilibrium. this is my problem ;/
0
reply
natninja
  • Community Assistant
Badges: 20
Rep:
?
#9
Report 5 years ago
#9
(Original post by MEPS1996)
I see what you mean, but it is my understanding that moments have to be zero about any point for rotational equilbirium. If you combine all the forces into one by vector addition, then by taking moments about any point which is not on the line of action of this combined force has a non-zero moment, contradicting my definition that moments have to be zero about any point for the rotational equilibrium. this is my problem ;/
for each individual point the moment would be non-zero but because of symmetry the total sum of moments from every point together would be zero
0
reply
Stonebridge
Badges: 10
#10
Report 5 years ago
#10
(Original post by MEPS1996)
I see what you mean, but it is my understanding that moments have to be zero about any point for rotational equilbirium. If you combine all the forces into one by vector addition, then by taking moments about any point which is not on the line of action of this combined force has a non-zero moment, contradicting my definition that moments have to be zero about any point for the rotational equilibrium. this is my problem ;/
It's not "any" point, it's the sum of moments about "all" points.
You can't pick one point at random and say there is a moment about that one so there is no equilibrium. You could pick a second point at random and find that the moment about that is equal and opposite to the first. The sum would be zero.
The centre of mass is the special place whereby if you did what you are doing for ALL points on the mass the sum of the moments would be zero.
reply
MEPS1996
Badges: 4
Rep:
?
#11
Report Thread starter 5 years ago
#11
(Original post by Stonebridge)
If you have a force acting on the body through the c of m and confine the body to rotate about some other point (this I assume to be what you call "a point on the line of action of the force") where this point is not the centre of mass, then the body is not in rotational equilibrium. This is because in order to do that you need to apply some sort of force at the other point in question in order to create this situation. These two forces create a couple which rotates the object.
If you are saying the object is not confined to rotate about this second point then taking moments about it is pointless,because you can now go on to choose any number of similar points on the object and make the same claim about moments about those points. The sum of all of the moments about all of the possible points will be zero.

If this isn't what you mean, please draw a diagram to explain.
I understood you could take moments about any number of points, inside or outside the object. In my notes it is said that "For an object to be in rotational equilibrium, the sum of clockwise moments = sum of anticlockwise moments about any given point" ie. resultant moment about any point (just a point arbitrarily chosen, not necessarily a pivot or hinge). This is obviously not the case when for example, one force acts through centre of mass which means it should be in rotatational equilibrium, BUT taking moments about a point not on the line of action gives a resultant non-zero moment about that point. I think the definition i quoted above is wrong. Do you think so?
0
reply
Stonebridge
Badges: 10
#12
Report 5 years ago
#12
(Original post by MEPS1996)
I understood you could take moments about any number of points, inside or outside the object. In my notes it is said that "For an object to be in rotational equilibrium, the sum of clockwise moments = sum of anticlockwise moments about any given point" ie. resultant moment about any point (just a point arbitrarily chosen, not necessarily a pivot or hinge). This is obviously not the case when for example, one force acts through centre of mass which means it should be in rotatational equilibrium, BUT taking moments about a point not on the line of action gives a resultant non-zero moment about that point. I think the definition i quoted above is wrong. Do you think so?
I think you are mixing up two different ideas.
-An object in rotational equilibrium
-The single force through the centre of mass.


The 1st idea.
If an object is in rotational equilibrium it means that there is no resultant moment acting on it. Moments cause rotation. No moment = no rotation.
It doesn't matter which point you take moments about.
We are talking about a number of forces (moments) acting on the object, and these are balanced. Clockwise = anticlockwise.
2nd idea
The centre of mass of an object is the special point through which a single resultant force acting produces no rotation.

In most moments questions you have a number of forces acting on a solid object and these are in equilibrium. This is nothing to do with centre of mass, except that one of these forces could be acting through it, eg the weight of the object. However, this is only one of a number of forces that are in equilibrium.
reply
MEPS1996
Badges: 4
Rep:
?
#13
Report Thread starter 5 years ago
#13
(Original post by Stonebridge)

The 1st idea.
If an object is in rotational equilibrium it means that there is no resultant moment acting on it. Moments cause rotation. No moment = no rotation.
It doesn't matter which point you take moments about.
We are talking about a number of forces (moments) acting on the object, and these are balanced. Clockwise = anticlockwise.
In order to find the resultant moment on an object, surely a point has to be picked first. How do you decide which?
0
reply
Stonebridge
Badges: 10
#14
Report 5 years ago
#14
(Original post by MEPS1996)
In order to find the resultant moment on an object, surely a point has to be picked first. How do you decide which?
I take it you've done problems on this. There are plenty on this forum.
When solving a moments problem you usually choose
-moments about the pivot or pivots
-moments about the support points
-moments about any point that allows you to eliminate one of the unknown forces

It depends on the question.

Post one here and I'll tell you which points to choose..
reply
MEPS1996
Badges: 4
Rep:
?
#15
Report Thread starter 5 years ago
#15
(Original post by Stonebridge)
I take it you've done problems on this. There are plenty on this forum.
When solving a moments problem you usually choose
-moments about the pivot or pivots
-moments about the support points
-moments about any point that allows you to eliminate one of the unknown forces

It depends on the question.

Post one here and I'll tell you which points to choose..
Actually you said resultant moment is zero, so you must have been referring to the sum of moments about all points, for the whole object, no? You said earlier that moments about a point do not have to be zero for rotational equilbrium. Could you explain this idea of the sum of moments about all the points?
0
reply
Stonebridge
Badges: 10
#16
Report 5 years ago
#16
(Original post by MEPS1996)
Actually you said resultant moment is zero, so you must have been referring to the sum of moments about all points, for the whole object, no? You said earlier that moments about a point do not have to be zero for rotational equilbrium. Could you explain this idea of the sum of moments about all the points?
We seem to be going round in circles.
We've done quite a bit of explaining and are getting nowhere.

I'm still not sure what the problem is here.
It's seems to be connected with your initial question about centre of mass. The statement in your first post about an object being in equilibrium with a single force through the c of m is not really helpful.
To summarise.
Equilibrium is normally about 2 or more forces and/or moments being in balance. A single force through the c of m of an object produces no rotation. That is not really a question of equilibrium. It is actually a definition of centre of mass. This has been dealt with in earlier posts.
Are we finished with that point now?

The second problem seems to concern rotational equilibrium, which is not really connected with the 1st problem. The sum of the moments of the forces on an object in such equilibrium is zero. You can take moments about any point but some are more helpful than others in solving the problem.

Are you having difficulties answering questions on moments where you have an object like a plank resting on two supports and you are asked to find the force at one of the supports? You seemed to suggest this in you recent post.

I also think there is confusion over the word moment.
The equilibrium problem is about moments of forces.

When talking about centre of mass, you are referring to so called mass moments.
It's the sum of these (mass x distance from pivot) that is equal to zero.
reply
MEPS1996
Badges: 4
Rep:
?
#17
Report Thread starter 5 years ago
#17
(Original post by Stonebridge)
A single force through the c of m of an object produces no rotation. That is not really a question of equilibrium. It is actually a definition of centre of mass
I understand you must be growing tired of our slightly circular (no pun intended) discussion. I am sorry about this but would like to thank you for your help. It is not questions it is just concepts i am struggling with. The root of my problem is this: an object with one force throught the c of m is in rotational equilibrium. Everywhere I look on the internet, the definition of rotational equlibrium is that the net moments of forces about any point are supposed to be zero. This is obviously not the case with the object with one force through the c of m. This is the heart of my problem.
0
reply
Stonebridge
Badges: 10
#18
Report 5 years ago
#18
(Original post by MEPS1996)
I understand you must be growing tired of our slightly circular (no pun intended) discussion. I am sorry about this but would like to thank you for your help. It is not questions it is just concepts i am struggling with. The root of my problem is this: an object with one force throught the c of m is in rotational equilibrium. Everywhere I look on the internet, the definition of rotational equlibrium is that the net moments of forces about any point are supposed to be zero. This is obviously not the case with the object with one force through the c of m. This is the heart of my problem.
Ok.
I think I may just be beginning to see what the problem is here. I'll need another think about this and reply tomorrow when I have more time to answer. Let's hope we stop going round in circles. Need some equilibrium.
reply
Stonebridge
Badges: 10
#19
Report 5 years ago
#19
(Original post by MEPS1996)


*******
I've had some time today to get an answer together.

Right’ let’s go back a bit first.


We are talking about rotation. What hasn't been made clear in this discussion is that to produce rotation in a rigid body you need to identify two forces, called a “couple”. A couple is two equal parallel forces acting in opposite directions but with different lines of action. The torque produced by that couple (its turning effect or moment) is equal to one of those forces times the perpendicular distance between the lines of action.

A single force produces translation, not rotation. The simplest case is a single force acting on a point mass. There is no rotation. The mass accelerates. F=ma

A rigid body acts as if all its mass is concentrated at a single point, the centre of mass. You see this with gravitation where the force acts as if the mass of the Earth was all at the centre. Applying a single force to a rigid body such that the line of action acts through the c of m has exactly the same effect as if all the mass of the body was at the centre and you were just applying it to a point mass. There is no rotation and the body just has linear motion like a point mass. There is no couple, no torque, no rotation.

In the typical questions where you have moments of forces, you only show one force producing rotation. This is strange as I just said you need two forces to produce rotation. What is happening in these questions using moments is that you are simplifying the diagram by removing half of the couple.

So on a beam where you have two supports (at A and B) applying upwards forces and the weight of the beam W downwards at the c of m as shown here in fig 1



only one of the forces in each couple is shown. The other force acts at the point you take the moment of the force about. To understand this you need to think of this beam floating in space and the only force on it is the one originally labelled W downwards. fig 2. (The beam wouldn't have weight in space so just imagine someone pushing it there.) To make the beam rotate about the point B you need to apply an equal and opposite force to it at that point and hold B steady. Then you have a couple and rotation is possible. The torque (turning effect) is W x L (the same as if you took the moment of the force W about B). If there is no force acting at B, there is no couple and no turning effect. The object doesn't rotate.
A free rigid object can only rotate about its centre of mass. This is what the c of m is by definition. It cannot rotate about some other arbitrary point unless you constrain it to do so by applying a force at that point and fixing that point somehow, and applying a couple with a moment about that point.

So in your original question, when you say that the force through the c of m has a moment about some other point in the mass, and therefore the mass can’t be in rotational equilibrium, the fallacy is that in order to have a moment about that point, and a turning effect, there must be a couple. But where is the other force? If there is only one force, there is no couple, no moment and no turning effect.
So the problem arose because you applied the techniques of moments questions (where they work just fine) to the situation where there is only one force. The moment technique is only half the story (literally!) so when you apply it to the case of the force through the centre of mass, you are missing half the story. A single force acting through the c of m doesn't have a moment about some other point in the mass if it’s the only force acting. You need a second force (at that point) and a couple, to provide a moment (and rotation) about that other point.

I ask that you take some time to read through this carefully a couple of times.
It would also be worth while looking up some of the theory on couple, torque and rotation of a rigid body.
I really don't have the time to go through all of that here.
reply
MEPS1996
Badges: 4
Rep:
?
#20
Report Thread starter 5 years ago
#20
(Original post by Stonebridge)
I've had some time today to get an answer together.

Right’ let’s go back a bit first.


We are talking about rotation. What hasn't been made clear in this discussion is that to produce rotation in a rigid body you need to identify two forces, called a “couple”. A couple is two equal parallel forces acting in opposite directions but with different lines of action. The torque produced by that couple (its turning effect or moment) is equal to one of those forces times the perpendicular distance between the lines of action.

A single force produces translation, not rotation. The simplest case is a single force acting on a point mass. There is no rotation. The mass accelerates. F=ma

A rigid body acts as if all its mass is concentrated at a single point, the centre of mass. You see this with gravitation where the force acts as if the mass of the Earth was all at the centre. Applying a single force to a rigid body such that the line of action acts through the c of m has exactly the same effect as if all the mass of the body was at the centre and you were just applying it to a point mass. There is no rotation and the body just has linear motion like a point mass. There is no couple, no torque, no rotation.

In the typical questions where you have moments of forces, you only show one force producing rotation. This is strange as I just said you need two forces to produce rotation. What is happening in these questions using moments is that you are simplifying the diagram by removing half of the couple.

So on a beam where you have two supports (at A and B) applying upwards forces and the weight of the beam W downwards at the c of m as shown here in fig 1



only one of the forces in each couple is shown. The other force acts at the point you take the moment of the force about. To understand this you need to think of this beam floating in space and the only force on it is the one originally labelled W downwards. fig 2. (The beam wouldn't have weight in space so just imagine someone pushing it there.) To make the beam rotate about the point B you need to apply an equal and opposite force to it at that point and hold B steady. Then you have a couple and rotation is possible. The torque (turning effect) is W x L (the same as if you took the moment of the force W about B). If there is no force acting at B, there is no couple and no turning effect. The object doesn't rotate.
A free rigid object can only rotate about its centre of mass. This is what the c of m is by definition. It cannot rotate about some other arbitrary point unless you constrain it to do so by applying a force at that point and fixing that point somehow, and applying a couple with a moment about that point.

So in your original question, when you say that the force through the c of m has a moment about some other point in the mass, and therefore the mass can’t be in rotational equilibrium, the fallacy is that in order to have a moment about that point, and a turning effect, there must be a couple. But where is the other force? If there is only one force, there is no couple, no moment and no turning effect.
So the problem arose because you applied the techniques of moments questions (where they work just fine) to the situation where there is only one force. The moment technique is only half the story (literally!) so when you apply it to the case of the force through the centre of mass, you are missing half the story. A single force acting through the c of m doesn't have a moment about some other point in the mass if it’s the only force acting. You need a second force (at that point) and a couple, to provide a moment (and rotation) about that other point.

I ask that you take some time to read through this carefully a couple of times.
It would also be worth while looking up some of the theory on couple, torque and rotation of a rigid body.
I really don't have the time to go through all of that here.
Thank you very much for your time, this answer is very​ helpful indeed
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

University open days

  • University of East Anglia
    All Departments Open 13:00-17:00. Find out more about our diverse range of subject areas and career progression in the Arts & Humanities, Social Sciences, Medicine & Health Sciences, and the Sciences. Postgraduate
    Wed, 30 Jan '19
  • Solent University
    Careers in maritime Undergraduate
    Sat, 2 Feb '19
  • Sheffield Hallam University
    City and Collegiate Campus Undergraduate
    Sun, 3 Feb '19

Brexit: Given the chance now, would you vote leave or remain?

Remain (553)
80.49%
Leave (134)
19.51%

Watched Threads

View All