Rotational Mechanics:Small Problems

[Doc Al]

Are frictional forces said to be electromagnetic because they are associated with heat?

No. All contact forces, which are interactions between atoms and molecules, are fundamentally electromagnetic--as opposed to nuclear or gravitational.

 

[Urmi Roy]

It took a while to accumulate all these facts together in my head, but I think I've finally come to a conclusion. Please confirm if I'm right.

From what I figured, if we have a wheel, on level ground,on which a force is being applied tangentially,this force 'F' serves to accelerate the CM of the wheel as well as to supply a torque to the wheel.
The effect of this torque is felt on all the individual particles of the wheel,of which one is the lowermost point 'P', at which the wheel is in contact with the ground.

'F' tries to push this lowermost point to an adjacent position,but just as Doc Al pointed out, this point behaves quite like feet, running on the ground. In pushing against the ground due to the effect of 'F', the lowermost point 'P' experiences a reactional force from the ground due to the ground's friction, which resists its pushing past,and hence accelerating.
(A pair of feet running on the ground first push the ground and receive a reactional force due to friction from the ground).
However, the 'F' keeps on acting and the net force on 'P' is zero (since F and friction at 'P' are opposite),so it moves past its original position,but at uniform velocity(consiering this from the point of view of the point 'P',it doesn't have zero velocity like it would appear to an observer at rest with respect to the ground).

From the perspective of the CM, there is a net force 'F' so CM of the wheel accelerates.

Its the same thing for a wheel on a ramp,but here, 'F' is actually the component of gravitational force acting down the ramp.

All this is applicable only if the force with which 'P' tries to push past is less than the limiting friction.
If the force is greater than limiting friction,the wheel spins at a certain angular velocity depending on the effective torque and the linear velocity is determined separately by the net linear force ( by the way, can we have a force which only has an effective torque,but doesn't cause any linear acceleration of the wheel its working on--as in "an automobile with its engine revved to even 12000 rpm on a frictionless surface, which will stay put with an enormous angular velocity (measured at its wheels) but zero linear velocity."??).I suppose we can find out the angular and linear velocities imparted here separately.

Its true that it is difficult to imagine there to be no frictional force for a wheel rotating without slipping,and upon which there is no other force acting,but in this case,I suppose we can say that the 'P' doesn't have any tendency to 'push past' the ground, so in turn, the ground doesn't have to give any reactional force.

 

[BobbyBear]

From what I figured, if we have a wheel, on level ground,on which a force is being applied tangentially,this force 'F' serves to accelerate the CM of the wheel as well as to supply a torque to the wheel.
The effect of this torque is felt on all the individual particles of the wheel,of which one is the lowermost point 'P', at which the wheel is in contact with the ground.

'F' would not produce torque at the points of the wheel on its line of action.
For the solid to have a net torque, the torque at its centre of mass must be non-zero.
If 'F' is applied at the centre of mass of the wheel and there is no friction, then the wheel will not rotate. But if 'F' is applied somewhere other than the centre of mass, it will produce a net torque about the centre of mass and set the wheel rotating on its own.
When there is a friction force at 'P', it too produces torque at the centre of mass of the wheel. If 'F' is applied at the centre of mass, the it is the friction force at 'P' that is responsible for setting the wheel in rotational motion.
At least I think so.

However, the 'F' keeps on acting and the net force on 'P' is zero (since F and friction at 'P' are opposite),so it moves past its original position,but at uniform velocity(consiering this from the point of view of the point 'P',it doesn't have zero velocity like it would appear to an observer at rest with respect to the ground).

From the perspective of the CM, there is a net force 'F' so CM of the wheel accelerates.

I'm not following you:( How is there no net force on 'P'? If that were so then P would move in a straight line.
I don't understand what you're saying about the movements from different perspectives. Please explain?

by the way, can we have a force which only has an effective torque,but doesn't cause any linear acceleration of the wheel its working on--as in "an automobile with its engine revved to even 12000 rpm on a frictionless surface, which will stay put with an enormous angular velocity (measured at its wheels) but zero linear velocity."??).I suppose we can find out the angular and linear velocities imparted here separately.

Yes, but if there's friction it will tend to convert some of the angular movement into linear movement . . . the reverse of what it tends to do when the agent provoking movement is a force applied at the cenre of mass instead of a moment.

Its true that it is difficult to imagine there to be no frictional force for a wheel rotating without slipping,and upon which there is no other force acting,but in this case,I suppose we can say that the 'P' doesn't have any tendency to 'push past' the ground, so in turn, the ground doesn't have to give any reactional force.

I like that way of thinking about it :) 'P' doesn't have any tendency to push past the ground because the wheel has no tendency to accelerate - 'P' pushing against the ground would be to accelerate (or deccelerate) the wheel.

 

[Urmi Roy]

'F' would not produce torque at the points of the wheel on its line of action.
For the solid to have a net torque,...At least I think so.

I was just trying to say that the 'F' does effect the state of motion of 'P',and tries to accelerate it.

I'm not following you:( How is there no net force on 'P'? If that were so then P would move in a straight line.

Concentrating upon the instant that 'P' is the lowermost point,it does move in a straight line,doesn't it?

I don't understand what you're saying about the movements from different perspectives. Please explain?

Well, actually, I found that in rotational mechanics,one tends to say 'torque with respect to CM' or 'torque with respect to any other point',...quite like different frames of references, really.I don't know how appropiate that is in this case,but I just tried it out.

Yes, but if there's friction it will tend to convert some of the angular movement into linear movement . . . the reverse of what it tends to do when the agent provoking movement is a force applied at the cenre of mass instead of a moment.

...meaning,basically that this is possible,but friction is not such an example...did I understand that right?

Please bear with me if I'm being a little stupid, but I've always had an ambiguity with this topic.

 

[sganesh88]

Static friction does work in certain situations when the surface itself accelerates with respect to the observer. Consider the example of two blocks placed one above the other on a frictionless surface. By pushing the lower block of mass M with a force F, the upper block of mass m, also accelerates because the static friction is doing work on it. (Assuming the value of $$\frac{Fm}{M+m}$$ is below the max static friction force)

However, the 'F' keeps on acting and the net force on 'P' is zero (since F and friction at 'P' are opposite),

Net force in the horizontal direction is zero,yes. Not in the vertical direction. Since you're considering the particles of the body-wheel- you have to consider the so-far-internal entity, namely the centripetal force which becomes an external force now. The rigidity of Newton's laws sometimes frightens me. :)

so it moves past its original position,but at uniform velocity(considering this from the point of view of the point 'P',it doesn't have zero velocity like it would appear to an observer at rest with respect to the ground).

No need to consider view points here. Just assume you're in an inertial frame and are provided with all facilities to measure the velocities and accelerations of individual particles as well as the wheel as a whole. Do check out what an inertial frame means if you don't already know. A final year Automobile engineering friend of mine once asked me, while i was explaining some stuff-and simultaneously doubting whether I'm right-, if such a frame called inertial frame really existed or if I'm just bluffing.
Back to our good ol wheel. The point P(the contact point of wheel with the ground) does come to zero velocity each time before being uplifted by the centripetal force. It cannot have the "uniform velocity" as you mentioned, because anything above zero isn't admissible to the Earth's surface. Its just like me. tooo lazy!. Watch the cycloid Urmi.

 

[BobbyBear]

Static friction does work in certain situations when the surface itself accelerates with respect to the observer. Consider the example of two blocks placed one above the other on a frictionless surface. By pushing the lower block of mass M with a force F, the upper block of mass m, also accelerates because the static friction is doing work on it. (Assuming the value of $$\frac{Fm}{M+m}$$ is below the max static friction force)

Ooooh tricky! that's something I need to ponder over:P:P However, yes, okay . . . but maybe now we need to redefine what 'doing work' is. The friction force would be doing work upon the top block, but it'd merely be transmitting part of the total work (the ratio m/M) beind done by the force F. It'd be acting like an internal force (so long as it's static friction). So maybe what we mean when we say that static friction cannot do work, is that we mean that it does not dissipate energy, thinking that friction in general (kinetic friction) is a dissipative force.
Yes?:P

 

[BobbyBear]

A final year Automobile engineering friend of mine once asked me, while i was explaining some stuff-and simultaneously doubting whether I'm right-, if such a frame called inertial frame really existed or if I'm just bluffing.

As far as I've been given to understand, it's Newton's first law that claims the existence of inertial frames.

 

[sganesh88]

Yes?:P

I'm afraid, no. Why are you bothered about static friction doing work anyway? Btw i was also irked when i heard gravity does work.

 

[sganesh88]

As far as I've been given to understand, it's Newton's first law that claims the existence of inertial frames.

Yes. So anyway first law created it. I didn't.

 

[BobbyBear]

Net force on individual particles

About the net force on 'P' (I'm assuming we are considering the situation that Urmi Roy described: the wheel subject to a tangential force 'F' (applied at its centre of mass?), and a static friction force at the point of contact 'P', thus the wheel has both a linear acceration and an angular acceleration (both of them compatible with the no slipping condition).

However, the 'F' keeps on acting and the net force on 'P' is zero (since F and friction at 'P' are opposite),so it moves past its original position,but at uniform velocity(consiering this from the point of view of the point 'P',it doesn't have zero velocity like it would appear to an observer at rest with respect to the ground).

Net force in the horizontal direction is zero,yes. Not in the vertical direction. Since you're considering the particles of the body-wheel- you have to consider the so-far-internal entity, namely the centripetal force which becomes an external force now. The rigidity of Newton's laws sometimes frightens me. :)
[...]
Back to our good ol wheel. The point P(the contact point of wheel with the ground) does come to zero velocity each time before being uplifted by the centripetal force. It cannot have the "uniform velocity" as you mentioned, because anything above zero isn't admissible to the Earth's surface. Its just like me. tooo lazy!. Watch the cycloid Urmi.

I'm not sure how you'd know what forces are acting on individual particles of the wheel. Are you working it out from the movement you know of the wheel, from which you know the movement of each particle of the wheel, as we're considering the solid to be rigid?
So! basically, the centre of mass has a linear acceleration (and no angular acceleration), so the net force upon it is a linear force to the right.
And all points, including point 'P', have the same linear acceleration as the centre of mass, plus an angular acceleration about the centre of mass (superposition of two movemets). Thus, they are all subject to the same force that the centre of mass is, plus a centripetal force directed toward the centre of mass equal to the 'mass of the particle' times the distance of the particle to the centre of mass and the square of the angular velocity of the wheel at each instant.

 

[BobbyBear]

I'm afraid, no. Why are you bothered about static friction doing work anyway? Btw i was also irked when i heard gravity does work.

Oh :(
but I'm correct in saying that in your example the friction force doesn't do 'it's own' work, just transmits part of the work done by the force 'F'. And its true that friction cannot provoke movement on its own, it cannot transform some other kind of energy into kinetic energy! that's all I'm trying to say. At least this is true?

 

[Doc Al]

Static friction does work in certain situations when the surface itself accelerates with respect to the observer. Consider the example of two blocks placed one above the other on a frictionless surface. By pushing the lower block of mass M with a force F, the upper block of mass m, also accelerates because the static friction is doing work on it. (Assuming the value of $$\frac{Fm}{M+m}$$ is below the max static friction force)

Good point. Whether a force does work on a system depends on the reference frame used to analyze the system. But the key point about static friction is that it's a passive force. For the "real" source of the energy used to accelerate the top block one must look to what's doing the work on the lower block.

 

[Urmi Roy]

Static friction does work in certain situations when the surface itself accelerates with respect to the observer. Consider the example of two blocks placed one above the other on a frictionless surface.

Thanks,this really helped to clear my concepts further!

Net force in the horizontal direction is zero,yes. Not in the vertical direction. Since you're considering the particles of the body-wheel- you have to consider the so-far-internal entity, namely the centripetal force which becomes an external force now. The rigidity of Newton's laws sometimes frightens me. :)

Will this have a serious bearing on the point of view I have formed about this entire event of roling without slipping?

No need to consider view points here. Just assume you're in an inertial frame and are provided with all facilities to measure the velocities and accelerations of individual particles as well as the wheel as a whole.
Back to our good ol wheel. The point P(the contact point of wheel with the ground) does come to zero velocity each time before being uplifted by the centripetal force. It cannot have the "uniform velocity" as you mentioned.

Just very cautiously,let me ask if this idea of considering view points,or rather frames of reference is actually wrong,even if I don't need it here.

Also, please tell me how I could modify my understanding of 'rolling' by stating what is wrong and what is right about my post(post 72 of this thread).

 

[BobbyBear]

Well, actually, I found that in rotational mechanics,one tends to say 'torque with respect to CM' or 'torque with respect to any other point',...quite like different frames of references, really.I don't know how appropiate that is in this case,but I just tried it out.

The moment produced by a force is not uniform in space, it depends upon the point you consider. Only the moment produced by a pair of forces of equal magnitude and opposite direction whose lines of action do not coincide produce a uniform moment in all of space.
That's why we talk of the moment (or torque) with respect to a point.

Maybe you're mixing this idea with different referece frames? I really don't think there's any relationship, but if you find there is let me know!

 

[sganesh88]

I'm not sure how you'd know what forces are acting on individual particles of the wheel. Are you working it out from the movement you know of the wheel, from which you know the movement of each particle of the wheel, as we're considering the solid to be rigid?

Assuming pure rolling;taking the mass of the particle to be some $$\Delta$$m, we can compute the instantanoues force acting on it at a particular position w.r.t COM.

Thus, they are all subject to the same force that the centre of mass is, plus a centripetal force directed toward the centre of mass equal to the 'mass of the particle' times the distance of the particle to the centre of mass and the square of the angular velocity of the wheel at each instant.

Considering pure rolling, the contact point P should have a zero horizontal velocity component, whatever the tangential force. The only unbalanced force acting on it is the centripetal force.

but I'm correct in saying that in your example the friction force doesn't do 'it's own' work, just transmits part of the work done by the force 'F'. And its true that friction cannot provoke movement on its own, it cannot transform some other kind of energy into kinetic energy! that's all I'm trying to say.

Going by the definition of work done which is, the dot product of the force and displacement, static friction can claim that it can do work. But yes, it can't cause motion on its own; i.e., without another force entering the picture.

 

[sganesh88]

Will this have a serious bearing on the point of view I have formed about this entire event of roling without slipping?

Rolling can be analysed without bothering about what happens at the "particle" level by just considering the wheel as a rigid body and computing the external forces and torques(w.r.t COM preferably) on it. But understanding the complex motions of particles is also quite fun.

Just very cautiously,let me ask if this idea of considering view points,or rather frames of reference is actually wrong,even if I don't need it here. Also, please tell me how I could modify my understanding of 'rolling' by stating what is wrong and what is right about my post(post 72 of this thread).

Reference frames are important when you describe motion. And you've come a long way from post #1. Much faster than me, i confess. Just keep thinking about it. Refer some books and internet articles. Anyway is this just for understanding purpose or are you giving some lecture on it?

 

[Urmi Roy]

Rolling can be analysed without bothering about what happens at the "particle" level ... But understanding the complex motions of particles is also quite fun.

I thought that perhaps analysing in this way might help in my understanding,but if you think it's not, I'll try in a different way.

Reference frames are important when you describe motion. And you've come a long way from post #1. Much faster than me, i confess.

BobbyBear said that the usage of 'frames of reference' in rotational mechanics isn't quite correct.Please tell me where I'm going wrong.

Just keep thinking about it. Refer some books and internet articles. Anyway is this just for understanding purpose or are you giving some lecture on it?

Pleeease don't leave it at that! I don't want to just improve in this,I want to finally remove this doubt that has been bothering me for so long,even though its just for the sake of my satisfaction,I don't have any lectures to give! I can only do it if you all help me!

 

[sganesh88]

I thought that perhaps analysing in this way might help in my understanding,but if you think it's not, I'll try in a different way.

Yes. it does. But to understand the motion of the wheel as a whole, the former approach would suffice.

BobbyBear said that the usage of 'frames of reference' in rotational mechanics isn't quite correct.Please tell me where I'm going wrong.

BobbyBear said reference frames aren't analogous to reference points w.r.t which moments are measured. Quantities like displacement,force, velocity and the whole lot of physical quantities lose their meaning in the absence of a reference frame. My initial doubt has been confirmed. Read about reference frames. Physicsforums and wikipedia would help you in it, I'm sure.

 

[sganesh88]

Pleeease don't leave it at that! I don't want to just improve in this,I want to finally remove this doubt that has been bothering me for so long,even though its just for the sake of my satisfaction,I don't have any lectures to give! I can only do it if you all help me!

What is the doubt that you presently have?

 

[Urmi Roy]

Actually I don't have anything new. As I said,this issue of 'rolling friction' has been bothering me for quite a while,the main difficulty of which stemmed from a particular extract of a book I read.

So,after getting together all that Doc Al, BobbyBear, vin300 and ofcourse, you said, I just tried to summarise the new picture in my head, in post 72.

Since then,you pointed out that viewing it from the aspect of individual particles isn't necessary, so apart from that, I just wanted to confirm my idea (as presented in #72) was basically right.

If you think I have any major problems, please modify my post and tell me the true 'picture'.

 

[sganesh88]

Ya. You have got the picture right, i think. But question yourself continuously. Doubts can never cease.
and regarding rolling friction, someone has already explained it, i see i.e., the offset of normal force towards the front, creating a braking torque.

 

[praveen kumar]

I've just got some small problems this time.
The first being,do the particles at the axis of a rotating body really stand still??Don't they spin about themselves??

No.I think that axis of rotation is dimensionless.So the particles on the axis of rotation won"t spin around themselves.

 

[Urmi Roy]

Ya. You have got the picture right, i think. But question yourself continuously. Doubts can never cease. .

Thanks,that's a load off my mind!

Also,I'm quite sure I'll have enough doubts to keep myself and physicsforums busy!

and regarding rolling friction, someone has already explained it, i see i.e., the offset of normal force towards the front, creating a braking torque.

Did you mean H.C Verma? Have you read this book?

 

[sganesh88]

Did you mean H.C Verma? Have you read this book?

No. I meant someone who participated in this discussion. And I've
just heard of the H.C.Verma book.. Seems it's a good one. But the explanation is the same and can be found in some good sites.
http://webphysics.davidson.edu/faculty/dmb/PY430/Friction/rolling.html
Wiki too has a good article on rolling friction.

 

[Urmi Roy]

Hi,I'm back with some more questions in rotational mechanics..I would love to get some more help!

Please explain these points giving necessary examples.

1.The moment induced by two equal and opposite forces is equal to the moment of one force about the point of action of the other. It doesn’t matter which force you use to do this calculation.

2.If an object translates in space in a circular trajectory, with constant speed,then a centripetal acceleration, and hence a centripetal force, is needed, so the object cannot be in equilibrium.
But if an extended object does not translate, it only rotates
around a fixed axis, in the absence of dissipative forces and with no other forces, it will continue to rotate.

3.Depending on the torques acting on an object, the angular momentum of it will remain conserved in two directions only.

4.Angular momentum and angular velocity vector need not be in the same direction

 

[sganesh88]

If you have any specific doubt regarding the above points, ask. I don't understand 3; 4 is wrong. Angular momentum and angular velocity have the same direction.

 

[Urmi Roy]

If you have any specific doubt regarding the above points, ask. I don't understand 3; 4 is wrong. Angular momentum and angular velocity have the same direction.

Thanks for your prompt reply, sganesh88.

Sorry for being unclear about the points,but you see, I got the last two points from my book, 'Pradeep's Physics' and they were just a part of the text (in the section 'angular momentum'),without any further clarification---so,because I couldn't understand what they meant,I thought of posting it as a question.

For the first two,well,they were from two websites that I was reading--if you realize which concept is being described,just explain it to me in short.

As far as the first point goes, it said that there was a rod,pivoted in the middle,and made to rotate about that point by applying a couple--the author was calculating the net torque on the rod,and he said that we could calculate it by either summing the individual torques on the arms,or by 'finding the moment of one force about the point of action of the other'.

But is this justified? I mean, we calculate the torque about the axis,but as he says,we are changing the axis to a line about which there is no rotation.

As for the 2nd point,it was also taken from my book,and I guess we'll just have to treat it as a general statement on rotation--which if you yourself understand,then I would like you to please explain.

 

[sganesh88]

But is this justified? I mean, we calculate the torque about the axis,but as he says,we are changing the axis to a line about which there is no rotation.

You could take moment w.r.t any point though in the above problem, the pivot point is preferable. And the first point can be proved mathematically from the expression for moment.

As for the 2nd point,it was also taken from my book,and I guess we'll just have to treat it as a general statement on rotation--which if you yourself understand,then I would like you to please explain.

The second point speaks first about a mass point and then about a rigid body-which contains a collection of mass points-. A mass point under no external force will, by Newton's second law, continue along a straight line with its original speed; it needs a centripetal force to revolve about an axis in space. But in case of the rigid body mentioned, the individual mass points not lying on the axis of rotation will be subjected to radial forces from other mass points though there is no external force on the body as a whole. So apply Second law for individual mass points and you see why they go along a circle around an axis passing through the COM of the rigid body; apply it for the entire rigid body and you see why the COM doesn't accelerate.

 

[D H]

4 is correct. Angular momentum and angular velocity do not have to point in the same direction. The angular momentum and angular velocity of a rigid body are related by

$$\boldsymbol L= \mathbf I\,\boldsymbol{\omega}$$

The angular momentum will point in the same direction as the angular velocity if and only if the angular velocity is an eigenvector of the inertia tensor.The answer to the second question lies in conservation of angular momentum. One way to look at conservation of angular momentum is that it is a consequence of the strong form of Newton's third law. The weak form states that if object A exerts a force F on object B, then object B exerts an equal but opposite force on object A. That linear momentum is conserved in an insolated system is a consequence of the Newton's second law and the weak form of his third law. (Alternately, the weak form of Newton's third law is a consequence of the conservation of linear momentum.)

The weak form of Newton's third law does not say anything about the orientation of the equal but opposite forces. The strong form of this law adds the constraint that the forces are directed along the line connecting the two particles. With this added constraint, the total angular momentum for an isolated collection of particles is conserved. The collection of particles do not need to form a rigid body. They just need to obey Newton's third law (strong form).

 

[vin300]

I had also read somewhere that the law of conservation of angular momentum alters the weight(gravitational acc) of the body and the cause remains unknown

 

[BobbyBear]

Please explain these points giving necessary examples.

1.The moment induced by two equal and opposite forces is equal to the moment of one force about the point of action of the other. It doesn’t matter which force you use to do this calculation.

That is because the moment produced by a couple (two forces of equal magnitude and opposing direction) is constant through all of space, and its value is F times d, where d is the separation between the lines of action of the two forces. Obviously, if you take the moment with respect to a point on the line of action of one of the forces, that force's contribution will be zero. But regardelss of the point about which you take it, the moment produced by the couple,
$$\vec{M} = \vec{r_1}\times \vec{F_1} + \vec{r_2}\times \vec{F_2}$$
is the same w.r.t. every point of space. Try it out and see that it's true! :P

2.If an object translates in space in a circular trajectory, with constant speed,then a centripetal acceleration, and hence a centripetal force, is needed, so the object cannot be in equilibrium.
But if an extended object does not translate, it only rotates
around a fixed axis, in the absence of dissipative forces and with no other forces, it will continue to rotate.

I'd like to simply lay down a couple of ideas first that might help. Thinking of a particle translating in a circular trajectory with constant speed, the centripetal force is always perpendicular to the trajectory (or velocity vector), and therefore is doing no work upon the particle. In the absence of any other force, the (kinetic) energy of the particle must be conserved (so I think this is simply an application of the law of conservation of energy, yes?)

In the case of the solid rotating without translating about an axis, you could apply this reasoning to each individual particle?
Or alternatively, you could apply Newton's second law for rotation to the solid:
$$\vec{M} = I \frac{d \vec{L}}{d t}}$$
: since there is no external moment, the angular momentum and thus the angular velocity of the solid is conserved.

Can someone comment on these ideas, please?

The second point speaks first about a mass point and then about a rigid body-which contains a collection of mass points-. A mass point under no external force will, by Newton's second law, continue along a straight line with its original speed; it needs a centripetal force to revolve about an axis in space. But in case of the rigid body mentioned, the individual mass points not lying on the axis of rotation will be subjected to radial forces from other mass points though there is no external force on the body as a whole. So apply Second law for individual mass points and you see why they go along a circle around an axis passing through the COM of the rigid body; apply it for the entire rigid body and you see why the COM doesn't accelerate.

Now I have a slight problem . . . firstly, with respect to sganesh88's comment above, does the rotation of the solid have to be about an axis passing through its COM? Urmi only talked about 'an axis', so why did you assume an axis passing through the COM? If the axis of rotation was not through the COM, would that mean that the solid would be translating as well as rotating (because the COM would be describing some trajectory)? And how do you apply Newton's law of rotation to a solid?? I mean:

what is meant by "net torque on a rod" for example? We always talk about torque of a force wrp a point, so what does the 'torque of a solid' mean? You don't always have a 'couple' acting upon a solid, do you?

Say, if you have a rod with a magnet at one end suspended in a magnetic field, there will be a force upon the rod but it will only be applied at one end of the rod. The rod will rotate until its length is aligned with the magnetic field, but about which point is it rotating? What is the torque on the rod? Since the torque (due to the magnetic force) w.r.t. evert point in space is different, how can we talk of "the torque" on the rod, and if we do, what exactly do we mean by it?

 

[D H]

I'd like to simply lay down a couple of ideas first that might help. Thinking of a particle translating in a circular trajectory with constant speed, the centripetal force is always perpendicular to the trajectory (or velocity vector), and therefore is doing no work upon the particle. In the absence of any other force, the (kinetic) energy of the particle must be conserved (so I think this is simply an application of the law of conservation of energy, yes?)

This is true only for an object rotating around one of its principle axes. In general, it is not true. The motion in general is not circular, even in the absence of external forces / torques.

Or alternatively, you could apply Newton's second law for rotation to the solid:
$$\vec{M} = I \frac{d \vec{L}}{d t}}$$
: since there is no external moment, the angular momentum and thus the angular velocity of the solid is conserved.

Can someone comment on these ideas, please?

BobbyBear, what you wrote is the freshman physics version of the rotational analog of Newton's second law. This expression is only true under special circumstances. Freshman physics rotational mechanics problems are carefully constructed so as to make this simple form applicable. Angular momentum is always conserved in the absence of external torques. Angular velocity is not.

For a rigid body, the inertia tensor is constant in a frame fixed with respect to the body. That does not necessarily mean it is constant in an inertial frame. The problem is that the inertia tensor is a second order tensor, not a simple scalar. Suppose no external forces act on a rigid body. Angular momentum is conserved along with linear momentum and energy. The angular momentum of the body is $\boldsymbol L = \mathbf I \boldsymbol{\omega}$. Because the inertia tensor is not necessarily constant in an inertial frame, angular velocity is not conserved. There is no such thing as a law of conservation of angular velocity.

Because the inertia tensor is constant in a frame fixed with respect to a constant mass rigid body, this is one of those cases where working in a rotating frame is easier than working in an inertial frame. The derivative of any vector quantity $\boldsymbol q$ from the perspectives of a rotating frame versus an inertial frame are related by

$$\left.\frac{d \boldsymbol q}{dt}\right|_{\text{inertial}} = \left.\frac{d \boldsymbol q}{dt}\right|_{\text{rotating}} + \boldsymbol{\omega}\times \boldsymbol q$$

Applying this to the angular momentum vector $\boldsymbol L = \mathbf I \boldsymbol{\omega}$ and taking advantage of the fact that the inertia tensor is constant in the rotating frame,

$$\left.\frac{d \boldsymbol L}{dt}\right|_{\text{inertial}} = \mathbf I \left.\frac{d \boldsymbol \omega}{dt}\right|_{\text{rotating}} + \boldsymbol{\omega}\times (\mathbf I \boldsymbol{\omega})$$

Back to Newton's second law. F=ma is also a freshman physics version of Newton's second law. A more general expression is $d\boldsymbol p/dt = \boldsymbol F_{\text{ext}}$, where $\boldsymbol p$ is the linear momentum of the object in question and $F_{\text{ext}}$ is the external force acting on the object. The rotational analog of this is $d\boldsymbol L/dt = \boldsymbol {\tau}_{\text{ext}}$, where $\boldsymbol {\tau}_{\text{ext}}$ is the external torque acting on the body. This expression is only true from the perspective of an inertial frame. Combining this with the rotating frame expression yields

$$\mathbf I \left.\frac{d \boldsymbol \omega}{dt}\right|_{\text{rotating}} + \boldsymbol{\omega}\times (\mathbf I \boldsymbol{\omega}) = \boldsymbol {\tau}_{\text{ext}}$$

In the absence of external torques, the angular velocity will be constant only if the angular velocity is zero or if it is parallel to the the angular momentum.

Now I have a slight problem . . . firstly, with respect to sganesh88's comment above, does the rotation of the solid have to be about an axis passing through its COM?

The motion of a rigid body can always be expressed as a translation of the center of mass plus a rotation about an axis through the center of mass. Moreover, when one models the motion in this way, the translational and rotational mechanics are decoupled. (This is not true for a non-rigid body.)

 

[BobbyBear]

*flails*
Okay, D.H. I think I'm a bit of a freshbear as far as solid dynamics goes :P I shall strive to brush up a little on the subject when I can - thanks for your insights, which I shall take some time to process :P

And as far as my other question (about the rod), you said:

The motion of a rigid body can always be expressed as a translation of the center of mass plus a rotation about an axis through the center of mass. Moreover, when one models the motion in this way, the translational and rotational mechanics are decoupled. (This is not true for a non-rigid body.)

Yes, I know that . . . so does that mean that if the solid is rotating about an axis that does not contain its centre of mass, its movement is not a pure rotation: it can be viewed as the superposition of a translation of its centre of mass and a rotation about an axis through the centre of mass. Is that correct then?

 

[D H]

BobbyBear,

What I didn't tell you is that the instantaneous motion of a rigid body can be expressed as the linear translation of any point in the body plus a rotation about an axis passing through that point.

A freely rotating object in space can be viewed as rotating about any axis parallel to the central axis. Asking which axis is the correct one implies that all others are somehow "incorrect". This is not the case. The equations of motion, when done correctly, will yield the same results regardless of which point is chosen as the central point. The reason for choosing the center of mass as the central point is because this choice makes the translational and rotational dynamics decouple. This makes the math easier.

An example of where a different choice might make more sense is a rocket. A rocket that launches something into orbit is oriented vertically at takeoff but is oriented horizontally when it reaches orbital velocity about ten minutes after takeoff. Meanwhile, the rocket has burned off 90% or so of if its mass. The center of mass of a rocket moves as the rocket consumes fuel. Some people prefer to model the rotational dynamics in a frame fixed with respect to a point on the rocket structure; others prefer to model the rotational dynamics with a center of mass frame. Which is correct? Wrong question. Both viewpoints are correct. Which is easier? Both viewpoints are messy. The rocket is not a constant mass rigid body. The translational and rotational equations of motion do not decouple in either a center of mass frame or a structural frame (or any other frame).

 

[sganesh88]

Now I have a slight problem . . . firstly, with respect to sganesh88's comment above, does the rotation of the solid have to be about an axis passing through its COM?

With no external force, rotation must be about an axis passing through the COM.

what is meant by "net torque on a rod" for example? We always talk about torque of a force wrp a point, so what does the 'torque of a solid' mean? You don't always have a 'couple' acting upon a solid, do you?

Yes. It doesn't make sense under the definition of torque. But if someone says so, it has to mean torque about the COM or a pivot point.