Rotational Mechanics:Small Problems

[Urmi Roy]

Static friction acting on the wheel does no work--the displacement of the point of contact is zero. However, Newton's laws still apply and that force does contribute to the wheel's acceleration. But don't confuse that with doing work--the ground is not an energy source.

I'm sorry, I still don't completely get it...please elaborate a little further.

 

[Urmi Roy]

Isn't the work done by a body to produce a displacement greater on rough sufaces than smoother sufaces because the forces of friction are different?
Yes, you do more work,but that doesn't go into the motion of the body. It appears on the underside of the body.

This is applicable in the case of,say for example a box being dragged across the floor, where friction plays a plain role of opposing relative motion between the box and the floor--here, it's converting mechanical energy into heat energy and hence dissipating it. In this case, friction does its expected role of preventing motion.

However, in case of rolling,it is seen that the friction is responsible for acceleration of the wheel,so it's not dissipating energy in this case,instead, it's speeding the wheel up,which kind of appears as if its 'providing' energy to the wheel.

This is where I'm having the problem.

And why are frictional forces basically said to be electromagnetic?
When a linearly moving body comes in contact with a frictional surface, what was previosly only linear motion now becomes rotational too, the total momentum remaining the same.maybe this could explain why no work is done by the force.

Well, I'm a little confused if we can consider the initial linear momentum being conserved by having the final sum of angular and linear momentum equal to it..can we sum angular and linear momenta?

 

[vin300]

I'm just having a little problem in understanding how the centre of mass changes position,since I read that the position of the centre of mass depends only on the structure of the body in concern and the individual positions of its infinitely small constituent masses--all of which remain unchanged in this case.

I didn't mean centre of mass changes position within the body, it changes position in the observer's frame.
Whatever internal changes take place in the body, its CM does not change position in the observer's frame but changes in its own frame, so the observer interprets "no linear motion"

 

[vin300]

As the centre of the sphere decelerates,the friction should be opposite to its velocity

Exactly, the friction must be opposite to the velocity, that is the velocity of the body at the point of contact, and not the velocity of its CM.
Whatever the motion, friction is always "resistive".

that is towards left (there is a simple diagram of a wheel rolling towards the reader's right).

It is towards the right.

But this friction will have a clockwise torque that should increase the angular velocity of the sphere.

Isn't that stupidity by a proffessor in a technical institute?

The contact is not a single point as we normally assume, rather there is an area of contact.

It forms a curve of contact.

 

[BobbyBear]

Okay, here is an extract from the book I'm referring to..."When a sphere is rolled on a horizontal table,it slows down and eventually stops.The forces acting on the sphere are a. weight mg, b. friction at the contact and c. the normal force by the table on the sphere.
As the centre of the sphere decelerates,the friction should be opposite to its velocity,that is towards left (there is a simple diagram of a wheel rolling towards the reader's right). But this friction will have a clockwise torque that should increase the angular velocity of the sphere.

The book is 'Concepts of Physics' part 1, by H.C Verma (proffessor of IIT Kanpur).

Urmi, I think the inconsistency lies in the assumption that the friction force is to the left. As DocAl has explained in this thread, you deduce the friction force that is compatible with Newton's laws. If there is no external force or torque upon the wheel prompting its movement, and the wheel is rolling and not slipping, then the friction force must be zero. This is the only compatible solution to applying both Newton's law of linear and rotational motion to the wheel in which you must consider the friction force as an unknown to be solved for (although under the assumption that static friction is capable of preventing slipping even if in the end the friction force is zero). The wheel will carry on its motion with uniform angular velocity indefinitely: it will neither accelerate nor deccelerate.

Although it seems a little strange to think that the friction force in this case is zero, remember that this is what the rigid-body model is predicting. And as the author of your book explained, it's not what exactly really happens, as we know by experience the wheel would tend to slow down if there was no external force (or torque) in favour of provoking its movement. But under the ideal assumptions this is what would happen, and what approximately happens when the two surfaces are almost indeformable.

 

[BobbyBear]

Well, I'm a little confused if we can consider the initial linear momentum being conserved by having the final sum of angular and linear momentum equal to it..can we sum angular and linear momenta?

You'd sum up the kinetic energies of translation and rotation.

 

[BobbyBear]

Exactly, the friction must be opposite to the velocity, that is the velocity of the body at the point of contact, and not the velocity of its CM.
Whatever the motion, friction is always "resistive".

friction is indeed in the direction opposing the relative velocity at the point of conctact, though if I may add, in the case of static friction, the relative velocity at the point of contact is nil, so the direction of the friction force is such that it opposes the impending relative velocity, if you get me (ie. the relative velocity that would be if the surfaces were frictionless).

 

[vin300]

I'd like assurance though that for perfectly rigid bodies, the concept of 'rolling friction' is meaningless . . . it would be zero always - which is corroborated by the fact that a rigid body that is rolling without slipping is not dissipating mechanical energy as heat due to friction.

Rigid bodies rolling without slipping might not dissipate mechanical energy as heat, the forces acting may weaken its intermolecular forces which break down at a later point of time. The energy is conserved.

 

[vin300]

friction is indeed in the direction opposing the relative velocity at the point of conctact, though if I may add, in the case of static friction, the relative velocity at the point of contact is nil, so the direction of the friction force is such that it opposes the impending relative velocity, if you get me (ie. the relative velocity that would be if the surfaces were frictionless).

When the body comes in contact with the surface in a direction different from the normal reaction or weight, it exerts a force on the suface and the surface exerts an equal force on the body but this force does not in any way assist the motion of the body, so it loses energy.

 

[BobbyBear]

Urminess,

adding to what I said in reference to the paradoxical scenario from the extract of your book about the friction force having to be zero, the fact that the friction force is zero doesn't mean that there could be a nonzero static friction force if so required (eg if another external force were to come into play). That is, the maximum static friction force is greater than zero. In this case, the friction force is zero, but the wheel is rolling without slipping, thus the linear velocity of the centre of mass is related to the angular velocity of the wheel.

But if the maximum friction force was zero, then even though we'd have exactly the same forces acting upon the wheel, and the wheel would be rotating with the same constant angular velocity (which would be whichever intial angular velocity you like), the wheel would be slipping without rolling (or rolling and slipping, depending on the initial movement - but in any case the rolling would not be due to the friction and the two movements would be uncorrelated), and the linear velocity of its centre of mass would be constant but arbitrary.

How would the system "know" whether there is a capacity for friction or not, if in both cases the friction force (by requirement of the conservation laws) is nil? Well you could think how the system would respond if you were to apply an inifinitesimal force to the wheel - in the first case, a nonzero friction force would appear, so that the movement provoked by this infinitesimal force would be a rolling without slipping movement, but not so in the second case, as that infinitesimal force would contribute to the system's linear movement but not affect its rotational movement. Since the case we've described is an ideal limiting case, in reality the system would never be faced with this quandary, as it would always be somewhat away from the ideal case, and it would know in which situation it stands.

 

[BobbyBear]

A rigid body that is rolling without slipping does dissipate mechanical energy as heat.

o:
This crushes all my notions:( Please explain? And, please tell me if you're considering there to be "rolling friction", maybe we're just thinking of different things? I'm assuming there is no "rolling friction", because the contact surfaces are not deforming, and contact occurs at a single point . . . although as I've discussed earlier, one could consider 'rolling friction' (for the sake of approximating reality better) as well as the bodies to be perfectly rigid (for the sake of applying Newton's laws using concentrated forces instead of having to consider internal stresses and what not), in which case I agree that there'd be energy dissipated as heat even though the wheel is not slipping, but due only to the 'rolling friction' which is really an attempt to quantify the energy that goes into deforming the surfaces.

 

[Doc Al]

Okay, here is an extract from the book I'm referring to...

It starts off with the big question: Why does a rolling sphere slow down.

Then,coming to the main issue,it says...

"When a sphere is rolled on a horizontal table,it slows down and eventually stops.The forces acting on the sphere are a. weight mg, b. friction at the contact and c. the normal force by the table on the sphere.
As the centre of the sphere decelerates,the friction should be opposite to its velocity,that is towards left (there is a simple diagram of a wheel rolling towards the reader's right). But this friction will have a clockwise torque that should increase the angular velocity of the sphere.
There must be an anticlockwise torque that causes the decrease in angular veocity.

Infact when, the sphere rolls on the table, both the sphere and the surface deform near the contact. The contact is not a single point as we normally assume, rather there is an area of contact.The front part pushes the table a bit more strongly than the back part. As a result, the normal force (by the table on the sphere) does not pass through the centre of the sphere, it is shifted towards the right of the centre of mass.
This force,then, has an anticlockwise torque. The net torque causes an angular deceleration."

OK, this is just a description of rolling friction. The deformable surface is "bunched up" a bit ahead of the rolling sphere, which changes the direction of the force it exerts on the sphere.

The book is 'Concepts of Physics' part 1, by H.C Verma (proffessor of IIT Kanpur).

I'm not familiar with that one.

Now, in reference to this, it seems that if we neglect rolling friction(which we should be doing in the study of rigid bodies), the sphere will accelerate,due to the torque of static friction even on level round--this is strange, as friction is supposed to be a dissipative force--it is not supposed to favour relative motion between the sphere and table.

Neglecting rolling friction, there are no friction forces acting on the rolling sphere. It would just keep rolling.

 

[vin300]

o:
This crushes all my notions:( Please explain? And, please tell me if you're considering there to be "rolling friction", maybe we're just thinking of different things? I'm assuming there is no "rolling friction", because the contact surfaces are not deforming, and contact occurs at a single point . . . although as I've discussed earlier, one could consider 'rolling friction' (for the sake of approximating reality better) as well as the bodies to be perfectly rigid (for the sake of applying Newton's laws using concentrated forces instead of having to consider internal stresses and what not), in which case I agree that there'd be energy dissipated as heat even though the wheel is not slipping, but due only to the 'rolling friction' which is really an attempt to quantify the energy that goes into deforming the surfaces.

Before I could edit it understanding the mistake, it said I've to login again and the statement remained.
The contact surface must break apart after some time if the friction is too much as I said earlier.

 

[Doc Al]

Static friction acting on the wheel does no work--the displacement of the point of contact is zero. However, Newton's laws still apply and that force does contribute to the wheel's acceleration. But don't confuse that with doing work--the ground is not an energy source.

I'm sorry, I still don't completely get it...please elaborate a little further.

As long as we are just talking about static friction (and ignoring rolling friction due to deformation of the surface), there is no relative motion between the contact point and the surface. No slipping means no displacement and thus no work being done. Work is done by kinetic friction (and rolling friction), not by static friction. This is why you can apply conservation of mechanical energy to the wheel rolling down the incline--there are no dissipative forces (the friction is static).

This is applicable in the case of,say for example a box being dragged across the floor, where friction plays a plain role of opposing relative motion between the box and the floor--here, it's converting mechanical energy into heat energy and hence dissipating it. In this case, friction does its expected role of preventing motion.

However, in case of rolling,it is seen that the friction is responsible for acceleration of the wheel,so it's not dissipating energy in this case,instead, it's speeding the wheel up,which kind of appears as if its 'providing' energy to the wheel.

This is where I'm having the problem.

When the wheel rolls down the incline without slipping, static friction acts up the incline. That friction does two things:
(1) It slows down the translational motion of the wheel. (The net force on the wheel down the incline is less than it would be on a frictionless surface, thus the wheel's acceleration is less.)
(2) It increases the rotational speed of the wheel, since it applies a torque.

The friction doesn't provide energy, it just allows some of the gravitational energy to be converted to rotational kinetic energy.

Another example: When you walk or run (without slipping), again friction propels you forward yet it does no work and provides no energy. The energy is provided by your muscles; ground friction allows you to convert your internal chemical energy into translational kinetic energy. (If friction provided the energy, you wouldn't get tired. )

 

[BobbyBear]

​

As long as we are just talking about static friction (and ignoring rolling friction due to deformation of the surface), there is no relative motion between the contact point and the surface. No slipping means no displacement and thus no work being done. Work is done by kinetic friction (and rolling friction), not by static friction. This is why you can apply conservation of mechanical energy to the wheel rolling down the incline--there are no dissipative forces (the friction is static).

O: if the friction is static, how would there be kinetic friction as well? There either is relative motion or not..

 

[BobbyBear]

ooh, I think you just mean in general . . .
obviously the friction cannot be static and kinetic at the same time:P Sorry, I misinterpreted your meaning

 

[BobbyBear]

When the body comes in contact with the surface in a direction different from the normal reaction or weight, it exerts a force on the suface and the surface exerts an equal force on the body but this force does not in any way assist the motion of the body, so it loses energy.

Ya, I agree, friction cannot, by its very nature, increase the overall motion of an object, though ideally, if there is only static friction (and no deformation), there would be no dissipation of energy either. By definition static friction cannot do work! (another issue is whether in reality you'd have whatever other dissipating phenomena taking place).

 

[Doc Al]

O: if the friction is static, how would there be kinetic friction as well? There either is relative motion or not..

That's true. At any given time there's either static or kinetic friction, not both. I was just pointing out that work is done by kinetic friction, not static friction.

 

[vin300]

Ya, I agree, friction cannot, by its very nature, increase the overall motion of an object, though ideally, if there is only static friction (and no deformation), there would be no dissipation of energy either. By definition static friction cannot do work! (another issue is whether in reality you'd have whatever other dissipating phenomena taking place).

The post by me you're talking about above makes sense only if it is not static friction.

 

[vin300]

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

 

[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 \Deltam, 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