Rotational mechanics


by Urmi Roy
Tags: mechanics, rotational
BobbyBear
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#55
Jul10-09, 02:46 AM
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Quote Quote by Urmi Roy View Post
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
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#56
Jul10-09, 02:50 AM
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Quote Quote by Urmi Roy View Post
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
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#57
Jul10-09, 02:58 AM
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Quote Quote by vin300 View Post
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
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#58
Jul10-09, 03:11 AM
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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
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#59
Jul10-09, 03:14 AM
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Quote Quote by BobbyBear View Post
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
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#60
Jul10-09, 03:20 AM
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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
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#61
Jul10-09, 03:31 AM
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Quote Quote by vin300 View Post
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
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#62
Jul10-09, 04:49 AM
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Quote Quote by Urmi Roy View Post
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
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#63
Jul10-09, 04:54 AM
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Quote Quote by BobbyBear View Post
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
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#64
Jul10-09, 05:11 AM
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Quote Quote by Doc Al View Post
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.
Quote Quote by Urmi Roy View Post
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).

Quote Quote by Urmi Roy View Post
This is applicable in the case of,say for example a box being dragged accross 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
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#65
Jul10-09, 05:25 AM
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Quote Quote by Doc Al View Post
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
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#66
Jul10-09, 05:28 AM
P: 162
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
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#67
Jul10-09, 05:33 AM
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Quote Quote by vin300 View Post
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
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Jul10-09, 05:37 AM
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Quote Quote by BobbyBear View Post
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
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#69
Jul10-09, 05:41 AM
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Quote Quote by BobbyBear View Post
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
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#70
Jul10-09, 05:51 AM
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Are frictional forces said to be electromagnetic because they are associated with heat?
Doc Al
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Jul10-09, 06:32 AM
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Quote Quote by vin300 View Post
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
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#72
Jul10-09, 01:16 PM
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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 aswell 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 recieve 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.


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