Solving Confusions on Pulling a Spool

[member 731016]

Homework Statement

Pls see below

Relevant Equations

Pls see below

For this problem,

I have a few confusions:

(a): Why can't we use the UAM equations? Is it because the spool cannot be approximated as a point particle as it is cylinder not a sphere?

(b): I don't understand their statement about that the point of application of the force of static friction dose not move though a displacement so dose no work. To me it looks like it the point of application of static friction moves though almost the same displacement as the point of application of tension (apart from the fact they are different distances from the AoR)

(c): Why do they not include the force of friction in the net work energy theorem statement? I thought that $W = F_{net} \times \vec r$ and you don't have to take into account whether each force acting on the object is part of what causes the displacement. It seems like they did not take into account the static friction for in their energy statement even though it is a force making up the net force acting on the spool

(d) Is there a way to prove that the hand dose indeed move a distance $l + L$?

Many thanks!

 

[erobz]

(d) Is there a way to prove that the hand does indeed move a distance $l + L$?

If it was just sliding and not rotating, how far would the hand be displaced?

 

[member 731016]

If it was just sliding and not rotating, how far would the hand be displaced?

Thank you for your reply @erobz !

$l$ I believe since the problem states that the spool rotates without slipping a distance $L$ not $l$

Many thanks!

 

[erobz]

Thank you for your reply @erobz !

$l$ I believe since the problem states that the spool rotates without slipping a distance $L$ not $l$

Many thanks!

If it was sliding and not rotating the string is not unwinding. Try again.

 

[haruspex]

Why can't we use the UAM equations?

I am unfamiliar with that abbreviation. What equation do you have in mind?

the point of application of the force of static friction dose not move though a displacement so dose no work

Static friction never does work. The force does not displace the surface it acts on. It doesn’t 'know' about the displacement of the cylinder's mass centre.

 

[member 731016]

If it was sliding and not rotating the string is not unwinding. Try again.

Thank you for your reply @erobz ! Oh sorry! The other way round, so the distance the spool moves would be little l $l$.

Many thanks!

 

[erobz]

Thank you for your reply @erobz ! Oh sorry! The other way round, so the distance the spool moves would be little l $l$.

Many thanks!

I’m not sure what is happening, I was looking for the answer $L$. You seemed to be saying $l$?

 

[member 731016]

I am unfamiliar with that abbreviation. What equation do you have in mind?Static friction never does work. The force does not displace the surface it acts on. It doesn’t 'know' about the displacement of the cylinder's mass centre.

Thank you for your reply @haruspex !

UAM stands for Uniformly Accelerated Motion. It is a term used in the AP physics for the kinematic equations.

Many thanks!

 

[member 731016]

I’m not sure what is happening, I was looking for the answer $L$. You seemed to be saying $l$?

Thank you for your reply @erobz!

Sorry how it it L? I though L is for rolling?

Many thanks!

 

[erobz]

Thank you for your reply @erobz!

Sorry how it it L? I though L is for rolling?

Many thanks!

I’m trying to build up to it in pieces.

If it was sliding without rotation the spools mass and hand both move through the distance $L$. Do you agree?

 

[member 731016]

Static friction never does work. The force does not displace the surface it acts on. It doesn’t 'know' about the displacement of the cylinder's mass centre.

Thank you for your reply @haruspex !

How do we prove that static friction from the surface dose not displacement spool?

Many thanks!

 

[member 731016]

If it was sliding without rotation the spools mass and hand both move through the distance $L$. Do you agree?

Thank you for your reply @erobz !

I guess I might agree, however, what was you reasoning behind that?

Many thanks!

 

[erobz]

Thank you for your reply @erobz !

I guess I might agree, however, what was you reasoning behind that?

Many thanks!

My next step is if it had been rotating without slipping as the COM translates a distance $L$, how much rope has been unwound? Where is the hand relative to where it started applying the force?

 

[member 731016]

My next step is if it had been rotating without slipping as the COM translates a distance $L$, how much rope has been unwound? Where is the hand relative to where it started applying the force?

Thank you for your reply @erobz !

I think the hand would have unwound the same amount since each part as the angular velocity.

Many thanks!

 

[erobz]

Thank you for your reply @erobz !

I think the hand would have unwound the same amount since each part as the angular velocity.

Many thanks!

What does this mean. What amount of rope is between the drum and the hand after the COM moves a distance $L$?

 

[haruspex]

How do we prove that static friction from the surface dose not displacement spool?

It is not sliding. That means there is no relative motion between the surfaces in contact. Displacement, in the context of $W=\vec F.\vec d$ (note, dot product, not cross product as you wrote in post #1), is relative motion between the body applying a force and the body to which the force is applied, at the point where it is applied.

So how is it that a car can drive up hill? The axle exerts a torque on the wheel; static friction exerts a force up the hill on the wheel where it contacts the road; the result is a force up hill from the wheel on the axle, and that is what does work on the car chassis.

 

[haruspex]

What does this mean. What amount of rope is between the drum and the hand after the COM moves a distance $L$?

May I suggest explaining it in the reference frame of the cylinder? How far does the hand move away from the cylinder, and how far does the original position of the cylinder move in the other direction?

 

[member 731016]

What does this mean. What amount of rope is between the drum and the hand after the COM moves a distance $L$?

May I suggest explaining it in the reference frame of the cylinder? How far does the hand move away from the cylinder, and how far does the original position of the cylinder move in the other direction?

Thank you for your replies @erobz and @haruspex !

In the reference frame attached to the cylinder I will analyze how far the hand has moved away when the cylinder has rotated with slipping and so the COM has translated a distance $L$.

From the diagram the distance looks like L

Many thanks!

 

[member 731016]

It is not sliding. That means there is no relative motion between the surfaces in contact. Displacement, in the context of $W=\vec F.\vec d$ (note, dot product, not cross product as you wrote in post #1), is relative motion between the body applying a force and the body to which the force is applied, at the point where it is applied.

So how is it that a car can drive up hill? The axle exerts a torque on the wheel; static friction exerts a force up the hill on the wheel where it contacts the road; the result is a force up hill from the wheel on the axle, and that is what does work on the car chassis.

Thank you for your reply @haruspex !

Oh I now see. Since the spool is rolling without slipping, the point of contact is at momentary at rest relative to the surface which means that there is no displacement in the direction of the force of static friction. Is there a proof from this (maybe froma a diagram)?

Many thanks!

 

[haruspex]

Since the spool is rolling without slipping, the point of contact is at momentary at rest relative to the surface which means that there is no displacement in the direction of the force of static friction.

Yes.

Is there a proof from this

Not sure what you are expecting me to prove. I've simply explained what displacement means in the equation.

Edit: I need to explain a bit more.
Your equation $W=\vec F_{net}\cdot \vec r$ (more correctly $W=\int\vec F_{net}\cdot d\vec r$) is valid, and the static friction force does contribute to $\vec F_{net}$. And yet it does no work. All the work on the cylinder is done by the other horizontal force.

 

[erobz]

From the diagram the distance looks like L

I can see how you are allowing yourself to be mislead by the picture. Forget the accompanying image. It’s not showing you the final position of the hand, it is only showing you a potential starting position of the hand. By the time the wheel gets to $L$ the hand is well out of the frame of that image. They are not showing you where it’s at, they are leaving you to figure it out where it’s at. I recommend drawing your own diagram. Label the starting position of the hand and the wheel. Now displace the wheel a distance $L$, where is the hand relative to it’s starting position? Could it possibly only moved a distance $L$ if the wheel is rotating, and the length of rope between the hand and the wheel is growing?

 

[kuruman]

Forget the accompanying image.

The accompanying image is useful as a free body diagram and can be used to examine the force of static friction that was brought up. We write the usual Newton's second law for the translational motion $$T-f_s=ma_{\text{cm}}.\tag{1}$$ For the rotational motion, we express the torque about the contact point P and note that $\alpha = a_{\text{cm}}/R.$ Then $$T(R+r)=mR^2[I/(mR^2)+1]\frac{a_{\text{cm}}}{R}.\tag{2}$$We solve for the unknown acceleration of the CM and the force of friction to find
$$a_{\text{cm}}=\frac{T}{m}\frac{(1+r/R)}{[I/(mR^2)+1]}\tag{3}.$$At this point the kinematic equation $~2a_{\text{cm}}L=v_{\text{cm}}^2~$ yields directly the speed of the center of mass without need to worry about the work done by static friction or the distance over which the disembodied hand exerts a force.

The force of static friction is interesting because its magnitude and direction depend on the radius of the axle. $$f_s=\frac{T[I/(mR^2)-r/R]}{[I/(mR^2)+1]}\tag{4}$$For example, if $I/(mR^2)=1/2$ then $f_s = 0$ when $r =R/2.$ When $r >R/2$, the force of static friction is in the same direction as the motion. In this case, the accompanying picture can be forgotten but that's OK because it has already served its purpose.

Another interesting case is $r=-R$, i.e. the string is wrapped around the wheel and comes out from its bottom. Then $f_s=T$ which means that the net force is zero and the wheel will not accelerate. This explains why bottles placed on the supermarket belt with their axes perpendicular to the direction of the belt's motion, roll in place and don't move closer to the cashier.

 

[erobz]

The accompanying image is useful as a free body diagram and can be used to examine the force of static friction that was brought up. We write the usual Newton's second law for the translational motion $$T-f_s=ma_{\text{cm}}.\tag{1}$$ For the rotational motion, we express the torque about the contact point P and note that $\alpha = a_{\text{cm}}/R.$ Then $$T(R+r)=mR^2[I/(mR^2)+1]\frac{a_{\text{cm}}}{R}.\tag{2}$$We solve for the unknown acceleration of the CM and the force of friction to find
$$a_{\text{cm}}=\frac{T}{m}\frac{(1+r/R)}{[I/(mR^2)+1]}\tag{3}.$$At this point the kinematic equation $~2a_{\text{cm}}L=v_{\text{cm}}^2~$ yields directly the speed of the center of mass without need to worry about the work done by static friction or the distance over which the disembodied hand exerts a force.

The force of static friction is interesting because its magnitude and direction depend on the radius of the axle. $$f_s=\frac{T[I/(mR^2)-r/R]}{[I/(mR^2)+1]}\tag{4}$$For example, if $I/(mR^2)=1/2$ then $f_s = 0$ when $r =R/2.$ When $r >R/2$, the force of static friction is in the same direction as the motion. In this case, the accompanying picture can be forgotten but that's OK because it has already served its purpose.

Another interesting case is $r=-R$, i.e. the string is wrapped around the wheel and comes out from its bottom. Then $f_s=T$ which means that the net force is zero and the wheel will not accelerate. This explains why bottles placed on the supermarket belt with their axes perpendicular to the direction of the belt's motion, they roll in place and don't move closer to the cashier.

I was saying toss it for figuring out how far the hand is displaced for their question (d). Its making @Callumnc1 confused. They think the hand is only moving a distance $L$ because of how the diagram makes it appear. Arguably, they are not digesting the problem correctly, but that diagram is not helping in that department. The artist should have made the rope a bit shorter IMO.

 

[kuruman]

Yes, the drawing is misleading in that regard and contradicts the statement of the problem which points out that the distance by which the hand moves is “different from L”. There must have been some loss of signal from the author to the illustrator to the proofreader.

 

[haruspex]

They think the hand is only moving a distance L because of how the diagram makes it appear.

Not really. It shows a before and after position for the cylinder, and the distance between them is labelled L. Only one position is shown for the hand, so there is no indication of how far the hand moves.
Yes, it would have been clearer to show the after position of the hand as well, at a distance visibly greater than L.

 

[erobz]

Not really. It shows a before and after position for the cylinder, and the distance between them is labelled L. Only one position is shown for the hand, so there is no indication of how far the hand moves.
Yes, it would have been clearer to show the after position of the hand as well, at a distance visibly greater than L.

I would have just preferred the hand was not at ( or indistinguishably near) where they have labeled the length $L$. Based on their answers, it seems like it's the source of their confusion.

 

[member 731016]

Yes.

Not sure what you are expecting me to prove. I've simply explained what displacement means in the equation.

Edit: I need to explain a bit more.
Your equation $W=\vec F_{net}\cdot \vec r$ (more correctly $W=\int\vec F_{net}\cdot d\vec r$) is valid, and the static friction force does contribute to $\vec F_{net}$. And yet it does no work. All the work on the cylinder is done by the other horizontal force.

Thank you for your reply @haruspex !

How do we write the tension and static friction force in our work equation then?

Many thanks!

 

[haruspex]

Thank you for your reply @haruspex !

How do we write the tension and static friction force in our work equation then?

Many thanks!

Consider a simpler case: the string is attached at the top of the cylinder.
$F_{net}=T-F_f=ma=mr\alpha$.
Torque about point of contact with ground: $2rT=(I+mr^2)\alpha$.
After some time, the cylinder has rotated through angle $\theta$ and is rotating at rate $\omega$. The cylinder has moved $r\theta$ and the string has moved $2r\theta$.
Work done by string =$2r\theta T$.
KE of cylinder = $\frac 12 mr^2\omega^2+\frac 12 I\omega^2$.
$\omega^2=2\alpha\theta$. (Analogous to $v^2=2as$.)
So work done on cylinder =$(mr^2+I)\alpha\theta=2rT\theta$= work done by string.

 

[member 731016]

I can see how you are allowing yourself to be mislead by the picture. Forget the accompanying image. It’s not showing you the final position of the hand, it is only showing you a potential starting position of the hand. By the time the wheel gets to $L$ the hand is well out of the frame of that image. They are not showing you where it’s at, they are leaving you to figure it out where it’s at. I recommend drawing your own diagram. Label the starting position of the hand and the wheel. Now displace the wheel a distance $L$, where is the hand relative to it’s starting position? Could it possibly only moved a distance $L$ if the wheel is rotating, and the length of rope between the hand and the wheel is growing?

Thank you for you reply @erobz !

I will do that!

Many thanks!

 

[member 731016]

The accompanying image is useful as a free body diagram and can be used to examine the force of static friction that was brought up. We write the usual Newton's second law for the translational motion $$T-f_s=ma_{\text{cm}}.\tag{1}$$ For the rotational motion, we express the torque about the contact point P and note that $\alpha = a_{\text{cm}}/R.$ Then $$T(R+r)=mR^2[I/(mR^2)+1]\frac{a_{\text{cm}}}{R}.\tag{2}$$We solve for the unknown acceleration of the CM and the force of friction to find
$$a_{\text{cm}}=\frac{T}{m}\frac{(1+r/R)}{[I/(mR^2)+1]}\tag{3}.$$At this point the kinematic equation $~2a_{\text{cm}}L=v_{\text{cm}}^2~$ yields directly the speed of the center of mass without need to worry about the work done by static friction or the distance over which the disembodied hand exerts a force.

The force of static friction is interesting because its magnitude and direction depend on the radius of the axle. $$f_s=\frac{T[I/(mR^2)-r/R]}{[I/(mR^2)+1]}\tag{4}$$For example, if $I/(mR^2)=1/2$ then $f_s = 0$ when $r =R/2.$ When $r >R/2$, the force of static friction is in the same direction as the motion. In this case, the accompanying picture can be forgotten but that's OK because it has already served its purpose.

Another interesting case is $r=-R$, i.e. the string is wrapped around the wheel and comes out from its bottom. Then $f_s=T$ which means that the net force is zero and the wheel will not accelerate. This explains why bottles placed on the supermarket belt with their axes perpendicular to the direction of the belt's motion, roll in place and don't move closer to the cashier.

Thank you for you reply @kuruman ! I will read into that!

Many thanks!

 

[member 731016]

I was saying toss it for figuring out how far the hand is displaced for their question (d). Its making @Callumnc1 confused. They think the hand is only moving a distance $L$ because of how the diagram makes it appear. Arguably, they are not digesting the problem correctly, but that diagram is not helping in that department. The artist should have made the rope a bit shorter IMO.

Yes, the drawing is misleading in that regard and contradicts the statement of the problem which points out that the distance by which the hand moves is “different from L”. There must have been some loss of signal from the author to the illustrator to the proofreader.

I would have just preferred the hand was not at ( or indistinguishably near) where they have labeled the length $L$. Based on their answers, it seems like it's the source of their confusion.

Thank you for your replies @erobz, @kuruman and @haruspex !

Not really. It shows a before and after position for the cylinder, and the distance between them is labelled L. Only one position is shown for the hand, so there is no indication of how far the hand moves.
Yes, it would have been clearer to show the after position of the hand as well, at a distance visibly greater than L.

 

[member 731016]

Consider a simpler case: the string is attached at the top of the cylinder.
$F_{net}=T-F_f=ma=mr\alpha$.
Torque about point of contact with ground: $2rT=(I+mr^2)\alpha$.
After some time, the cylinder has rotated through angle $\theta$ and is rotating at rate $\omega$. The cylinder has moved $r\theta$ and the string has moved $2r\theta$.
Work done by string =$2r\theta T$.
KE of cylinder = $\frac 12 mr^2\omega^2+\frac 12 I\omega^2$.
$\omega^2=2\alpha\theta$. (Analogous to $v^2=2as$.)
So work done on cylinder =$(mr^2+I)\alpha\theta=2rT\theta$= work done by string.

Thank you for your reply @haruspex ! I will consider that simpler case!

Many thanks!

 

[member 731016]

Consider a simpler case: the string is attached at the top of the cylinder.
$F_{net}=T-F_f=ma=mr\alpha$.
Torque about point of contact with ground: $2rT=(I+mr^2)\alpha$.
After some time, the cylinder has rotated through angle $\theta$ and is rotating at rate $\omega$. The cylinder has moved $r\theta$ and the string has moved $2r\theta$.
Work done by string =$2r\theta T$.
KE of cylinder = $\frac 12 mr^2\omega^2+\frac 12 I\omega^2$.
$\omega^2=2\alpha\theta$. (Analogous to $v^2=2as$.)
So work done on cylinder =$(mr^2+I)\alpha\theta=2rT\theta$= work done by string.

Thank you for your reply @haruspex!

I might add a bit more explanation to your working as it a good exercise and it makes sure that my understanding is correct.

The net force on the CM is $F_{net}=T-F_f=ma_{CM}=mr\alpha$. (Assuming rolling without slipping)

And the torque is $\tau = 2rT + (0)(f_s)=(I_{CM}+mr^2)\alpha$ since the lever arm for the static friction is zero.

Where $I_{CM}+mr^2$is the rotational inertia cylinder about the point of contact from the parallel axis theorem where r is the perpendicular distance between from the CM to point of contact and $I_{CM}$ is the moment of inertia of the cylinder about its COM.

Since the string is a distance $r$ further than the CM, then it will moves though twice the distance that the COM travels, so it moves so if the COM moves though $r\theta$ then the string will have moved $2r\theta$.

These results can be found from the arc length formula my imagining the string getting put around the cylinder (this is only true for rolling with slipping for some reason).

The cylinder can be imagined as rotating about the point of contact with an angular velocity $\omega$ such that the bottom of the cylinder has zero tangential velocity assuming the rolling without slipping condition is satisfied.

Therefore, the rotational kinetic energy is given by the formula $KE = \frac 12 mr^2\omega^2+\frac 12 I_{CM}\omega^2$

Therefore, as the string moves parallel to the displacement of the cylinder then the work done is $W = 2Tr\theta$

And from the work energy theorem since the cylinder can be modelled as a point particle, then $\Delta KE = (mr^2+I)\alpha\theta=2rT\theta = W$ which is the work done by the string on the cylinder

Many thanks!

 

[member 731016]

Consider a simpler case: the string is attached at the top of the cylinder.
$F_{net}=T-F_f=ma=mr\alpha$.
Torque about point of contact with ground: $2rT=(I+mr^2)\alpha$.
After some time, the cylinder has rotated through angle $\theta$ and is rotating at rate $\omega$. The cylinder has moved $r\theta$ and the string has moved $2r\theta$.
Work done by string =$2r\theta T$.
KE of cylinder = $\frac 12 mr^2\omega^2+\frac 12 I\omega^2$.
$\omega^2=2\alpha\theta$. (Analogous to $v^2=2as$.)
So work done on cylinder =$(mr^2+I)\alpha\theta=2rT\theta$= work done by string.

If I try to calculate the work done by the net force again using the method that is confusing me then,

$W = F_{net}(2r\theta)$
$W = (T - F_f)(2r\theta)$
$W = 2Tr\theta - 2F_fr\theta$ Which is a different result than above.

They should give the same results, correct?

EDIT:

I don't think we need to do the dot product here because we have already accounted for the direction in Newton II?Many thanks!

 

[member 731016]

Consider a simpler case: the string is attached at the top of the cylinder.
$F_{net}=T-F_f=ma=mr\alpha$.
Torque about point of contact with ground: $2rT=(I+mr^2)\alpha$.
After some time, the cylinder has rotated through angle $\theta$ and is rotating at rate $\omega$. The cylinder has moved $r\theta$ and the string has moved $2r\theta$.
Work done by string =$2r\theta T$.
KE of cylinder = $\frac 12 mr^2\omega^2+\frac 12 I\omega^2$.
$\omega^2=2\alpha\theta$. (Analogous to $v^2=2as$.)
So work done on cylinder =$(mr^2+I)\alpha\theta=2rT\theta$= work done by string.

However how dose friction do not work?

To me if the cylinder is moving to the right, then the force of static friction is to the left and what causes the cylinder to rotate (I rote memorized that, I'm not actually saw how it dose) which means that the force of static friction should do negative work since the force is to the left but the displacement of the cylinder is to the right so $\cos180 = -1$

Many thanks!

 

[haruspex]

Thank you for your reply @haruspex!

I might add a bit more explanation to your working as it a good exercise and it makes sure that my understanding is correct.

The net force on the CM is $F_{net}=T-F_f=ma_{CM}=mr\alpha$. (Assuming rolling without slipping)

And the torque is $\tau = 2rT + (0)(f_s)=(I_{CM}+mr^2)\alpha$ since the lever arm for the static friction is zero.

Where $I_{CM}+mr^2$is the rotational inertia cylinder about the point of contact from the parallel axis theorem where r is the perpendicular distance between from the CM to point of contact and $I_{CM}$ is the moment of inertia of the cylinder about its COM.

Since the string is a distance $r$ further than the CM, then it will moves though twice the distance that the COM travels, so it moves so if the COM moves though $r\theta$ then the string will have moved $2r\theta$.

These results can be found from the arc length formula my imagining the string getting put around the cylinder (this is only true for rolling with slipping for some reason).

The cylinder can be imagined as rotating about the point of contact with an angular velocity $\omega$ such that the bottom of the cylinder has zero tangential velocity assuming the rolling without slipping condition is satisfied.

Therefore, the rotational kinetic energy is given by the formula $KE = \frac 12 mr^2\omega^2+\frac 12 I_{CM}\omega^2$

Therefore, as the string moves parallel to the displacement of the cylinder then the work done is $W = 2Tr\theta$

And from the work energy theorem since the cylinder can be modelled as a point particle, then $\Delta KE = (mr^2+I)\alpha\theta=2rT\theta = W$ which is the work done by the string on the cylinder

Many thanks!

All correct.

If I try to calculate the work done by the net force again using the method that is confusing me then,

$W = F_{net}(2r\theta)$

No, the net force does not act at height 2r.

the force of static friction is to the left and what causes the cylinder to rotate … which means that the force of static friction should do negative work

Yes, the static friction, together with the string tension, provides the torque leading to rotation. But that does not mean the friction does work. It merely redirects some of the work done by the string into rotational energy.

 

[member 731016]

All correct.

Thank you for your reply @haruspex !

No, the net force does not act at height 2r.

Whoops, the net force is the mass x acceleration of the CM, which means that it acts at the height of the CM so at r, is that reasoning correct please?

Yes, the static friction, together with the string tension, provides the torque leading to rotation. But that does not mean the friction does work. It merely redirects some of the work done by the string into rotational energy.

Sorry, I don't understand yet. So I understand that the friction produces a clockwise torque about the CM as with the tension. Maybe I should try to understand the origin of the static friction force. So at the moment when the cylinder is pulled by a tension force T, at the point of contact with the ground, the force of static friction oppose the relative motion by exert a force $F_s$ to the left which stops the point of contact translating with respect to the ground and allows the cylinder to roll.

I think the thing I don't understand it how the point of application of the static friction force dose not move and therefore has no displacement so not work is done (That is my textbooks explanation).

I did find this on Physics SE https://physics.stackexchange.com/q...n-friction-do-no-work-in-case-of-pure-rolling

So it looks like there are two ways analyzing the situation.

The first looks purely at the point of contact. It appears that actually the point of contact dose moves with respect to the surface, but perpendicular to the force as shown below.

Therefore, as the point of application moves perpendicular for the differential time $dt$ that static friction for is acting on the point, then the work done is $dW = F_s~\dot~\vec {ds} = -F_s\hat {i}~\dot~ds\hat j = 0$

Many thanks!

 

[erobz]

I think the thing I don't understand it how the point of application of the static friction force dose not move and therefore has no displacement so not work is done (That is my textbooks explanation).

I did find this on Physics SE https://physics.stackexchange.com/q...n-friction-do-no-work-in-case-of-pure-rolling

So it looks like there are two ways analyzing the situation.

The first looks purely at the point of contact. It appears that actually the point of contact dose moves with respect to the surface, but perpendicular to the force as shown below. View attachment 322778
Therefore, as the point of application moves perpendicular for the differential time $dt$ that static friction for is acting on the point, then the work done is $dW = F_s~\dot~\vec {ds} = -F_s\hat {i}~\dot~ds\hat j = 0$

Many thanks!

We consider the point of contact to be instantaneously at rest. Although the points of contact change as the wheel rolls along, "while" in contact with the surface they have no velocity, thus no displacement.

 

[haruspex]

how the point of application of the static friction force dose not move

Picture it instead as a 'rack and pinion'. The cylinder is now a cog wheel with its teeth engaging similar teeth along flat ground.
As each tooth on the wheel engages a tooth on the track it becomes stationary. While two such teeth are in contact, neither moves horizontally.

 

[haruspex]

the net force is the mass x acceleration of the CM, which means that it acts at the height of the CM so at r

No. To find where a resultant force acts you have to see where it would exert the same torque.
Taking moments about the point of contact, if the net force acts at height h then $(T-F_f)h=2rT$. We already found $T-F_f=mr\alpha$ and $2rT=(I+mr^2)\alpha$, so $h=(r+\frac I{mr})$.
You can check that multiplying the net force by the displacement at height h gives the correct work done.

 

[member 731016]

We consider the point of contact to be instantaneously at rest. Although the points of contact change as the wheel rolls along, "while" in contact with the surface they have no velocity, thus no displacement.

Thank you for you reply @erobz!

I agree that it is instantaneously at rest

Maybe for a few nanometers even, the point of contact of the cylinder will move perpendicular to the surface while the force of fraction is still acting. However, as the dot product is still zero, then the work done by friction from this microscopic view of friction is still zero.

Correct?

Many thanks!

 

[haruspex]

Maybe for a few nanometers even, the point of contact of the cylinder will move perpendicular to the surface while the force of fraction is still acting. However, as the dot product is still zero, then the work done by friction from this microscopic view of friction is still zero.

Correct?

Many thanks!

Yes.

 

[member 731016]

Picture it instead as a 'rack and pinion'. The cylinder is now a cog wheel with its teeth engaging similar teeth along flat ground.
As each tooth on the wheel engages a tooth on the track it becomes stationary. While two such teeth are in contact, neither moves horizontally.

Thank you for your reply @haruspex !

 

[member 731016]

Yes.

Thank you for your reply @haruspex !

 

[member 731016]

No. To find where a resultant force acts you have to see where it would exert the same torque.
Taking moments about the point of contact, if the net force acts at height h then $(T-F_f)h=2rT$. We already found $T-F_f=mr\alpha$ and $2rT=(I+mr^2)\alpha$, so $h=(r+\frac I{mr})$.
You can check that multiplying the net force by the displacement at height h gives the correct work done.

Thank you for your reply @haruspex ! I will check that out!

 

[member 731016]

I'm still confused why

$W_{net} = (\vec T +\vec F_f) \cdot \vec r$
$W_{net} = (T\hat j - F_f\hat i) \cdot (r\theta\hat i)$

is giving that friction contributes to the net work done on the cylinder.

Many thanks!

 

[haruspex]

I'm still confused why $W_{net} = (\vec T +\vec F_f) \cdot (r\theta\hat i)$ is not working here thought.

Many thanks!

Because the net force does not act at height r, as I showed in post #40.

 

[member 731016]

Because the net force does not act at height r, as I showed in post #40.

Thank you for your reply @haruspex !

I will plug that it then

Many thanks!

 

[member 731016]

Because the net force does not act at height r, as I showed in post #40.

Also don't you think it strange that the net force dose not act at the same height as the CM? I guess maybe the only special case where the net force acts at same place as the CM is for a point particle, correct?

Many thanks!

 

[member 731016]

Because the net force does not act at height r, as I showed in post #40.

Is this please correct @haruspex?

$W_{net} = (T\hat j - F_f\hat i) \cdot [(r+\frac I{mr})\theta\hat i]$

However, this means that friction contributes to the net work done by the cylinder?

It should not from this?

Many thanks!