Solving Confusions on Pulling a Spool

[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!

 

[haruspex]

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

No, I already answered that in post #20. It contributes to the net force, but that does not imply it contributes to the work.

 

[member 731016]

No, I already answered that in post #20. It contributes to the net force, but that does not imply it contributes to the work.

Thank you for your reply @haruspex!

Sorry I am still confused.

How does the net force not contribute to the net work? The net work is made up of the net force x displacement of the object under the net force correct? This means that object if the friction force dose indeed do no work on the cylinder then it should cancel in the net work equation, correct?

But from that equation the force of friction dose negative work.

Are you suggesting that I just have $F_{net} = T$ (which is what the textbook did) instead of $F_{net} = T - F_f$in the net work equation, but this would not be true since if we are trying to find the net work of the entire object then we have to include all the forces acting on the total displacement of the object, correct?

Here is the net work equation I am talking about $W_{net} = F_{net}dcos\theta$

Where d is the total displacement of the object under the net force
Many thanks!

 

[haruspex]

How does the net force not contribute to the net work?

No, the net force does the net work; but each force that contributes to the net force acts at a different point, so you cannot tell from the magnitude of each force how much each contributes to the net work.

 

[member 731016]

No, the net force does the net work; but each force that contributes to the net force acts at a different point, so you cannot tell from the magnitude of each force how much each contributes to the net work.

Thank you for your reply @haruspex!

Oh is the net work equation only valid for point particles not a rigid object like a cyclinder? Is it possible to apply it to the cyclinder correctly?

Many thanks!

 

[haruspex]

Oh is the net work equation only valid for point particles not a rigid object

No, it's still valid.
Two forces $\vec F_1,\vec F_2$ act at $\vec r_1, \vec r_2$, displacing them by $\Delta \vec r_1, \Delta\vec r_2$.
The net force is $\vec F_1+\vec F_2$, the net work is $\vec F_1\cdot\Delta \vec r_1+ \vec F_2\cdot\Delta \vec r_2$.
Clearly, $\Delta \vec r_2$ could be zero, meaning the second force does no work.

If you want to think of it as a resultant force $\vec F'$ acting at some point $\vec r'$ and moving through some displacement $\Delta\vec r'$, then we have to find the place where this resultant force acts:
$\vec F'=\vec F_1+\vec F_2$,
$\vec r'\times\vec F'=\vec r_1\times\vec F_1+\vec r_2\times\vec F_2$.
That is not enough to determine $\vec r'$, but it is enough to find the line of action of the resultant.
I'd like to continue that, but not enough time now.

 

[member 731016]

No, it's still valid.
Two forces $\vec F_1,\vec F_2$ act at $\vec r_1, \vec r_2$, displacing them by $\Delta \vec r_1, \Delta\vec r_2$.
The net force is $\vec F_1+\vec F_2$, the net work is $\vec F_1\cdot\Delta \vec r_1+ \vec F_2\cdot\Delta \vec r_2$.
Clearly, $\Delta \vec r_2$ could be zero, meaning the second force does no work.

If you want to think of it as a resultant force $\vec F'$ acting at some point $\vec r'$ and moving through some displacement $\Delta\vec r'$, then we have to find the place where this resultant force acts:
$\vec F'=\vec F_1+\vec F_2$,
$\vec r'\times\vec F'=\vec r_1\times\vec F_1+\vec r_2\times\vec F_2$.
That is not enough to determine $\vec r'$, but it is enough to find the line of action of the resultant.
I'd like to continue that, but not enough time now.

Thank you for your reply @haruspex!

Ahh, thank you for mentioning that displacements! But if I think the cylinder rolling again, isn't the displacement for force of static friction and tension the same (if we are analyzing the whole cylinder instead of just the point of applications)

Please don't worry about this if you don't have time :)

Many thanks!

 

[jbriggs444]

Ahh, thank you for mentioning that displacements! But if I think the cylinder rolling again, isn't the displacement for force of static friction and tension the same (if we are analyzing the whole cylinder instead of just the point of applications)

There is room for confusion when we talk about "work".

We can talk about the work done by individual forces on individual points of application. This appears to be the sense that @haruspex is using when he speaks of "net work" -- the total of the works done by all of the individual forces.

i.e. $$\sum_i \vec{F_i} \cdot \vec{s_i}$$where each $F_i$ is one force and $s_i$ is the displacement for the point of application of that force.

This is distinct from $$( \sum_i \vec{F_i} ) \cdot \vec{s_\text{cm}}$$ where each $F_i$ is the same as above but $s_\text{cm}$ is the displacement of the center of mass.

The latter formula yields something that can be called "center of mass work".

Together with the two definitions for work there are two versions of the work-energy theorem.

If you are using the first definition of work ("net work" as @haruspex has been using the term), then the net work is equal to the change in total energy of the object. This includes the bulk kinetic energy associated with the movement of the object as a whole, rotational kinetic energy of the object as a rigid whole, any further kinetic energy associated with the relative motion of the parts (such as movement of pistons in an engine block or of whirlpools in a pool of water), vibrations, thermal energy, electrical energy (if the object is a generator, for instance), chemical energy (one can drive some endothermic chemical reactions by doing work on an object). In the case at hand, all we have is linear kinetic energy and rotational kinetic energy. All of those other places where energy could show up are irrelevant for purposes of this problem.

This version of the work-energy theorem is all about conservation of energy.

If you are using the second definition of work ("center of mass work"), then the work is equal to the change in linear kinetic energy of the object. This is just the kinetic energy associated with the motion of the object as a whole: $\frac{1}{2}m_\text{tot}v_\text{cm}^2$.

This version of the work-energy theorem is about conservation of momentum.

I apologize if you were already well aware of this and I am pontificating for no good reaason.

 

[haruspex]

any further kinetic energy associated with the relative motion of the parts (such as movement of pistons in an engine block or of whirlpools in a pool of water), vibrations, thermal energy, electrical energy (if the object is a generator, for instance), chemical energy (one can drive some endothermic chemical reactions by doing work on an object)

Yes, counting the total work done by the forces will include internal energy gained. e.g. compressing a spring, where the net force and net torque are both zero.
But a third formula (I believe) captures all added KE, linear and rotational, so would equal total work done by the forces on a rigid body, as here. This is $$( \sum_i \vec{F_i} ) \cdot \vec{s_\text{eff}}$$where $\vec{s_\text{eff}}$ is an effective point of application of the net force, i.e. a point about which the net force would have the same torque as that of the applied forces.

 

[jbriggs444]

But a third formula (I believe) captures all added KE, linear and rotational, so would equal total work done by the forces on a rigid body,

Yes, I've used that sort of approach in a few problems. Never with any firm theoretical foundation. It always seemed like something that should be obviously correct -- similar to ascribing an effective mass to a bicycle by doubling up the rim weight of the wheels.

 

[member 731016]

There is room for confusion when we talk about "work".

We can talk about the work done by individual forces on individual points of application. This appears to be the sense that @haruspex is using when he speaks of "net work" -- the total of the works done by all of the individual forces.

i.e. $$\sum_i \vec{F_i} \cdot \vec{s_i}$$where each $F_i$ is one force and $s_i$ is the displacement for the point of application of that force.

This is distinct from $$( \sum_i \vec{F_i} ) \cdot \vec{s_\text{cm}}$$ where each $F_i$ is the same as above but $s_\text{cm}$ is the displacement of the center of mass.

The latter formula yields something that can be called "center of mass work".

Together with the two definitions for work there are two versions of the work-energy theorem.

If you are using the first definition of work ("net work" as @haruspex has been using the term), then the net work is equal to the change in total energy of the object. This includes the bulk kinetic energy associated with the movement of the object as a whole, rotational kinetic energy of the object as a rigid whole, any further kinetic energy associated with the relative motion of the parts (such as movement of pistons in an engine block or of whirlpools in a pool of water), vibrations, thermal energy, electrical energy (if the object is a generator, for instance), chemical energy (one can drive some endothermic chemical reactions by doing work on an object). In the case at hand, all we have is linear kinetic energy and rotational kinetic energy. All of those other places where energy could show up are irrelevant for purposes of this problem.

This version of the work-energy theorem is all about conservation of energy.

If you are using the second definition of work ("center of mass work"), then the work is equal to the change in linear kinetic energy of the object. This is just the kinetic energy associated with the motion of the object as a whole: $\frac{1}{2}m_\text{tot}v_\text{cm}^2$.

This version of the work-energy theorem is about conservation of momentum.

I apologize if you were already well aware of this and I am pontificating for no good reaason.

Thank you for your reply @jbriggs444 ! That is helpful!

I would like to find the net work from on the cylinder from $W_{net} = F_{net} \cdot vec r = (T\hat j - F_f\hat i) \cdot [(r+\frac I{mr})\theta\hat i]$, would you please give me some guidance for how to do that?

@haruspex said the displacement was different for each force (which means that I will have to do sperate dot products for each force), however I'm not sure how find the displacement of each point of application when the cylinder is rotating.

Many thanks!