Solving Confusions on Pulling a Spool

[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!

 

[member 731016]

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.

Thank you for your reply @haruspex !

 

[member 731016]

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.

Thank you for your reply @jbriggs444 !

 

[haruspex]

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!

$T\hat j$?

 

[member 731016]

$T\hat j$?

Thank you for your reply @haruspex ! Oh whoops sorry, should be

$T\hat i$

Many thanks!

 

[member 731016]

$T\hat j$?

Here is the new equation $W_{net} = F_{net} \cdot \vec r = (T\hat i - F_f\hat i) \cdot [(r+\frac I{mr})\theta\hat i]$

Many thanks!

 

[member 731016]

I think it should actually be:

$W_{net} = F_{net} \cdot \vec r = T\hat i \cdot [(r+\frac I{mr})\theta\hat i] - F_f\hat i \cdot r_f\hat i$ where $r_f$ is the displacement of the point of application of the friction force

Many thanks!

 

[member 731016]

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

Thank you for your reply @erobz !

Back to the original problem, I think I am still struggling to understand how the hand moves an amount $l + L$ relative to the spool.

Many thanks!

 

[haruspex]

I think it should actually be:

$W_{net} = F_{net} \cdot \vec r = T\hat i \cdot [(r+\frac I{mr})\theta\hat i] - F_f\hat i \cdot r_f\hat i$ where $r_f$ is the displacement of the point of application of the friction force

Many thanks!

Since $F_{net}=T\hat i - F_f\hat i$, $W_{net} =( T\hat i - F_f\hat i)\cdot (r+\frac I{mr})\theta\hat i=(T-F_f) (r+\frac I{mr})\theta$.

Btw, just noticed that there’s a false assumption we all may have been making implicitly, that the friction acts to the left. In fact, it will act to the right. But the equations are fine, it's just that the value of F_f will be negative.

 

[member 731016]

Since $F_{net}=T\hat i - F_f\hat i$, $W_{net} =( T\hat i - F_f\hat i)\cdot (r+\frac I{mr})\theta\hat i=(T-F_f) (r+\frac I{mr})\theta$.

Btw, just noticed that there’s a false assumption we all may have been making implicitly, that the friction acts to the left. In fact, it will act to the right. But the equations are fine, it's just that the value of F_f will be negative.

Thank you for your reply @haruspex!

Sorry what do you mean friction will act to the right?

Many thanks!

 

[haruspex]

Thank you for your reply @erobz !

Back to the original problem, I think I am still struggling to understand how the hand moves an amount $l + L$ relative to the spool.

Many thanks!

No, no! $L+l$ relative to the ground.
The spool moves $L$ relative to the ground and the hand moves $l$ relative to the spool.

what do you mean friction will act to the right?

What it says. If there were no friction, the spool would rotate too fast for rolling contact, so the force of friction from the ground acts in the forward direction (right, in the diagram).

 

[member 731016]

Since $F_{net}=T\hat i - F_f\hat i$, $W_{net} =( T\hat i - F_f\hat i)\cdot (r+\frac I{mr})\theta\hat i=(T-F_f) (r+\frac I{mr})\theta$.

Btw, just noticed that there’s a false assumption we all may have been making implicitly, that the friction acts to the left. In fact, it will act to the right. But the equations are fine, it's just that the value of F_f will be negative.

Thank you for your reply @haruspex!

So from that equation static friction contributes to the net work (so dose do work actually?). Assuming that $T > F_f$ then the $W_{net} > 0$

Many thanks!

 

[haruspex]

Thank you for your reply @haruspex!

So from that equation static friction contributes to the net work (so dose do work actually?). Assuming that $T > F_f$ then the $W_{net} > 0$

Many thanks!

No!! It does no work whichever way it acts, for all the reasons given before.

 

[member 731016]

No, no! $L+l$ relative to the ground.
The spool moves $L$ relative to the ground and the hand moves #l## relative to the spool.

Thank you for your reply @haruspex!

How dose it the hand move $L + l$ relative to the ground?

What it says. If there were no friction, the the spool would rotate too fast for rolling contact, so the force of friction from the ground acts in the forward direction (right, in the diagram).

How would the spool rotate if there was no friction? Oh I guess the tension would provide a torque and therefore an angular acceleration and there would be other torque to counter it I guess.

Many thanks!

 

[member 731016]

No!! It does no work whichever way it acts, for all the reasons given before.

Thank you for your reply @haruspex!

How sorry, the force of static friction is in the net work equation?

$W_{net} =(T-F_f) (r+\frac I{mr})\theta$

Many thanks!

 

[member 731016]

If the force of friction is to the right then actually the net work equation should be

$W_{net} =(T + F_f) (r+\frac I{mr})\theta$ (since both T and F_f act in positive x-direction)

Is this please correct @haruspex ?

Many thanks!

 

[haruspex]

How dose it the hand move L+l relative to the ground?

The centre of the spool moves $L$ and rotates an angle $\theta$. So the top of the spool moves further than the centre. A length $l$ of string that was wrapped around the spool has been payed out and is now horizontal.

How sorry, the force of static friction is in the net work equation?

Suppose you want to lift up a heavy plank, weight W, at its middle. The plank lies on two supports height H. You position a block, height H, at X from its midpoint. You go round to the other side and insert a rod under the midpoint of the plank to rest on the block. You hold the rod Y from the plank and lift distance L. The plank rises R.
Work done =WR
Lift force = F_L, normal force from block =F_B.
F_L+F_B=W
F_LY=F_BX
R/X=L/(X+Y)
Which leads to, work done by you = LF_L=WR
You did all the work, the block did none.
Yet, both your lifting force and the normal force from the block feature in the net force on the plank.
You will just have to accept that a force that contributes to the net force does not necessarily do work.

If the force of friction is to the right then actually the net work equation should be

$W_{net} =(T + F_f) (r+\frac I{mr})\theta$ (since both T and F_f act in positive x-direction)

No, as I wrote, the equations are fine. The sign you put on a force in an equation only has to match the convention you are adopting for which way is positive for that force. When you wrote -F in your force balance equation that just meant you were taking left to be positive for that force. When you later discover that F has a negative value that tells you that it acts to the right. There is no need to change any algebra.

 

[erobz]

@Callumnc1

Look, you have to take a minute to draw a diagram if you are going to learn to physics...

 

[member 731016]

The centre of the spool moves $L$ and rotates an angle $\theta$. So the top of the spool moves further than the centre. A length $l$ of string that was wrapped around the spool has been payed out and is now horizontal.

Suppose you want to lift up a heavy plank, weight W, at its middle. The plank lies on two supports height H. You position a block, height H, at X from its midpoint. You go round to the other side and insert a rod under the midpoint of the plank to rest on the block. You hold the rod Y from the plank and lift distance L. The plank rises R.
Work done =WR
Lift force = F_L, normal force from block =F_B.
F_L+F_B=W
F_LY=F_BX
R/X=L/(X+Y)
Which leads to, work done by you = LF_L=WR
You did all the work, the block did none.
Yet, both your lifting force and the normal force from the block feature in the net force on the plank.
You will just have to accept that a force that contributes to the net force does not necessarily do work.

No, as I wrote, the equations are fine. The sign you put on a force in an equation only has to match the convention you are adopting for which way is positive for that force. When you wrote -F in your force balance equation that just meant you were taking left to be positive for that force. When you later discover that F has a negative value that tells you that it acts to the right. There is no need to change any algebra.

Thank you very much for your reply @haruspex !

I'll ponder over this for a while before replying :)

Many thanks!

 

[member 731016]

@Callumnc1

Look, you have to take a minute to draw a diagram if you are going to learn to physics...

View attachment 322838

Thank you very much for your diagram @erobz !

That is very helpful :) I was not quite sure how to go about drawing the diagram

Many thanks!

 

[erobz]

Thank you very much for your diagram @erobz !

That is very helpful :) I was not quite sure how to go about drawing the diagram

Many thanks!

Well, it's something that take some practice...If you're not even going to try, stumble, and maybe fail... you aren't going to get better at it. Sometimes a picture is worth a thousand words. This thread kind of proves that point.

 

[member 731016]

Well, it's something that take some practice...If you're not even going to try, stumble, and maybe fail though you aren't going to get better at it. Sometimes a picture is worth a thousand words...This thread kind of proves that point.

Thank you for your reply @erobz !

I agree with everything you said, I will try to draw a diagram next time. Perhaps I should draw my own diagram for the spool?

Many thanks!

 

[erobz]

Thank you for your reply @erobz !

I agree with everything you said, I will try to draw a diagram next time. Perhaps I should draw my own diagram for the spool?

Many thanks!

Almost always there is a useful diagram that can be drawn for a problem.

Have a good night(or day).

 

[member 731016]

Almost always there is a useful diagram that can be drawn for a problem.

Thank you @erobz , I agree!

Many thanks!

 

[member 731016]

Have a good night(or day).

You too! Thanks for your help!

 

[member 731016]

The centre of the spool moves $L$ and rotates an angle $\theta$. So the top of the spool moves further than the centre. A length $l$ of string that was wrapped around the spool has been payed out and is now horizontal.

Thank you for your reply @haruspex !

I guess it makes sense intuitively that the distance the hand has moved with respect to the ground is the distance moved by the CM (L) + the amount of string unraveled (l)

Many thanks!

 

[member 731016]

No, as I wrote, the equations are fine. The sign you put on a force in an equation only has to match the convention you are adopting for which way is positive for that force. When you wrote -F in your force balance equation that just meant you were taking left to be positive for that force. When you later discover that F has a negative value that tells you that it acts to the right. There is no need to change any algebra.

Thank you for your reply @haruspex !

Ok but if I was taking left to be positive in this equation $W_{net} =(T - F_f) (r+\frac I{mr})\theta$ then tension should be $-T \hat i$ correct?

However, if we choose right to be positive then $F_{net} = T \hat i + F_f \hat i$ and $W_{net} =(T + F_f) (r+\frac I{mr})\theta$

Is that please correct?

Many thanks!

 

[haruspex]

Thank you for your reply @haruspex !

Ok but if I was taking left to be positive in this equation $W_{net} =(T - F_f) (r+\frac I{mr})\theta$ then tension should be $-T \hat i$ correct?

However, if we choose right to be positive then $F_{net} = T \hat i + F_f \hat i$ and $W_{net} =(T + F_f) (r+\frac I{mr})\theta$

Is that please correct?

Many thanks!

You took right as positive for F_net and T, left as positive for F_f: $F_{net}=T-F_f$, and with clockwise as positive for torque about ground level, $\tau_{net}=2rT$.
With right as positive for all forces, $F_{net}=T+F_f$, torque as before.
With left as positive for all forces, $F_{net}=T+F_f$, and with clockwise still as positive for torque about ground level, $\tau_{net}=-2rT$.

 

[member 731016]

Suppose you want to lift up a heavy plank, weight W, at its middle. The plank lies on two supports height H. You position a block, height H, at X from its midpoint. You go round to the other side and insert a rod under the midpoint of the plank to rest on the block. You hold the rod Y from the plank and lift distance L. The plank rises R.
Work done =WR
Lift force = F_L, normal force from block =F_B.
F_L+F_B=W
F_LY=F_BX
R/X=L/(X+Y)
Which leads to, work done by you = LF_L=WR
You did all the work, the block did none.
Yet, both your lifting force and the normal force from the block feature in the net force on the plank.
You will just have to accept that a force that contributes to the net force does not necessarily do work.

Thank you for your reply @haruspex !

That seems like the most non-intuitive physics fact I've ever heard of!

Do you know where to learn that fact from?

BTW, I am going to draw a diagram and post it sometime so that I make sure I understand your analogy correctly

Many thanks!

 

[member 731016]

You took right as positive for F_net and T, left as positive for F_f: $F_{net}=T-F_f$, and with clockwise as positive for torque about ground level, $\tau_{net}=2rT$.
With right as positive for all forces, $F_{net}=T+F_f$, torque as before.
With left as positive for all forces, $F_{net}=T+F_f$, and with clockwise still as positive for torque about ground level, $\tau_{net}=-2rT$.

Thank you for your reply @haruspex !

What I did was wrong, correct?

I am not allowed to take right as positive for one force and left as positive for another force as you have to define one coordinate system and use it to for determining the direction of the forces

Many thanks!

 

[haruspex]

What I did was wrong, correct?

Yes,

I am not allowed to take right as positive for one force and left as positive for another force as you have to define one coordinate system and use it to for determining the direction of the forces

No, you are free to define positive separately for each force, each displacement, each acceleration. Just be clear about it and do it consistently through the equations.

If all such variables are along the perpendicular axes, it is a good idea to pick one convention for them all. If not, you could decompose each oblique variable into such components or define the conventions individually.

Many prefer to guess which way each will act and define positive accordingly. That's ok as long as you handle each consistently and accept that some may turn out to have negative values.

That said, there may be cases where the equation does change if you guessed wrongly, but I can only think of quite contrived ones.

 

[member 731016]

Yes,

No, you are free to define positive separately for each force, each displacement, each acceleration. Just be clear about it and do it consistently through the equations.

If all such variables are along the perpendicular axes, it is a good idea to pick one convention for them all. If not, you could decompose each oblique variable into such components or define the conventions individually.

Many prefer to guess which way each will act and define positive accordingly. That's ok as long as you handle each consistently and accept that some may turn out to have negative values.

That said, there may be cases where the equation does change if you guessed wrongly, but I can only think of quite contrived ones.

Thank you for your reply @haruspex! That is very interesting!

Many thanks!