# Solving Confusions on Pulling a Spool

• member 731016
In summary: Thank you for your reply @erobz!I think the hand would have unwound the same amount since each part as the angular velocity.
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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!

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Callumnc1 said:
(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
erobz said:
If it was just sliding and not rotating, how far would the hand be displaced?

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

Many thanks!

Callumnc1 said:

##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.

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member 731016
Callumnc1 said:
Why can't we use the UAM equations?
I am unfamiliar with that abbreviation. What equation do you have in mind?

Callumnc1 said:
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
erobz said:
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!

Callumnc1 said:
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
haruspex said:
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.

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

Many thanks!

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

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

Many thanks!

Callumnc1 said:

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
haruspex said:
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.

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

Many thanks!

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

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

Many thanks!

Callumnc1 said:

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
erobz said:
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?

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

Many thanks!

Callumnc1 said:

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##?

member 731016
Callumnc1 said:
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.

member 731016
erobz said:
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 and erobz
erobz said:
What does this mean. What amount of rope is between the drum and the hand after the COM moves a distance ##L##?
haruspex said:
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!

haruspex said:
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.

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!

Last edited by a moderator:
Callumnc1 said:
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.
Callumnc1 said:
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.

Last edited:
member 731016
Callumnc1 said:
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?

Last edited:
member 731016
erobz said:
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.

Last edited:
member 731016 and nasu
kuruman said:
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.

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member 731016
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.

member 731016 and erobz
erobz said:
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.

member 731016
haruspex said:
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
haruspex said:
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.

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

Many thanks!

Callumnc1 said:

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
erobz said:
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!

kuruman said:
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!

erobz said:
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 said:
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.
erobz said:
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 !
haruspex said:
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.

haruspex said:
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!

haruspex said:
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.

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!

haruspex said:
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!

haruspex said:
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!

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