Solving Confusions on Pulling a Spool

[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.