Kinda hard problem involving friction and a spool.

[charmedbeauty]

Homework Statement

A string is attached to the drum (radius r) of a spool (radius R). (A spool is device for storing string, thread etc.) A tension T is applied to the string at angle θ above the horizontal. The coefficients of kinetic and static friction between floor and spool are μk and μs respectively. We are interested in whether and when the spool will move left or right, and how this depends on the nature of the floor.

Showing all working, calculate the critical value of θ (call it θc) at which the condition goes from rolling to skidding.


Homework Equations

The Attempt at a Solution

ok so I did this...

the point at which its starts skidding is when Nμ_s=Tcosθ_c

solving for θ_c=cos^-1(Nμ_s/T).

but this is wrong I can tell why... because you can't really define an angle by a force... but I don't understand why the expression is wrong,. help please I've been at it for a while. Thanks.

The right answer should be θ_c=cos^-1r/R

but I don't know how they get this?

 

[Infinitum]

I don't see anything wrong with your answer. You only need to express it in a different way.

You'll have an equation,

$$f_s-Tcos\theta = ma$$

Now you should also be able to make a torque equation, about the center,

$$\tau = I\frac{a}{r}$$

Write the torque in terms of the forces on the spool, and then derive an expression for acceleration, a, substituting from the first. Now apply the limiting condition that a=0.

 

[charmedbeauty]

I don't see anything wrong with your answer. You only need to express it in a different way.

You'll have an equation,

$$f_s-Tcos\theta = ma$$

Now you should also be able to make a torque equation, about the center,

$$\tau = I\frac{a}{r}$$

Write the torque in terms of the forces on the spool, and then derive an expression for acceleration, a, substituting from the first. Now apply the limiting condition that a=0.

Where does the I come from?

it is an exam question so I don't think it should be in terms of any I, since the only I on the data sheet is for a sphere and a rod.

with the torque how do I express it if it's the sum of the forces?, than shouldn't it be...
$\tau$=Nf_s+Tcosθ ?

 

[Infinitum]

Where does the I come from?

it is an exam question so I don't think it should be in terms of any I, since the only I on the data sheet is for a sphere and a rod.

with the torque how do I express it if it's the sum of the forces?, than shouldn't it be...
$\tau$=Nf_s+Tcosθ ?

Noo! 'I' will not be a term in the answer. You only need it to solve for the critical value of θ. Or, if it makes you feel better, just write I in terms of mass and radii(it won't bother you in getting the critical value)

Also, torque is not exactly the sum of forces, but sum of moments. What is the torque due to the tension force?? and frictional force??

 

[azizlwl]

1.. The coefficients of kinetic and static friction between floor and spool are μk and μs respectively. We are interested in whether and when the spool will move left or right, and how this depends on the nature of the floor.

Showing all working, calculate the critical value of θ (call it θc) at which the condition goes from rolling to skidding.



The right answer should be θ_c=cos^-1r/R

but I don't know how they get this?

I think the answer has a factor of μk or μs?

 

[Infinitum]

I think the answer has a factor of μk or μs?

It does not. The answer is, as charmedbeauty said,

$$\theta = cos^{-1}\frac{r}{R}$$

An alternate answer as solved by charmedbeauty can also be considered right, though, where

θc=cos^-1(Nμs/T).

Just need to find what N would be here.

 

[charmedbeauty]

Noo! 'I' will not be a term in the answer. You only need it to solve for the critical value of θ. Or, if it makes you feel better, just write I in terms of mass and radii(it won't bother you in getting the critical value)

Also, torque is not exactly the sum of forces, but sum of moments. What is the torque due to the tension force?? and frictional force??

so $\tau$=r×Fcosθ_c

but shouldn't r be the distance of the rope to the pulley not 'r' as in radius?

Im not to sure with $\tau$ in how it works so I just wikie'd it quickly... but from what I gather this should be the $\tau$ equation.

Im really lost in how to relate this to r, R?

 

[Infinitum]

so $\tau$=r×Fcosθ_c

but shouldn't r be the distance of the rope to the pulley not 'r' as in radius?

Im not to sure with $\tau$ in how it works so I just wikie'd it quickly... but from what I gather this should be the $\tau$ equation.

Im really lost in how to relate this to r, R?

PF library : https://www.physicsforums.com/library.php?do=view_item&itemid=175

Basically, torque about a point is the product of the force times the perpendicular distance between the point and line of action of the force. So, the torque due to tension about the center is given by,

$$\tau = Tr$$

As T and r are perpendicular. It is taken to be positive because it is in the anticlockwise direction. A clockwise torque would be considered negative.

Can you say what the torque due to frictional force would be?

 

[charmedbeauty]

PF library : https://www.physicsforums.com/library.php?do=view_item&itemid=175

Basically, torque about a point is the product of the force times the perpendicular distance between the point and line of action of the force. So, the torque due to tension about the center is given by,

$$\tau = Tr$$

As T and r are perpendicular. It is taken to be positive because it is in the anticlockwise direction. A clockwise torque would be considered negative.

Can you say what the torque due to frictional force would be?

is r the distance of the rope of the inner radius 'r'?

in any case I don't see why T and r are perpendicular?

the only thing I could say is perpendicular is Tsinθ and F_f

? am I missing something...also what is meant by 'cross'? I thought it was just times?

Thanks.

 

[Infinitum]

is r the distance of the rope of the inner radius 'r'?

in any case I don't see why T and r are perpendicular?

the only thing I could say is perpendicular is Tsinθ and F_f

? am I missing something...also what is meant by 'cross'? I thought it was just times?

Thanks.

r is the inner radius, R is the outer radius. Well, that's what the question said, I think.

Cross denotes the cross product.

 

[charmedbeauty]

r is the inner radius, R is the outer radius. Well, that's what the question said, I think.

Cross denotes the cross product.

Ohh now I see yes they are so it does not matter what θ the tension is applied it will always be perpendicular to inner radius.

so having said that... I think i need to consider some kind of ratio I'm guessing that's where r/R comes from...

I have a feeling it might be something like this,

cosθ_c=Tr/Nμ_sR

but I'm not exactly sure

 

[Infinitum]

Ohh now I see yes they are so it does not matter what θ the tension is applied it will always be perpendicular to inner radius.

Yep!

so having said that... I think i need to consider some kind of ratio I'm guessing that's where r/R comes from...

I have a feeling it might be something like this,

cosθ_c=Tr/Nμ_sR

but I'm not exactly sure

For that, why don't you try out what I hinted you in post #2? You won't get the relation that you've stated above, btw.

 

[ehild]

This problem is rather difficult. The given answer - cos(θ_c)=r/R - means that angle where the spool changes direction of rolling supposed the static friction is large enough with respect the applied tension.

ehild

 

[charmedbeauty]

Yep!





For that, why don't you try out what I hinted you in post #2? You won't get the relation that you've stated above, btw.

from this I get

$\tau$=I((f_s-Tcosθ)/mr)=ma

does the I cancel out with the m?

I'm still lost how does R come into this?

 

[Infinitum]

from this I get

$\tau$=I((f_s-Tcosθ)/mr)=ma

does the I cancel out with the m?

I'm still lost how does R come into this?

I think you aren't using the definition of torque correctly yet. The total torque about center is,

$$Tr-f_sR$$

Do you see how this comes?

And you have the force equation in f_s. Substitute f_s from this.

 

[charmedbeauty]

I think you aren't using the definition of torque correctly yet. The total torque about center is,

$$Tr-f_sR$$

Do you see how this comes?

And you have the force equation in f_s. Substitute f_s from this.

ok so...

Tr=f_sR

f_s=Tr/R

Nμ_s=Tr/R

unless Nμ_s cancels with T I don't know how to get this in the right ratio...

since Trcosθ will yield θ_c= R/r of that kind how would you get r/R

doesn't that imply Nμ_scosθ_c is needed.?

 

[Infinitum]

ok so...

Tr=f_sR

This applies only when the torque is zero. Torque = I(a/r), as i mentioned in an earlier post. You then have to put a=0. All this is just to get the 'type' of answer you require. Your result is correct, as it is. But finding N would require you to know the mass of the spool, which is not given in the problem, and hence getting the answer in terms of m would not be right.

You are also not using the other equation $f_s - Tcos\theta = ma$

 

[charmedbeauty]

This applies only when the torque is zero. Torque = I(a/r), as i mentioned in an earlier post. You then have to put a=0. All this is just to get the 'type' of answer you require. Your result is correct, as it is. But finding N would require you to know the mass of the spool, which is not given in the problem, and hence getting the answer in terms of m would not be right.

You are also not using the other equation $f_s - Tcos\theta = ma$

ohh I think I get it from the eqn above solve for f_s

f_s=ma+Tcosθ

Nμ_s=ma+Tcosθ

wait

f_sN-Tcosθ=ƩF=ma

torque=f_sR-Tr=ma

f_sN-Tcosθ=Tr-f_sR

if I collect like terms I have

f_s(N+R)=T(r+cosθ)

I'm still confused I have a feeling things should be cancelling but I don't know why? ahh I hate torque!

but thanks for the help so far infinitum.!

 

[azizlwl]

.

You are also not using the other equation $f_s - Tcos\theta = ma$

I think f_s would never be greater than applied force. Equal or less.

 

[ehild]

It is not clear what the problem asks. If you play with a spool (that for a sewing thread for example) you will find that you can pull it so as it rolls without slipping with a small force, and it can roll both forward and backward, according to the angle of the thread with respect to the horizontal. At angles less than θ_c, the spool rolls forward, at greater angles it rolls backward. It is interesting what happens when pulling at the critical angle. The spool can move without rotation: it only skids. But this is a singular behaviour.

Pure rolling means that the linear velocity of the CM is the same as the linear velocity at the perimeter of the spool V_CM=Rω. The same relation holds between linear acceleration of the CM and angular acceleration.
The point of contact with the ground is the instantaneous axis of rotation. You can write up the torque with respect to it. If the line of the thread goes through that axis, the torque is zero. Pulling at different angles, the torque changes sign according to the direction of the "arm of force". See picture.
Pure rolling is possible only when the static friction is great enough to keep the lowest point of the spool motionless with respect to the ground. That means a relation between tension, angle, and coefficient of static friction. If this relation is not fulfilled the spool can both skid and rotate, V_CM=ωR is not valid any more. You can write up equations for both the angular acceleration and for the acceleration of the CM under the effect of the tension and kinetic friction, they are not coupled any more. The answer given refers to the singular case, and the question
"calculate the critical value of θ (call it θc) at which the condition goes from rolling to skidding" is confusing as skidding happens only when cosθ=r/R, neither at smaller nor at greater angles.

ehild

 

[Infinitum]

I think f_s would never be greater than applied force. Equal or less.

True. It doesn't matter as a=0 for the limiting case, I believe. Anyway, a correct equation would rather be,

$$Tcos\theta - f_s = ma$$

Thanks for the correction!

It is interesting what happens when pulling at the critical angle. The spool can move without rotation: it only skids. But this is a singular behaviour.

"calculate the critical value of θ (call it θc) at which the condition goes from rolling to skidding" is confusing as skidding happens only when cosθ=r/R, neither at smaller nor at greater angles.

This is exactly the critical(and only) angle that is asked for, I think. According to the way I solved it, it doesn't give me any other result other than $cos\theta=r/R$ (no inequality)

 

[azizlwl]

Here example from Schaum series, 3000 problem in physics

If the origin for calculating torques is taken at P, point of contact between spool and the tabletop, the torques due to gravity, friction, and the normal force are all zero.

The force colored blue the torque about P is zero. The spool will not rotate if pulled gently and does not slip. It must be that the spool remain motionless

From the configuration,

Cosθ=r/R

 

[ehild]

Here example from Schaum series, 3000 problem in physics

If the origin for calculating torques is taken at P, point of contact between spool and the tabletop, the torques due to gravity, friction, and the normal force are all zero.

The force colored blue the torque about P is zero. The spool will not rotate if pulled gently and does not slip. It must be that the spool remain motionless

From the configuration,

Cosθ=r/R

Both the linear and angular accelerations are zero at angle θc. It does not mean that the spool is motionless.
It will not rotate but slips if pulled hard enough at that angle. Try. I just pull a spool on the table and it slips without rolling. Somehow it is set into motion without rotation. It might be because of the neglected rolling resistance. Once it moves, it can be kept in motion.

ehild

 

[ehild]

True. It doesn't matter as a=0 for the limiting case, I believe. Anyway, a correct equation would rather be,

$$Tcos\theta - f_s = ma$$

Thanks for the correction!

What about fs when the spool rolls backwards (T is positive, a is negative)?

This is exactly the critical(and only) angle that is asked for, I think. According to the way I solved it, it doesn't give me any other result other than $cos\theta=r/R$ (no inequality)

Try an other way to solve for other angles when pulled with a certain force T assuming the spool does not perform pure rolling, but it both skids and rotates. There is a limit for the angle when pure rolling changes to skidding and rolling: at that angle when the tension is not enough to overcome the maximum static friction.

Imagine a frictionless surface. How will the spool move if pulled?

ehild

 

[Infinitum]

The force colored blue the torque about P is zero. The spool will not rotate if pulled gently and does not slip. It must be that the spool remain motionless

Both the linear and angular accelerations are zero at angle θc. It does not mean that the spool is motionless.

I tried out the spool experiment too, it only skids at that particular angle, without rolling, as ehild says. But that is only considering that we pull it hard enough. Azizlwl's example involves a gentle force, hence the motionless spool. Considering we can pull it quite hard, it will not have any particular angle where it can stay motionless, and has to skid. But skidding at this point or not(zero linear acceleration being the limiting case), this is that exact angle where applied force will never cause a rotational acceleration, and it is what is asked in the problem.

 

[Infinitum]

What about fs when the spool rolls backwards (T is positive, a is negative)?

It really shouldn't matter. 'a' is the variable here, it can be negative too, so it includes that case. As for a limiting case, 'a' would be 0, which is considered in the question.

Try an other way to solve for other angles when pulled with a certain force T assuming the spool does not perform pure rolling, but it both skids and rotates. There is a limit for the angle when pure rolling changes to skidding and rolling: at that angle when the tension is not enough to overcome the maximum static friction.

I made an assumption that the friction is sufficiently enough, as with most rotational motion problems. I'll try it without that assumption too.

Imagine a frictionless surface. How will the spool move if pulled?

ehild

Non-roll skidding at the critical value, and -never- zero linear acceleration

 

[schaefera]

Ok, think like this:
Sum of forces in x: Tcos(u)-f=ma_x
Sum of forces in y: Tsin(u)+N-mg=ma_y
Sum of torques: Tr-fR=I(alpha)

The sum in y doesn't help us, but at u_c (the critical angle), so try from there.

 

[Infinitum]

Ok, think like this:
Sum of forces in x: Tcos(u)-f=ma_x
Sum of forces in y: Tsin(u)+N-mg=ma_y
Sum of torques: Tr-fR=I(alpha)

The sum in y doesn't help us, but at u_c (the critical angle), Tcos(u_c)=f and Tr=fR because acceleration is zero (limiting case) and therefore alpha is zero.

*snip*

This is exactly what I suggested, but my assumption, like yours, was that acceleration is zero because there's sufficient friction. There is this different case, suggested by ehild where linear acceleration is not zero, but there is no rotational motion. Hence, instead of having the statement as

because acceleration is zero (limiting case)

it should probably remodeled to only $\alpha = 0$ and not linear acceleration

PS: Please don't post full solutions

 

[schaefera]

My bad I apologize! But how do we solve the case with linear acceleration since we can't use the same algebra to cancel everything we don't know?

 

[ehild]

Remember the original text of the problem:

A tension T is applied to the string at angle θ above the horizontal. The coefficients of kinetic and static friction between floor and spool are μk and μs respectively. We are interested in whether and when the spool will move left or right, and how this depends on the nature of the floor.

So given T and the coefficients of friction. At what angles will the spool roll without skidding and at what angle does it start to skid?
Note that skidding does not mean pure translation. The wheels of a car rotate on ice or in mud when the car skids...

θc you arrived at, does not depend on the magnitude of tension neither is it influenced by the nature of the floor.

Applying a tension T, what is the acceleration in terms of the angle? What about the effect of non-infinite static friction? For what angles does the condition of pure rolling hold? And what happens if the angle is outside that range?

ehild

 

[ehild]

My bad I apologize! But how do we solve the case with linear acceleration since we can't use the same algebra to cancel everything we don't know?

In case of skidding, the friction is kinetic, instead of being static. You know that kinetic friction is μN. You can express N with the weight and T and the angle.

There are two equations, one for linear acceleration, and the other for angular acceleration. You can solve both in terms of tension, angle, mass and moment of inertia and see how they are related at different pulling angles.

ehild

 

[schaefera]

So, the case where it's static friction is the one case in which there is no acceleration and then we turn the cranks with kinetic friction there to find forward and backwards dependencies?

 

[ehild]

So, the case where it's static friction is the one case in which there is no acceleration and then we turn the cranks with kinetic friction there to find forward and backwards dependencies?

No, there is static friction in case of pure rolling. During rolling, there can be acceleration, both linear and angular, related by the equation a=R dω/dt.
The rolling can happen both forward and backwards.

When the spool skids, the friction is kinetic. There can be both linear and angular acceleration.

ehild.

 

[Infinitum]

No, there is static friction in case of pure rolling. During rolling, there can be acceleration, both linear and angular, related by the equation a=R dω/dt.
The rolling can happen both forward and backwards.

When the spool skids, the friction is kinetic. There can be both linear and angular acceleration.

ehild.

We're asked when the spool switches from rolling to skidding. So rolling with skidding shouldn't be considered...

 

[ehild]

calculate the critical value of θ (call it θc) at which the condition goes from rolling to skidding.

It is not the spool that switches from rolling to skidding.
Applying the same tension, the spool will not switch from rolling to skidding at the critical angle, but from rolling to standing...
But it can skid at the critical angle without rotating.

Do you only want to answer to the problem writer? Are you not interested in the phenomenon?

ehild