# Kinda hard problem involving friction and a spool.

1. Jun 4, 2012

### charmedbeauty

1. The problem statement, all variables and given/known data

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.
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2. Relevant equations

3. The attempt at a solution

ok so I did this...

the point at which its starts skidding is when Nμs=Tcosθc

solving for θc=cos-1(Nμs/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 θc=cos-1r/R

but I don't know how they get this?

2. Jun 4, 2012

### 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.

3. Jun 4, 2012

### charmedbeauty

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...
$\tau$=Nfs+Tcosθ ?

4. Jun 5, 2012

### Infinitum

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 wont 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??

5. Jun 5, 2012

### azizlwl

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

6. Jun 5, 2012

### Infinitum

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.

7. Jun 5, 2012

### charmedbeauty

so $\tau$=r×Fcosθc

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?

8. Jun 5, 2012

### Infinitum

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?

9. Jun 5, 2012

### charmedbeauty

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 Ff

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

Thanks.

10. Jun 5, 2012

### Infinitum

r is the inner radius, R is the outer radius. Well, thats what the question said, I think.

Cross denotes the cross product.

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11. Jun 5, 2012

### charmedbeauty

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 thats where r/R comes from...

I have a feeling it might be something like this,

cosθc=Tr/NμsR

but I'm not exactly sure

12. Jun 5, 2012

### Infinitum

Yep!!

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

13. Jun 5, 2012

### ehild

This problem is rather difficult. The given answer - cos(θc)=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

14. Jun 5, 2012

### charmedbeauty

from this I get

$\tau$=I((fs-Tcosθ)/mr)=ma

does the I cancel out with the m?

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

15. Jun 5, 2012

### Infinitum

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 fs. Substitute fs from this.

Last edited: Jun 5, 2012
16. Jun 5, 2012

### charmedbeauty

ok so...

Tr=fsR

fs=Tr/R

s=Tr/R

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

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

doesn't that imply Nμscosθc is needed.?

17. Jun 5, 2012

### Infinitum

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 $f_s - Tcos\theta = ma$

Last edited: Jun 5, 2012
18. Jun 5, 2012

### charmedbeauty

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

fs=ma+Tcosθ

s=ma+Tcosθ

wait

fsN-Tcosθ=ƩF=ma

torque=fsR-Tr=ma

fsN-Tcosθ=Tr-fsR

if I collect like terms I have

fs(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.!

19. Jun 6, 2012

### azizlwl

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

20. Jun 6, 2012

### 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 θc, 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 VCM=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, VCM=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

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Last edited: Jun 6, 2012