Friction on Pulleys: Axle or Surface?

[andyrk]

When we say of a pulley having friction, do we mean its axle about which it rotates has friction or the surface over which the string is wound has friction? Either way, if their is friction in a pulley, then when a string passes over the pulley, why does the string have different tension over the two sides of the pulley, when the two ends are connected by some weights?

 

[Tanya Sharma]

When we say of a pulley having friction, do we mean its axle about which it rotates has friction or the surface over which the string is wound has friction?

There are two types of friction acting on the pulley.One at the axle,other between the pulley and the string.Generally the pulley is considered frictionless which means absence of friction at the pulley axle.There is sufficient friction present between the pulley and the string such that the string doesn't slip over the pulley.It is static friction which is responsible for the rotation of pulley along with the string .

 

[andyrk]

Ok so if we say that the pulley is frictionless does it mean:
(i)The string slips over the top and the axle is frictionless.
(ii)The pulley rotates with the string (sufficient friction is there between the string and the pulley) and the axis has friction.
(iii)The pulley rotates with the string (enough friction between the pulley and the string)
but the axle is frictionless.
(iv)Both the surface of the pulley and the axle have friction.

Which of the four cases would be true if we say the pulley is frictionless and the pulley has friction?

And if the pulley has friction why does it affect the tension on the two hanging sides of the string?

 

[Tanya Sharma]

We will discuss tension later but first understand the motion of the pulley and the string.

Do you understand what causes the pulley to rotate ?

It is the static friction between the pulley and the string which causes the pulley to rotate.The friction at the axle opposes this rotation.

The torque equation of the pulley about the axle is FR-fr =Iα .
Here ,
F is the static friction between the pulley and the string.
f is the friction at the axle.
R is the radius of the pulley
r is the radius of the axle.

Note:If there is no friction between the pulley and the string,the string simply slips over the pulley and the pulley doesn't rotate.

Does this makes sense ?

 

[andyrk]

Yes it does.

 

[Tanya Sharma]

Did you get the answer to the question you posed in post #3? If yes ,can you explain briefly what you have understood so that we can move forward ?

 

[andyrk]

Yup, it would be
"(iii)The pulley rotates with the string (enough friction between the pulley and the string)
but the axle has friction"

 

[Tanya Sharma]

Yup, it would be
"(iii)The pulley rotates with the string (enough friction between the pulley and the string)
but the axle has friction"

You haven't understood.I have already explained it to you in post#2 that a frictionless pulley means absence of friction at the axle.

Ok so if we say that the pulley is frictionless does it mean:
(i)The string slips over the top and the axle is frictionless.
(ii)The pulley rotates with the string (sufficient friction is there between the string and the pulley) and the axis has friction.
(iii)The pulley rotates with the string (enough friction between the pulley and the string)
but the axle has friction
(iv)Both the surface of the pulley and the axle have friction.

None of the options is correct.

 

[andyrk]

Oh I meant to say that the axle is frictionless. I corrected that in #3 in (iii).
Yup..I agree with that.

 

[Tanya Sharma]

Okay ...

Now,we will try and understand tensions on the two sides of the pulley.To keep things simple,the string is considered massless.So,for the time being assume string to be massless.

Reconsider the torque equation of the pulley FR-fr=Iα .Now what happens is that the torque on the pulley due to friction(FR) is equal to the difference in the torques due to tensions on the two sides (T2R-T1R).

So,we can rewrite the torque equation of the pulley about the axle as (T2-T1)R -fr =Iα.Keep this equation in mind.

Now we will simplify things one step at a time.

If the pulley is frictionless ,'f' in the second term on the L.H.S becomes zero and we have (T2-T1)R=Iα.

Case 1) Pulley has mass , I≠0 .So,T2≠T1 i.e tensions on the two sides are unequal.So,if the question says the pulley is frictionless but has mass then the tensions on the two sides are different.

Case 2) Pulley is massless , I=0 .So,T2=T1 i.e tensions on the two sides are equal. So,if the question says the pulley is frictionless and massless then the tensions on the two sides are same.

 

[andyrk]

Okay ...

Now,we will try and understand tensions on the two sides of the pulley.To keep things simple,the string is considered massless.So,for the time being assume string to be massless.

Reconsider the torque equation of the pulley FR-fr=Iα .Now what happens is that the torque on the pulley due to friction(FR) is equal to the difference in the torques due to tensions on the two sides (T2R-T1R).

So,we can rewrite the torque equation of the pulley about the axle as (T2-T1)R -fr =Iα.Keep this equation in mind.

Now we will simplify things one step at a time.

If the pulley is frictionless ,'f' in the second term on the L.H.S becomes zero and we have (T2-T1)R=Iα.

Case 1) Pulley has mass , I≠0 .So,T2≠T1 i.e tensions on the two sides are unequal.So,if the question says the pulley is frictionless but has mass then the tensions on the two sides are different.

Case 2) Pulley is massless , I=0 .So,T2=T1 i.e tensions on the two sides are equal. So,if the question says the pulley is frictionless and massless then the tensions on the two sides are same.

Yup got it now thanks! but one more thing:
SO the pulley doesn't rotate in case 2) does it?

 

[Enigman]

Nope.

 

[andyrk]

Thanks for the replies. Just one last thing, that, if we want a pulley having mass to not rotate, then we will consider the surface of the pulley to be frictionless, right? So that the string slips upon it? And here the fr term that we have taken, the r stands for the radius of the axle right? And the R for the radius of the pulley..i.e (R>r). Right?

 

[Tanya Sharma]

Yup got it now thanks! but one more thing:
SO the pulley doesn't rotate in case 2) does it?

No .The pulley rotates in both the cases.

In case 2) since the pulley is massless it requires zero torque to rotate it.Reconsider torque equation
(T2-T1)R = Iα . As the value of I becomes smaller i.e → 0 ,the difference in the tensions →0.It will still rotate with tensions on both side same.

Anyways, an ideal pulley with mass 0 is an assumption made in physics to simplify problems just like frictionless surface,massless string etc

Thanks for the replies. Just one last thing, that, if we want a pulley having mass to not rotate, then we will consider the surface of the pulley to be frictionless, right? So that the string slips upon it? And here the fr term that we have taken, the r stands for the radius of the axle right? And the R for the radius of the pulley..i.e (R>r). Right?

Correct

 

[andyrk]

Ok. You mean the tending to zero makes it still rotate. But I am talking about the exact situation when the mass is actually zero. It can't happen but let's say. Then would the pulley be rotating? I don''t think it should rotate then. As I=0...Then the net torque from Newton's second law for rotational motion is also 0 means rotation isn't possible.

 

[Tanya Sharma]

Dont go by intuition.Check your reasoning with the maths .

As I=0...Then the net torque from Newton's second law for rotational motion is also 0 means rotation isn't possible.

Let T=Iα .If I=0 ,then T=0 .It doesn't mean α =0.

You require zero net torque to rotate an object having zero moment of inertia(I=0),just like you don't need force to accelerate a massless object .Remember massless objects are ideal cases which do not exist in real world.They are considered to simplify the situation.

 

[andyrk]

So we don't require any net torque to rotate a massless body. Does that mean the body rotates on its own?

 

[Tanya Sharma]

Please reread post#14 and post#16.You would understand things better if instead of taking M=0, consider what happens when M → 0.

 

[andyrk]

Didn't understand it from reading the posts you mentioned. The posts hold true for the cases when I→0 not when I=0. As for I=0, net angular acceleration may not be 0 and the body may rotate but how? You said that for rotating a body with no mass we don't require any torque but how does it rotate then I am just not able to understand. It can't be rotating on its own? A body rotating without any torque, just doesn't make sense to me :/

 

[Enigman]

Didn't understand it from reading the posts you mentioned. The posts hold true for the cases when I→0 not when I=0. As for I=0, net angular acceleration may not be 0 and the body may rotate but how? You said that for rotating a body with no mass we don't require any torque but how does it rotate then I am just not able to understand. It can't be rotating on its own? A body rotating without any torque, just doesn't make sense to me :/

A body having an angular acc. without any torque makes just about as much sense as it not having any mass both are just ideal cases. These assumptions are used to make approximate measurements by uncomplicating the math. If you want to analyse the pulley or get more accurate measurements you will have to take into consideration the mass and friction involved.
Also bodies rotate without torque all the time, they just don't change their angular momentum without any external torque.

 

[voko]

@andyrk: didn't you ask this same question here about three months ago? I believe you were given all the answers, many times over.