# Homework Help: Massless and Frictionless Pulley

1. Jun 18, 2013

### andyrk

Can a massless and frictionless pulley rotate if a string goes over it connecting two blocks of unequal weight? I think it shouldn't rotate and the string just skids/slides over the top of the pulley. Is it correct?

2. Jun 18, 2013

### PeterO

I think "a massless and frictionless pulley" is just the description used in a problem where you are intended to ignore the effects of mass and friction in the pulley. I have always assumed the frictionless part refers to the axle/bearing rather than the groove the string runs in.

3. Jun 18, 2013

### dreamLord

Yes, the frictionless part doesn't mean that the surface of the pulley is frictionless - instead it's basically saying that there are no additional torques in the system other than the ones provided by the strings.

4. Jun 18, 2013

### andyrk

Then of agreeing with # frictionless means frictionless axle. And for #3 you are saying that no external torque except that provided by the strings. So in such a case we take tension in both the strings to be same. So the net torque at the highest point should be 0 and infact on any other point because at every point the net force due to tension is 0. So still how does the pulley rotate?

5. Jun 18, 2013

### PeterO

Can a massless and frictionless pulley rotate if a string goes over it connecting two blocks of unequal weight? I think it shouldn't rotate and the string just skids/slides over the top of the pulley. Is it correct?

I addressed the frictionless reference by saying it is the "axle" that is frictionless.

The massless nature removes the only other possible complication in a real-world example of this.

First: The pulley will rotate!

If the masses are unequal, the large one will descend - accelerate down even - and the small one will accelerate up.
While it is easy to calculate that acceleration with a massless/frictionless pully [ie. ignoring the effects of the pully], it becomes very complicated if friction and the pulley mass are taken into account. For the frictionless/massless situation, the tension in the string on both sides of the pulley is the same - and not zero.

With friction the tensions would be different - and if it was large enough, and the two masses were not too different, the masses might not move at all.

If the pulley has considerable mass, then as the masses accelerate up/down, the system also has to start this pulley spinning. For a very heavy pulley that might take a lot of energy form the system, so the masses will accelerate very slowly. Accurately taking account of the pulley mass is not impossible, but is beyond the scope of the calculations expected to be done when these problems are first encountered.

Consider this example:
Stand a bicycle upside down [resting on seat and handle-bars] then by dragging/whipping on the front tyre, set the wheel spinning. Pretty easy - takes little energy.

Now replace the wheel with a 6cm thick steel disc the same diameter as the bicycle wheel.
Even without actually attempting this I think you can see it will be much harder to set that steel disc spinning. Even more-so of there was a lot of friction in the axle of the steel disc.

That shows the effect there would be if the pulley had mass.

6. Jun 19, 2013

### andyrk

Thanks for this great explanation. But I still have a doubt pertaining, is the surface/grooves where the string is moving frctionless or not? Or is that just never mentioned and taken to be frictionless? About the massless and frictionless pulley case the tensions are same so for the pulley to rotate there needs to be a torque provided at the top most point, i.e where the strings pass over the pulley tangentially. But at that point the left and right tensions totally cancel of as a result so does the anticlockwise and clockwise torque. So as there is no torque being provided at the topmost point how will the pulley rotate? Or is the torque coming from somewhere else? And by tension is 0 I meant that the next force due to tension at the top most point is 0. Not the strings of course.

Last edited: Jun 19, 2013
7. Jun 19, 2013

### PeterO

If the pulley has no mass, you don't need any force to cause it to accelerate!!

Suppose the surface was frictionless. The force on the surface would then be 0 N

Let's apply Newton's second law: F = ma

Thus 0 = 0 x a so a = 0/0 which is undefined - but I reckon will probably match the acceleration of the string, so the string won't slip.

Now if there is friction, and a tendency for the string to move in one direction.

F = m x a

Friction = 0 x a so a = Friction / 0

Now since that acceleration is not infinity, Friction must be zero, as only the undefined 0/0 can possibly give the correct acceleration. So even when we assume there is friction, the maths says the value of the friction is zero!!

A "massless/frictionless pulley" is just another way of saying "ignore the effects of the pulley while doing this problem"

If you took real measurements with a real pulley, you may be able to work backwards from the measure acceleration and masses to find out the mass of the pulley and the friction involved.

8. Jun 19, 2013

### ehild

See attachment. The surface of the pulley is rough. The (blue) string moves from left to right, the right piece downward, the left piece upward. The tension of the piece wound over the pulley adds up to some normal force and causes friction. If the coefficient of friction is great enough, the friction is static, the surface of the pulley moves together with the string.

The torque comes from the forces acting at the points where the vertical parts of the string are tangent to the pulley. The torque = (T1-T2)R (R is the radius, and T1 is the tension in the right piece of string, T2 is the tension in the left one).

If the pulley is massless the torque has to be zero and the tensions are equal.

ehild

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9. Jun 19, 2013

### andyrk

So according to what you said, we cannot define acceleration of a massless object. But you said that it would probably match the acceleration of the strings. So you did DEFINE the acceleration of the pulley. Isn't it so?

So does that mean that a massless pulley doesn't rotate at all? In all the problems involving massless pulleys? So it would have no angular accelration and no tangential acceleration at any point on its boundary. Right?
So we assume massless pulleys to be not rotating and pulleys having mass to be rotating always? This is an assumption right? That's why we take different tensions in the string passing over a pulley having mass, so that it can rotate which we assume it does?

10. Jun 19, 2013

### andyrk

I guess I come to 2 conclusions then:
(1)A massless pulley is always assumed to be NOT rotating.
(2)A pulley having mass is always assumed to be rotating.
In all the problems that one faces upto highschool level for ease of calculations.

Last edited: Jun 19, 2013
11. Jun 19, 2013

### ehild

The massless pulley has zero moment of inertia. So it needs zero torque for angular acceleration. It will rotate.

ehild

12. Jun 19, 2013

### Tanya Sharma

Hi andyrk

The conclusions you have drawn are wrong .

13. Jun 19, 2013

### andyrk

But then what will it rotate because of? Ok, I agree according to you that it will rotate WHEN a string goes over the pulley having unequal weights attached across its ends. Torque is zero even in this case and the pulley is frictionless both on the axle and the surface of grooves. So it is similar to saying that the pulley is present just there without any string passing over it. As in a totally isolated pulley. Even here no torque is being given to the pulley, but still it will rotate, as it doesn't have any moment of inertia. It will start rotating just by itself right?

14. Jun 19, 2013

### ehild

You have to consider that zero mass or zero moment of inertia as a limit. The pulley has mass, but very small compared to the masses hanging on the ropes. It needs some non-zero torque to start rotating. That torque comes from unequal tensions. And the edge of the pulley moves with the same acceleration and speed as the string, so the angular acceleration is β=a/R
If you draw the FBD-s and write up the equations ma=ƩF for the hanging bodies and Iβ=Ʃτ for the rotation of the pulley, you get

m1a=T1-m1g,
m2a=m2g-T2,
R(T2-T1)=Iβ=Ia/R--> (I/R2)a=T2-T1

(m1+m2+I/R2)a=m2g-m1g.
The whole system moves with acceleration a

a=(m2-m1)g/(m1+m2+0.5I/R2).

If the pulley is a solid disk, I=0.5mR2, and

a=(m2-m1)g/(m1+m2+0.5m).

If m<<m1+m2 a≈(m2-m1)g/(m1+m2)

Substituting back into the two first equations, you get that T1=T2=2g(m1m2)/(m1+m2).

ehild