Ideal Vs Real Mass Pulley system

  • #1
vcsharp2003
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Homework Statement:
How does a real mass pulley system differ from an ideal mass pulley system when applying Newton's Second of Law of motion?
Relevant Equations:
F = ma and Torque = r X F
Below is an ideal mass pulley system that we encounter in many problems under Newton's Second Law of Motion questions.
Its supposed to be massless and frictionless i.e. string slips over the pulley and pulley does not rotate.

  1. In a real system, the pulley is assumed to be massless, whereas in reality it does have some mass and therefore some moment of inertia.
  2. In a real system, the pulley string contact surface would not be frictionless, so the string will not slip over the pulley causing the pulley to rotate. Perhaps, the string could also slip in some cases where coefficient of static friction is very low between string and pulley?
  3. In a real system, the friction force on the pulley ( not on the string) would be acting downwards on left side of the pulley and upwards on the right side of the pulley. On the top part of the pulley this friction force would act in leftwards direction. It is this friction force that would cause the pulley to rotate counter clockwise.
  4. In a real system, the reason why pulley would rotate is because the tensions are unequal on left side and right sides of the string over the pulley. The left side tension would be greater than the right side tensions resulting in a net counter clockwise torque that would cause the pulley to rotate counter clockwise.
Is bullet#3 the correct explanation for why the pulley rotates or bullet#4 is the correct explanation?

I think bullet#3 is the correct explanation because tension does not act on the pulley but on the string. We are looking at the pulley only as the system and therefore, we are interested in forces acting on the pulley and not on the string.
IdealPulleySystem.png


IMG_20210418_132319__01.jpg
 
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Answers and Replies

  • #2
haruspex
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Homework Statement:: How does a real mass pulley system differ from an ideal mass pulley system when applying Newton's Second of Law of motion?
Relevant Equations:: F = ma and Torque = r X F

Its supposed to be massless and frictionless i.e. string slips over the pulley and pulley does not rotate.
Nonsense.
When a question states that the pulley is frictionless it means there is no friction at the axle. Often a pulley is given as frictionless but massive, and you have to assume no slip between pulley and string and take the inertia of the pulley into account.
If there were no friction between pulley and string there would be no point in making it a pulley; it could just as well be a smooth fixed cylinder.

I am not sure what bullet 3 means by "the friction force on the pulley ( not on the string)". Perhaps that is referring to axle friction. If it means friction between pulley and string then I don't see how it differs from bullet 2.

Why are you ruling out bullet 2?
 
  • #3
vcsharp2003
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If there were no friction between pulley and string there would be no point in making it a pulley; it could just as well be a smooth fixed cylinder.
Would the tension be the same on both sides when a smooth fixed cylinder is used?
I am not sure what bullet 3 means by "the friction force on the pulley ( not on the string)".
Friction acts between two surfaces and also on both surfaces of contact. For example, friction from a table top acts on a sliding mass placed on it, and at the same time due to Newton's 3rd Law of Motion an equal and opposite friction force will act on the table. In the same way, I have described the frictional force between string's and pulley's contact surface, but only looking at the friction force on the pulley's surface.
Why are you ruling out bullet 2?
I am discounting bullet#4 ( not 2 as you mentioned) as an explanation for pulley's rotation. I feel bullet#3 is the explanation for rotation of pulley. In bullet#3 I am mentioning the friction direction which is not mentioned in bullet#2.
 
  • #4
haruspex
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Would the tension be the same on both sides when a smooth fixed cylinder is used?
Yes.
I am discounting bullet#4 ( not 2 as you mentioned)
Of the four, you appeared to have narrowed it down to just 3 or 4, so implicitly you had discounted 2.
So 2 and 3 are the same explanation, but in 3 you go into more detail.
I would consider both correct.
 
  • #5
vcsharp2003
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I would consider both correct.
I think bullet#4 is correct as it looks at pulley plus the section of string over pulley as the system, whereas in bullet#3 only the pulley is the system. In bullet#4, the friction between string and pulley would be an internal force and therefore ignored.

Another difference in real mass pulley system that I forgot is that the string will have some mass per unit length, which will cause the tension in the string to be different at different points even on the same side of the pulley.
 
  • #6
Lnewqban
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When you use a system of pulleys to lift a heavy load in real life, you quickly understand how important the following things are:

1) A rope that stretches little as you pull.

2) A diameter of rope that allows a comfortable and strong hand grip.

3) A high coefficient of static friction of rope that allows a comfortable and strong hand grip, as well as zero slip around the pulleys.

4) Pulleys that add little weight and inertia to the natural weight and inertia of the load to be lifted (hence, the mass-less pulleys of theoretical problems).

5) Pulleys of big diameters and with properly lubricated axles (hence, the friction-less pulleys of theoretical problems).

As things deviate from the above, your input work increases respect to the useful work of lifting the load certain height.
 
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  • #7
haruspex
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I think bullet#4 is correct
I tejected #4 because it can be that the tensions are unequal yet it does not rotate.
 
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  • #8
vcsharp2003
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I tejected #4 because it can be that the tensions are unequal yet it does not rotate.

How can that be? Is it the torque due to friction at the pulley's axle that balances the torque due to tensions?

If tensions are unequal then there is going to be a net torque and therefore, rotation of the pulley. Unless, some other forces are involved besides tensions on both sides.
 
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  • #9
vcsharp2003
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Pulleys that add little weight and inertia to the natural weight and inertia of the load to be lifted

How does this make it better when lifting a load?

I am thinking that rotational inertia means a greater lifting force has to be applied on one end of the pulley.
 
  • #10
haruspex
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How can that be? Is it the torque due to friction at the pulley's axle that balances the torque due to tensions?
Yes, static friction on the axle.
 
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  • #11
Lnewqban
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How does this make it better when lifting a load?

I am thinking that rotational inertia means a greater lifting force has to be applied on one end of the pulley.
You are correct.
Big diameter pulleys are preferred because they have better grip of the rope and rotate slower, generating less friction in the axle.
Also, bending the ropes consume energy, more for small diameters.

At the same time, they should be as light as possible; otherwise, you will have excessive rotational and linear inertia.
The small weight becomes important also when you need several pulleys for more mechanical advantage, as you will be lifting the weight of the pulleys closer to the load.

6-detail-of-ropes-and-pulleys-of-a-wooden-sailboat.jpg
 
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