Find the Smallest Possible Tension in a Massless String Supporting a Disk

[Jzhang27143]

A disk of mass M and radius R is held up by a massless string. (The two ends of the string are connected to a ceiling and the disk rests on the bottom of the string.) The coefficient of friction between the disk is μ. What is the smallest possible tension in the string at its lowest point?

This is from "Introduction to Classical Mechanics" by David Morin. I am confused as to how T(∏/2) = Mg/2. T(∏/2) refers to the tension in the rightmost point of the disk where the string does not touch the disk anymore.)

 

[Simon Bridge]

What does the $\pi/2$ refer to?

I take it, by "in the bottom" you mean the disk rests in the loop of the string?

If the disk were just held up by two vertical lengths of string - what would the tension in each string be?

 

[Jzhang27143]

They would be mg/2. In the problem, why do the tensions in the vertical lengths have to be equal? The tension in the string increases opposite the direction of friction so from this argument, I see that the vertical lengths have different tensions. What am I missing?

 

[Simon Bridge]

How do you see that the vertical lengths have different tensions?
Are they hanging at different angles?
Is the disk spinning?

Remember - I cannot see any diagram you may be looking at.

 

[PJC01]

I would approach this problem by first considering the forces acting on the disk. The weight of the disk, given by the product of its mass (M) and the acceleration due to gravity (g), exerts a downward force on the disk. This force must be balanced by the upward force provided by the tension in the string.

To find the smallest possible tension in the string, we need to consider the point at which the tension is at its lowest. This occurs at the point where the string is tangent to the disk, also known as the contact point. At this point, the tension in the string is equal to the weight of the disk, as there is no friction acting on the disk yet.

However, as the string rotates around the disk, friction between the disk and the string will come into play. The coefficient of friction (μ) determines the strength of this frictional force. As the string continues to rotate, the frictional force will increase and the tension in the string will decrease until it reaches a minimum at the point where the string is tangent to the disk again.

Using the equation for the minimum tension at the contact point (T(∏/2) = Mg/2), we can see that the tension is directly proportional to the mass of the disk and the acceleration due to gravity, and inversely proportional to the radius of the disk. This means that as the mass or the radius of the disk increases, the minimum tension in the string will also increase.

In conclusion, the smallest possible tension in the string supporting the disk will occur at the point where the string is tangent to the disk, and it is equal to half the weight of the disk. This minimum tension can be affected by the mass and radius of the disk, as well as the coefficient of friction between the disk and the string.