David Morin classical mechanics Problem 2.6: Disk held up by a massless string

In summary: So the normal force is the same at any point on the arc (assuming the string is tensioned equally in all directions).The tension in the string exerts a pressure on the cylinder. Over a small (effectively almost straight) section of arc the force is the pressure times the length. So the normal force is the same at any point on the arc (assuming the string is tensioned equally in all directions).
  • #1
John100861
3
0
Homework Statement
(a) A disk of mass M and radius R is held up by a massless string, as
shown in Fig. 2.12. The surface of the disk is frictionless. What is
the tension in the string? What is the normal force per unit length
that the string applies to the disk?
Relevant Equations
ΣF=0
Screenshot 2020-07-05 at 4.58.03 PM.png

The first part is easy, we have 2T= Mg
T= 0.5 Mg
Now for the second part where I'm having trouble understanding Morin's solution:
I take the normal force on a small circle arc to be N, we know that the y component of the normal force must be balance with Mg for the whole disk, therefore
Ny = Nsin(θ)
dNy= Ncos(θ)dθ
Ncos(θ)dθ= Mg
And this is where I have trouble, I end up with Mg= 0 when plugging in the limits [0, π]
Morin's solution suggests that the normal force in the arc should be written as Ndθ but I don't understand why. Please point out what's wrong with my approach, and help me understand the solution, thanks.
.
 
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  • #2
You cannot define N in that way since it depends on the length of arc. Better to consider N as a force per unit length.
Pinpointing your error is not possible until you say how you are defining theta.
If it is the angle between the dashed line and the radius to an element ##rd\theta## of arc, the normal force is ##Nrd\theta##, and its Y component is ##Nr\cos(\theta)d\theta##.
 
  • #3
haruspex said:
You cannot define N in that way since it depends on the length of arc.
Hi thanks for your reply! Yes I'm defining that to be theta. Could you please elaborate on this? It's still not clear to me why I can't define N that way.
 
  • #4
John100861 said:
Hi thanks for your reply! Yes I'm defining that to be theta. Could you please elaborate on this? It's still not clear to me why I can't define N that way.
The normal force on a small circle of arc will depend on the length of the arc. So you need it as (some variable) x length of arc, i.e. as a force per unit length.
 
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  • #5
haruspex said:
The normal force on a small circle of arc will depend on the length of the arc. So you need it as (some variable) x length of arc, i.e. as a force per unit length.
Why does it depend on the length of the arc?
 
  • #6
John100861 said:
Why does it depend on the length of the arc?
The tension in the string exerts a pressure on the cylinder. Over a small (effectively almost straight) section of arc the force is the pressure times the length.
 

Related to David Morin classical mechanics Problem 2.6: Disk held up by a massless string

1. What is the setup of David Morin classical mechanics Problem 2.6?

The problem involves a disk held up by a massless string, with the string attached to the center of the disk and hanging vertically. The disk has a mass of 2kg and a radius of 0.5m.

2. What is the mass of the disk in this problem?

The disk in this problem has a mass of 2kg.

3. What is the radius of the disk in this problem?

The disk in this problem has a radius of 0.5m.

4. How is the string attached to the disk in this problem?

The string is attached to the center of the disk in this problem.

5. What is the role of the massless string in this problem?

The massless string is used to hold up the disk and maintain its vertical position. It also allows for the calculation of the tension force acting on the string and the disk.

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