Solving Stress in a Ring of Radius 1/10 m, Mass 1 kg/m

  • Thread starter Thread starter altamashghazi
  • Start date Start date
  • Tags Tags
    Stress String
Click For Summary
SUMMARY

The discussion focuses on calculating the stress in a thin wire ring of radius 0.1 m and mass per unit length of 1 kg/m, rotating with an angular velocity of 2 rad/s. The relevant equations include stress (s = f/a) and centripetal force (f = mv²/r). The tension force acting on a small mass element dm is derived, leading to the equation 2Tdθ = (dm * r * ω²). By substituting dm in terms of the total mass M and simplifying, the tension force can be determined, which is essential for calculating the stress in the ring.

PREREQUISITES
  • Understanding of basic mechanics, specifically centripetal force.
  • Familiarity with stress and tension concepts in materials.
  • Knowledge of angular motion and its equations.
  • Ability to manipulate calculus concepts, particularly in relation to small angles.
NEXT STEPS
  • Study the derivation of centripetal force in rotating systems.
  • Learn about stress-strain relationships in materials science.
  • Explore the implications of uniform density in circular motion problems.
  • Investigate the effects of varying angular velocities on tension in rotating rings.
USEFUL FOR

Students in physics or engineering disciplines, particularly those studying mechanics and material science, as well as educators looking for practical examples of stress calculations in rotating systems.

altamashghazi
Messages
42
Reaction score
0

Homework Statement



a thin wire of crosssection area 10^-2 m^2 is used to make a ring of radius 1/10 m. it is given ω=2 rad/s on a smooth table about its centre. find stress in ring?( mass/length of wire is 1 kg/m.)

Homework Equations


s=f/a
f=mv^2/r


The Attempt at a Solution


my only difficulty is the area on which i should take the force to be acting.
 
Physics news on Phys.org
On a small mass dm at an angle d(theta) tension force acts which is tangential to circle.

The component of this force towards the centre is 2Tsin(d(θ)). For small angles sin(dθ) is equal to dθ so centripetal force is 2Tdθ.

What should this centripetal force be equal to in terms of mass dm radius and ω. (Hint : you have written an equation in velocity.substitute velocity with rω.[Why?])

2Tdθ=(dm*r*ω^2)

Now all that is left is to calculate value of small mass dm in terms of mass M of ring and dθ (assuming uniform density) What will the equation be?

If you can write it you can cancel dθ on both sides and obtain value of tension force.

Using stress equation you wrote i.e Force/Area its acting upon

The force is tension.
What will be the area.?

What is the final answer
 

Attachments

  • tension.jpg
    tension.jpg
    5.6 KB · Views: 555

Similar threads

  • · Replies 21 ·
Replies
21
Views
2K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
Replies
2
Views
2K
Replies
3
Views
2K
  • · Replies 12 ·
Replies
12
Views
5K
Replies
14
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K