Elasticity of a Wire spinning in a circle

In summary, at the lowest point of the path, the elongation is 0.014 cm2 and at the highest point of its path, the elongation is 1380,000,000*Δl.
  • #1
HoodedFreak
30
0

Homework Statement


A 12.0-kg mass, fastened to the end of an aluminum wire with an unstretched length of 0.70 m, is whirled in a vertical circle with a constant angular speed of 120 rev>min. The cross-sectional area of the wire is 0.014 cm2. Calculate the elongation of the wire when the mass is (a) at the lowest point of the path and (b) at the highest point of its path.

Homework Equations


F⊥ = mv^2/R
F = m*a
Y = (F⊥/A)/(Δl/lo)
v = wR

The Attempt at a Solution


[/B]
a) So if we consider the mass at the end of the string. We get that T - mg = m*v^2/R
T - 12g = 12*(wR)^2/R
T - 12g = 12*w^2/R

R = Δl + lo = 0.7 + Δl

w = 120 * 2π/ 60 = 4π

T - 12g = 12*16π / (0.7 + Δl)

I'm not sure where to go from here
 
Physics news on Phys.org
  • #2
What are the material properties of Aluminium ?
 
  • #3
Nidum said:
What are the material properties of Aluminium ?

Young's modulus for aluminum, according to google is 69 N/m^2
 
  • #4
HoodedFreak said:
Young's modulus for aluminum, according to google is 69 N/m^2
Look more closely at the web page. I think you'll find you've omitted quite a large power of 10.
Indeed, I would be surprised if the elongation in this case made enough difference to the length that you need to take it into account in the expression for centripetal force.
 
  • #5
Ok Young's Modulus is the property you need to help solve this problem . You got the magnitude wrong though - should be 69 GN/m^2 or 69 GPa .
 
  • #6
haruspex said:
Look more closely at the web page. I think you'll find you've omitted quite a large power of 10.
Indeed, I would be surprised if the elongation in this case made enough difference to the length that you need to take it into account in the expression for centripetal force.

Nidum said:
Ok Young's Modulus is the property you need to help solve this problem . You got the magnitude wrong though - should be 69 GN/m^2 . Often written Gpa .

Right, yeah, so that would be 69*10^9 then.

Okay, so if I plug in 69*10^9 * Δl/lo = T/0.014,

then T = 69*10^9*0.014 * Δl / 0.7 = 1380,000,000*Δl

SO

1,380,000,000*Δl - 12g = 12*16π / (0.7 + Δl)

(1,380,000,000*Δl - 12g) * (0.7 + Δl) = 603

And I solve this quadratic?

It seems like I'm doing something wrong
 
  • #7
Also, to harupex's point, should I ignore the elongation and just treat the radius of the circle as R = lo = 0.7m since it should be small in comparison to the length.
 
  • #8
HoodedFreak said:
Also, to harupex's point, should I ignore the elongation and just treat the radius of the circle as R = lo = 0.7m since it should be small in comparison to the length.

Yes .
 
  • #9
It is a bit difficult to follow your calculations as written .

For the wire under tension T what is :

The stress ?
The strain ?
The change in length ?
 
  • #10
Nidum said:
It is a bit difficult to follow your calculations as written .

For the wire under tension T what is :

The stress ?
The stress should be T/A = T/0.014

The strain ?
The strain should be Δl/lo = Δl/0.7

The change in length ?

The change in length should be Δl

Using the two equations I had:

T - 12g = 12*16π / (0.7 + Δl)

and

Y * Strain = Stress ⇒ 6.9*10^9 * Δl/0.7 = T/0.014

Solving for T from the first equation to get T = 12*16π / (0.7 + Δl) + 12g

Plugging it into the second, to get

12*16π / (0.7 + Δl) + 12g = 6.9*10^9 * Δl/0.7 * 0.014

Now in order to solve this equation I would have to multiply out the (0.7 + Δl) and get a quadratic, which seems unnecessarily complicated, assuming I've done everything right to this point. So if I take your advice and ignore the Δl, I get:

12*16π / 0.7 + 12g = 6.9*10^9 * Δl/0.7 * 0.014

solving this huge mess gives me: 7.094 * 10^-6.

Which doesn't seem right to me
 
  • #11
HoodedFreak said:
T - 12g = 12*(wR)^2/R
T - 12g = 12*w^2/R
Check that step.
HoodedFreak said:
= 12*16π / (0.7 + Δl)
What happened to the power of 2 on w?
 
  • #12
haruspex said:
Check that step.

What happened to the power of 2 on w?

I was rushing everything and made a lot of mistakes. I corrected those mistakes since then and got the correct answer. Thanks for the help!
 

1. What is the concept of the "Elasticity of a Wire spinning in a circle"?

The elasticity of a wire spinning in a circle refers to the ability of a wire to stretch and deform when subjected to forces while rotating in a circular motion. This phenomenon is also known as centripetal force, which is responsible for keeping the wire in its circular path.

2. How does the elasticity of a wire affect its spinning in a circle?

The elasticity of a wire plays a crucial role in its spinning in a circle. The wire needs to have a certain level of elasticity to withstand the centrifugal force generated by its spinning motion. If the wire is too stiff, it may break under the force, and if it is too elastic, it may lose its circular path and become unstable.

3. What factors affect the elasticity of a wire spinning in a circle?

The elasticity of a wire spinning in a circle can be affected by various factors such as the material of the wire, its thickness, and the speed of its rotation. Thicker wires tend to be less elastic than thinner ones, and materials with higher elasticity, such as rubber, will be more suitable for spinning in a circle.

4. How can the elasticity of a wire be measured?

The elasticity of a wire can be measured by calculating its Young's modulus, which is a measure of its stiffness. This can be done by applying a known force to the wire and measuring the amount of deformation it undergoes. The higher the Young's modulus, the stiffer the wire will be.

5. What are some practical applications of understanding the elasticity of a wire spinning in a circle?

Understanding the elasticity of a wire spinning in a circle is crucial in many practical applications such as roller coasters, bungee jumping, and amusement park rides. It is also essential in the design and construction of suspension bridges, where wires are used to support the weight of the bridge and withstand the forces of wind and traffic.

Similar threads

  • Introductory Physics Homework Help
Replies
6
Views
238
  • Introductory Physics Homework Help
Replies
16
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
7K
  • Introductory Physics Homework Help
Replies
5
Views
2K
Replies
2
Views
3K
  • Introductory Physics Homework Help
Replies
5
Views
10K
  • Introductory Physics Homework Help
Replies
2
Views
681
  • Introductory Physics Homework Help
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
12
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
782
Back
Top