Angular Motion and Young's Modulus

In summary: Glad I could help. Just remember to always double check your calculations and make sure you're using the correct values and equations.
  • #1
fajoler
11
1

Homework Statement



A mass of 13.0 , fastened to the end of an aluminum wire with an unstretched length of 0.50 is whirled in a vertical circle with a constant angular speed of 130 . The cross-sectional area of the wire is 1.3×10−2 . The Young's modulus for aluminum is Pa.Calculate the elongation of the wire when the mass is at the lowest point of the path.

Calculate the elongation of the wire when the mass is at the highest point of its path.

Homework Equations



Y=(F/A)*(L/[tex]\Delta[/tex]L)

Ac=rw^2

The Attempt at a Solution

Alright so first thing first what I did was I converted all my units so I could work them into the equations. I have:

Y = 7 * 10^10 Pa
w = 13.6 rad/sec
A = 1.3*10^-6 m squared
radius = 6.4*10^-4 m
mass = 13.0 kg
L = 0.5mNow I need to find the forces acting on the wire, right? So I can plug that into the Young Modulus equation.At the lowest point of the path, I found that the sum of the forces = mg - mrw^2
If you make it so downwards direction is positive.

I calculated that the sum of the forces = (13 kg)*(9.8m/s) - (13 kg)*(6.4*10^-4m)*((13.6 rad/sec)^2) = 125.85 NI plug the sum of the forces into the Young's modulus equation so that:

[tex]\Delta[/tex]L = (F*L)/(A*Y) = [(125.85 N)*(0.5m)]/[(1.3*10^-6 m squared)*(7.0 * 10^10 Pa)]This comes out to 6.9*10^-4 m.

Is this answer correct? Did I do everything I was supposed to do?I used the same procedure for when the mass is at it's highest point, except the sum of the forces should now equal mg + mrw^2right?

So if I do that, the sum of the forces = (13 kg)*(9.8 m/s^2) + (13 kg)*(6.4*10^-4 m)*((13.6 rad/sec)^2)This comes out to 128.95 N

If I plug this into Young's Modulus equation, I get

[tex]\Delta[/tex]L = (F*L)/(A*Y) = (128.95 N)*(0.5m) / (1.3*10^-6 m^2)*(7.0 * 10^10 Pa)this comes out to 7.1*10^-4 m.Yet Mastering Physics continues to tell me it's wrong, so what am I doing wrong? I hope you guys can help. Thanks!
 
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  • #2
Hi fajoler,

fajoler said:

Homework Statement



A mass of 13.0 , fastened to the end of an aluminum wire with an unstretched length of 0.50 is whirled in a vertical circle with a constant angular speed of 130 . The cross-sectional area of the wire is 1.3×10−2 . The Young's modulus for aluminum is Pa.


Calculate the elongation of the wire when the mass is at the lowest point of the path.

Calculate the elongation of the wire when the mass is at the highest point of its path.



Homework Equations



Y=(F/A)*(L/[tex]\Delta[/tex]L)

Ac=rw^2


The Attempt at a Solution




Alright so first thing first what I did was I converted all my units so I could work them into the equations. I have:

Y = 7 * 10^10 Pa
w = 13.6 rad/sec
A = 1.3*10^-6 m squared
radius = 6.4*10^-4 m

You found this using A = pi r2, right? So this is the radius of the wire itself.

mass = 13.0 kg
L = 0.5m


Now I need to find the forces acting on the wire, right? So I can plug that into the Young Modulus equation.


At the lowest point of the path, I found that the sum of the forces = mg - mrw^2

No, this is not correct. The idea behind Newton's law here is:

(sum of the physical forces) = (mass) (acceleration)

Since the acceleration is centripetal in this case, we know that a=r w2, so:

(sum of physical forces) = (mass) (r w2)

But on the left hand side, you need the two forces that are actually pushing or pulling on the object. What two forces are that? And do you end up adding or subtracting them, based on the direction?

If you make it so downwards direction is positive.

I calculated that the sum of the forces = (13 kg)*(9.8m/s) - (13 kg)*(6.4*10^-4m)*((13.6 rad/sec)^2) = 125.85 N

In terms of the numerical values, here you are using r=radius of the wire. But what radius is used in the centripetal acceleration formula

a = r w2?

I plug the sum of the forces into the Young's modulus equation so that:

[tex]\Delta[/tex]L = (F*L)/(A*Y) = [(125.85 N)*(0.5m)]/[(1.3*10^-6 m squared)*(7.0 * 10^10 Pa)]


This comes out to 6.9*10^-4 m.

Is this answer correct? Did I do everything I was supposed to do?


I used the same procedure for when the mass is at it's highest point, except the sum of the forces should now equal mg + mrw^2


right?

So if I do that, the sum of the forces = (13 kg)*(9.8 m/s^2) + (13 kg)*(6.4*10^-4 m)*((13.6 rad/sec)^2)


This comes out to 128.95 N

If I plug this into Young's Modulus equation, I get

[tex]\Delta[/tex]L = (F*L)/(A*Y) = (128.95 N)*(0.5m) / (1.3*10^-6 m^2)*(7.0 * 10^10 Pa)


this comes out to 7.1*10^-4 m.


Yet Mastering Physics continues to tell me it's wrong, so what am I doing wrong? I hope you guys can help. Thanks!
 
  • #3
thanks so much. it turns out the problem i was having was the miscalculation of the radius. stupid mistakes like misreading directions can lead to a lot of trouble...

thanks again!
 
  • #4
fajoler said:
thanks so much. it turns out the problem i was having was the miscalculation of the radius. stupid mistakes like misreading directions can lead to a lot of trouble...

thanks again!

You're welcome!
 
  • #5


Your calculations seem to be correct. However, it's possible that there is a rounding error or a small mistake in your units. I would recommend double checking your calculations and making sure all units are consistent throughout. It's also possible that the online system is not accepting your answer due to a specific format it requires. I would suggest trying different formats for your answer (e.g. scientific notation, rounding to a certain number of significant figures, etc.) to see if that makes a difference. If you are still having trouble, I would recommend reaching out to your instructor or a classmate for further assistance.
 

Related to Angular Motion and Young's Modulus

1. What is angular motion and how is it different from linear motion?

Angular motion is the movement of an object around a fixed point or axis. It is different from linear motion because it involves a circular or rotational movement rather than a straight-line movement.

2. How is Young's Modulus related to angular motion?

Young's Modulus, also known as the elastic modulus, is a measure of the stiffness or rigidity of a material. It is related to angular motion because it determines the amount of force needed to produce a certain amount of deformation in a material.

3. What is the formula for calculating angular velocity?

The formula for calculating angular velocity is ω = Δθ/Δt, where ω is angular velocity, Δθ is the change in angle, and Δt is the change in time.

4. How does angular acceleration affect an object's motion?

Angular acceleration is the rate of change of angular velocity. It affects an object's motion by causing it to either speed up or slow down its rotation around a fixed point or axis.

5. What factors affect Young's Modulus?

The factors that affect Young's Modulus include the type of material, its composition, temperature, and the presence of any external forces or stresses.

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