Angular Motion and Young's Modulus

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Homework Help Overview

The discussion revolves around a problem involving angular motion and Young's modulus. A mass is attached to an aluminum wire and is whirled in a vertical circle, with the goal of calculating the elongation of the wire at both the lowest and highest points of the path.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the forces acting on the wire and how to apply Young's modulus to find elongation. There are attempts to clarify the correct approach to calculating the sum of forces at different points in the motion.

Discussion Status

Some participants have provided guidance on the correct interpretation of forces and the application of equations. There is acknowledgment of miscalculations and the importance of careful reading of problem statements.

Contextual Notes

Participants note potential issues with unit conversions and the radius used in calculations, indicating that these factors may have contributed to discrepancies in results.

fajoler
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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/\DeltaL)

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:

\DeltaL = (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

\DeltaL = (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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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/\DeltaL)

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:

\DeltaL = (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

\DeltaL = (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!
 
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!
 
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!
 

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