- #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!