Mass hanging on a steel wire

[giodude]

Homework Statement

(a) An object of mass 0.5 kg is hung from the end of a steel wire of length 2 m and of diameter 0.5 mm. (Young's modulus = $2 \times 10^{11} \frac{N}{m^{2}}$). What is the extension of the wire?

(b) The object is lifted through a distance $h$ (thus allowing the wire to become slack) and is then dropped so that the wire receives a sudden jerk. The ultimate strength of steel is $1.1 \times 10^{9} \frac{N}{m^{2}}$. What is the largest possible value of $h$ if the wire is not to break?

Relevant Equations

$A = \frac{\pi \times d^{2}}{4}$
$k = \frac{AY}{l_{0}}$
$F = mg = kx$
$stress = \frac{\Delta P}{A}$
$strain = \frac{\Delta l}{l_{0}}$
$Y = \frac{stress}{strain}$

Hi, I solved part (a) and will provide my solution below. However, I've been working on part (b) for quite a bit and reviewed the provided, relevant text a few times now but haven't been able to find what I'm missing:

Solution (a):
Using $A = \frac{\pi \times d^{2}}{4}$, $k = \frac{AY}{l_{0}}$, and $F = mg = kx$ we have:

$x = \frac{m \times g \times l_{0} \times 4}{\pi \times d^{2} \times Y} = 0.25 mm$

Solution (b):
We can use the cross sectional area and the given ultimate strength of steel to solve for the maximum allowed force:
$1.1 \times 10^{9} = \frac{F}{A}$
$F = 216 N$

I'm also aware that the potential energy at the peak just before the rod is dropped is $PE = mgh$ and that this potential energy is converted entirely to kinetic energy during the sudden jerk such that:

$PE = mgh = F \times \Delta l = KE$

However, I get stuck dealing with the $\Delta l$ term which I'm using to denote the wire extension that occurs during the sudden jerk.

Any suggestions on where I'm going wrong would be much appreciated!

 

[haruspex]

What type of motion will occur when the wire becomes taut?

 

[giodude]

Simple harmonic motion?

 

[haruspex]

Simple harmonic motion?

Right, so you cannot use $F\times\Delta l$ since F is not constant.
Instead of trying to find the velocity when it becomes taut, think about the conservation of energy from start to finish. What is the change in GPE?

 

[giodude]

The change in GPE will be mgh from start to finish. Which means KE will have changed from 0 to mgh at the moment of maximum tension.

 

[haruspex]

The change in GPE will be mgh from start to finish. Which means KE will have changed from 0 to mgh at the moment of maximum tension.

Where do you think the maximum tension occurs in the motion?

 

[giodude]

It'd occur just before the sudden jerk? So, at PE = 0 and KE = mgh?

 

[giodude]

I think I'm starting to get somewhere. We know, due to the starting state of lifting and dropping that the initial speed, $v_{0} = 0$. Additionally, we know that $KE = mgh$ at the point of peak tension. Therefore,

$\Delta KE = \frac{1}{2} \times m \times v_{f}^{2} - \frac{1}{2} \times m \times v_{0}^{2} = m \times g \times h$

Plugging in $v_{0} = 0$, we have:
$\frac{1}{2} \times m \times v_{f}^{2} = m \times g \times h$
$v_{f} = \sqrt{2 \times g \times h}$

 

[giodude]

We also know that in SHM $x = A cos(\omega t + \alpha)$, where $\omega^{2} = \frac{k}{m}$. So,

$v = \frac{dx}{dt} = - \omega A sin(\omega t + \alpha)$
$a = \frac{d^{2}x}{dt^{2}} = - \omega^{2} A cos(\omega t + \alpha)$

KE is maximum where $a = 0$ at $cos(\omega t + \alpha) = 0$ and $sin(\omega t + \alpha) = 1$. Therefore, $\sqrt{2gh} = \omega A$

 

[haruspex]

We also know that in SHM $x = A cos(\omega t + \alpha)$, where $\omega^{2} = \frac{k}{m}$. So,

$v = \frac{dx}{dt} = - \omega A sin(\omega t + \alpha)$
$a = \frac{d^{2}x}{dt^{2}} = - \omega^{2} A cos(\omega t + \alpha)$

KE is maximum where $a = 0$ at $cos(\omega t + \alpha) = 0$ and $sin(\omega t + \alpha) = 1$. Therefore, $\sqrt{2gh} = \omega A$

But you don't want max velocity. What motion variable is maximised when the tension is maximum?

 

[giodude]

Oh, I see. If $T = kx - mg = ma$ is at its maximum then acceleration needs to be maximum. As a result, we want $cos(\omega t + \alpha) = 1$ and $sin(\omega t + \alpha) = 0$.

This makes sense because this sets $x = Acos(\omega t + \alpha)$ to its max $x = A$.

 

[haruspex]

Oh, I see. If $T = kx - mg = ma$ is at its maximum then acceleration needs to be maximum. As a result, we want $cos(\omega t + \alpha) = 1$ and $sin(\omega t + \alpha) = 0$.

This makes sense because this sets $x = Acos(\omega t + \alpha)$ to its max $x = A$.

Right.
You still have one small inexactitude, but it's probably insignificant. The loss in GPE will be a little more than mgh.

 

[giodude]

Whats the inexactitude?

Also how does this reconcile with $v_{f} = \sqrt{2gh}$ if $v = 0$ at time $t$?

 

[giodude]

We also know $T = 216 N$ from the original post, given the ultimate strength of the wire. So,

$T = kx - mg$
$216 N = kx - (0.5 kg)(9.8 \frac{m}{s^{2}})$
$kx = 220.9 N$
$x = \frac{220.9 N}{k} = 0.0113 m$

Dimensionally this resolves properly. However, the solution in the textbook is $h = 0.23m$. Hmmm.

 

[giodude]

Oh, think I figured it out. This might be where the inexactitude you mentioned is coming in:

The $x$ in the previous message is the amplitude. Now we can plug it into $mgh = \frac{1}{2}kA^{2}$ which we derive from the conservation of energy equation:
$h = \frac{kA^{2}}{2mg} = 0.2536m$

Hopefully this is correct! Either way, have learned a bunch here. Thank you for your help!

 

[haruspex]

$T = kx - mg$

The total extension is x+(mg)/k.
The small inexactitude is that the lost GPE is mg(h+x), but x is tiny compared to h.