Falling mass, additionally accelerated by a rope

[greypilgrim]

Hi.

The following situation:

The pulleys are fixed to the floor/ceiling and massless, as are the ropes, and there is no friction. At $t=0$, the masses are released from rest.

For the moment, I'll assume $m_2\gg m_1$. So $m_2$ will accelerate at $g$ and the red rope at $\frac{d_1}{d_2}g$. How do I now find the accelerations of motion for $m_1$?

At the very beginning, as far as I can see $m_1$ experiences a downward force $F_1=m_1 g+\frac{d_2}{d_1}m_2 g$. But I'm quite sure the tension in the red rope decreases, but how exactly?

 

[BvU]

Hi,

you want to write down a few equations. choose/define relevant variables and work towards $N$ equations with $N$ unknowns.
What about the tensions in the red rope at the (massless!) lower pulley ?
And tensions in the red and blue ropes at the upper pulley ?
And the in the blue rope is related to the acceleration of $m_2$

 

[jbriggs444]

For the moment, I'll assume $m_2\gg m_1$. So $m_2$ will accelerate at $g$ and the red rope at $\frac{d_1}{d_2}g$. How do I now find the accelerations of motion for $m_1$?

At the very beginning, as far as I can see $m_1$ experiences a downward force $F_1=m_1 g+\frac{d_2}{d_1}m_2 g$. But I'm quite sure the tension in the red rope decreases, but how exactly?

If $m_2 \gg m_1$ then we can treat the downward acceleration of $m_2$ as being fixed. You already realize this.

This means that the downward acceleration of $m_1$ is also fixed. (Assuming $d_2 > d_1$ so that the rope does not go slack).

What is the resulting acceleration of $m_1$?
What forces act on $m_1$?
Which of those forces do you already know?
Does this allow you to calculate the remaining force?

If $m_2$ is not vastly greater than $m_1$ then you will have some simultaneous equations to solve as @BvU suggests.

 

[greypilgrim]

This means that the downward acceleration of is also fixed.

By "fixed", do you mean constant?

 

[jbriggs444]

By "fixed", do you mean constant?

Constant yes. But I mean that it is "determined". Given what you already know, there is only one value it could possibly be.

 

[greypilgrim]

But if the blue rope accelerates at $g$, then the red one does so at $\frac{d_1}{d_2}g$. And if my $F_1$ from #1 is correct and constant, then $m_1$ accelerates at $g+\frac{d_2}{d_1}\frac{m_2}{m_1}g$ which is bigger than $\frac{d_1}{d_2}g$ given that $\frac{m_2}{m_1}\gg 1$.

This would mean that the rope goes slack immediately. But then again $m_1$ falls freely at $g$, allowing the rope to tension again and so forth... What's wrong?

 

[jbriggs444]

But if the blue rope accelerates at $g$, then the red one does so at $\frac{d_1}{d_2}g$. And if my $F_1$ from #1 is correct and constant, then $m_1$ accelerates at $g+\frac{d_2}{d_1}\frac{m_2}{m_1}g$ which is bigger than $\frac{d_1}{d_2}g$ given that $\frac{m_2}{m_1}\gg 1$.

This would mean that the rope goes slack immediately. But then again $m_1$ falls freely at $g$, allowing the rope to tension again and so forth... What's wrong?

But if the blue rope accelerates at $g$, then the red one does so at $\frac{d_1}{d_2}g$. And if my $F_1$ from #1 is correct and constant

Huh? You have given no valid computation for $F_1$.

What you know is how $m_1$ moves. You can use that to calculate what $F_1$ must be.

 

[greypilgrim]

I used the law of the lever on the wheel and axle and added this force to the gravitational force acting on $m_1$ (but I see now that this can't be right).

What you know is how m1 moves. You can use that to calculate what F1 must be.

Huh? That seems backwards to me. How would I know how $m_1$ moves without knowing $F_1$ first and calculate the acceleration out of that?

EDIT: Ah, I see. If there's no slack, $m_1$ obviously has to accelerate at the same rate as the blue rope.

 

[jbriggs444]

EDIT: Ah, I see. If there's no slack, m1 obviously has to accelerate at the same rate as the blue rope.

Yep.

 

[Baluncore]

Things are not always as simple as they seem. When I see an analysis of the Indian Rope Trick, or a released balloon, I immediately fear the speed of sound in rope.

 

[greypilgrim]

Okay, that makes my initial question trivial. I was way too focused on finding the forces first to realize that the acceleration can be calculated right away.

So $a_1=\frac{d_1}{d_2}g$ and since $F_1=m_1 a_1=m_1 g+T_r$, the tension in the red rope is
$$T_r=\frac{d_1}{d_2}m_1 g-m_1 g=\left(\frac{d_1}{d_2}-1\right)m_1 g\ .$$

However, now the law of the lever
$$T_r\cdot d_1=T_b\cdot d_2=m_2 g\cdot d_2$$
seems to fail completely. Is this an artifact of $m_2\rightarrow\infty$ or does that generally not hold for accelerated systems?

 

[haruspex]

$$T_b\cdot d_2=m_2 g\cdot d_2$$

Since $m_2$ is accelerating there must be a net force on it, so $T_b\neq m_2g$.

 

[Lnewqban]

I used the law of the lever on the wheel and axle and added this force to the gravitational force acting on $m_1$ (but I see now that this can't be right).

Falling m2 pulls down on also falling m1 via the lever (mechanical advantage of d2/d1).
Therefore, m2 must accelerate at a rate lesser than g, while m1 must accelerate at a rate greater than g.

As the height difference between both masses decrease as they fall, the combined (m1+m2) center of mass must be descending at a value of acceleration that is within the range of values at which m1 and m2 are accelerating.

 

[greypilgrim]

Since $m_2$ is accelerating there must be a net force on it, so $T_b\neq m_2g$.

Ah, I see. I think I'm running into trouble here because of my assumption that if $m_2\gg m_1$, then $m_2$ is "practically" free falling, which would mean that $T_b\approx 0$ (right?), but on the other hand still having $m_1>0$ which means that $T_b=\frac{d_1}{d_2}T_r>0$ (but still small, so so it kind of works out).

I guess I should try and tackle the full problem without approximations.

 

[haruspex]

I guess I should try and tackle the full problem without approximations.

Good idea.

 

[greypilgrim]

Okay. For consistency, let $T_1=T_r$ and $T_2=T_b$. I'll assume no slack, which means $a_1>g$ (otherwise, both masses are free falling). Then I get the four equations:
$$\begin{aligned}
m_1 a_1&=m_1 g+T_1 \\
m_2 a_2&=m_2 g-T_2 \\
\frac{a_1}{d_1}&=\frac{a_2}{d_2} \\
T_1 d_1&=T_2 d_2
\end{aligned}$$
I used a CAS to solve this for me:
$$\begin{aligned}
a_1&=d_1 g\cdot\frac{d_1 m_1+d_2 m_2}{d_1^2 m_1+d_2 ^2 m_2} \\
a_2&=d_2 g\cdot\frac{d_1 m_1+d_2 m_2}{d_1^2 m_1+d_2 ^2 m_2} \\
T_1&=d_2 m_1 m_2 g\cdot\frac{d_1-d_2}{d_1^2 m_1+d_2 ^2 m_2} \\
T_2&=d_1 m_1 m_2 g\cdot\frac{d_1-d_2}{d_1^2 m_1+d_2 ^2 m_2}
\end{aligned}$$
Now in the limit $\frac{m_1}{m_2}\rightarrow 0$, this simplifies to:
$$\begin{aligned}
a_1&=\frac{d_1}{d_2}g \\
a_2&=g \\
T_1&=\left(\frac{d_1}{d_2}-1\right)m_1 g \\
T_2&=\frac{d_1}{d_2^2}\left(d_1-d_2\right)m_1 g=\frac{d_1}{d_2}\left(\frac{d_1}{d_2}-1\right)m_1 g=\frac{d_1}{d_2}T_1
\end{aligned}$$
That looks good so far. Going back to the general solution, the condition $a_1>g$, is:
$$d_1 g\cdot\frac{d_1 m_1+d_2 m_2}{d_1^2 m_1+d_2 ^2 m_2}>g$$
And this simplifies to $d_1>d_2$, which was to be expected.

Is all that correct?

 

[greypilgrim]

Oh and another question, more mathematical: If I take the limit $m_1\rightarrow 0$ instead of $\frac{m_1}{m_2}\rightarrow 0$, I get the same accelerations, but $T_1=T_2=0$. Is this some zeroth vs. first order approximation thing? But why do I get the same accelerations then?

 

[jbriggs444]

Oh and another question, more mathematical: If I take the limit $m_1\rightarrow 0$ instead of $\frac{m_1}{m_2}\rightarrow 0$, I get the same accelerations, but $T_1=T_2=0$. Is this some zeroth vs. first order approximation thing? But why do I get the same accelerations then?

If $m_1$ approaches zero while $m_2$ remains finite and approximately constant then $\frac{m_1}{m_2}$ also approaches zero. So naturally you get the same limiting accelerations either way.

If you get a finite limiting acceleration and you are considering a mass $m_1$ that approaches zero then of course the tension $T_1$ will approach zero. A non-zero limiting tension would produce an infinite limiting acceleration of $m_1$. Which you've calculated is not the case.

 

[kuruman]

I solved the equations for $a_1$ first and then saw what happens when $m_2>>m_1$. To do this, I simplified the algebra (no need for a CAS) by replacing $m_2=\mu m_1$ and $d_1=d_2 \delta.$ Note that scaling parameters $\mu$ and $\delta$ are both greater than 1. The bottom two of the starting equations in post #16 reduce to $~T_2=T_1\delta~$ and $~a_2=a_1/\delta.$ These can be substituted back into the top two equations to yield a system of two equations and two unknowns. One then gets $~a_1=f_1(\mu,\delta)g~$ and $~T_1=f_2(\mu,\delta)m_1g$.

One can then investigate the behavior of dimensionless numerical constants $f_1(\mu,\delta)$ and $f_2(\mu,\delta)$ in the limit $\mu >>1$ at fixed $\delta$. The result may be surprising.