Can Different Pulley Radii Achieve Accelerations Greater than Gravity?

[cragar]

If I wanted to accelerate a mass faster than g would I just take a rod and connect two pulleys
on each end but they would have different radii. And then I could wrap one with rope and tie it to a mass. And the other pulley would have a rope and a mass attached to it, and the rod would rotate on a bearing and because they have different radii I should be able to get different accelerations. Are their other ways with ropes and pulleys and masses just under the influence of gravity to accelerations larger than g?

 

[Jorriss]

Have you taken an intro mechanics course? If so, I would recommend drawing a force diagram for the system you describe.

 

[Mmm_Pasta]

I don't see how the system you described could accelerate faster than g. Each mass experiences a tension that prevents it from accelerating faster than g.

 

[kurros]

I am sure there are zillions of ways. If you allow me a lever, for simplicity: put the fulcrum very close to one end, put some gigantic mass on that end, and a light mass on the long arm of the lever, and you can generate some enormous acceleration in the light mass. You can of course do the same sort of thing in many different ways with pulleys.

edit: or am I being stupid. The other replies now make me doubt myself and I will have to go and think again :p.

edit 2: No it's fine, this is the whole principle of mechanical of advantage of levers and pulleys, they can act as force multipliers. Oh wait but the same force is applied at both ends. Oh damn. I haven't thought about mechanics in a while :p.

edit 3: No wait, I am right. I calculated it, and for such a lever, if the large mass totally dominates the torque on the lever, the acceleration of the smaller mass is
$$ a = g \frac{l}{x} $$
where x is the short arm of the lever and l is the long arm.

 

[Simon Bridge]

edit 3: No wait, I am right. I calculated it, and for such a lever, if the large mass totally dominates the torque on the lever, the acceleration of the smaller mass is
$$ a = g \frac{l}{x} $$
where x is the short arm of the lever and l is the long arm.

And x < l so a > g ... is that the reasoning?
Could you demonstrate the calculation you did one step at a time, using Mx > ml and l > x and start from: $\sum\tau = I\alpha$

 

[Simon Bridge]

If I wanted to accelerate a mass faster than g would I just ...

What exactly is the objective here?
To accelerate a mass faster than g just fire it from a cannon - it will have a very brief but very high acceleration.
There are zillions of ways - to get something to accelerate more, you push it harder.

But if your aim is to have a free-fall mass accelerate by more than, say 9.8m/s/s, then you need a higher gravitational field.
You'll start to get a feel for this sort of thing when you start drawing the free-body diagrams.
I am interested where you got the idea from though.

 

[kurros]

And x < l so a > g ... is that the reasoning?
Could you demonstrate the calculation you did one step at a time, using Mx > ml and l > x and start from: $\sum\tau = I\alpha$

Yes that was the idea. There is actually a faster way than I originally used, I realize now:

So the large mass $M$ has a lever arm of length $x$, and the smaller mass $m$ has a arm of length $l$. Let the lever start out horizontal and at rest.

If the system is dominated by the larger mass, then the angular acceleration (about the fulcrum) of the system is $\alpha = g/x$, since the larger mass falls under gravity as normal and barely notices the presence of the lever.

The smaller mass just goes along for the ride, and being part of the system experiences the same angular acceleration as the rest of it, i.e $\alpha = a/l$. Since these things must be equal we immediately get $a = g\frac{l}{x}$

 

[WannabeNewton]

cragar I'm not sure you worded your question in the way you intended. If gravity $g$ is the only force influencing an individual massive particle then by definition the particle is in free fall with acceleration $g$. When you constrain its path using things like rods and/or ropes, at least for some finite interval of time, you are introducing constraint forces (e.g. tension) on the particle; it is no longer only under the influence of gravity.

 

[Simon Bridge]

Yes that was the idea. There is actually a faster way than I originally used, I realize now:

So the large mass $M$ has a lever arm of length $x$, and the smaller mass $m$ has a arm of length $l$. Let the lever start out horizontal and at rest.

If the system is dominated by the larger mass, then the angular acceleration (about the fulcrum) of the system is $\alpha = g/x$, since the larger mass falls under gravity as normal and barely notices the presence of the lever.

The smaller mass just goes along for the ride, and being part of the system experiences the same angular acceleration as the rest of it, i.e $\alpha = a/l$. Since these things must be equal we immediately get $a = g\frac{l}{x}$

You have applied the approximation too early ... start out without assuming the domination of the large mass ... work out the equation of motion, then see what happens when you increase the large mass.

 

[kurros]

You have applied the approximation too early ... start out without assuming the domination of the large mass ... work out the equation of motion, then see what happens when you increase the large mass.

Ok maybe I am being stupid... but try this:

Total torque on the lever is
$$\tau = Mgx - mgl = I\alpha$$
Moment of inertia of the system is
$$I = Mx^2 + ml^2$$
Angular acceleration is therefore
$$\alpha = \frac{Mgx-mgl}{Mx^2 + ml^2} $$
The magnitude of the linear acceleration of the smaller mass is as before:
$$a = \alpha l = \frac{(Mgx-mgl)l}{Mx^2 + ml^2} = g\frac{(Mx-ml)l}{Mx^2 + ml^2} $$
And applying $Mx>>ml$ and $l>x$ we get my answer as before, since they render the $ml$ and $ml^2$ terms irrelevant:
$$a = g\frac{l}{x} $$
You still disagree?

edit: Err wait, maybe the ml^2 term is not irrelevant, one second...
edit 2: No, I stand by it's irrelevance.
edit 3: To be more convincing of this:

$$ \frac{(Mx-ml)l}{Mx^2 + ml^2} \rightarrow \frac{Mxl}{Mx^2 + ml^2} = \frac{1}{Mx^2/Mxl + ml^2/Mxl} = \frac{1}{x/l + ml/Mx} \rightarrow l/x $$

edit 4: I might believe you if you say my first limit messes up the second one here though.
edit 5: Ok no I won't, the -ml term goes to zero by itself just fine.

 

[Simon Bridge]

Hmmm ... I have not said that I don't believe you, it's just that you kept getting worried about your math and I was trying to show how you can get certain ... the worry crept in each place where you jumped a few steps so you had to go do a lot of back-and-fill... which makes you sound confused.

Your were looking for the instantaneous acceleration when the lever is horizontal - and we are using a massless lever in the coordinate system of the pivot point.

$$Mgx - mgl = (ml^2+Mx^2)\frac{a}{l}$$ ... the mgl torque is chosen negative so the anticipated m acceleration will be positive.

This rearranges to: $$a=\frac{(Mx-ml)l}{ml^2+Mx^2}g$$ ... so, in order for a > g, you need $$Mlx-ml^2 > ml^2 + Mx^2 \\ \Rightarrow M-m\frac{l}{x} > m\frac{l}{x}+M\frac{x}{l}\\ \Rightarrow M\left (1-\frac{x}{l}\right ) > 2m\frac{l}{x}$$

R=l/x is down to the lever - the question is, can we find an M that will make the LHS > RHS ... well, measure M in units of m, and let R be something simple like 2, then the relation works for M > 8m ... only need to demonstrate one instance to prove the point.

Notice how I didn't need to go to some hypothetical overwhelming mass?
Limiting cases are often confusing.

You can also do it by conservation of energy:
$(Mgx-mgl)\sin\theta = \frac{1}{2}(ml^2+Mx^2)\dot{\theta}^2 : \theta(0)=0$ ... differentiate both sides and use $a=l\ddot{\theta}$

I still wonder what OP is trying to do.

 

[kurros]

Hmmm ... I have not said that I don't believe you

Oh. Well alright then :). I started to become unsure due the character of the other responses to this thread, although when I made my first claim based on intuition I was feeling pretty sure about it. I then convinced myself with a quick consideration of the hypothetical overwhelming mass case which I reported the result of, and then the way you put your comment started to make me worry that I made some mistake in my haste :p. So the doubt went up and down, with each new doubt forcing me to be more thorough :p.

Notice how I didn't need to go to some hypothetical overwhelming mass?
Limiting cases are often confusing.

Yes I agree this is better. I was just out for speed before :). Plus the simplicity of the result was appealing.

I still wonder what OP is trying to do.

*shrugs*

 

[cragar]

What exactly is the objective here?
To accelerate a mass faster than g just fire it from a cannon - it will have a very brief but very high acceleration.
There are zillions of ways - to get something to accelerate more, you push it harder.

But if your aim is to have a free-fall mass accelerate by more than, say 9.8m/s/s, then you need a higher gravitational field.
You'll start to get a feel for this sort of thing when you start drawing the free-body diagrams.
I am interested where you got the idea from though.

My physics professor told us a while back that you could have a pulley set up with masses, and he had like 4 pulleys, that you could get accelerations faster than g with the setup. I realize now my original set up won't work, but I was just trying to figure out how you would make one.

 

[kurros]

Ahh, right. Well you should be able to do it with a block and tackle type of setup.

 

[WannabeNewton]

Consider the following system: http://s16.postimg.org/5jqfsup9x/atwood.png

Let $T$ be the tension across the main string (the one connected to $m_1$). Note that the net force on the lower pulley must be zero (otherwise it would have infinite acceleration since its a massless pulley) so this tells us that the tension on the string attached to $m_2$ is $2T$; note that just because the net force on the lower pulley is zero doesn't mean it won't accelerate-think of it as saying that because it is massless it takes no force to accelerate it. The lower pulley will accelerate with some magnitude whilst following $m_2$ as is obvious from the diagram but it's actual dynamics we don't care for because it is massless so it's dynamics won't contribute anything to the dynamics of the rest of the system.

The equations of motion then become $T - m_1g = m_1a_1, 2T - m_2g = m_2a_2$. Because the length of the string is constant, we find that $a_1 = -2a_2$ and using this we can solve for the unknowns. Do this and consider the case $m_2 >> m_1$; you will find that $a_1 = 2g$ because $m_2$ is approximately in free fall.