Atwood’s machine is the name of a device that looks like this:
Also known as “two masses on a pulley”. Surprisingly, this simple device comes up a lot in intro physics texts. It also brings up some interesting issues. I will go over the basic way to solve a problem like this (as an example) and then talk about the other interesting issues it brings up.
Problem: A small, low mass, pulley has a light string over it connected to two masses, m1 and m2. If released from rest, what is the acceleration of the two masses.
Where to start? This is actually a very difficult question for introductory students. When in doubt, start with a picture – at least that is my recommendation. If you have a physics text, nearly all of the new ones have some type of problem solving strategy. I suggest you try one of these (at least when you are stuck). It is surprisingly difficult to get students to solve problems in the same way that experts do. I think that I started solving problems in a more expert manner when I started teaching as a graduate student. This may be why it is useful to work in groups, it is kind of like teaching. Nonetheless, I digress. Since you already have a picture of the atwoods machine, I will draw two free body (force) diagrams.
Notice that the tension on the two masses is the same. This will not always be true. In order for the tension to be the same the mass of the rope must be negligible (massless rope is available from PASCO). Also, the mass of the pulley must be small (technically the moment of inertia of the pulley must be small). These two things are not too difficult to achieve so I will proceed with the magnitudes of the tension forces being the same.
The next thing to think about is what strategy to use. There some basics to consider. Would work-energy be good here? How about Newton’s second law? What about plain old kinematics? The kinematics approach will not work because the acceleration is not known. There is probably a way to make work-energy work (get it?), but in general, the work-energy approach is good if you know or a looking for forces, distance and velocities. That leaves Newton’s second law. Here is a review if you missed it before. There are several forms of this strategy, but since I am looking for acceleration, I will use:
But wait! There are two objects, what to do? Simple, I will just use Newton’s second law twice. If I call the vertical direction the y-direction, I can write for the two masses:
Here, this is a scalar equation (only in the y-direction). Also, I have assumed that mass 1 will accelerate in the negative y direction and mass 1 will accelerate in the positive y direction. If the two masses are connected by a non-stretchable rope, then the magnitudes of the acceleration must be the same (which I call “a”). From here, I want to solve for the acceleration. Although everything looks like a variable, really only T and a are variables. I assume that I would know the two masses and g. Notice that there are two variables and two equations. This is a situation I like to call “two equations and two unknowns”. I am surprised by how many students only try to solve these equations by multiplying one of the equations by a constant and adding it to the other. This can work, but not always. I suggest solving one of the equations for T and plugging that solution into the other equation. I will start by solving the first equation for T:
Now I will use this expression in the second equation. That will create an equation with only the variable “a”
Now I need just solve this for “a”
- Does this result have the correct units? Yes. The fraction has kg/kg and g has units of N/kg which is equivalent to m/s2. It is always a good idea to check and see if your answer has the correct units. It doesn’t mean your answer is correct, but if it is the wrong units you can be sure the answer is wrong.
- Does this result seem reasonable? Yes. The fraction in front of g has a smaller value on the top (since it is the difference in the two masses). This will make the acceleration smaller than the acceleration of a free falling object. Makes sense. Also, I found a positive value for a assuming m1 > m2. This also makes sense as it would accelerate in the direction of the heavier mass (which is what I assumed).
- What could go wrong? The common mistake that I see (and that I made as an undergrad – I remember this) is to look at mass m1 and say it has two forces (gravity and the tension). Then say hey, look. The tension (T) is just the weight of m2. This is not true. If mass m2 had a tension equal to m2g on it, its acceleration would be 0 m/s2. Clearly that does not happen. Instead, mass 2 is accelerating up. The tension must be greater than its weight. You could solve for the value of the tension force and check this for yourself.
There are two key assumptions. First, that the mass of the pulley is small. Second that the mass of the string is small. What if the mass of the pulley is NOT small? If there is also friction between the pulley and the string, then the tension on the two masses will NOT be the same. Maybe this picture will help:
Here I drew the tensions not vertical so that it would look a little better. The tension on the left is greater than on the right. The result is that there is a net torque on the pulley. This torque increases the angular speed of the pulley. If the mass is small, this difference in tensions is not noticeable. I know what you are saying. If there is a difference in tensions, shouldn’t the pulley change its momentum also? No. These are not the only forces on the pulley. There is also a force from the axle where the pulley is connected. This is also the case for a “massless” pulley. Both of those tension forces were the same magnitude, but down. This means there must be an upward force from the axle or the pulley would accelerate downward.
Can I model this in Fantastic Contraption?
So, I set up a simple situation. Here is a video:
Does this work as it is supposed to? Clearly, you can see that the mass of the “string” is not zero. Also, the mass of the pulley is not zero. Let me proceed anyway. Using Tracker Video Analysis I obtained vertical position data for one of the masses. Here is a plot of that data:
I fit a quadratic function to the data to see if the acceleration is constant. Looks close enough to constant. From the fit, the acceleration of the mass is 0.302 U/s2. Remember that U is the distance across one of the ball. According to this site, the size of a ball is 40 units and the acceleration of a free falling object is 300 units/s2. So, my U is one of their U’s. I know that is confusing. Let me just say that my gravity should be 300/40 = 7.5 U/s2. Now, if I use the results from above (ignoring the mass of the string and pulley), I should get an acceleration of:
Notice that I called the mass of the mass, m (which canceled). This gives a result much greater than that measured from the video. Ok, so it didn’t work. I have some other ideas to try with fantastic contraption. One would be to measure the moment of inertia of a ball by letting it roll down an incline. That would be a post for another day.