## The Asymmetric Atwood Machine.Some physics problems can be solved by several different methods. Often textbook problems are idealized by assuming that friction and other energy dissipative processes have negligible effect, all bodies behave as rigid bodies, any connecting cords have negligible mass, are perfectly flexible but do not stretch. Very often these problems yield to any one of several methods.
- Dynamic analysis using only Newton's laws and geometric constraints.
- Energy analysis.
- Momentum analysis.
The pulley at the top consistes of two or more pulleys of different radii fastened together on a common axle. We treat it as frictionless and of negligible mass. Being massless, its moment of inertia is zero. The supporting strings are weightless, unstretchable, yet perfectly flexible.
Let the tension in the left string be labeled T
## Dynamic analysis.We will adopt the convention that upward on the left is positive, while downward on the right is positive. Clockwise rotation is positive. Clockwise torques are positive. (We have very good reason to guess that the pulley rotation will be clockwise, but any consistent sign convention would do as well and give the same physical answers.)We'll first do this "by the book", using the "divide and conquer" approach. We mentally subdive the system into parts. We then apply Newton's second law to each part of the system, one at a time, generating as many equations as there are parts in the system. Then these equations are solved simultaneously, eliminating unwanted quantities to obtain the answers of interest. Then if you really want to know values for the quantities that you eliminated, you can put the answers back into the equations to generate them.
At each step be very careful to identify all forces acting upon the part of the system you have chosen, and There are situations where you can shorten the process by lumping several parts of the system together, if they have the same motion and their mutual forces of interaction are not of interest to you. Those forces, being equal in size and oppositely directed are then "internal forces" and "drop out" of the calculation of net force, net torque, and net impulse. We will see an example of that as we go along. So let's begin. This will not only be fun, but also instructive.
By Newton's second law, the left mass m has two forces acting on it. The upward force T
The right mass 2m has upward force T
Newton's first law for torques on the pulley: -T
Therefore T
T So:
2 (T Multiply the second one by 2.
4mg - 2 (T
Add 1 and 3 to eliminate T
- mg + 4mg = m(a
(T
(T This is greater than mg, so m accelerates upward.
(T
This is half of T
## Energy analysis.What can energy equations tell us?Suppose the left side moves upward a distance y. Then the right side moves down a distance 2y. The change in potential energy plus the change in kinetic energy is zero for a closed system.
mg2y - mgy = ½mv
mgy = ½mv
gy = (3/2)v
v It is unnecessary to explicitly use the concept of work in this example. Doing so would make the derivation messier. The gravitational force does work on the masses, the tension forces do also, but by use of the potential energy concept we can treat this system as if it were closed, and the work done by the tension forces is all internal. So the increase of kinetic energy may be thought of as due to the net work the forces due to gravity give to the system. What is the kinetic energy of the objects after a given time?? If m moves up a vertical distance y, then 2m moves down 2y by the geometric constaints.
(2m)g(2y) - mgy = (1/2m(v (Gravitational potential change = kinetic energy change, from rest position.)
3mgy = (1/2)m(v (The change in kineteic energy is the final kinetic energy of m plus the final kinetic energy of 2m.)
3mgy = (1/2)m(v
But by geometric constraints, (v
3mgy = (9/2)m(v
gy = (3/2)(v
v
v The kinetic energy of m is (1/2)m(2/3)gy = (1/3)mgy The kinetic energy of 2m is (1/2)2m4(2/3)gy = (8/3)mgy The total kinetic energy of both together is (1/3 + 8/3)mgy = (9/3)mgy = 3mgy, which equals the total change in potential energy. This serves as a check on the math. It is unnecessary to explicitly use the concept of work in this example. Doing so would make the derivation messier. The gravitational force does work on the masses, the tension forces do also, but by use of the potential energy concept we can treat this system as if it were closed, and the work done by the tension forces is all internal. So the increase of kinetic energy may be thought of as due to the net work the forces due to gravity give to the system. Gravitational potential energy changes as the bodies move. It decreases because the center of mass of the system lowers. This is no different than when a body freely falls, and its gravitational potential energy decreases in the same amount its kinetic energy increases, thus the total energy of the system remains constant. Notice that I never considered the system as a whole when using F = ma. How could I, for m and 2m have different accelerations, so what meaning could one attach to the "acceleration of the whole system"? As my physics profs used to say, the method of solving such problems is "divide and conquer", that is, look at each part of the system and apply F = ma to each in turn.
Now one Finally, one can clean up the loose ends by using the kinematic laws to find other quantities. For example, the acceleration (which we found so easily by the dynamic analysis).
Use v
(2/3)gy
a
This raises an important question. How do we know that this system is one where the acceleration of each part is constant, therefore allowing us to use the "Galileo" kinematic equations that were derived on the assumption of constant acceleration? This might be considered a "hidden assumption" in most textbook analyses of such dynamic systems. There is, in fact, a brief period of time, just as the system is released, during which the tension forces quickly change and the acceleration rapidly changes. For example, when the system is held at rest (acceleration is zero), the tension on m is just mg, but when the system is released, this tension increases very rapidly to (4/3)mg, and during this time its acceleration quickly increases from zero to g/2. We are, in fact, ignoring this brief (though interesting) process, during which brief time the system moves very little, and are doing the analysis on the much larger duration of the process during which the acceleration is nearly constant. Qustion: Can you describe a situation where this shortcut would not be justified for this system? ## Momentum analysis.There is still another way to do this problem. Use Impulse = change in momentum. This is seldom done in textbooks. All answers should come out to be the same, however.Impulse is ∫ F(t) dt = Δ(mv) over the duration of time during which the force acts. The time need not be short. Let the system accelerate at a constant rate during time t. The net gravitational impulse on the system equals the system's change of momentum.
(2m)gt - mgt) = 2mv
mgt = (2m2 - m)v To compare with the previous results, we have in our toolkit the kinematic equations for constant acceleration (sometimes called "Galileo's equations of motion"). Use this one.
y and
v
So gt = 3v
v
y
1 = 9(a
Finally, a
## Angular momentum method.This problem also can be analyzed by considering the system's angular momentum. It is a trivial exercise for the student. But let's use a creative shortcut, possible because we have assumed the pulley moment of inertia is negligibly small.Take the center of the pulley as the center of torques. Take clockwise torques as positive, counter-clockwise negative. This is not the usual textbook convention, but you could reverse the signs without harm so long as you are consistent. The torque due to the gravitational force on mg is mgR. The torque due to the gravitational force on 2m is (2m)g(2R). So long as the weights on the strings do not reach the pulley, the system behaves the same as if the weights were affixed at R and 2R on the pulley itself, and the motion was through a very small angle.
Use τ We are treating the system as a whole, with two gravitational torques acting upon it.
-mgR + 4mgR = (mR
So 3mR = 9mR And 3g = 9Rα
But by geometry, αR = a
So a
No surprise, is it? This method is the easiest of all, because of that negligible mass pulley. If the pulley mass is negligible, so is its moment of inertia, and therefore the net torque on it is zero even if it is rotating, and we can confidently say that the tensions of the left and right string are in ratio of 1:2. These forces are internal to the system, and provide no work and no impulse to the system. Therefore we don't need to use the tensions in the calculation. However, now that we have the accelerations of each mass, we can easily calculate the tensions using F
ma
mg/3 = T
T As we found by previous methods.
## Non-negligible pulley mass.This suggests that we can just as easily analyze this system when the pulley mass is significant. Just modify our previous equation:
-mgR + 4mgR = (mR By adding a term for the angular momentum of the pulley (I):
^{2} + 2m(2R)^{2} + I)α Or, in more general form:
_{1}gR_{1} + m_{2}gR_{2} = (m_{1}R_{1}^{2} + m_{2}R_{2}^{2} + I)α This equation is the basis of several instructive exercises. The pulley's moment of inertia can be measured directly. The pulley is removed and weighed. Suspend the pulley from some point. These pulleys usually have a flange at the rim which is a convenient suspension point. Then let it oscillate in simple harmonic motion, measuring the period of its motion with a stopwatch. From this data its moment of inertia about the suspension point can be calculated. The parallel-axis theorem may then be used to determine its moment of inertia about its own rotation axis. This may be roughly checked from dimension measurements if its geometry is simple. This result is inserted into Eq. 2, which is solved as before, predicting the acceleration of the asymmetric Atwood machine, which can then be confirmed experimentally. Or, the acceleration of the system may be measured, and Eq. 2 used to calculate I.
Some students don't realize that a body moving in a straight line can have angular velocity, angular acceleration and angular momentum about a point. The vector form of the definition of angular velocity is ## Historical note.This problem example illustrates why it took so long in the history of physics to develop the concepts of energy and momentum. The issue was "What is the proper measure of motion?" Candidates were "vis viva" (mv^{2}) and momentum (mv). Some physical situations could be analyzed using either one. But some other problems required both used together. We now realize that we need both to adequatly deal with all possible interactions between material bodies. We now also recognize that energy is a scalar quantity, while momentum is a vector quantity. They aren't interchangeable.
## Comparison of linear motion and angular motion.Typically, physics textbooks introduce straight line motion with constant acceleration first, in a chapter all its own. Then motion with constant acceleration in a plane is considered with examples of trajectories of projectiles. This requires vector algebra. Finally the special case of circular motion is taken up, and it may seem like a repeat of the process and derivations already learned to deal with straight line motion. But there are traps for the unwary. Straight line motion begins with the variables time, distance, velocity and acceleration. Circular motion begins with time, angle, angular velocity and angular acceleration. Though the process of derving results from these seems identical for kinematic equations, surprises are in store when one moves on to work, energy, impulse and momentum.I will not do the derivations here, for they can be found in any good physics textbook. But the textbook may not summarize the results and critically compare them as I will do here.
Note that angular position, speed and acceleration are The angular quantities and their equations need to be independently derived from first principles. This table is only a summary of the results to serve as a roadmap to the equations and a memory aid. We are not at all surprised at the analogy holding up when we derive the angular counterparts of "Galileo's" kinematic equations for motion with constant acceleration.
But remember, this is
In two dimensions, with velocity and acceleration expressed as vectors, these patterns are still analogous, but the terms of the first two are all vectors, so they add by vector addition. The last equation has scalar terms, and the term as becomes
In two dimensions, with angle, angular velocity and angular acceleration expressed as vectors, these patterns are still analogous. The terms of the first two are all vectors, so they add by vector addition. The last equation has scalar terms, and αθ becomes
## Inertia.Students hear the word inertia often, but how many know what it means? What is the measure of inertia? Inertia is a measure of the "reluctance" or "sluggishness" of a body to be set into motion when a force is applied. Newton's law F = ma in the rearranged form a = F/m shows that the acceleration of a body of mass m resulting from application of a force is inversely proportional to the body's mass. Mass is the measure of translational inertia. The equation α = τ/I shows that the rotational acceleration of a body is inversely proportional to its moment of inertia. The moment of inertia is the measure of its rotational inertia.
In classical physics, mass does not depend on how fast the body accelerates. Even if it is at rest is has mass. Mass is a property of the body. Its moment of inertia depends only its mass distribution with respect to a potential axis of rotation. But even if it is not rotating, its moment of inertia may be calculated from I = ∫ r
- Donald E. Simanek, September 3, 2010.
Go to the next chapter, The Mechanical Universe. |