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EN4 : Dynamics and Vibrations

4.5 Oblique Central Impact

In the last section we discussed impacts that took place when the two particles in question were moving along the same straight line. Such an impact is called direct central impact, and in this case the problem can be completely solved by considering it in a one-dimensional setting. In this section, we shall generalize this to include impact that occurs when the particles are not moving along the same straight line. We shall, however, assume that the particles are still confined to the same plane. Such impacts are called oblique central impacts. We shall make another assumption that the during impact, i.e., when the particles are in contact, there are no frictional effects. This is, of course, an unrealistic assumption but it simplifies our calculations.

figure253

Consider, then, two particles (assumed spherical) of mass tex2html_wrap_inline638 and tex2html_wrap_inline640 approaching each other with velocities tex2html_wrap_inline642 and tex2html_wrap_inline644 as shown in the figure. The global tex2html_wrap_inline646 and tex2html_wrap_inline648 axis are also shown. The question in oblique central impact problems, like that in the direct central impact problems, is to determine the velocities tex2html_wrap_inline650 and tex2html_wrap_inline652 after impact.

We shall use the impulse-momentum relationship to look at this problem just as in the case of direct central impact. During the time that the spheres are in contact, they assert equal and opposite forces on each other. The direction of this force is along the line joining the centers of the spheres, in other words in the direction normal to the contacting surfaces. We denote this direction with the unit vector tex2html_wrap_inline654 ; and the tangential direction by tex2html_wrap_inline656 . As you will see, working this problem out in this new basis defined by tex2html_wrap_inline656 and tex2html_wrap_inline654 will be much easier. Using the impulse-momentum relationship one can easily show that the total momentum of the system is conserved, i.e.,

eqnarray548

Writing components in the basis tex2html_wrap_inline654 and tex2html_wrap_inline656 we get that

eqnarray554

which is just a mathematical way of stating that total momentum in the normal and tangential direction is conserved. How many unknowns are there in this problem? If you look at the equations carefully, you will see that there are four(4) unknowns; they are tex2html_wrap_inline674 and tex2html_wrap_inline676 . The principle of conservation of momentum gives us two equations. We need two additional equations for a complete solution. We may, of course, bring into action the idea of the coefficient of restitution, which is the ratio of the magnitude of the relative velocity of separation to that of the relative velocity of approach in the normal direction, in symbols it reads

eqnarray556

Ok, that gives us one more equation making the total number of equations so far three. We need one more equation to solve for the four unknowns. Where do we get this from? Here is where we assume that there is no friction between the spheres, i.e., there is no tangential force on the spheres. Well, then we know from the restated-Newton's Law that tangential component of the momentum of each ball is conserved. But that will give us two equations,

eqnarray558

But now we have five equations! Not really, since the last two equations that we just derived would automatically imply that the total momentum in the tangential direction is conserved. Thus we drop the second equation of obtained from the conservation of the total momentum. Collecting all the equations, we get

eqnarray560

which is a well-posed linear algebraic system for four unknowns. What we shall do now is to apply these equations some oblique central impact problems through some examples.

Example 1: A ball of mass m moving with speed v bounces on a floor at an angle tex2html_wrap_inline682 . If the coefficient of restitution is e, find the angle tex2html_wrap_inline686 at which the ball emerges from the floor.

Let us assume that the ball is the mass `` tex2html_wrap_inline638 '' and that the earth is mass `` tex2html_wrap_inline640 ''. In this case the tex2html_wrap_inline654 and tex2html_wrap_inline656 are parallel to tex2html_wrap_inline648 and

figure292

tex2html_wrap_inline646 . Thus we can write the initial velocity of the ball as tex2html_wrap_inline700 . The velocity of the ball after impact can be written as tex2html_wrap_inline702 . We know that the velocity of earth is unaffected in our frame before and after impact. Now, from the third equation that we had collected before, we have

eqnarray565

From the definition of the coefficient of restitution, we get

eqnarray567

It therefore follows that

eqnarray569

Note that the maximum value of tex2html_wrap_inline686 is tex2html_wrap_inline682 and this occurs when e = 1. The minimum value of tex2html_wrap_inline686 is zero, and this happens when e = 0.

Example 2: A pool player has to decide the distance d at which the cue ball is to placed so that the 8-ball is potted. Both the balls have equal mass m and radius R. Neglect friction.

figure318

It is clear that for the 8-ball to be potted, its velocity vector has to be inclined at an angle tex2html_wrap_inline720 . Our task will be to determine d such that this happens. To analyze this. construct a coordinate system with tex2html_wrap_inline656 and tex2html_wrap_inline654 as shown. It is clear that the angle of inclination of tex2html_wrap_inline654 with the tex2html_wrap_inline646 direction is tex2html_wrap_inline732 . Thus in this coordinate system we can write the initial velocity of the cue ball (mass `` tex2html_wrap_inline638 '') as tex2html_wrap_inline736 . The velocity of the 8-ball is tex2html_wrap_inline738 . The final velocities may be written as tex2html_wrap_inline740 and tex2html_wrap_inline742 . Let us obtain all unknowns. From conservation of total momentum we know that

eqnarray574

figure341

The definition of coefficient of restitution gives us

eqnarray576

From the conservation of tangential component of momentum of each of the balls we get that

eqnarray578

From the first two equations it follows that

eqnarray580

Thus the final velocity of the 8-ball is

eqnarray582

i.e., it is exactly along tex2html_wrap_inline654 . But we know that tex2html_wrap_inline654 is inclined at an angle tex2html_wrap_inline750 to the tex2html_wrap_inline646 axis, and this must be equal to tex2html_wrap_inline682 . Thus, imposing that tex2html_wrap_inline756 gives

eqnarray586

4.6 Angular Momentum and Angular Impulse

This section will define ``rotational'' analogs to linear momentum and linear impulse called ``angular momentum'' and ``angular impulse''. If you have not seen these before they might appear rather strange! You should not find this surprising - these concepts are indeed hard to grasp and to explain! With that warning, lets get to the definitions of angular momentum.

Consider a particle of mass m moving with a velocity tex2html_wrap_inline760 and whose position vector tex2html_wrap_inline762 is measured from a point O with respect to a fixed basis. The angular momentum of the particle about the point O is defined as

eqnarray590

Thus angular momentum is defined as the cross product of the position vector with linear momentum. Several features may be noted:

What is the significance of angular momentum? To see this consider a particle of mass m executing uniform circular motion with angular velocity tex2html_wrap_inline792 along a circle of radius r. The velocity at any instant of time is given by tex2html_wrap_inline796 and the position vector is given by tex2html_wrap_inline798 . The angular momentum of the particle about the origin is given by tex2html_wrap_inline800 . This is amazing! Note that the angular momentum about the origin is a constant for uniform circular motion just as linear momentum was constant for uniform straight line motion!

Now we pop the all important question: for linear motion we can state Newton's Law as ``the rate of change of momentum is equal to the applied force'' - is there an analogue for angular momentum? In other words is there an ``Angular Newton's Law''. Lets see. To this end, compute the time derivative of angular momentum

eqnarray604

Note that the tex2html_wrap_inline818 vanishes because velocity is parallel to linear momentum. Amazing! We find a very interesting result that the rate of change of angular momentum is the applied moment which is the ``angular Newton's Law'' that we were after. And again, playing the same game, we see that if the moment of the force acting on a particle about a point is zero, then the angular momentum of the particle about that point is conserved. You will see through examples later that this is an extremely useful result.

Using ``Angular Newton's Law'' we can now develop the angular impulse-angular momentum relationship. To this end, we define the angular impulse tex2html_wrap_inline820 (about point O) as

eqnarray616

By integrating ``Angular Newton's Law'' we come to an analogous conclusion that

eqnarray618

Example 1: A ball of mass tex2html_wrap_inline638 attached to a string undergoes uniform circular motion at speed v along a circle of radius r. A second ball of mass tex2html_wrap_inline640 is located along the circular path. The mass tex2html_wrap_inline638 collides with mass tex2html_wrap_inline640 . The two particles stick together after collision. Compute the final speed of circular motion.

The initial angular momentum of particle 1 about the point O is

eqnarray620

figure402

The angular momentum of the particle 2 before impact is

eqnarray622

After impact the particles stick together and therefore the angular momenta are

eqnarray624

The total angular momentum is conserved, thus

eqnarray628

Example 2: Problem 3, Midsemster Exam I (conducted on Tuesday, March 9, 1999).

This example will concern part 3.4 of the question where the angular speed of the Habitat/rocket stage system is to be determined as a function of time if the length of the cable as a function of time is tex2html_wrap_inline846 . Note that from the free body diagram that you obtained in part 3.1, that the net force acting on both the Habitat and the rocket stage only the tension in the cable. But this does not produce a moment about the point O, and thus the angular momentum of the system is conserved. Let us compute the total angular momentum of the system any given time:

eqnarray632

But by conservation of angular momentum this quantity must be equal to to the angular momentum at time t=0 or

eqnarray634

whence it follows that

eqnarray636

The simplicity afforded by the principle of conservation of angular momentum in solving this problem is indeed impressive!



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