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All of the work we have done on Newton's Laws has prepared us to deal with a great many problems in dynamics. However, a large class of problems are still not "possible" for us to do and an even larger class can be made easier if we complete our discussion of rotational dynamics by finding the complete analog of Newton's Second Law of Motion for rotations. By virtue of our success with analogy so far, we could just suggest that we can define a vector quantity, L, such that the torque follows the equation Tnet = dL/dt. This is analogous to Fnet = dp/dt. The quantity L is therefore referred to as the angular momentum. To find the appropriate form for L, we first have to define the direction of the torque. The convention that has been found to be useful is to define the direction of T to be along the axis of rotation. Since there are two possible directions along any line, the convention is to use the right-hand rule to define the direction of T, i.e. you curl the fingers of your right hand along the direction of rotation and your thumb indicates the direction of T along the axis of rotation. This is just a convention, but it is one you will see often in the study of electricity and magnetism. Given this convention, we should define the torque as being T = r x F where the cross product of the two vectors has a magnitude of rF*sin(theta) with theta being the angle between the two vectors and the direction of the cross-product is perpendicular to both r and F.
According to the textbooks in calculus and physics, the cross-product
represents the area of a parallelogram defined by sides which have the lengths
of r and F. To make use of the cross-product, we have to keep in
mind the method for specifying direction in three-dimensional space: with unit
vectors along the three orthogonal directions.
You can play around with an interactive, graphical picture of the cross-product by looking at a resource provided by Syracuse University, the vector cross-product tutorial. For our purposes, we only need to know a few things about the cross-product. First, if we define our axes so that two vectors, A and B lie in a plane (which we are always free to do), then if C = A X B, then C is perpendicular to the plane containing A and B. Thus if A and B lie along the plane containing the unit vectors i and j, then C lies only along k.
We also need to know that C is zero if A and B are
parallel or anti-parallel. One other thing we need to know is the time
derivative of a cross-product. For a general vector, r, the definition of
the derivative is exactly what you'd expect,
It is easy to prove that, for any two vectors dependent on time, u(t)
and v(t) that d/dt[u(t) X v(t)] = du(t)/dt X
v(t) + u(t) X dv(t)/dt. Therefore, if we want to have a
cross-product definition of the angular momentum L that matches our
definition of torque, we need L to be r X p where r is the
vector from the axis of rotation to the place where the force acts to produce a
torque or where the particle having momentum p is located. To see that
our choice for the definition of L is consistent with our definition of
torque, let's work through the definition.
| dL/dt | = d/dt[r(t) X p(t)] |
| = dr(t)/dt X p(t) + r(t) X dp(t)/dt | |
| = v(t) X p(t) + r(t) X F(t) | |
| = 0 + r(t) X F(t) ==> | |
| dL/dt | = r(t) X F(t) |
where we note that the velocity v and the momentum, mv must always be parallel so the cross product of velocity and momentum is always zero. In terms of our original definition of torque from the work-energy theorem, T = I*alpha (where the direction of alpha is defined the same way as the direction of the torque is defined), we have L = I*omega since we must define alpha = domega/dt to be consistent with our earlier definitions of angular velocity and angular acceleration.
The importance of the angular momentum comes into play when we attempt to do problems for which the net external force can not be usefully defined to be zero. A typical example is the following problem:
Solution: This problem can not be solved by any of our previous
methods of dynamics since, at the instant of collision, the hinge exerts a
force on the rod just as the particle imparts a force to it. If the rod is
rigid, then the application of this force from the hinge is instantaneous.
Momentum for the rod and particle is therefore not conserved since the net
external force on the system of rod and particle is not zero at the moment of
collision. Including the hinge does not help since we do not know exactly how
it transmits forces to the ceiling and therefore how the ceiling affects the
hinge (and the rod and particle through the hinge). The forces are impractical
to calculate so we are stuck. That is, until we notice that the external
torque on the system of rod and particle remains zero if we calculate
the torque about an axis that goes through the hinge. Since the unknown forces
from the ceiling and the rod go through this point, their lever arm is zero no
matter what their magnitude or direction. According to our definition of
angular momentum, if the external torque is zero, then dL/dt = 0 ==>
angular momentum is conserved. Hence, we can relate the angular momentum
before the collision to the angular momentum after the collision. Before the
collision, the angular momentum is defined by the angular momentum of the
particle and the rod relative to the axis of rotation through the hinge,
thus
| L0 | = Lm + LM |
| = r X p + 0 | |
| = mr X v0 |
since h is the distance of the particle from the axis of rotation through
the hinge at the time of impact. To be consistent with our definition of the
direction of omega, we need to relate the angular velocity and the
tangential velocity through the relationship, v0, perp = omega X r, i.e. only
the part of the velocity that is perpendicular to the position vector,
r, contributes to the angular velocity around the axis of rotation
(this is a generalization of our understanding of rotational velocity and
tangential velocity. See page 342 of the Lerner text). From the figure below,
we see that r*sin(phi) = h.
This is a constant that does not change as m
approaches the rod. Thus the angular momentum is constant as we expect before
the collision and has a magnitude which equals mv0h. Since there are no external torques
until the rod begins to swing upward after the collision, we can apply angular
momentum conservation by finding the angular momentum just after the
collision. Here the I*omega form of the angular momentum is more useful.
The rotational inertia after impact is (1/3)Mh2 + mh2. Since the initial and final angular
momenta must be equal, we have the magnitudes as
| L0 | = Lf |
| mv0h | = [(1/3)Mh2 + mh2]omega ==> |
| omega | = mv0/ (h[(1/3)M + m]) |
Since energy is conserved after the collision and the motion is purely
rotational, we can find the height to which the center of mass of the rod and
the particle rises. The center of mass for both is located at a distance
of
down from the hinge. The center-of-mass for the system rises to a height y
above the point d below the hinge where y is determined by energy conservation
as follows
| (M + m)gy | = ½ I*omega2 |
| = ½ [(1/3)Mh2 + mh2] [m2v02/ (h2[(1/3)M + m]) | |
| = m2v02/ (2[(1/3)M + m]) ==> | |
| y | = m2v02/ (2(M + m) [(1/3)M + m]) |
From our original picture, we see that, if the center-of-mass rises by distance y from an original distance d below the hinge, then cos(theta) = (d - y)/d ==> theta = arccos[(d-y)/d] where we have found the values of d and y in terms of masses, h, and the initial velocity.
Solution: Seems like a tough problem, but, considering the system of barbell and projectile, we note that there are no external forces acting during the collision, therefore, both angular momentum and linear momentum are conserved during the collision. Let's consider what each tells us. First, note that the center-of-mass (CofM) of the barbells is in the middle of the rod connecting them. Therefore, it makes sense to consider the rotation about an axis through the center-of-mass of the barbell for the elastic collision case. In this case, we can calculate the initial angular momentum of the system, assuming the projectile strikes the system at a distance l + R from the CofM of the barbell and strikes at a right angle to the rod, as L0 = mv0(l + R). Linear momentum conservation tells us that, for a coordinate system with one axis parallel to the projectile's initial velocity, v0, the momentum of the projectile and CofM of the barbell after an elastic collision can only be along this original direction (parallel or antiparallel) since there is no initial momentum along the perpendicular direction (i.e. along the direction of the rod) and linear momentum is conserved. Therefore, remembering that the projectile must come off the collision moving in the opposite direction,
| linear momentum: | p0 = pf ==> | mv0 = -mvf + 3MVf |
| K.E.: | K0 = Kf ==> | = ½ mv02 = ½ mvf2 + ½ (3M)Vf2 |
We do our usual trick of squaring the momentum equation, dividing that by m, and subtracting the kinetic energy equation from the result to get
We can use this in the original linear momentum equation to find the final speed of the CofM of the barbell as
| mv0 | = mvf + 3MVf ==> |
| = m(3M/m - 1)Vf + 3MVf ==> | |
| = (6M - m)Vf ==> | |
| Vf | = mv0/(6M - m) |
The velocity of the projectile is
Now angular momentum conservation gives us, assuming the final angular velocity of the barbell about an axis of rotation through the CofM is w:
| L0 | = Lf |
| mv0(l + R) | = -mvf(l + R) + I*w ==> |
| w | = mv0(l + R)( (vf+ vf)/I |
The rotational inertia of the barbell is derived using the parallel-axis theorem and the definition of rotational inertia
| I | = Ispheres + Irod |
| = 2[(2/5)MR2 + M(l/2 + R)2] + (1/12)Ml2 |
There's not much point to combining terms in the expression for I. We have our solution.
For the inelastic case, you should evaluate on your own using the method just described.
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larryg@upenn5.hep.upenn.edu