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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