Archive for March, 2009

Monday Math 65

March 30, 2009

Divisibility Tests Part 5: Tests for 13

Divisibility tests for 13 are also quite rare, however we can create a test analogous to our first for 7.

We note that 9*10+1=91=7*13, so with n=10a+b and q=a-9b, we see
so 9n+q is a multiple of 13, which means that n is divisible by 13 if and only if q is divisible by 13. This gives us the test: mulitply the ones digit of our number by 9, and subtract from the number formed by the rest of the digits. We see that this doesn’t work well for two-digit numbers, but a little better for three- or four-digit numbers.
For example:
so all of these numbers are divisible by 13.

As with 7, this procedure may be laborious for very large numbers. However, we can see that
This means 1000=13*77-1, and so we see our large-number divisibility test for 7 also works for 13.
Thus, we split our large number into sets of three digits, and then take the alternating sum of these numbers. For example:
so 348,970,129,352 is divisible by 13
149-2*9=131=13*10+1, which is not divisible by 13,
so 4,626,145,317,014,712 is not divisible by 13.


Physics Friday 65

March 27, 2009

Consider a loop of wire, forming a simple closed plane curve. We let the plane of the loop be the xy plane. Then let us parametrize the curve via (x(u),y(u)), 0≤uumaxwith counterclockwise orientation (so that the normal vector to the loop via the right-hand rule is in the positive z direction).

Suppose we then have a uniform magnetic field of magnitude B directed at an angle α from the positive z axis; we can choose our x axis so that is in the xz plane with positive x component.

Now, let us have a current I in the loop (positive I indicates counterclockwise current). What then, is the force on an element du at
(x(u),y(u)), and what is the net effect of this force on the loop?

The magnetic force due to field on a length of wire carrying a current I is . Thus, for an element du, the length vector is . We also have . Thus, we have force
To find the total force, we integrate over u:

(here we have used the fact that , as our curve is closed, and similarly for ).

So there is no net force on the loop. However, let us pick a point (xc,yc,0) in the plane of the loop, and find the torque about that point. For an element of the curve, the torque element is

Integrating this, and using again the fact that , we get

and similarly, , so we see that the x and z components of the torque are zero, and
. Note that the torque is independent of the plane point (xc,yc,0) chosen, so we can choose it to be the center of mass of the loop.

For the remaining integral in our torque, we note that
, and to this integral over our closed curve (c), we apply Green’s Theorem (see also here), which says
, where D is the plane region bound by the simple closed curve ∂D. Here, f(x,y)=0, g(x,y)=x, so
, and so
where A is the area enclosed by our loop. Our torque is thus:
Now, using the area vector , we see that , so
Noting that the magnetic moment of an object can be defined by the torque it experiences from an external magnetic field via
, we see that the magnetic moment of any planar current loop is
. Lastly, notice that the direction of our torque is such that the loop rotates to align the magnetic moment (and thus the area vector) with the magnetic field.

Monday Math 64

March 23, 2009

Divisibility Tests Part 4: Tests for 7

One generally does not see tests for divisibility by 7, as these aren’t as simple or quick as the more common tests. However, tests do exist.

First, let us consider two numbers a and b, and examine n=10a+b, q=a-2b. In particular,
, so 2n+q is a multiple of 7, which means that n is divisible by 7 if and only if q is divisible by 7. This gives us the following test: mulitply the ones digit of our number by 2, and subtract from the number formed by the rest of the digits.
For example:
so all of these numbers are divisible by 7.
For large numbers, we can repeat the test:
so 41,237 is divisible by 7 (41,237=7*5891)

Note that for very large numbers, the procedure may be laborious. However, we can develop a second test by noting that
This means 1000=7*143-1, and so:

and so on, so that
is divisible by 7 for all non-negative integers k. This creates a test by analogy to our divisibility test for 11.
Thus, we split our large number into sets of three digits, and then take the alternating sum of these numbers. For example:
so 348,967,129,356,874 is divisible by 7.
We can also combine the tests:
so 1,626,145,217,013,712 is divisible by 7


March 21, 2009

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Physics Friday 64

March 20, 2009

[part 3 of 3]

In part 2, we showed how if the distant point mass M orbits our oblate spheroid (or our ellipsoid orbits it) in a circular orbit, with an orbital frequency Ω much faster than the torque produced, one may time-average the torque to get a result where the rotation axis of our ellipsoid, the z’ axis of the body frame, maintains the same angle with, and preccesses about, the axis perpendicular to the plane of the orbit, the z axis of the inertial frame. In particular, we found that
, where is the angular frequency of the precession, ωz is the angular velocity of the ellipsoid’s rotation, θ0 is the angle between z’ and z, and d is the distance between our ellipsoid and the mass M.

So far we’ve kept all this in general, abstract terms. Now we will move this to a concrete example: the shape of the planet Earth may be approximated as an oblate spheroid of equatorial radius a=6378.1 km and polar radius c=6356.8. More specifically, if we use either pair of values given here for the moment of inertia of the earth, we get a value for of about 0.00323.

The other major parameters for the Earth are θ0 and ωz. The tilt of the earth’s axis, θ0, is approximately 23.44°=0.4091 radians, so that cosθ0=0.9175. The angular frequency of the earth’s rotation about its axis, ωz is found to be 7.2921×10-5 s-1, using as rotational period one sidereal day.

Now, we consider first the effect of the sun. The mass of the sun is 1.99892×1030 kg, so GM for the sun is 1.3272×1020 m3 s-2. The average distance between the earth and the sun is d=1.496×1011 m. Plugging these into our precession formula,
gives a precession frequency of about 2.417×10-12 s-1, which corresponds to a period of 2.600×1012 s, or about 82,000 years.

Now, for the moon. The mass of the moon is is 7.3477×1022 kg, so GM for the moon is 4.903×1012 m3 s-2. The average earth-moon distance is 3.844×108 m. Ignoring the tilt of the moon’s orbit relative to the ecliptic, we can use our precession formula as before, giving a precession frequency of about 5.262×10-12 s-1, which corresponds to a period of 1.194×1012 s, or about 38,000 years.

As noted before, we have made many simplifying assumptions and approximations; the actual rate of the precession of Earth’s axis, historically called the “precession of the equinoxes,” has a period of about 25,700 years. Further, there are added complications, such as nutation, due to factors we have ignored (such as eccentricity of the orbits of the moon and Earth, the tilt of the moon’s orbit relative to the ecliptic, the time-varying nature of the torque, and so on). See here for a deeper treatment.

Monday Math 63

March 16, 2009

Divisibility tests Part 3: 11

Last time, we proved the tests for divisibility by 3 and 9. Now, for 11, we again use the factoring rule . We also use the rule that if n is odd.
The first tells us , and so is divisible by 11 (as 99=11*9) for even k.
The second tells us that
So is divisible by 11 for all non-negative integers k.
So we consider the alternating sum . The difference between n and a is then
, and as each term of the series is divisible by 11, the difference is a multiple of 11. Thus, n is divisible by 11 if and only if a is a multiple of 11.

An few examples of this test:
n=297: 7-9+2=0=0*11, so 297 is divisible by 11 (297=27*11)
9-2+5-2+1=11 (12,529=1139*11)
6-3+8-6+9-5+4=13, so 4,596,836 is not divisible by 11.
0-9+6-9+4-9+4-9=-22=-2*11, so 94,949,690 is divisible by 11 (94,949,690=8,631,790*11).

Physics Friday 63

March 13, 2009

[Part 2 of ?]

In the previous part, we introduced a spinning oblate spheroid, and showed that a distant point mass M at displacement d will exert a torque on the spheroid that can be approximated as
. Now, suppose that d is in the xy-plane of the inertial frame. Further, suppose that our point mass is orbiting our spheroid in a circular orbit of angular frequency Ω (or that our spheroid is orbiting our point mass in a circular orbit of angular frequency Ω; the model will turn out the same). Then in the inertial frame. Suppose then at a time t the x and x’ axes coincide. Then, if we let the angle between z and z’ be θ0, we see that d has components in the body coordinates of .

Now, if the density of our spheroid is sufficiently symmetric about its axis, then we will have Ix’=Iy’, and the moment of inertia tensor in the body coordinates will be . Using our results from part one, we find the torque in this situation;
Now, we presently have and , so we can rewrite the above in a way independent or our choice of x’ and y’ axes:
giving torque:

Supposing that this torque is small enough that any precession produced is of frequency much slower than Ω, we can then average the torque over time; recalling the time average of trigonometric functions and their products, we get average torque

Recalling that our object has angular momentum along the z’ axis, we see then that our average torque is perpendicular to our angular momentum, and to our z axis (as it is along the cross product ). Thus, as in here, we have precession of our spheroid’s rotation about the z axis (so θ0 is constant), and from our previous work on torque-driven precession, we see that the precession has angular frequency
(The negative sign indicates that the direction here is opposite the sense of the rotation ωz.)
Now, supposing our spheroid has total mass ME, then Kepler’s third law for our spheroid-point mass orbit tells us that the period T of the orbit is . Since , we find , which lets us rewrite the precession frequency in terms of the orbital frequency and the ratio of the masses :

Monday Math 62

March 9, 2009

Divisibility Tests
Part 2: 3 and 9

One may know that the tests for divisibility by 3 and 9 involve adding the digits of a number: a number is divisible by 3 if and only if the sum of its digits is divisible by 3, and divisible by 9 if and only if the sum of its digits is divisible by 9.
The key is to recall the factoring rule , used to prove the geometric series formulas. This means that . (This just means 9, 99, 999, 9999, etc. are all multiples of 9, which should be quite obvious).
Let us consider an integer n. Let a0 be the ones digit, a1 the tens digit, a2 the hundreds digit, and so on, with am the highest digit, so that n is an (m+1)-digit number. This means
. The sum of the digits, in turn, is . The difference between these two is
As this is a multiple of 9, we see that n is divisible by 9 if and only if s is divisible by 9, and as a multiple of 9 is also a multiple of 3, n is divisible by 3 if and only if s is divisible by 3.


March 8, 2009

This is hilarious:

Physics Friday 62

March 6, 2009

[Part 1 of ?]

Let us consider a rotating oblate spheroid with equatorial radius a and polar semi-axis c. We choose an inertial frame (x,y,z) and body coordinate system (x’,y’,z’), both with origin at the ellipsoid’s center of mass. Suppose we have a rotation about a principal axis with angular momentum , where Iz’ is the moment of inertia along the z’ principal axis and is the z’ unit vector).

Now, let us put a point mass M at a displacement d from the center of our ellipsoid. Then a mass element of our ellipsoid at position r ( is the density at that point in the object, not necessarily uniform). experiences a gravitational force due to the mass M of

Integrating the torque element due to this force over the volume V of the ellipsoid gives a total torque of

Now, suppose our point mass is far from our body, compared to its size, so that , where and . Now using
, we see that approximating to first order in r/d,
, so we can in this situation approximate our torque by:

Now, to find the first integral on the right, we note that
For a an object in volumeV of total mass m and density , the center of mass has position (vector)

(see here); as our origin is the center of mass, we thus see that the integral , and the first integral in our approximation is the zero vector, and

Consider the moment of inertia tensior I of our object (see here):
, where E3 is the 3×3 identity matrix and is the outer product of r with itself. Applying the tensor to d, we use and to get
Next, consider the cross product of d and this vector:
which is our remaining integral in the torque approximation: