Posts Tagged ‘Torque’

Friday Physics 146

December 10, 2010

Consider a (uniform) rod of mass m and length l leaning against a wall at an angle θ from the horizontal; both the wall and floor are made of the same material, with the same coefficient of static friction μ. What is the minimum value of μ for the rod to remain static, without slipping down?

Drawing a diagram, we see that there are five forces:

  • Nw, the normal force exerted by the wall on the upper end of the rod
  • Nf, the normal force exerted by the floor on the lower end of the rod
  • fw, the friction force exerted by the wall on the upper end of the rod
  • ff, the friction force exerted by the floor on the lower end of the rod
  • W=mg, the weight of the rod, acting on the center of the rod

[Fig. 1]
To have equilibrium we need forces and torques to cancel.
Cancellation of forces in the horizontal direction gives us Nw=ff, and cancellation in the vertical gives us fw+Nf=mg.
Now, for torque, we consider the torque about the lower end. The torque exerted by the weight is thus , while the forces exerted by the wall produce torques in the opposite direction of
and .
Equating, we have
and multiplying by , we get

and combining with the vertical equation,
Now, by the definition of coefficient of static friction, we have
and . Thus, we have the minimum value when these ratios, and are equal; thus, the relation between them gives us
Solving for positive μ,

(Here, we use that for 0<θ<π/2, the secant of θ is positive, and so .)
Note that μ goes to zero as θ approaches π/2, and increases with decreasing angle. Thus, leaning the rod in a near-vertical position requires less friction, and so is easier, than leaning in a shallower angle; as one should expect from practical experience.


Physics Friday 133

August 27, 2010

Part 13: Electromagnetic FIelds and Angular Momentum

Last week, we considered momentum conservation in electromagnetism, finding that the momentum density is equal to , where is the Poynting vector; and that the momentum flux density tensor is , where is the Maxwell stress tensor, a symmetric rank-two tensor with components . Today, we consider angular momentum.
Recalling that for a particle of momentum p at displacement x from the origin, the angular momentum about the origin is
, and the force on a particle of charge q is , so the torque about the origin on the particle is
Converting to charge and current densities, if is the total (mechanical) angular momentum of the charges in a volume V, then
However, we showed last week that
where is the divergence of the Maxwell stress tensor , which is a vector with ith component .
Thus, we plug this in to get:
Note that since , ,
and the order can be interchanged between time derivatives and volume integrals, so the above becomes:
and so we see is the angular momentum of the fields in the volume, and so is the angular momentum density of the electromagnetic fields.
Now, let us consider the flux term. The ith component of the cross product of vectors a and b is , where is the Levi-Civita symbol. Thus, the ith component of is
Now, using the product rule, , and , so
and so
since due to the symmetry of and the antisymmetry of the Levi-Civita symbol (exchange the labels on the dummy indices j and k, then reverse orders on , with no sign change, and on , with sign change, to find that the sum is its own opposite, and therefore zero).
Note that the cross product of vectors a and b is the vector with ith component . Similarly, the cross product of the vector a with the tensor is the tensor with components
In this vein, we see then that , and therefore, that
the ith component of the divergence of the tensor (or, more accurately, pseudotensor) . Thus, since the corresponding components are equal,
and so
giving us our integral continuity equation. Just as is the linear momentum flux density tensor of electromagnetic fields, so we see that is the angular momentum flux density tensor of electromagnetic fields. Thus, the differential form for our continuity equation is
And just as one may find the total force the fields exert on the charges in a volume by integrating the total force via , one can do the same with to find the total torque.

Physics Friday 100: Another Classic

December 4, 2009

Consider a spool of string with radius r, and ends capped by disks of radius R>r. The spool is placed on its side on a horizontal, so that it can roll on end rims. Suppose one pulls on an unwound end of the string, at some angle θ from the underside horizontal, as pictured below.

It’s clear from basic physical intuition that if θ=90°, so that one pulls straight up, the spool will tend to unwind, rolling to the left in our picture.

Similarly, if we pull horizontally, θ=0°, then the spool will roll to the right, and will wind up the string.

So, then, somewhere between these two angles should be a transition between these behaviors, a critical angle θc where the spool drags without rotating, neither winding nor unwinding. What is this angle?

Physics Friday 97

November 13, 2009

Suppose we have a cube, of edge length L, with mass M symmetrically distributed so that the geometric center of the cube is the center of mass. Suppose we have a fluid with density ρ, such that (the average density of the cube is less than that of the fluid), so that the cube will float in the fluid. If the cube is floating such that the upper face is entirely outside the fluid, and is tipped toward one edge by an angle θ from the horizontal, then what is the torque on the cube about its center?

Physics Friday 96

November 6, 2009

Suppose, as in this post, we have a fluid of density ρ, with some arbitrary three-dimensional object immersed in the fluid, occupying a volume V. This object has surface ∂V. We saw that for any point on the surface, the pressure is , and the force on an area element is:
. Thus, the torque on that element is
where r is the coordinate vector to the element. Recalling that for vectors v and w and scalar a,
, so since P(x,y,z) is a scalar, we see

and so the total torque (about the origin) is
and one form of the divergence theorem tells us that for vector field A,

Now, the product rule for the curl of the product of a scalar field ψ and a vector field a is
Now, applying that to , and noting that (as it is a radial vector field), and , we see
Now, since is a constant vector, it can be “factored out” of the integral:
We recall that the total buoyant force on the object is ; plugging this into the above, we see

Now, recall that the center of mass of a region with density function ρ(x,y,z) is . If the density is a constant, it factors out of the integrals, and one gets . Thus, if we consider the volume as if it were filled with the fluid; that is to say, the volume of fluid displaced, its center of mass would be , and then we see that
which is equivalent to the torque if the buoyant force acted entirely on the point rb; this point is called the center of buoyancy. Just as the force of gravity on an extended object can be treated as if it acts entirely on the center of mass, the buoyant force can be treated as if it acts entirely on the center of buoyancy.

Physics Friday 66

April 3, 2009

Last time, we showed that a plane loop carrying current I placed in a uniform magnetic field experiences a torque , where is the magnetic moment of the loop, and is the area vector for the loop, as given using the right-hand rule.

Now, let us consider an electron moving in a circular orbit of radius r and angular velocity ω0. Then the period of the orbit is , and so the charge passing through any point on the orbit per unit time is , and so we can treat the circular orbit as a current loop of current (the minus sign indicates that the current is opposite in direction to the motion of the electron, as the electron has a negative charge). Thus, the orbit has a magnetic moment, .

Unlike the previous case of a current loop, here we also have to consider the angular momentum of our electron’s orbit as well. As our orbit is circular, the angular momentum is just , where me is the mass of the electron. Thus, we can rewrite the magnetic moment in terms of the angular momentum:
and the magnetic moment is proportional to the angular momentum of the orbit, with the constant of proportionality dependent only on the properties of the electron.

Note here that since the magnetic moment is proportional to the angular momentum, the torque due to an external magnetic field, is perpendicular to both and . Thus, as seen in previous work, the orbit, if not perpendicular to the magnetic field, will precess about the field. This is an example of Larmor precession, which occurs whenever there is a magnetic moment proportional to angular momentum exposed to an external magnetic field. The frequency of this precession, the Larmor frequency, for this problem is . In the more general form, the constant of proportionality between the magnetic moment and angular momentum is called the gyromagnetic ratio (or sometimes the magnetogyric ratio), and usually denoted by γ: . Then the Larmor frequency is . These frequencies, when applied to a charged particle with spin, are important in spin transitions, and play an important role in systems such as nuclear magnetic resonance

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.

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.

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 :

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: