Centripetal e.g.
when a mass on the end of a string is swung in a circle, the
inward centripetal acceleration causes an outward centrifugal force on the mass which
causes a matching tensile force in the string. Of course, it is the
tensile force of the string which causes the curvature of the
trajectory in the first place, and thus, the centripetal acceleration. (Attempts to unravel
causality often lead to this The equation for centripetal acceleration can be derived from the following vector diagram. Let R1 and V1 be the radius and velocity vectors at some time, t1. (V1 is a free vector, i.e. it may be drawn anywhere as long as its direction and magnitude is preserved.) Assume a circular trajectory with constant tangential velocity and radius of curvature. Some short duration of time, dt, later, the radius and velocity vectors have the directions of R2 and V2. Since R1 equals R2 and V1 equals V2, the two triangles are isosceles. Since V1 is perpendicular to R1, and V2 is perpendicular to R2, the acute angles of both triangles are equal. Since the two triangles are similar, the ratios of corresponding sides are equal. The subscripts have been dropped since the magnitudes of the vectors are not changing. The above equation can be rearranged: Note that if the velocity vector triangle were multiplied by the scalar, mass, it would become a momentum vector triangle. The centripetal acceleration can also be expressed as a function of the rate of change in direction. Letting d-Theta represent the small angle (in radians) that the radius vector travels through in differential time, dt. The angular velocity is represented by Omega. Squaring: Multiplying both sides by R yields the two equivalent expressions for centripetal acceleration. The above expressions for centripetal acceleration are adequate for circular trajectories with constant velocity. ## Aerodynamic lift from centripetal accelerationExplaining aerodynamic lift to students, whether aeronautical engineers or pilots, presents a dilemma for instructors. All of the popular explanations invoke at some point the Bernoulli equation. or, For an idealized gas flow with constant density, (incompressible), no friction, (inviscid), and no heat transfer, (adiabatic) the total energy density must remain constant, since no work is being done on or by the gas, and no heat is being added or lost. Total energy density is the sum of the random internal kinetic energy density and the kinetic energy density of the flow, known as dynamic pressure, q. Total energy density is: or, Giving the correct Bernoulli equation as: Another problem with Bernoulli is that, even if we attribute Consider the flow above the wing. The downward centripetal acceleration of the airflow at some height, h, above the wing is: Where the radius of curvature is a function of camber, (R For a given combination of camber and angle of attack the radius of curvature is a complex function of height above the wing. The vertical pressure gradient at each height is equal to the density of the air times the downward acceleration: The above pressure gradient is analogous to the tension in a string attached to a swinging mass - it is the pressure gradient required to maintain the flow curvature. You can think of the pressure gradient as causing the flow curvature, or the flow curvature as causing the pressure gradient through centripetal acceleration - ishkabibble. The two effects are in equilibrium. The upward centrifugal pressure gradient is balanced by the product of density and downward centripetal acceleration. Since the curvature of the flow is convex on top, the pressure gradient must be positive, i.e. increasing pressure with increasing height. At some height above the wing, the radius becomes infinite for all practical purposes, and the pressure is equal to ambient. (The flow becomes straight.) At the top surface of the wing, the total pressure drop from ambient becomes: The same argument can be made for the flow beneath the wing, The radius of curvature function will be different, however, there will still be some finite distance from the wing at which the influence of the wing is negligible, the flow becomes straight, and the pressure is equal to ambient. For the bottom surface, the equation for the total pressure difference from ambient has the same form as above except for the sign, i.e. the pressure at the wing is greater than ambient. The total pressure difference between the bottom of the wing and the top of the wing is: We have insufficient knowledge of the radius of curvature distributions to allow solving the non-dimensional integrals inside the brackets; instead we rely on wind-tunnel test data. The dimensionless value of the bracketed expression is, by convention, evaluated as one half the coefficient of lift. Total lift on a wing of area, S is: The radius of flow curvature, R, decreases with increasing camber or increasing angle of attack, which explains why increasing camber (flaps) increases lift for a given angle of attack and why increasing angle of attack increases lift for a given camber. ## Centrifugal force on the hydrogen electronModern
physics has evolved its explanation of the atomic structure well beyond
the
simplistic and naive idea of classical electron orbits. It is of
historical interest, however, to occasionally revisit this original
intuitive model, since many of its results are still valid today. The
inward Coulomb force of attraction to the proton nucleus is inversely
proportional to radius squared. From the definition of the classical
electron radius, r _{e}, it can be written as:Setting these two forces in equilibrium: (Divide both forces by m _{e}c^{2}, and multiply by r.)One
of the great mysteries of quantum physics is the experimental evidence
that the ratio v/c of the hydrogen electron is always an integer
fraction of the constant alpha, originally known as Sommerfeld's
constant. (Approximately 1/137) At N=1, (The ground state), the radius defines the Bohr radius. Some down to Earth examples of centrifugal force. ## How much lighter is an object at the Equator Vs. a pole?Only the centripetal acceleration of a point on the Equator must be considered. The
determination of the Earth's circumference facilitates the
determination of both the tangential velocity at the Equator and the
radius of the Earth. ^{8} ft.Tangential velocity of the Equator is: The radius of the Earth at the equator is: Centripetal acceleration at the Equator is: The centrifugal force acts outward (up), partially canceling gravitational weight. (As a point of interest, the above ratio is approximately the same as the flattening ratio of the Earth - 0.003353.) A fully loaded 747-400 (875,000lb) on the ground in Singapore weighs approximately 3,000lb less than it would in Thule Greenland (N76.5 ^{o}) with the same mass. (Both centrifugal force and its vertical
component decrease with the cosine of latitude. The vertical
component of centrifugal force, therefore, decreases by the cosine of
latitude squared.)## How much lighter is an airplane flying east at 600 knots over the Equator?This is a problem of the third kind. The aircraft is flying along an arc which has the same radius of curvature as the Earth. The total centripetal acceleration can be found by adding the aircraft's ground speed to the tangential velocity of Earth. The total tangential velocity of the aircraft is, therefore, 2535 ft/sec. Total centripetal acceleration is: The fractional reduction in weight is: A 747-400 at maximum gross weight, (875,000lb), flying east over the equator at 600 knots (ground speed) is approximately four tons lighter than its one-g gravitational weight. ## What is the altitude of a geostationary satellite? For a satellite to remain stationary relative to the surface of the earth, it must be in a circular equatorial orbit with an angular rate matching that of the earth. The centripetal acceleration must also equal the acceleration of gravity at the orbit radius. Since the acceleration of gravity is inversely proportional to radius-squared, the product of radius-squared and acceleration of gravity is constant. Using the known equatorial radius of the earth, R Substituting from the first equation for g: Now it's just a matter of plugging in the following known quantities on the rhs. The radius of the earth at the Equator:
The acceleration of gravity at the Equator: The rotation rate of the earth relative to the inertial reference frame: This is the radius of the geosynchronous orbit from the center of the earth. For the altitude, we must subtract the radius of the earth. ## Will PANSTARRS C/2011 L4 ever return? If only the gravitational field of the Sun is considered, the answer appears to be no, since the velocity of the comet at perihelion exceeded the escape velocity of the Sun. The velocity at perihelion has been reported to be 172,000 mph. The specific kinetic energy (kinetic energy per unit mass) is merely one half velocity squared. Dimensioning specific energy as velocity squared is perfectly valid regardless of the units of distance and time. For example, we could choose an arbitrary unit of mass as one earth mass. We might then define some force (I'll call it "sum" - short for "some force".) as the force required to accelerate one earth mass at the rate of one astronomical unit per year per year. A natural unit of energy would then be "au-sum". (Note that the above equation has the equivalent dimension of "au-sum per earth mass".) Specific potential energy can also be expressed in au ^{2}/yr^{2}.Since Earth's orbit is nearly circular, the Sun's gravitational acceleration at 1au can be estimated by setting it equal to the centripetal acceleration of the earth. ( V ^{2}/R would yield the same result.)The gravitational acceleration is also equal to the product of the universal gravitational constant, G, and the mass of the Sun, M _{s}, divided by radius squared.Therefore, the product , GM _{s}, can be defined as:Gravitational force per unit mass is equal to gravitational acceleration. With radius in astronomical units: The specific potential energy of the sun at any radius can be found by integrating the force per unit mass from the radius to infinity. Since it takes positive work to completely remove a mass, the potential energy is considered negative. The comets radius from the Sun at perihelion was: Since the total energy per unit mass is greater than zero, the kinetic energy is sufficient to escape the gravitational field of the Sun. This agrees with the JPL data which lists the eccentricity of the comet's trajectory as greater than one. (An eccentricity of zero is a circle, less than one an ellipse, one a parabola, and greater than one a hyperbola.) Since the Sun contains over 99.8% of the entire mass of the solar system, it appears that the comet would escape even if the entire mass of the solar system were concentrated in the Sun. (Dividing the specific potential energy by 0.998 gives a potential energy for the entire solar system of -131.15au ^{2}/yr^{3}.) |