Monday, April 2, 2018
Poggendorff in 1853 designed an apparatus to demonstrate the changes in the tensions in Atwood's Machine that result from the acceleration of the masses. Mach gives a brief description in his Science of Mechanics. A compound Atwood's Machine is easier to analyze since the suspension points from the upper pulley don't change while the ends of the beam in Poggendorff's apparatus do. Only two parameters are required to specify the configuration of the compound machine, the rotation angles of the two pulleys or the arcs s ant t.
The positions of the masses are specified by the vertical coordinates, xi. From the balance of forces for each mass we obtain expressions relating the tensions and torques to the two accelerations, s̈ and ẗ.
One can then find the formulas for the individual tensions and by using the difference in tensions acting on the two pulleys we can derive a pair of linear equations for the unknown accelerations.
If we have cylindrical pulleys we can relate their moment of inertia to their masses as follows.
We are now in a position to compute the accelerations and tensions for the compound pulley by solving the system of linear equations.
All the tensions are reduced except for that of the first mass when the masses are accelerating.
Thursday, March 29, 2018
Consider a variation of Atwood's machine with a mass hanging from a string wrapped around a pulley.
The expressions for the height of the mass and balance of forces are the same as before. For the pulley we need to balance the torque, the product of tension and its distance from the center of the pulley, with the inertial force due to the acceleration of the pulley. It is assumed that the quantity of motion for the pulley is proportional to the rate of change of the angle θ or P=Iω=Iθ̇ where I is the moment of inertia. Note that the resistance to angular acceleration is directed upwards. This time the inertial factor, m+I/r2, includes the inertia of the pulley.
And the tension in the string is again less than the gravitational force acting on the mass m. Barton's Analytic Mechanics derives the formulas for Atwood's machine including the inertia of the pulley.
Wednesday, March 28, 2018
Atwood's machine gives a unique insight into forces acting on a body. One can find a description of it's use in Comstock's System of Natural Philosophy. The machine consists of two weights hanging from a pulley. One only needs one variable, s, to represent the position of the apparatus.
The value s is the distance along the perimeter of the pulley corresponding to its rotation through an angle θ. From the formulas for the positions we can express the acceleration of the masses and ignoring the inertia of the pulley we can look at the balance of forces acting on the masses due to gravity, the tension in the string and the "inertial force" associated with resistance of a mass to a change in its state of motion. After solving for the acceleration of the pulley we can obtain the tension in the string.
The balance forces acting on the pulley gives us the tension in the line holding it in position. The net force acting on the apparatus is the difference in gravitational force and an inertial factor equal to the sum of the masses and more generally including the moment of inertia of the pulley. The tension in the string depends on the distribution of the masses and is maximum when both masses are equal and the system is in equilibrium.
The units of the vertical scale are T/g and the horizontal scale is the ratio of one mass to their sum.
A difference in tension would be needed to accelerate the pulley so in general one cannot say that the tension in the string on both sides of the pulley would be the same.
Tuesday, March 20, 2018
Lagrange introduced the Method of Multipliers in his Analytical Mechanics of 1811 but when one first encounters this method it's not very clear why it works. One can deduce the procedure starting with a set of condition equations, Φ, and use least squares.
One ends up with a linear combination of the condition equations with arbitrary coefficients, the dΦ, set equal to zero. Division by dΦ1 removes some of the arbitrariness since the linear combination equals a constant, -Φ1.
Edit (Mar 20): Add dΦ and last sentence.
Monday, March 19, 2018
The gradient correction equation procedure is not free from error and can be forced to fail as seen in the following example.
The procedure works correctly for an initial rough estimate of x=2, y=6 for the zeroes of the Condition Equations Φ.
But the procedure fails along the x=y diagonal.
Saturday, March 17, 2018
This iterative correction for the fit of the parameters for the catenary using the least squares gradient formula works quite well too. The equations of condition for xmin and L are fairly smooth over a wide range as these grid plots show.
And the iteration of the corrections converges rapidly to the optimal values of xmin and L using an initial rough estimate of their values.
Using the formulas for the catenary we can now compute exact values for ymin and y(b).
Friday, March 16, 2018
Using smin and σ=A/(gλ) to rewrite the parametric equations for x and y one can eliminate the mass density λ and show that the curve is in fact a catenary.
Supplemental (Mar 16): One can also show that the tension in the string is,