Homework 6 - APPM 2350 HOMEWORK 06 Due Thursday, October 7,...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: APPM 2350 HOMEWORK 06 Due Thursday, October 7, 2010 at the start of your recitation FALL 2010 1. You know the routine for the first problem. 2. If you have not finished problem 1, go back and finish it. 3. Suppose you have just washed your favorite 100% cotton turtleneck sweater and have placed it in the dryer. At a particular time t0 , the neck of the sweater is 18 inches in circumference, while the height of the neck is 3 inches. Sadly, you have forgotten that you are supposed to air dry the sweater. At time t0 , the circumference of the neck is shrinking at a rate of 0.2 in/min, while the twisting and tumbling action of the dryer is causing the neck to lengthen at a rate of 0.1 in/min. How is the volume of the space inside the neck changing at time t = t0 ? Is the volume increasing or decreasing? 4. Suppose you have two particles with positions x1 (t) and x2 (t) and velocities x1 (t) and x2 (t). They ˙ ˙ have total kinetic energy m1 m2 T= (x1 (t))2 + ˙ (x2 (t))2 , ˙ (1) 2 2 where mj denotes the mass of the j th particle, which we assume is constant with respect to time. We further assume the particles are interacting in a potential V depending only upon their separation. Thus potential energy is given by V = V ( | x 1 ( t ) − x 2 ( t) | ) . (2) Common potentials like k , (3) | x1 − x2 | where depending on our choice of k we are talking about a gravitational or electric potential, are of this form. Now, we introduce a very special function called the Lagrangian, which is V ( | x1 − x2 | ) = L ( x 1 , x2 , x 1 , x 2 ) = T ( x 1 , x 2 ) − V ( | x 1 − x 2 | ) . ˙˙ ˙˙ ∂L d ∂L − =0 ∂ xj dt ∂ xj ˙ for j = 1, 2. What makes this equation special is that it gives us back Newton’s second law Fj = mj xj , ¨ where Fj is the force acting on particle xj . (a) Using Lagrange’s equation (5), find F1 and F2 . The point here is that the Lagrangian gives us a way to describe our two particle system that is the same as if we had used Newton’s second law, except now we are only working with energies. Another energy based approach in physics is to work with what are called Hamiltonians. (b) Using Lagrange’s equation (5), show that 2 d ￿ ∂L x j ( t) ˙ − L(x1 (t), x2 (t), x1 (t), x2 (t)) = 0 . ˙ ˙ dt ∂ xj ˙ j =1 (4) The Lagrangian, L, satisfies a very particular equation called the Euler-Lagrange equation, which is (5) (6) (7) 2 ￿ j =1 This means we can define a constant with respect to time, say H , such that H = xj ˙ ∂L − L. ∂ xj ˙ (c) Now, suppose we define the momentum of particle j to be pj (t) = mj xj (t) . ˙ ∂L ∂L , and pj = ˙ . ∂ xj ˙ ∂ xj (d) Also, we can define the Hamiltonian, H , as Show that pj = H (x1 , x2 , p1 , p2 ) = p1 x1 + p2 x2 − L(x1 , x2 , x1 , x2 ) . ˙ ˙ ˙˙ (9) (8) Rewrite the right hand side of (9) in terms of only xj and pj using your results from problem dH (c). Express H in terms of the kinetic energy T and potential energy V . If = 0, what have dt you shown? You should now see that the Hamiltonian is the total energy while the Lagrangian was the difference between kinetic and potential energy. The last issue though is if Lagrange’s equation (5) could get us back to Newton’s second law, is there a system of equations for the Hamiltonian that can get us to Newton’s second law? The answer is yes, and the equations are called Hamilton’s equations of motion. They are xj ˙ pj ˙ for j = 1, 2. (e) Using the definition of momentum, pj , your Hamiltonian H (x1 , x2 , p1 , p2 ), and your results from problem (c), prove that Hamilton’s equations of motions are true. Have you gotten Newton’s second law back? Explain why. ∂H ∂ pj ∂H =− ∂ xj = (10) (11) ...
View Full Document

Ask a homework question - tutors are online