Class #10 Energy Applications Rolling down a ramp Pendulum Simple Solid Potential wells 2nd derivative as a spring constant

L8-2 Energy problems - Rolling h x O x A hoop and a cylinder of equal radius “R” and mass M roll down equivalent ramps What is velocity “v” at ramp bottom in each case? Which shape “wins” the ramp race.

Simple and Solid Pendula Approach is same for solid pendulum If replace z with z of CM and If replace with appropriate moment of inertia

L10-1 Solid Pendulum L m1 L is distance from pivot to CM of m1 R is radius of spherical pendulum bob. What is correction to ordinary pendulum frequency if R=L/2?

Math Physics ODE Summary

Analogy of 1-D system to roller coaster x K(x)=E-U(x) <- General 1-D system K(x)=E-mgx <- Roller Coaster

Potential Wells K < 0 K > 0 Mass m Spring constant k

Series Expansion of potential around a critical point Taylor series -- Generic Taylor series -- Potential Can be ignored or set to zero … “gauge invariance” Is already zero for potential evaluated about a critical point

L10-2 Potential Wells I’ve fitted equivalent parabolas to this function. Roots are 1.3637, -0.8863 Value of U 8.3, 26.2 2nd derivs are 34.5869, -30.6684 Solid state --- Semiclassical … Bandgap What is equation of parabola of equivalent curvature? What is resonant frequency of a 1 kg mass operating in this potential?

L 10-3 Potential Wells Calculate the Equilibrium separation of two water molecules Calculate the vibration frequency assuming M=18 amu 1 amu=

Taylor 7-50 A mass m1 rests on a frictionless horizontal table. Attached to it is a string which runs horizontally to the edge of the table, where it passes over a frictionless small pulley and down to where it supports a mass m2. Use as coordinates x and y, the distances of m1 and m2 from the pulley. Find the tension forces on the two masses. Now repeat including friction on m2 The df d whatever gives the direction of the force. If lambda is positive and df/dq is, then constraint is in direction of increasing q. m2 m1

Atwood’s Machine Let’s talk about directions of forces. Because g is assumed to be down, so Force of gravity is down, thus U=mgy. Df/dy is positive. It wouldn’t have to be. Thus direction of y enters into constraint as well. If direction of y2 were different than direction of y1 then force would be different. m1 m2