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Class #10 Energy Applications Rolling down a ramp Pendulum

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Presentation on theme: "Class #10 Energy Applications Rolling down a ramp Pendulum"— Presentation transcript:

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

2 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.

3 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

4 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?

5 Math Physics ODE Summary

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

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

8 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

9 L10-2 Potential Wells I’ve fitted equivalent parabolas to this function. Roots are , Value of U 8.3, 26.2 2nd derivs are , 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?

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

11 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

12 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


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