#### Transcript Lecture 24 - McMaster University

Energy • • • Potential energy Examples with rotation Force and Potential Energy Serway 8.4-8.6; 10.8 Physics 1D03 - Lecture 24 Mechanical Energy E = K + U = K + Ugravity + Uspring + ... Mechanical energy is conserved by “conservative” forces; the total mechanical energy does not change if only conservative forces do work. Remember, for a conservative system: ΔK+ΔU=0 Physics 1D03 - Lecture 24 Non-Conservative Forces Non-conservative forces cause the mechanical energy of the system to change. Divide work W into work by conservative forces and work by “other” forces.: DK = ΣW = Wc + Wother Wc can be replaced by potential energy terms: Wc = − DU , so and, since E=K+U, DK = - DU + Wother DE = Wother The change in mechanical energy is equal to the work by “other” forces. “Other” means any force not represented by a term in the potential energy. It includes non-conservative forces, but also externallyapplied forces, conservative or not, that transfer energy into or out of the system. Physics 1D03 - Lecture 24 Conservation of Energy Other types of energy: -electrostatic P.E. -chemical P.E. -nuclear P.E. -etc., etc., and thermal energy (actually just kinetic and potential energy at a microscopic scale) Then: The total energy of the universe is conserved. Physics 1D03 - Lecture 24 Rotation about a fixed axis Kinetic Energy: Work: Power: K = ½ Iw 2 dW d (infinitesimal rotation) W D (for a constant torque) P w Physics 1D03 - Lecture 24 Example: Pole vault Jumping animals, from a frog (10-2 kg) to a horse (103 kg) can all jump about 1 or 2 metres vertically. Why can a pole-vaulter jump 6 m? Guess: Initial kinetic energy built up during several seconds of effort can be converted to elastic and then gravitational potential energy. The pole-vaulter doesn’t need to get all his mechanical energy from a single muscle contraction. Kinetic energy Elastic potential energy Gravitational potential energy Physics 1D03 - Lecture 24 To test our guess, see if the numbers fit: If he runs at 9 m/s, and can convert all his kinetic energy into gravitational potential energy, how high would he get? Ki = (1/2) mv2 = Uf = mgy so y = v2/2g = 4.1 m for v = 9 m/s ( for v = 10 m/s, y = 5.1 m) This is enough to explain a 6-m vault. His mechanical energy actually increases as the pole straightens, as his muscles (e.g., arms) do work. Physics 1D03 - Lecture 24 Quiz: Bungee cord m A mass attached to a 50-m elastic cord is dropped. The mass falls 80 m (stretching the cord 30 m) before the block stops momentarily. How far has it fallen when it reaches its maximum speed ? 50 m a) 50 m b) less than 50 m c) more than 50 m 30 m m Physics 1D03 - Lecture 24 Quiz A thin (uniform) pole of length L and mass m, initially vertical and stationary, is allowed to fall over. Assuming it falls as if hinged at the base, what is its angular velocity just before it hits the ground? How much potential energy does it lose as it falls? A) mgL B) 2mgL C) mgL/2 Physics 1D03 - Lecture 24 Gravitational potential energy depends on the height of the center of mass. As the stick falls, gravitational potential energy is converted to kinetic energy of rotation; the total mechanical energy is constant. Initial: yCM = L/2, U = mgL/2, K = 0 so E = mgL/2 CM L/2 Final: yCM = 0, U =0, so E = K = ½ I w 2 Initial energy = final energy: ½ I w 2 = mgL/2 For a uniform thin rod, I = 1/ mL2, 3 3g so (after a little work), w L Physics 1D03 - Lecture 24 Questions: What is the (linear) speed of the tip and of the centre of mass? How do these compare with a particle that falls a distance L/2? Linear velocities: v = rw At the tip: r = L, g vtip wL 3 L 3gL L At the CM: r = L/2, vCM L 1 w 2 3gL 2 -------------------------------------------------------------------------------For a particle falling from a height L/2, mgL/2 = ½ m v2, and v gL (faster than the CM of the rotating stick). Physics 1D03 - Lecture 24 Example The end of the string is glued to the rim of the pulley. After the mass reaches its lowest point, the pulley continues to turn, and raises the mass back to ¾ of its original height. What is the frictional torque at the axle of the pulley? R m Physics 1D03 - Lecture 24 Initial Energy = Final Energy + Work Done There is no kinetic energy at (1) and (2) MgL 0 W 4 W 14 MgL (1) R (2) y 0 For the full problem. . . frictional torque W f total y 1 4 L total angle rotation total f L 34 L 7L R 4R W total 1 4 MgL MgR L 7 7 4 R y L Physics 1D03 - Lecture 24 Forces and Potential Energy One dimension, motion along the x axis: Given U(x), find F. For a small displacement Dx, the work done is W = Fx Dx If the force is conservative, then W DU as well. So, DU Fx Dx , or dU Fx dx (Extra) Similarly, dU = -dW = -Fx dx, so x2 U ( x2 ) U ( x1 ) Fx ( x)dx x1 Physics 1D03 - Lecture 24 Quiz U The potential energy of a particle, acted on by conservative forces only, is shown as a function of position. At which of the points on the graph is the net force on the particle: i) A E B x D C zero? – QUIZ (A,B,C,D,E) ii) in the positive x direction? iii) a maximum? Physics 1D03 - Lecture 24