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This announcement provides information about a physics lecture covering chapters 9-14 of the textbook. It also includes details about an upcoming exam and a physics colloquium.
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Announcements • Remember -- Tuesday, Oct. 28th, 9:30 AM – Second exam (covering Chapters 9-14 of HRW) – Bring the following: • 1 equation sheet • Calculator • Pencil • Clear head • Note: If you have kept up with your HW, you may drop your lowest exam grade • Today --Thursday, Oct. 23th, 4 PM – Physics Colloquium by Professor Bernd Schüttler, Dept. of Physics, U. Ga – will discuss the analysis of biological systems in terms of a physical and mathematical model • Today’s lecture – review Chapters 9-14, problem solving techniques PHY 113 -- Lecture 14R
Gravitational forces and energy m r v0 Energy needed to escape Earth’s gravitational field, assuming an initial velocity : v0 PHY 113 -- Lecture 14R
Energy needed to go from one stable circular orbit to another: R1 R2 PHY 113 -- Lecture 14R
Energy needed to go from one stable circular orbit to another -- Example: How much energy is needed to take a satellite of mass m=100kg from the international space station (R1=RE+390 km) to its usual orbit (R2=RE+600 km)? R1 R2 PHY 113 -- Lecture 14R
Problem solving skills Equation Sheet Math skills • Advice: • Keep basic concepts and equations at the top of your head. • Practice problem solving and math skills • Develop an equation sheet that you can consult. PHY 113 -- Lecture 14R
Problem solving steps • Visualize problem – labeling variables • Determine which basic physical principle applies • Write down the appropriate equations using the variables defined in step 1. • Check whether you have the correct amount of information to solve the problem (same number of knowns and unknowns. • Solve the equations. • Check whether your answer makes sense (units, order of magnitude, etc.). PHY 113 -- Lecture 14R
Center of mass ri rj PHY 113 -- Lecture 14R
Position of the center of mass: Velocity of the center of mass: Acceleration of the center of mass: PHY 113 -- Lecture 14R
Physics of composite systems: Center-of-mass velocity: Note that: PHY 113 -- Lecture 14R
A new way to look at Newton’s second law: Define linear momentum p = mv • Consequences: • If F = 0 p = constant • For system of particles: PHY 113 -- Lecture 14R
Statement of conservation of momentum: If mechanical (kinetic) energy is conserved, then: PHY 113 -- Lecture 14R
Snapshot of a collision: Pi Impulse: Pf PHY 113 -- Lecture 14R
Angular motion angular “displacement” q(t) angular “velocity” angular “acceleration” s “natural” unit == 1 radian Relation to linear variables: sq = r (qf-qi) vq = r w aq = r a PHY 113 -- Lecture 14R
r1 v1=r1w w r2 v2=r2w Special case of constant angular acceleration: a = a0: w(t) = wi + a0 t q(t) = qi + wi t +½ a0 t2 (w(t))2 = wi2 + 2 a0 (q(t) - qi) PHY 113 -- Lecture 14R
Newton’s second law applied to center-of-mass motion Newton’s second law applied to rotational motion ri mi di Fi PHY 113 -- Lecture 14R
Object rotating with constant angular velocity (a = 0) w R v=Rw v=0 Kinetic energy associated with rotation: “moment of inertia” PHY 113 -- Lecture 14R
Kinetic energy associated with rolling without slipping: Distance to axis of rotation Rolling: PHY 113 -- Lecture 14R
Torque and angular momentum Define angular momentum: For composite object: L = Iw Newton’s law for torque: If ttotal = 0 then L = constant In the absence of a net torque on a system, angular momentum is conserved. PHY 113 -- Lecture 14R
Center-of-mass Torque on an extended object due to gravity (near surface of the earth) is the same as the torque on a point mass M located at the center of mass. mi ri rCM PHY 113 -- Lecture 14R
Notion of equilibrium: Notion of stability: T- mg cos q = 0 -mg sin q = -maq F=ma r q T t=I a r mg sin q = mr2 a = mraq Example of stable equilibrium. mg(-j) PHY 113 -- Lecture 14R
Analysis of stability: PHY 113 -- Lecture 14R