PH 105-003/4 ----Monday, Oct. 8, 2007 Homework: PS8 due Wednesday: conservation of energy Chapter 9: momentum Review Wednesday & Friday: define p = mv Why? Then dp/dt=0 for isolated system Dp = F dt = impulse Elastic collision: KE conserved Completely inelastic collision: stick together (v1f = v2f = vf) – solve C of Mom. eqn. for vf. 3/24/2016 A. PH 105 Elastic Collisions in 1D Equations: m1v1i + m2v2i = m1v1f + m2v2f (Cons . of px) ½m1v1i2 + ½ m2v2i2 = ½ m1v1f2 + ½ m2v2f2 (Conservation of E) Known: v1i & v2i Unknown: v1f & v2f – can solve 2 equations. But quadratic! (Wednesday, Oct. 10) Trick (only works for 1D elastic collisions): can show relative velocity reverses, (v1f - v2f) = - (v1i - v2i) 3/24/2016 A. PH 105 Ungraded question (related to PS 7, Chapter 8) A mass m strikes a spring with spring constant k. There is a friction force of magnitude f = mKmg. In a process in which the spring moves a distance d, conservation of mechanical energy E = K + U can be written as A. Ef = Ei B. Ef = Ei + f d C. Ef = Ei - f d Answer is (c) -- the mass must lose energy to friction. Example: Consider two equal masses m1 = m2 = m, m1 moving to the right at velocity v and m2 stationary. Find the velocities after an elastic collision. Solution: We know conservation of momentum, m1v1i + m2v2i = m1v1f + m2v2f and (ONLY because this is 1D elastic) relative velocity reverses, (v1f - v2f) = - (v1i - v2i) .... 3/24/2016 A. PH 105 Clicker question: A small mass moving at +8 m/s collides elastically with a stationary larger mass, which moves away at +2 m/s. What is the velocity of the smaller mass after the collision? -6 0.1 Clicker question: In the previous question, how much heavier was m2? That is, find the ratio m2/m1. 7 0.1 PH 105-003/4 ----Monday, Oct. 15, 2007 Homework: PS9 (short) due Wednesday: conservation of momentum Rest of Chapter 9: Center of Mass Why? If a system isn’t just a point mass, but we want to treat it as one (use F=ma, for example) what is its position x? Answer: use xcm where (for a simple system of 2 masses m1 and m2) xcm m1 x1 m2 x2 m1 m2 m1 x x1 Check: x1=x2 case, m1=0 case, … 3/24/2016 m2 A. PH 105 xcm x2 Example: Consider a system of two masses m1 m2 x 0 xcm 6m If they are equal (m1 = m2) then m1 x1 m2 x2 m1 x1 m1 x2 x1 x2 0 6 m xcm 3m 2 m1 m2 m1 m1 2 3/24/2016 A. PH 105 Clicker question (not recorded): In the system of two masses, suppose m1=1 kg, m2=2 kg. Find the center of mass coordinate xcm. m1 m2 x 4 0.1 0 6m A. 9.7 Deformable bodies (skip) 9.8 Rockets: Serway derives vf vi=0 mi ln( ) v0 mf mi exhaust v0 3/24/2016 mf vf Clicker question (didn’t have time): Use the rocket velocity equation to calculate the final velocity vf (in m/s) , given that mf = 1000 kg, mi = 7389 kg, and the rocket exhaust velocity v0 = 200 m/s, 400 20 3/24/2016 vf mi ln( ) v0 mf A. PH 105-003/4 ----Wednesday, Oct. 17, 2007 Please turn in filled-out evaluations in box at front. Talk Thursday: “Energy for All in the 21st Century ” Homework: PS9 (short) due tonight: conservation of momentum Test next Friday Recoil problems: rocket, gun, hose, …. Linear vs. Rotational motion: x <--> q, etc. Torque: F=ma t = I a where I = S m r2 Kinetic energy: ½ m v2 ½ I w2 Clicker Question: Two cylindrical drums (on axles through their centers) have string wound around them. One string has an object of mass m attached to it, the other is pulled on by a constant force T = mg. Which drum will turn faster? A T = mg A. The one with constant tension mg. B. The one with the mass m attached C. They will turn at the same rate. B m Clicker Q [not done]: Find the acceleration of the falling mass in the case mdisk = m = 2 kg, r = 0.2 m. r 6.7 0.4 m A. PH 105-003/4 ---- Friday, Oct. 19, 2007 Homework (PS 9, Ch 9&10) is on WebAssign: Test next Friday Did: Linear vs. Rotational motion: x <--> q, etc. F=ma t = I a, where I = S m r2 Torque: t = F r More generally, t = F r sin f Line of action Calculating I for various shapes (make table) Parallel Axis Theorem Kinetic energy: ½ m v2 ½ I w2 A. PH 105-003/4 ---- Monday, Oct. 22, 2007 Homework (PS 9, Ch 9&10) is due Wednesday 11PM Test this Friday Review: F=ma t = I a, where I = S m r2 Torque: t = F d = F r sin f (d = r sin f = moment arm) Line of action Calculating I for various shapes (made table) Parallel Axis Theorem: I = Icm + Mr2 A. Example: An airplane has wingspan 20 m, length 20 m, and mass 20,000 kg. It snags a weather balloon at the end of its wing, which exerts a drag force 105 N on the wing. Will the plane crash? Assume it crashes if it rotates > 1 radian in 1 second (the pilot’s reaction time). To answer this, answer (a) what is the torque about the center of mass? (aa) where is the center of mass? model: 2 rods, each 10,000 kg (ab) what is the torque? (b) What is the moment of inertia? (c) What is the angular acceleration? (d) How far does it turn in 1 second? Clicker question: What is the moment of inertia of a rod of length L about a point on the rod, L/8 from the center? For simplicity, use m = 1 kg and L = 20 m, and give the answer in kg m2. Hint: for a rod of length L rotating about its center of mass, I/mL2 = 1/12. 39.6 1 A. Monday, continued Kinetic energy: Ktotal = Ktrans + Krot = ½ m vcm2 + ½ I w2 Rolling motion -- calculate v at bottom of incline Conservation of energy: mgh = ½ mv2 + ½ I w2 & v = w r give v2 = 2gh/(1+I/MR2) A. PH 105-003/4 ---- Wednesday, Oct. 24, 2007 Homework (PS 9, Ch 9&10) is due tonight 11PM Test this Friday Review: Parallel Axis Theorem: I = Icm + Mr2 Kinetic energy: Ktotal = Ktrans + Krot = ½ m vcm2 + ½ I w2 Rolling motion -- calculate v at bottom of incline Conservation of energy: mgh = ½ mv2 + ½ I w2 & v = w r give v2 = 2gh/(1+I/MR2) vsphere>vdisk > vhoop Angular momentum