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Welcome back to Physics 215 Today’s agenda: • Circular motion • Impulse and momentum Physics 215 – Fall 2014 Lecture 07-1 1 Current homework assignment • HW6: – Knight textbook Ch.8: 36, 46, 48; – Ch.9: 28, 48, 70 – due Friday, Oct. 10th in recitation Physics 215 – Fall 2014 Lecture 07-1 2 Forces in circular motion • Motion around circular track, constant speed (for now): arad = v2/r Physics 215 – Fall 2014 Lecture 07-1 3 Two identical balls are connected by a string and whirled around in circles of radius r and 2r at angular speed. The acceleration of ball B is 1. 2. 3. 4. four times as great twice as great equal to one half as great as the acceleration of ball A. Physics 215 – Fall 2014 Lecture 07-1 4 The two identical balls are whirled around in a circle as before. Assume that the balls are moving very fast and that the two strings are massless. The tension in string P is 1. 2. 3. less than equal to greater than the tension in string R. Physics 215 – Fall 2014 Lecture 07-1 5 Car on circular track with friction • Motion around circular track, constant speed (for now): arad = v2/r Physics 215 – Fall 2014 Lecture 07-1 6 Motion of car on banked circular track (no friction) car N R q a= W Speed v Horizontal forces: Vertical: Physics 215 – Fall 2014 Lecture 07-1 7 Motion on loop-the-loop What is normal force on car at top and bottom of loop? Neglect friction; assume moves with speed vB at bottom and vT at top At bottom Physics 215 – Fall 2014 car At top Lecture 07-1 8 Impulse • Constant force F12 acting on object 1 due to object 2 for a time Dt yields an impulse I12 = F12 Dt • In general, for a time varying force need to use this for small Dt and add: I = F(t)Dt = Physics 215 – Fall 2014 Lecture 07-1 9 Impulse for time varying forces F(t) * area under curve equals impulse t1 Physics 215 – Fall 2014 t2 t Lecture 07-1 10 Impulse change in momentum • Consider first constant forces ... • Constant acceleration equation: vf = vi + at mvf - mvi = mat = • If we call p = mv momentum we see that Dp = Physics 215 – Fall 2014 Lecture 07-1 11 Impulse demo • Cart equipped with force probe collides with rubber tube • Measure force vs. time and momentum vs. time • Find that integral of force curve is precisely the change in p! Physics 215 – Fall 2014 Lecture 07-1 12 Definitions of impulse and momentum Impulse imparted to object 1 by object 2: I12 = F12 Dt Momentum of an object: p = mv Physics 215 – Fall 2014 Lecture 07-1 13 Impulse-momentum theorem Inet = Dp The net impulse imparted to an object is equal to its change in its momentum. Physics 215 – Fall 2014 Lecture 07-1 14 Consider the change in momentum in these three cases: A. A ball moving with speed v is brought to rest. B. The same ball is projected from rest so that it moves with speed v. C. The same ball moving with speed v is brought to rest and immediately projected backward with speed v. In which case(s) does the ball undergo the largest change in momentum? 1. 2. 3. 4. Case A. Case B. Case C. Cases A and B. Physics 215 – Fall 2014 Lecture 07-1 15 Throwing ball at student on cart (Neglecting vertical forces for short time interval Physics 215 – Fall 2014 Lecture 07-1 while student catches ball.) 16 Notice • By Newton’s 3rd law, the force on the ball is equal and opposite to the force on the student • Acts for same time interval equal and opposite changes in momentum Physics 215 – Fall 2014 Lecture 07-1 17 Newton’s 3rd law and changes in momentum F1,2 = -F2,1 If all external forces (weight, normal, etc.) cancel: Physics 215 – Fall 2014 Lecture 07-1 18 Conservation of momentum • Assuming no net forces act on bodies there is no net impulse on composite system • Therefore, no change in total momentum D(p1+ p2) = 0 1 Physics 215 – Fall 2014 F 2 system Lecture 07-1 19 Conservation of momentum (for a system consisting of two objects 1 and 2) If the net (external) force on a system is zero, the total momentum of the system is constant. Whenever two or more objects in an isolated system interact, the total momentum of the system remains constant Physics 215 – Fall 2014 Lecture 07-1 20 Conservation of momentum with carts • Two carts initially at rest with compressed spring between them -- track with motion detectors pi = 0 pf = Physics 215 – Fall 2014 Lecture 07-1 21 A cart moving to the right at speed v collides with an identical stationary cart on a low-friction track. The two carts stick together after the collision and move to the right. What is their speed after colliding? 1. 2. 3. 4. 0.25 v 0.5 v v 2v Physics 215 – Fall 2014 Lecture 07-1 22 A student is sitting on a low-friction cart and is holding a medicine ball. The student then throws the ball at an angle of 60° (measured from the horizontal) with a speed of 10 m/s. The mass of the student (with the car) is 80 kg. The mass of the ball is 4 kg. What is the final speed of the student (with car)? 1. 2. 3. 4. 0 m/s 0.25 m/s 0.5 m/s 1 m/s Physics 215 – Fall 2014 Lecture 07-1 23 Momentum is a vector! pA,initial + pB,initial = pA,final + pB,final • Must conserve components of momentum simultaneously • In 2 dimensions: Physics 215 – Fall 2014 Lecture 07-1 24 At the intersection of Texas Avenue and University Drive, a blue, subcompact car with mass 950 kg traveling east on University collides with a maroon pickup truck with mass 1900 kg that is traveling north on Texas and ran a red light. The two vehicles stick together as a result of the collision and, after the collision, the wreckage is sliding at 16.0 m/s in the direction 24o east of north. Calculate the speed of each vehicle before the collision. The collision occurs during a heavy rainstorm; you can ignore friction forces between the vehicles and the wet road. Physics 215 – Fall 2014 Lecture 07-1 25 Reading assignment • More impulse and momentum, collisions • Remainder of Chapter 9 in textbook Physics 215 – Fall 2014 Lecture 07-1 26