Physics 1501: Lecture 18, Pg 1 Physics 1501: Lecture 18 Today’s Agenda l Announcements çHW#7: due Oct. 21 l Midterm 1: average = l Topics çImpulse çCenter of Mass çRotational Kinematics

Physics 1501: Lecture 18, Pg 2 Momentum Conservation l The concept of momentum conservation is one of the most fundamental principles in physics. l This is a component (vector) equation. çWe can apply it to any direction in which there is no external force applied. l You will see that we often have momentum conservation even when (mechanical) energy is not conserved.

Physics 1501: Lecture 18, Pg 3 Elastic vs. Inelastic Collisions l A collision is said to be elastic when energy as well as momentum is conserved before and after the collision. K before = K after çCarts colliding with a spring in between, billiard balls, etc. vvivvi l A collision is said to be inelastic when energy is not conserved before and after the collision, but momentum is conserved. K before  K after çCar crashes, collisions where objects stick together, etc.

Physics 1501: Lecture 18, Pg 4 Force and Impulse F t titi tftf tt I l The diagram shows the force vs time for a typical collision. The impulse, I, of the force is a vector defined as the integral of the force during the collision. I Impulse I = area under this curve ! Impulse has units of Ns.

Physics 1501: Lecture 18, Pg 5 Force and Impulse F t tt l Using the impulse becomes: impulse = change in momentum !

Physics 1501: Lecture 18, Pg 6 Force and Impulse I l Two different collisions can have the same impulse since I depends only on the change in momentum, not the nature of the collision. tt F t F t tt same area F  t big, F small F  t small, F big

Physics 1501: Lecture 18, Pg 7 Force and Impulse tt F t F t tt F  t big, F small F  t small, F big soft spring stiff spring

Physics 1501: Lecture 18, Pg 8 Lecture 18, ACT 1 Force & Impulse l Two boxes, one heavier than the other, are initially at rest on a horizontal frictionless surface. The same constant force F acts on each one for exactly 1 second. çWhich box has the most momentum after the force acts ? (a) (b) (c) (a) heavier (b) lighter (c) same FF light heavy

Physics 1501: Lecture 18, Pg 9 Average Force and Impulse tt F t F t tt F av  t big, F av small F av  t small, F av big soft spring stiff spring F av

Physics 1501: Lecture 18, Pg 10 System of Particles: l Until now, we have considered the behavior of very simple systems (one or two masses). l But real life is usually much more interesting ! l For example, consider a simple rotating disk. l An extended solid object (like a disk) can be thought of as a collection of parts. The motion of each little part depends on where it is in the object!

Physics 1501: Lecture 18, Pg 11 System of Particles: Center of Mass l How do we describe the “position” of a system made up of many parts ? Center of Mass l Define the Center of Mass (average position): çFor a collection of N individual pointlike particles whose masses and positions we know: (In this case, N = 2) y x rr2rr2 rr1rr1 m1m1 m2m2 R R CM

Physics 1501: Lecture 18, Pg 12 System of Particles: Center of Mass l If the system is made up of only two particles: y x rr2rr2 rr1rr1 m1m1 m2m2 R R CM where M = m 1 + m 2 So: (r 1 - r 2 )

Physics 1501: Lecture 18, Pg 13 System of Particles: Center of Mass l If the system is made up of only two particles: y x rr2rr2 rr1rr1 m1m1 m2m2 R R CM r r r 2 - r 1 where M = m 1 + m 2 + If m 1 = m 2 the CM is halfway between the masses. Active Figure

Physics 1501: Lecture 18, Pg 14 System of Particles: Center of Mass l The center of mass is where the system is balanced ! çBuilding a mobile is an exercise in finding centers of mass. (Calder was a great applied physicist!) m1m1 m2m2 + m1m1 m2m2 +

Physics 1501: Lecture 18, Pg 15 Example Calculation: l Consider the following mass distribution: (24,0) (0,0) (12,12) m 2m m R CM = (12,6)

Physics 1501: Lecture 18, Pg 16 R rr CM dm M     System of Particles: Center of Mass l For a continuous solid, we have to do an integral. y x dm r where dm is an infinitesimal mass element.

Physics 1501: Lecture 18, Pg 17 Example: Astronauts & Rope l A male astronaut and a female astronaut are at rest in outer space and 20 meters apart. The male has 1.5 times the mass of the female. The female is right by the ship and the male is out in space a bit. The male wants to get back to the ship but his jet pack is broken. Conveniently, there is a rope connected between the two. So the guy starts pulling in the rope. Does he get back to the ship? Does he at least get to meet the woman?

Physics 1501: Lecture 18, Pg 18 Example: Astronauts & Rope 1. Find position relative to the male astronaut where they meet. Use Center of Mass M = 1.5m m

Physics 1501: Lecture 18, Pg 19 Example: Astronauts & Rope 2. Find x cm F ext =0 so v cm is constant at zero, x cm is constant. M = 1.5m m x 0 20m

Physics 1501: Lecture 18, Pg 20 Example: Astronauts & Rope Solve symbolically,

Physics 1501: Lecture 18, Pg 21 Example: Astronauts & Rope Put in numbers,

Physics 1501: Lecture 18, Pg 22 Example: Astronauts & Rope They end up 12m away from the ship, So he doesn’t get back to the ship, but he does get to the woman.

Physics 1501: Lecture 18, Pg 23 Lecture 18, ACT 2 Center of Mass Motion l A woman weighs exactly as much as her 20 foot long boat. l Initially she stands in the center of the motionless boat, a distance of 20 feet from shore. Next she walks toward the shore until she gets to the end of the boat. çWhat is her new distance from the shore. (There no horizontal force on the boat by the water). 20 ft ? ft 20 ft (a) (b) (c) (a) 10 ft (b) 15 ft (c) 16.7 ft before after

Physics 1501: Lecture 18, Pg 24 Center of Mass Motion: Review l We have the following law for CM motion: l This has several interesting implications: l It tell us that the CM of an extended object behaves like a simple point mass under the influence of external forces: FA çWe can use it to relate F and A like we are used to doing. F l It tells us that if F EXT = 0, the total momentum of the system does not change. çAs the woman moved forward in the boat, the boat went backward to keep the center of mass at the same place. Active Figure

Physics 1501: Lecture 18, Pg 25 Chap. 10: Rotation l Up until now we have gracefully avoided dealing with the rotation of objects. çWe have studied objects that slide, not roll. çWe have assumed wheels are massless. l Rotation is extremely important, however, and we need to understand it ! l Most of the equations we will develop are simply rotational versions of the ones we have already learned when studying linear kinematics and dynamics.

Physics 1501: Lecture 18, Pg 26 Rotational Variables l Rotation about a fixed axis: çConsider a disk rotating about an axis through its center: l First, recall what we learned about Uniform Circular Motion: (Analogous to )  

Physics 1501: Lecture 18, Pg 27 Rotational Variables... Now suppose  can change as a function of time: l We define the angular acceleration:    Consider the case when  is constant.  We can integrate this to find  and  as a function of time: constant

Physics 1501: Lecture 18, Pg 28 Rotational Variables... l Recall also that for a point a distance R away from the axis of rotation:  x =  R  v =  R And taking the derivative of this we find  a =  R   R v x  constant

Physics 1501: Lecture 18, Pg 29 Summary (with comparison to 1-D kinematics) AngularLinear And for a point at a distance R from the rotation axis: x = R  v =  R  a =  R

Physics 1501: Lecture 18, Pg 30 Example: Wheel And Rope l A wheel with radius R = 0.4m rotates freely about a fixed axle. There is a rope wound around the wheel. Starting from rest at t = 0, the rope is pulled such that it has a constant acceleration a = 4m/s 2. How many revolutions has the wheel made after 10 seconds? (One revolution = 2  radians) a R

Physics 1501: Lecture 18, Pg 31 Wheel And Rope... Use a =  R to find  :  = a / R = 4m/s 2 / 0.4m = 10 rad/s 2 l Now use the equations we derived above just as you would use the kinematic equations from the beginning of the semester. = 0 + 0(10) + (10)(10) 2 = 500 rad a R 