Quick Review: Work Done by Friction A non-conservative Force = KE + U. Friction takes Energy out of the system treat it like thermal energy

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2 Quick Review: Work Done by Friction A non-conservative Force Mechanical energy: Conservative System E mech = KE + U Friction takes Energy out of the system treat it like thermal energy ΔE mech 0 Lose Mechanical energy If there is no work done by an external force ΔE mech + ΔE Thermal = 0 E mech ( final) = E mech (initial) E th ( final) + E th (initial) K f + U f = K i + U i F f displacement

3 Quick Review: Work done by External Force Conserv Force Other Forces F gravitation F applied F spring F friction F air (drag) F tension F normal Conservative Force in isolated system ΔE mech = 0 ΔKE = ΔU If External force: (NO FRICTION) W ext,net = ΔE mech = ΔKE + ΔU Positive if energy is transferred to system Negative if energy is transferred from system

4 Quick Review: Work done by External Force + Friction Example: Ext Force with Friction W ext, fric = ΔE mech = ΔKE + ΔU W ext = F ext d W friction = f k d positive or negative always negative W ext, fric = F ext d f k d ( ) = ΔE mech Define: ΔE therm = f k d (increase in thermal energy by sliding) W ext,net = ΔE mech + ΔE therm = ( ΔKE + ΔU) + ΔE therm

5 From Potential Energy Curves to Force Curves

6 Finding a conservative force from potential energy F (x) U(x) In 1-D: ΔU(x) = W cons = Fdx x 2 x 1 then where F is a slowly varying, internal force acting on a particle in system ΔU(x) F(x)Δx Now go backwards, say you know the change in potential energy at some point and you want to know the force at that point (in the differential limit) F(x) = du(x) dx F grav (y) = du grav (y) dy F spring (x) = du spring (x) dx ( = mgy ) dy = = mg 1 ( kx 2 2 ) = kx dx

7 Potential Energy Curve The potential energy as a function of the a system with 1-D movement along x-axis while gravitational force does work on it: U(x) E mech = KE(x) + U(x) KE(x) = E mech U(x) F(x) = du(x) dx

8 Potential Energy Curve Plot of U(x), the potential energy as a function of the a system with 1-D movement along x-axis while a conservative force does work on it: F(x) = du(x) dx F(x) is negative slope of tangent to U(x)

9 Potential Energy Curve Plot of U(x), the potential energy as a function of the a system with 1-D movement along x-axis: E mech = KE(x) + U(x) = constant Equilibrium Points Turning point KE(x) = E mech U(x) F(x) = du(x) dx E mech = 3 J E mech = 4 J E mech = 1 J Equilibrium positions: where slope of U(x) curve is zero [i.e. F(x) = 0 ; NO FORCE] -> Neutral vs Unstable vs Stable Equilibrium KE=0 ; F=0 & if move left or right move back KE=0 ; F=0 but if move left or right force to move away KE=0 ; E mech =U stationary particle

10 Problem #121: A conservative force F(x) acts on a 2.0 kg particle. The potential energy U(x) is shown in the figure. When the particle is at x=2.0 m it has a velocity of -1.5m/s. (a) find the magnitude and direction of the force at this position. (b) What is the range of the allowed motion? (c) What is the velocity at x=7.0 m? (a) F(x=2)=-dU/dx=-( )/(4-1)=5 N (b) K i = mv2 2 = 2.25J X=2 m X=7 m E mec = K i + U i = 2.25J 7.5J = 5J F(x) 5 N Range given by green arrows. (c) K i = mv2 2 = J = 12J What is direction and magnitude of the force at 2, 7, 12 m?

11 Supplemental Homework #4: Look at the figure for potential vs x. The force acts on a 0.50 kg particles and U A =3 J, U B =7 J, U C =9 J. (a) Calculate the magnitude and direction of the force in all five regions. (b) Draw the plot. 9 J 3 J 7 J What happens at x=2, x=4, x=5, and x=6? 2N -6N

12 Chapter 9: Center of Mass + Linear Momentum

13 The Center of Mass Up to now, we ve taken boxes, balls, pigs, penguins to be particles. We know how to apply Newton s laws to determine dynamics of a point particle. Now we have a real mass (system of individual particles). How do we do this? 1) We find the Center of Mass (COM) : A point that moves as though all mass of a system were concentrated at that point (doesn t have to be on the object)

14 How to determine COM (Center Of Mass)? Experimentally: Find the point where it balances (as if all external forces are applied there) Mathematically: Find the effective position Simplest example: two particles x com = m 1 x 1 + m 2 x 2 m 1 + m 2 y com = m 1 y 1 + m 2 y 2 m 1 + m 2 = m 1 (0) + m 2 (0) m 1 + m 2 = 0 Note: Use symmetry!

15 Sample problem 9-3 What is the COM of the 3-particle system shown in the figure. 1) What is total mass M = 2) What are the components x com = 1 M y com = 1 M z com = 0 i=1 3) Write in vector notation 3 m i = ( ) = 15 kg N i=1 N i=1 m i x i m iy i = 1 M = 1 M ( 3kg 0m + 8kg 1m + 4kg 2m) = 1.07m ( 3kg 0m + 8kg 2m + 4kg 1m) = 1.33m r com [m] = x comˆ i + y com ˆ j + z ˆ com k = 1.07ˆ i ˆ j

16 In General: How to determine COM? N particles with total mass M M = N i=1 m i x com = 1 M y com = 1 M z com = 1 M N i=1 N i=1 N i=1 m ix i m iy i m iz i In Vector notation: Position of i th particle: Center of Mass : r i = x iˆ i + y i ˆ j + z ˆ i k r com = x comˆ i + y com ˆ j + z com ˆ k = 1 M N i=1 m ir i Here N is large but still countable

17 Checkpoint 1: The figure shows a uniform square plate from which four identical squares at the corners will be removed. (answer all in terms of quadrants, a) axes, or points) Where is the COM of the plate originally? a) b) Where is the COM of the plate originally? Where is the COM after removal of square 1? b) c) Where Where is is the the COM COM after after removal removal of of square square 1? 1 & 2? c) d) Where is the COM after removal of square 1 & 2? Where is the COM after removal of square 1, 2,& 3? d) e) Where is the COM after removal of square 1, 2, & 3? Where is the COM after removal of all four squares? e) Where is the COM after removal of all squares? Checkpoint com 1 COM com 2

18 CheckPoint #1: The figure shows a uniform square plate from which four identical squares at the corners will be removed. (answer all in terms of quadrants, axes, or points) a) Where is the COM of the plate originally? Checkpoint b) Where is the COM after removal of square 1? c) Where is the COM after removal of square 1 & 2? d) Where is the COM after removal of square 1, 2,& 3? e) Where is the COM after removal of all four squares?

19 COM: Solid Bodies What happens if N gets VERY large? For example a kilogram of material has atoms. Counting is impossible. Treat as a continuum of material x com = 1 M N i=1 m ix i x com = 1 M xdm x com = 1 V xdv y com = 1 M N i=1 m iy i y com = 1 M ydm z com = 1 M N i=1 m iz i z com = 1 M zdm M = N i=1 m i ρ = dm dv = M V Here mass density replaces mass

20 0 m 2 m 1 m 3 From symmetry x cm = L 2 = 9cm y cm = y 1m 1 + y 2 m 2 + y 3 m 3 m 1 + m 2 + m 3 = ( L 2 ) 14 + (0) 44 + L = 3.5cm

21 Rules of Center of Mass (1) Center of mass of a symmetric object always lies on an axis of symmetry. (2) Center of mass of an object does NOT need to be on the object.

22 Newton s 2 nd Law for a System of Particles z F m 1 1 F 2 m 2 F 2 m 3 O x y com Newton's Second Law for a System of Particles : Consider a system of n particles of masses m1, m2, m3,..., m and position vectors r, r, r,..., r, respectively The position vector of the center of mass is given by Mr = m r + m r + m r m r. We take the time derivative of both sides i n n d d d d d M rcom = m1 r1 + m2 r2 + m3 r mn rn dt dt dt dt dt Mvcom = m1v 1 + m2v2 + m3v mnvn. Here vcom is the velocity of the com and v is the velocity of the ith particle. We take the time derivative once more d d d d d M vcom = m1 v1 + m2 v2 + m3 v mn vn dt dt dt dt dt Macom = m1a 1 + m2a2 + m3a mnan. Here acom is the acceleration of the com and a is the acceleration of the ith particle. i n n

23 z F m 1 1 F 2 m 2 F 2 m 3 O x Ma = m a + m a + m a m a. com We apply Newton's second law for the ith particle: miai = Fi. Here Fi is the net force on the ith particle, y Macom = F1 + F2 + F Fn. The force Fi can be decomposed into two components: applied and internal: app int Fi = Fi + Fi. The above equation takes the form: Macom = F1 + F1 + F2 + F2 + F3 + F3 + + F + F Ma = app int app int app int app int ( ) ( ) ( )... ( n n ) ( F app F app F app... F app ) ( int int int... int n F F F Fn ) com The sum in the first set of parentheses on the RHS of the equation above is just F. net The sum in the second set of parentheses on the RHS vanishes by virtue of Newton's third law. The equation of motion for the center of mass becomes In terms of components we have: F = Ma F = Ma net, x com, x net, y com, y F net, z = Ma com, z n n Ma com = F net.

24 For a single particle, Newton s law For each particle in a system of particles, For a system of particles, F = Net forces! ma F = m i iai F = Ma ext com Internal forces that cancel each other: tension; explosions; electromagnetic attractions or repulsions,

25 Checkpoint Two skaters on frictionless ice hold opposite ends of a pole of negligible mass. An axis runs along the pole, and the origin of the axis is at the COM. On skater, Fred, weighs twice as much as the other skater, Ethel. Where do the skaters meet if (a) Fred pulls hand over hand along the pole so as to to draw himself to Ethel, (b) Ethel pulls hand over hand to draw herslef to Fred, and (c) both skaters pull hand over hand? Center of Mass

26 0 x Step I: choose a system boat(m b )+ dog(m d ) x Step II: Find net force F net = 0 COM remains unchanged Step III: choose coordinating system Step IV: D L b m b + Dm d 2 x cm = = m b + m d x + 2 L b m b + xm d 2 m b + m d

27 a) Where is the center of mass? b) Transfer 20 g from one to other--cm now? c) Release--How does CM move? d) What is the acceleration? (a) x cm = 0, y cm = 0 ( (b) Assume part of 1 transferred to 2 x cm = 25 ) y cm = 0 (c) y cm < 0 (d) F y = m 1 g m 2 g = ( m 1 + m 2 )a cm