z F 3 = = = m 1 F 1 m 2 F 2 m 3 - Linear Momentum dp dt F net = d P net = d p 1 dt d p n dt - Conservation of Linear Momentum Δ P = 0

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3 F 1 m 2 F 2 x m 1 O z F 3 m 3 y Ma com = F net F F F net, x net, y net, z = = = Ma Ma Ma com, x com, y com, z p = mv - Linear Momentum F net = dp dt F net = d P dt = d p 1 dt d p n dt Δ P = 0 - Conservation of Linear Momentum

t i F L R (t)dt F net (t)dt p i = Δp = J Impulse Change in Momentum is

4 P R,after Collision What changes momentum of each? P R,before Definition p f d p R = F L R (t)dt dp R = p i p f J t f t i t f t i F L R (t)dt F net (t)dt p i = Δp = J Impulse Change in Momentum is equal to Impulse acting on it Impulse Vector! Must satisfy for each direction! Impulse-momentum theorem

5 Elastic collision : TOTAL KE is conserved (~ Conservative forces ) AND if system is closed and isolated, the total linear momentum P cannot change (whether the collision is elastic or inelastic!). ΔKE = K f K i = ( K 1f K nf ) ( K 1i K ni ) = 0 Δ P = P f Pi = P1f P nf P1i+...+ P ni = 0 Inelastic collision : KE is not conserved (~ thermal energy ) However, if system is closed and isolated, the total linear momentum P cannot change (whether the collision is elastic or inelastic!). ΔKE = K f K i = ( K 1f K nf ) ( K 1i K ni ) 0 Δ P = P f Pi = P1f P nf P1i+...+ P ni = 0

6 Pure Inelastic Collision Ballistic pendulum: A bullet of mass m and initial velocity v 0 collides and sticks to a pendulum of mass M supported on a rope of length L. How high does the pendulum go before coming to rest? Conservation of momentum p i = mv 0 = p f = (m + M )v Kinetic Energy to Potential Energy (M + m)v 2 2 = ( M + m)gh v = mv 0 m + M mv 0 m + M 2g 2 = h

7 Inelastic Collision P. # 54: Ch. 9: A bullet of mass m is moving directly upward at a velocity v 0. It strikes a block of mass M initially at rest. It passes through the block but comes out with a velocity v 0 /4. How high does the block rise above its initial position?

8 D Elastic Collision: 2 particles 1 2 m 2 1 1v 1f + 2 m 2 1 2v 2 f = 2 m 2 1 1v 1i + 2 m 2 2v 2i m 1 v 1f v 2 f = m 1 v 1i v 2i v 1f = m 1 m 2 m 1 v 1i + 2m 2 m 1 v 2i v 2 f = 2m 1 m 1 v 1i m 1 m 1 v 2i 6 variables and 2 Equations 2 mass & 4 velocity

9 1D Elastic Collision: 2 particles with v 2i =0 v 1f = m 1 m 2 m 1 v 1i v 2 f = 2m 1 m 1 v 1i 5 variables and 2 Equations 2 mass & 3 velocity Special Cases: 1) Equal masses If m 1 = m 2 then v f 1 = 0 and v f 2 = v o1 v 1f = 0 v 2 f = v 1i

10 Examples of Elastic Collisions

11 1D Elastic Collision: 2 particles with v 2i =0 v 1f = m 1 m 2 m 1 v 1i v 2 f = 2m 1 m 1 v 1i 5 variables and 2 Equations 2 mass & 3 velocity Special Cases: 2) Massive target If m 1 << m 2 then f 1 v o1 and f 2 v 1f v 1i v 2 f 2m 1 m 2 m 2 v o1 v 1i

12 1D Elastic Collision: 2 particles with v 2i =0 v 1f = m 1 m 2 m 1 v 1i v 2 f = 2m 1 m 1 v 1i 5 variables and 2 Equations 2 mass & 3 velocity Special Cases: 3) Massive Projectile If m 1 >> m 2 then v 1f f 1 v o1 1i and v f2 f 2v 2 v o1 1i

13 Problem 10-4 A ball of mass m is fastened to a cord of length L. The ball is released when cord is horizontal. At bottom of path, the ball elastically strikes block of mass M initially at rest on frictionless floor. a) What is the speed of the ball right after the collision? b) What is the speed of the block right after the collision? c) Use Conservation of Mechanical Energy to find the velocity of the ball right before collision: E mech = 0 ( KE i + U i ) = KE f + U f ( 0 + mgl) = 1 mv 2 bottom ( ) ( 2 + 0) v bottom = 2gL = v mi Because collision is elastic, use COLM+COKE to find the velocity of the ball right after collision: Eqn# 9-67 v mf = m M m + M v mi if M>m, negative? d) What is the speed of the block right after the collision? Because collision is elastic, use COLM+COKE to find the velocity of the block right after collision: Eqn# 9-68 v Mf = 2m m + M v mi

14 Velocity of COM In a closed, isolated system the COM velocity ( ) of the system is CONSTANT. Why? v com F net = d P dt tot = 0 v com = P tot = Mv com = (m 1 ) v com P tot = p 1i + p 2i p 1i + p 2i (m 1 ) = p 1 f + p 2 f (m 1 ) P tot (m 1 ) = Constant!! Exploding things from Chapt. 9:

15 1-D Collisions If system is closed and isolated, the total linear momentum P cannot change Inelastic collision : KE is not conserved (~ thermal energy ) P before = P after 1- D 2 particles ( ) ( m 1 v 1i v 2i ) = m 1 v 1 f v 2 f Elastic collision : TOTAL KE is conserved (~ Conservative forces ) v 1 f = m 1 m 2 m 1 v 1i + 2m 2 m 1 v 2i (Eqn. 9-75) P before = P after 1- D KE before = KE after 2 particles v 2 f = 2m 1 m 1 v 1i m 1 m 1 v 2i (Eqn. 9-76) 1 m 2 1v 2 1i + 1 m 2 ( 2 2v 2i ) = 1 m v m v 2 ( f ) f

16 Chapter 10: Rotation

17 Chapter 10: Rotation In this chapter we will study the rotational motion of rigid bodies about a fixed axis. To describe this type of motion we will introduce the following new concepts: -Angular displacement -Average and instantaneous angular velocity (symbol: ω ) -Average and instantaneous angular acceleration (symbol: α ) -Rotational inertia, also known as moment of inertia (symbol I ) -Torque (symbol τ ) apple We will also calculate the kinetic energy associated with rotation, write Newton s second law for rotational motion, and introduce the work-kinetic energy theorem for rotational motion.

18 Here we assume (for now): Rigid body Axis of rotation Reference line Rotation Variables - all parts are locked together ( e.g. not a cloud) - a line about which the body rotates (fixed axis) - rotates perpendicular to rotation axis θ(t) = s r looking down Rotation axis ω(t) = dθ(t) dt Angular position (displacement) of line at time t: (measured from : ccw) Angle is in units of radians (rad) : 1 rev = 360 = 2πr = 2π rad r Angular velocity of line at time t: ω(t) (positive if ccw) α(t) θ(t) All points move with same angular velocity α(t) = dω(t) dt = arc length radius Δθ = θ 2 θ 1 = d 2 θ(t) dt 2 ( Δθ(t) ) + x ˆ Units of rad/sec Angular acceleration of line at time t: (direction: depends on ω(t) ) All points move with same angular acceleration Units of rad/sec 2

19 Are angular quantities Vectors? Some! Once coordinate system is established for rotational motion, rotational quantities of velocity and acceleration are given by right-hand rule (angular displacement is NOT a vector) Note: the movement is NOT along the direction of the vector. Instead the body is moving around the direction of the vector

20 Rotation with Constant Rotational Acceleration Recall: 1-D translational motion v = v 0 + at x = x 0 + v 0 t at 2 Rotational motion ω = ω 0 + αt θ = θ 0 + ω 0 t αt 2 Problem 10-10: A disk, initially rotating at 120 rad/s, is slowed down with a constant acceleration of magnitude 4 rad/s 2. a) How much time does the disk take to stop? ω = ω 0 + αt t = ( 0 120rad /s) 4rad /s 2 = 30s b) Through what angle does the disk rotate during that time? θ = θ 0 + ω 0 t + 1 αt 2 Δθ = (120rad /s)(30s) + 1 ( 4rad 2 2 /s2 )(30s) 2 = 1800rad = 286.5rev

(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.

(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. 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