I 2 comω 2 + Rolling translational+rotational. a com. L sinθ = h. 1 tot. smooth rolling a com =αr & v com =ωr

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2 com tot a com KE tot = KE trans + KE rot = KE trans 1+ I mr 2 slowest v COM = 2gL

2 Rolling translational+rotational smooth rolling a com =αr & v com =ωr Equations of motion from: - Force/torque -> a and α - Energy -> v and ω 1 I 2 comω Mv 2 = KE 2 com tot a com KE tot = KE trans + KE rot = KE trans 1+ I mr 2 slowest v COM = 2gL sinθ 1+ I mr 2 h L sinθ = h θ Hoop Disk Sphere Block fastest Δt bottom = 2L gsinθ 1+ I mr 2

Rolling Down a Ramp: Accelera2on by f s - Acceleration downwards - Friction provides torque - Static friction points upwards Newton s 2 nd Law Linear version x ˆ : f s F g sinθ = ma com y ˆ : N F g

3 Rolling Down a Ramp: Accelera2on by f s - Acceleration downwards - Friction provides torque - Static friction points upwards Newton s 2 nd Law Linear version x ˆ : f s F g sinθ = ma com y ˆ : N F g cosθ = 0 Angular version z ˆ : I com α = τ = Rf s a com = αr a com = R 2 ( ) mgsinθ ma com I com Question: 1. How much work is done by f s? A: 0 2. Is mechanical energy conserved? A: YES α = a com R = Rf s I com gsinθ a com = 1+ I com mr 2

4 Problem 11-HW7: A constant horizontal force is applied to a uniform solid cylinder by fishing line wrapped around the cylinder. The mass of the cylinder is M, its radius R and it rolls on a smooth surface. (a) What is the magnitude of α? (b) What is the acceleration of the COM? (c) What is the fric2onal force vector nota2on. P Find Torque about point P τ =Iα τ = 2RF app I = MR2 2 (com) + MR2 = 3MR2 2 (a) find a. 2RF app = a = 4F app 3M 3MR2 2 α (c) Find F f F = F app+ Ff = m a 4F F f = (M app 3M F app ) i = F app 3 i (b) find α. α= a R α = 4F app 3MR

5 Physics gone Wrong??

Angular momentum with respect to point O the angular momentum is defined

momentum: p = mv with components in x-y plane Units kg m 2 /s components

6 Angular momentum with respect to point O the angular momentum is defined as: = r p = m ( ) r v where r is the position vector of the particle with respect to O. Linear momentum when particle passes through point A it has linear momentum: p = mv with components in x-y plane Units kg m 2 /s components in z-direction only! ( to x-y plane) Just as in τ, has meaning only with respect to a given axis ccw + angular momentum cw - angular momentum

8 Newton s 2 nd Law in Angular Form (single par>cle) Relationship between force and linear momentum Newton s 2 nd Law in linear form F net = F i = ma = d p Relationship between torque and angular momentum Proof: Then d = m dr τ net = d x v + r x dv d = m ( v x v + r x a ) The vector sum of all the torques acting on a particle is equal to the time rate change of the angular momentum of that particle. But v x v ( ) = 0 d = m ( r x a ) = r x ma d == r x ma = r x F = τ

9 Angular Momentum of a Rigid Body Rota>ng about a Fixed axis dl τ net = I α = = I d ω L = I ω Example #3 (Problem 11-39) Three particles are fastened to each other with massless string and rotate around point O. What is the total angular momentum about point O? I tot = m( d) 2 + m( 2d) 2 + m( 3d) 2 = 14md 2 L tot = Iω = 14mωd 2 z ˆ [kg m 2 /s]

10 Conserva>on of Angular Momentum Conservation Law of Linear Momentum (closed, ΔP = 0 isolated system = no net external forces) Δ L = 0 L = L i f If the net external torque acting on a system is zero ( τ net = d ), the angular momentum of the system remains constant, no matter what happens within the system - Total angular momentum of system at all times is equal - Vector {conserved in all three directions, x-y-z} I i ω i = I f ω f - Initially rigid body redistributes mass relative to rotational axis

11 Demo: Conserva>on of Angular Momentum No external torques act L i = L f I i ω i = I f ω f ( I person + 2MR 2 ) i ω i = ( I person + 2Mr 2 ) f ω f if Δr then ω f if Δr then ω f > ω i < ω i

12 Demo: Conserva>on of Angular Momentum No external torques act L initial = L final = + =

Newton s 2 nd Law in Angular Form System of par-cles Total angular momentum L = l 1 + l 2 + l 3 +.

13 Newton s 2 nd Law in Angular Form System of par-cles Total angular momentum L = l 1 + l 2 + l l n = Relationship between torque and angular momentum for system of particles τ net = d L n i=1 τ net l i represents the net external torque.

14 Linear and angular rela>ons Force Linear momentum (one) Linear momentum (system) Linear momentum (system) Newton s second law (system) Conservation Law (closed,isolated) F v = p m p = P Mv com = P dp = F net Ma com = F net 0 = ΔP τ = r F l = r p L = l L = Iω τ net = d L τ net = Iα Δ L = 0 Torque Angular momentum (one) Angular momentum (system) Angular momentum (system, fixed axis) Newton s second law (system) Conservation Law (closed,isolated) τ net = dl has no unless the net torque has no meaning net, and the total unless the net torque τ net, and the total rotational momentum L,, are are defined defined with with respect respect to to the the same same origin origin

16 L = Iω τ = dl v = rω = I dω I(disk) = MR2 2 I(Sphere) = 2MR2 5 I(Hoop) = MR 2

17 Example: Problem #1 Two particles are moving as shown. a) What is their total moment of inertia about point O? b) What is their total angular momentum about point O? c) What is their net torque about point O? #2 a) The total moment of inertia about point O is found from: 2 I tot = = m i r i = (6.5 kg)(1.5 m) 2 + (3.1 kg)(2.8 m) 2 = 38.9 kg m 2 I i b) The total angular momentum about point 0 is: L tot = i = m i ( ri v i ) = m 1 ( r 1 v 1 sinθ 1 )( ẑ) + m 2 ( r 2 v 2 sinθ 2 )(+ẑ) L tot = [ (6.5 kg)(1.5 m)(2.2 m /s) + (3.1 kg)(2.8 m)(3.6 m /s)]ˆ z = 9.8 kg m 2 /s ˆ z c) The net torque about point 0 is: τ net = d L tot = 0 closed, isolated system

Example #3 : Clutch A wheel is rotating freely with angular speed ω i on a shaft whose rotational inertia is negligible.

$What is the angular speed of the the resultant combination of the shaft and two wheels What fraction of the original$ rotational kinetic energy is lost?

18 Example #3 : Clutch A wheel is rotating freely with angular speed ω i on a shaft whose rotational inertia is negligible. A second wheel, initially at rest and with twice the rotational inertia of the first, is suddenly coupled to the same shaft. What is the angular speed of the the resultant combination of the shaft and two wheels What fraction of the original rotational kinetic energy is lost? L i = L f ( Iω i )small wheel ( ) small + ( 2I i (0))large = Iω f wheel Iω i = 3Iω f Before wheel + ( 2Iω f ) large wheel After ω f = 1 3 ω i Fraction of KE lost : KE i KE f % = KE i 1 I 2 initialω i2 1 I 2 2 finalω f = I 2 2 initialω i ( ) 1 3 ( ) 2 % = % = ( 2 3)% = 66.7% 1 Iω 2 1 3I 2 i 2 ( )( 1 ω 3 i) 2 1 Iω 2 2 i %

19 Problem A particle of mass m slides down the frictionless surface through height h and collides with the uniform vertical rod (of mass M and length d), sticking to it. The rod pivots about point O through the angle θ before momentarily stopping. Find θ. What do we know? Where are we trying to get to? What do we know about the ramp? What does this give us? Conservation of Mechanical Energy going down ramp! Now what? After the collision, what holds? ( ) 2 ( + 0) ( KE init + U init ) = KE final + U final ( 0 + mgh) = 1 mv 2 final This gives us the speed of the mass right before hitting the rod: v m,bottom = 2gh There s a collision! What kind? What do we know about this collision? What is constant? Only Conservation of Angular Momentum holds! L initial = L final (choose a point wisely) ( r p m,0 + 0) = ( r p m, f + I rod ω) ( d(mv m,0 )) = md 2 ( ω bottom ) + ( 1 Md 2 3 )ω bottom This gives the angular speed of the rod/ mass just after hitting: ω bottom = Now Conservation of Mechanical energy holds again! Solve for θ: ( ) mv m,0 dm Md = m 2gh dm Md ( KE init + U init ) = ( KE final + U final ) 1 ( I 2 totω 2 bottom + 0) = ( 0 + mgd(1 cosθ final ) + Mg( 1 d)(1 cosθ 2 final )) θ = cos 1 1 m 2 h d m + 1 M 2 ( )( m + 1 M 3 )

20 Problem 11-40: The figure plots the torque τ that acts on a ini2ally sta2onary disk that can rotate about its center. Τ s =4.0 N. m. What is the angular momentum of the disk at (a) t=7.0 s, (b) t=20s? τ = dl L(t) = t 0 τ At 7.0 Sec L(7) = 2( 4) + 2( 3) ( 3) L(7) = 24kgi m2 s At 20.0 Sec ( ) ( ) L(20) = = 7.5 ( ) 3 2 2

Forces of Rolling. 1) Ifobjectisrollingwith a com =0 (i.e.no netforces), then v com =ωr = constant (smooth roll)

Forces of Rolling. 1) Ifobjectisrollingwith a com =0 (i.e.no netforces), then v com =ωr = constant (smooth roll) Physics 2101 Section 3 March 12 rd : Ch. 10 Announcements: Mid-grades posted in PAW Quiz today I will be at the March APS meeting the week of 15-19 th. Prof. Rich Kurtz will help me. Class Website: http://www.phys.lsu.edu/classes/spring2010/phys2101-3/