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# November Notes AP Physics 101

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This 216 page Bundle was uploaded by Kevin Kudah Vitalis on Monday August 15, 2016. The Bundle belongs to AP Physics 101 at University of Pennsylvania taught by Dr. Kristen Pacada in Fall 2016. Since its upload, it has received 9 views. For similar materials see Phys 101 in PHYSICS (PHY) at University of Pennsylvania.

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Date Created: 08/15/16

Momentum! Reminder: If two objects move at a constant velocity for the same amount of time, the distance they travel is proportional to their velocity. Δx vt = Δx v = t Both carts start at the same velocity and travel the same distance in the start in the middle of the track.th hit the barriers at the same time if they If I add weight to cart #2, enough to double its mass, then: 1. Cart #1 and Cart #2 still get to the ends at the same time. 2. Cart #2 arrives ﬁrst. 3. Cart #1 arrives ﬁrst. 4. Cart #1 moves but Cart #2 doesn’t move. Where should I put the carts to make sure that #1 and #2 arrive at the same time? 2m =m1. 2 1. Cart #1 should travel half the distance of #2. 2. Cart #2 should travel half the distance of #1. 3. Cart #1 should travel one third the distance of #2. 4. Cart #2 should travel one third the distance of #1. Where should I put the carts to make sure that #1 and #2 arrive at the same time? 3m =m1. 2 1. Cart #1 should travel half the distance of #2. 2. Cart #2 should travel half the distance of #1. 3. Cart #1 should travel one third the distance of #2. 4. Cart #2 should travel one third the distance of #1. What happens if cart #1 discharges without touching cart #2? 1. It still travels to the end of the track. 2. It travels in the opposite direction as when it was touching cart #2. 3. It doesn’t move. You just learned almost everything you need to know about momentum! But let’s look at the mathematical description of momentum anyhow... Why would we want to do this? •Notice that this is the ﬁrst time we’ve been able to predict the behavior of two objects simultaneously... •Also very useful for describing collisions ! ! Momentum is a vector quantity, the variable we use is ‘p’. p = mv Deﬁned as mass times velocity. F21 F12 1 2 Free body diagram of the cart experiment. ! ! p = mv Momentum is a vector quantity, the variable we use is ‘p’. Deﬁned as mass times velocity. F21 1 2 F12 ! kg⋅m [p]= s For an object of ﬁxed mass: ! ! Δv Δp ! Fnetma = m = Fnet = Δp Δt Δt Momentum is a little bit like kinetic energy: ! Work-kinetic energy theorem ΔK = F net=Work ! ! Δp = F net≡ J = impulse Impulse-momentum theorem ! ! ! p = mv F = ma = m Δv = Δp ! ! net Δt Δt Δp = F net≡ J = impulse F21 F12 1 2 ! ! ! ! F = Δp F = Δp 21 Δt 12 Δt p = m v ! p = m v ! 1 1 1 2 2 2 F = −F Newton’s third law 21 12 Δp 1 = −Δp 2 Δt Δt Δp 1 Δp =20 The carts had zero momentum before the spring, and zero total momentum after. If the net force acting on a system is zero, the total momentum of the system is conserved. ! ! ! ! Δv Δp p = mv F = ma = m = ! ! net Δt Δt Δp = F ⋅net J = impulse F21 1 2 F12 Δp 1 Δp =20 If the net force acting on a system is zero, the total momentum of the system is p − p + p − p = 0 conserved. ( 1f 1i) ( 2 f 2i) Note the similarities to conservation of p1i p =2i + p1f 2 f energy. ! ! ! ! Δv Δp p = mv F neta = m = ! ! Δt Δt Δp = F net≡ J = impulse Δp 1 Δp =20 p 1ip =2i + 1f 2 f F21 F12 1 2 Conservation of momentum in the cart system: m =m Before: v1=0, v2=0 1 2 So psystem. P = p + p = 0 initial 1i 2i P = p + p = m v + m v final 1f 2 f 1 f 2 f m 1 1fm v 2 2 f= 0 When masses are equal, they v = −v travel at the same speed in 1f 2 f opposite directions after an equal and opposite impulse. ! ! ! ! Δv Δp p = mv Fnetma = m = ! ! Δt Δt Δp = F net≡ J = impulse Δp 1 Δp =20 p1i p 2ip + 1f 2 f F21 F12 1 2 Conservation of momentum in the cart system: Before: v1=0, v2=0 2m =m 1 2 So psystem. Pinitialp1i p =2i Pfinal p 1fp 2 f= m 1 f m v 2 f m 1 1fm v 2 2 f= 0 m 1 1fm v 1 2 f= 0 m 1 1f−2m v 1 2 f 1 − v =1f 2 f cart 2 travels half the distance in the same time. 2 It also travels in the opposite direction. ! ! ! ! Δv Δp p = mv Fnetma = m = ! ! Δt Δt Δp = F net≡ J = impulse Δp 1 Δp =20 p1i p 2ip + 1f 2 f F21 F12 1 2 Conservation of momentum in the cart system: Before: v1=0, v2=0 3m =m 1 2 So psystem. Pinitial 1ip = 2i Pfinal p 1fp 2 f= m 1 f m v 2 f m 1 1fm v 2 2 f 0 m 1 1f3m v 1 2 f= 0 m 1 1f−3m v 1 2 f 1 − v =1f 2 f cart 2 travels one third the distance in the same time. 3 It also travels in the opposite direction. Collisions: Categorized into two types: Elastic Inelastic “Elastic” means energy is conserved - the energy in the collision is stored in the objects colliding as if they were covered in little springs - elastic potential energy “INelastic” means energy is -not- conserved - the objects do not behave like little springs - they behave more like blobs of clay. “Inelastic collisions” - the two objects stick together after they collide. •In order for them to stick, the two materials must deform/heat up. •This process requires energy, and the only energy source here is the kinetic energy of the moving objects. •In an inelastic collision, momentum is conserved but kinetic energy is not conserved. m 1 m 2 v1i v2i m 1 m2 vf “Inelastic collisions” - two objects stick together after they collide. Momentum is conserved. Kinetic energy is not conserved. m1 m2 v1i v2i m1 m 2 vf Pinitialp 1ip = 2iv + m1 1i 2 2i P final p 1+2= m(+ 1 v 2) f m 1 1im v 2 2i+ m(v 1 2) f Velocity of the m 1 1im v 2 2i coupled objects after = v f the collision. Signs (m 1 m 2) matter - velocity is a vector! “Elastic collisions” - the two objects move independently after collision. •Mechanical energy is stored during the collision •In an elastic collision, momentum is conserved and kinetic energy is also conserved. By combining conservation of energy and conservation of momentum, we can prove that: v −v = −(v −v ) v1i v 2i 1i 2i 1f 2 f v1fv 2 f m 1 m2 v 1i 2i v 1i v2i m m 2 v1fv 2 f 1 v1f v2f After the collision, the magnitude of the relative velocity is the same, but relative velocity vector points in the opposite direction. Two cars have a head-on collision and their bumpers lock. The ﬁrst car has a mass of 1000 kg and an initial velocity of 10 m/s, and the second car has a mass of 1500 kg and an initial velocity of 15 m/s and was headed in the opposite direction. Find the work done on the two-car system during the collision. m m m1=1000 kg 1 2 m2=1500 kg v v v1i10 m/s 1i 2i v =15 m/s m 1 m 2 2i vf=? v f Work = ΔKE m v + m v = m + m v ( ) 1 2 ⎛ 1 2 1 2 ⎞ 1 1i 2 2i 1 2 f KE −fKE = i (m 1 m v2) f ⎜ m 1 1i+ m v 2 2i⎟ 2 ⎝ 2 2 ⎠ m 1 1im v 2 2i 2 ⎛ 2 2⎞ = v f KEf− KEi=1(1000kg+1500k⎜5 m⎟ −⎜11000kg⋅⎜10m⎟ + 1500⋅⎜15m⎟ ⎟ (m +1m 2 ) 2 ⎝ s⎠ ⎝2 ⎝ s⎠ 2 ⎝ s⎠ ⎠ Work = −1.87⋅10 J 5 1000kg⋅10m /s+1500kg⋅−15m /s = −5m /s (1000kg+1500kg ) The force of the collision did work on the cars in the form of deformation, and acted to slow the system down, reducing kinetic energy. However, momentum is still conserved. How to measure the speed of a speeding bullet (or fastball)? A ballistic pendulum. Use conservation of momentum in an inelastic collision paired with conservation of energy to ﬁnd the initial speed of the bullet/fastball. m1 m 1 2 Δh v1 v2 ! ! p i m v1 1 pf= m(+ 1 v 2) 2 Conservation of energy - kinetic en1rgy of 2 pendulum immediately after collision(bm1+ m v2) m2+ m(gh1 2 ) gravitational potential energy. 2 Relationship between velocity of the pendulum v = 2gh immediately after the collision and its maximum 2 height. ! ! pi= p f Conservation of momentum between projectile object and its inelastic collision with another 1 1ect. 1 v 2) 2 and the ﬁnal height of the pendulum after an proje(m 1 m 2 ) 2gh inelastic collision. v1= m 1 Elastic collision in a Basketball/tennisball drop - why’s the tennis ball do that? What is its ﬁnal velocity? m b1 kg vbiv = −ti −v ) bf tf Relative velocities are maintained but inverted m=0.05 kg after the collision. t vbi3.8m/s v = v −v bf tf bi vti0m/s m v = m v + m v Conservation of momentum if tennis ball has ~0 b bi b bf t tf velocity to start. vti m vb bi (v −vb)+ tfv bi t tf Substitute in result from relative velocities maintained but in opposite direction. vbi m v = m v − m v + m v Solve for the velocity of the lighter ball. b bi b tf b bi t tf 2m v b bi = v tf m b m t 2⋅1kg⋅3.8m /s = v = 7.2m /s Smaller ball will shoot into the air! tf 1kg+0.05kg When predatory birds catch prey out of midair, they hunt by chasing them from behind, not by sneaking up on them by ﬂying in the opposite direction - how come? m h0.6 kg m p0.4 kg vhi12 m/s H vpi7 m/s θ=-20º θ +y v P H hi +x P v pi Φ v h+p The hawk grabbing the pigeon out of the air is an inelastic collision with conservation of momentum - we can ﬁnd the pair’s velocity and direction after the catch: m vh hi v = (m p pi)v h p h+p videos we watched in class: Momentum is a vector so: https://youtu.be/Lfq8K3qE7q8 https://youtu.be/cvH_Qa5svSw In x: m h hisθ + m v = (p p m )v h p h+p cosφ https://youtu.be/cvH_Qa5svSw 0.6kg⋅12 m cos−20º+0.4kg⋅7 m https://youtu.be/xNYCldtpiW0 mh hiosθ + m ppi s s m vh+p,x vh+pcosϕ = = = 9.56 m h m p 0.6kg+0.4kg s In y: m h hinθ = (m +m hv p h+p sinφ m 2 m 2 m vh+p= v h+p,x vh+p,y ⎜9.6 ⎟ + ⎜.46 ⎟ = 9.87 ⎝ s ⎠ ⎝ s ⎠ s 0.6kg⋅12 m ⋅sin−20º m h hinθ s m vh+p,y= vh+dinφ = = = −2.46 φ = tan12.46= −14º m h m p 0.6kg+0.4kg s 9.6 What happens in the other orientation? m h0.6 kg m p0.4 kg H vhi12 m/s vpi-7 m/s θ=-20º θ +y v hi P H +x P v pi Φ v h+p In x: m v cosθ +m v = (m +m )v cosφ h hi p p h p h+p m m m v cosθ + m v 0.6kg⋅12 cos−20º+0.4kg⋅−7 m vh+p,x vh+pcosϕ = h hi p pi= s s = 3.9 m h m p 0.6kg+0.4kg s In y: m v sinθ = (m + m )v sinφ h hi h p h+p m m v sinθ 0.6kg⋅12 ⋅sin−20º m vh+p,y vh+dsinφ = h hi = s = −2.46 mh+ m p 0.6kg+0.4kg s m 2 m 2 m vh+p= v h+p,x+ h+p,y= ⎜3.9 ⎟ +⎜2.46 ⎟ = 4.6 ⎝ s⎠ ⎝ s ⎠ s −1−2.46 φ = tan = −0.6º 3.9 What does energy loss look like in each scenario? H H +y θ vhi P v P H +x vpi HP P Φ v Φ v 2 2 2 2 2 2 ⎛ m ⎞ ⎛ m ⎞ m 2 2 ⎛ m ⎞ ⎛ m ⎞ m vh+p = v h+p,x +v h+p,y = ⎜9.6 ⎟ + ⎜.46 ⎟ = 9.87 vh+p = v h+p,x +v h+p,y = ⎜ 3.9 ⎟ + ⎜.46 ⎟ = 4.6 ⎝ s ⎠ ⎝ s⎠ s ⎝ s ⎠ ⎝ s ⎠ s 1 2 ⎛ 1 2 1 2⎞ 1 2 ⎛ 1 2 1 2 ΔKE = 2(m h m )p h+p− ⎝ 2m h hi+ 2 m p pi⎠ = ΔKE = (m h m )p h+p −⎜ m h hi+ m ppi⎟ = 2 ⎝ 2 2 ⎠ ⎛ 2 ⎛ 2 2⎞⎞ ⎛ 2 ⎛ 2 2⎞⎞ 1 (0.6kg+0.4kg )⎜ 9.9m ⎟ − 0.6kg⋅ ⎜2 m ⎟ +0.4kg⋅ ⎜ m ⎟ = −4.3 J 1 (0.6kg+ 0.4kg)⎛4.6 m⎞ − 0.6kg⋅ 12 m ⎞ + 0.4kg⋅ 7 m ⎞ = −42.4 J 2⎝ ⎝ s ⎠ ⎝ ⎝ s ⎠ ⎝ s ⎠ ⎠⎠ 2 ⎝ ⎝ s⎠ ⎝ ⎝ s ⎠ ⎝ s ⎠ ⎠⎠ What does this mean for the hawk? F ⋅t = impulse = m v − m v = Fhawk:xcrashimpulsehawk m h h f:xmh hi:x hawk:xcrash hawk h h f:x h hi:x m m 0.6kg(9.87m cos(−14 )−12 m cos(−20º))= −1.02N ⋅s 0.6kg(4.6 cos−0.6º−12 cos−20º)= −4.00 N ⋅s s s s s Fhawk:ycrash impulshawk= mh h f:ymh hi:y Fhawk:crash impulsehawk:ymh h f:ym h hi:y m m m m 0.6kg(9.87 sin(−14 )−12 sin(−20º)) = −1.03N ⋅s 0.6kg(4.6 sin− 0.6º−12 sin− 20º) = 2.4 ⋅s s s s s 2 2 2 2 Jnet (1.02Ns ) ( 1.03Ns) =1.45Ns Jnet (.00Ns ) ( 2.4Ns ) = 4.66Ns Estimate that in the collision it takes 0.01 s for momentum transfer between hawk and pigeon to occur: impulse 1.45N ⋅s impulse 4.66N ⋅s hawk = Force = =145N hawk= Force = = 466N timecrash hawk 0.01s time crash hawk 0.01s 466N = 3.2 times more force is exerted on the hawk in the 145N head-on collision than in the chase. How to measure the speed of a speeding bullet (or fastball)? A ballistic pendulum. Use conservation of momentum in an inelastic collision paired with conservation of energy to ﬁnd the initial speed of the bullet/fastball. m1 m 1 2 Δh v1 v2 ! ! p i m v1 1 pf= m(+ 1 v 2) 2 Conservation of energy - kinetic en1rgy of 2 pendulum immediately after collision(bm1+ m v2) m2+ m(gh1 2 ) gravitational potential energy. 2 Relationship between velocity of the pendulum v = 2gh immediately after the collision and its maximum 2 height. ! ! pi= p f Conservation of momentum between projectile object and its inelastic collision with another 1 1ect. 1 v 2) 2 and the ﬁnal height of the pendulum after an proje(m 1 m 2 ) 2gh inelastic collision. v1= m 1 Elastic collision in a Basketball/tennisball drop - why’s the tennis ball do that? What is its ﬁnal velocity? m b1 kg vbiv = −ti −v ) bf tf Relative velocities are maintained but inverted m=0.05 kg after the collision. t vbi3.8m/s v = v −v bf tf bi vti0m/s m v = m v + m v Conservation of momentum if tennis ball has ~0 b bi b bf t tf velocity to start. vti m vb bi (v −vb)+ tfv bi t tf Substitute in result from relative velocities maintained but in opposite direction. vbi m v = m v − m v + m v Solve for the velocity of the lighter ball. b bi b tf b bi t tf 2m v b bi = v tf m b m t 2⋅1kg⋅3.8m /s = v = 7.2m /s Smaller ball will shoot into the air! tf 1kg+0.05kg Youare watching a hawk chase a pigeon over the train tracks behind DRL.The two birds are ﬂying at a constant height of about 11 m above the tracks and the hawk is ﬂying straight south while the pigeon ﬂies straight north. You remember from physics class that a pigeon has a mass of about 0.5 kg and a hawk has a mass of about 0.6 kg. Strangely, the hawk ends up catching the pigeon in a head-on attack at the instant when both animals are ﬂying relatively slowly,2.1 m/s for the hawk and 2.5 m/s for the pigeon. Find the velocity of the hawk and pigeon together just after the pigeon is caught in the hawk’s claws,then write a sentence or two describing the motion of the hawk and pigeon together about half a second after the hawk catches the pigeon. m h0.6 kg +y m p0.5 kg +x vhi2.1 m/s H v =-2.5 m/s P P pi H θ=0º θ Φ v hi v pi vh+p In x: m v + m v = (m + m )v h hi p p h p h+p In y: no momentum in y mh hi m vp p = vh+p (m h m )p m m The momentum of the pigeon and the hawk are equal and opposite 0.6kg⋅2.1 +0.5kg⋅−2.5 m before the collision so the pair have zero velocity after the collision. vh+p = s s ≈ 0 Soon as the hawk catches the pigeon, the two will literally fall out of (0.6+0.5)kg s the sky in free fall. Further reason why hawks generally prefer to hunt following their prey - it is dangerous the other way! (25 points).You are watching a hawk chase a pigeon over the train tracks behind DRL.The two birds are ﬂying at a constant height of about 10 m above the tracks and the hawk is ﬂying straight south while the pigeon ﬂies straight north. You remember from physics class that a pigeon has a mass of about 0.4 kg and a hawk has a mass of about 0.6 kg. Strangely,the hawk ends up catching the pigeon in a head-on attack at the instant when both animals are ﬂying relatively slowly,which is about 3 m/s for the pigeon and 2 m/s for the hawk Find the velocity of the hawk and pigeon together just after the pigeon is caught in the hawk’s claws,then write a sentence or two describing the motion of the hawk and pigeon together about half a second after the hawk catches the pigeon. mh=0.6 kg +y mp=0.4 kg +x hi=2 m/s H v =-3 m/s P P pi H θ=0º θ Φ v hi v pi v h+p In x: m v + m v = (m + m )v h hi p p h p h+p In y: no momentum in y m h him v p p = vh+p (m h m )p m m The momentum of the pigeon and the hawk are equal and opposite 0.6kg⋅2 +0.4kg⋅−3 m before the collision so the pair have zero velocity after the collision. vh+p = s s ≈ 0 Soon as the hawk catches the pigeon, the two will literally fall out of (0.6+0.5)kg s the sky in free fall. Further reason why hawks generally prefer to hunt following their prey - it is dangerous the other way! You are working for a small bicycle company. Your current project is in selecting the best wheels for two upcoming bicycle models. Your boss wants you to around the wheel hub, which has a diameter of 3 cm, and the weight is released.As gravity pulls the weight down, the wheel undergoes angular acceleration until the entire string is unwound.At that instant, you measure the angular velocity of the wheel to be 7 radians/sec. What is the moment of inertia of the wheel? m=1 kg T I + ω f7 rad/s ω=i rad/s m r h=1 m r=0.015 m + T mg h ΣF = mg−T = ma vf=ω r f Στ = T ⋅r = Iα v 2 = 2aΔh f T = mg− ma 2 v f = a (mg−ma r = )α 2Δh a (mg− ma r =)I (7 rad ⋅0.015m) 2 2 r s −3 m (mg− ma r ) a = = 5.5⋅10 2 = I 2⋅1m s a ⎛ m −3 m ⎞ 2 ⎝1kg⋅9.8 s2−1kg⋅5.5⋅10 s 2⎠(0.015m ) I = = 0.4kg⋅m 2 5.5⋅10−3m s2 You are working for a small bicycle company. Your current project is in selecting the best wheels for two upcoming bicycle models. Your boss wants you to experimentally determine the moment of inertia of a bicycle wheel and tire. You do this by tying a 1-kg weight to one end of a 1-m string. The string is wrapped around the wheel hub, which has a diameter of 3 cm, and the weight is released.As gravity pulls the weight down, the wheel undergoes angular acceleration until the entire string is unwound.At that instant, you measure the angular velocity of the wheel to be 7 radians/sec. What is the moment of inertia of the wheel? m=1 kg ωf=7 rad/s ωi0 rad/s h=1 m r=0.015 m h 0m 1 2 1 2 vf=ω ⋅rf mgh i mv f+ Iω 2 2 rad m 1 2 vf= 7 ⋅0.015m = 0.1 2(mgh −imv ) f s s I = 2 ⎛v ⎞2 f ⎝ r ⎠ 2 ⎛ m 1 ⎛ m ⎞ ⎞ 2⋅ ⎜kg⋅9.8 2⋅1m− ⋅1kg⋅ 0.1 ⎜ ⎟ ⎟ ⎝ s 2 ⎝ s ⎠ ⎠ 2 I = 2 = 0.4kg⋅m ⎛ 7 rad ⎞ ⎝ s ⎠ Exam covers: Circular motion a=v /r Conservation of energy Momentum Torque (includes angular kinematics) Energy with rotation •No rolling per se, covered in pre-Tday quiz Will post a longer problem set before the end of the day to work in class on Wednesday - strongly encourage study groups between now and class on Wednesday. Moments of inertia for common shapes: 1 ML2 12 L R 1 2 MR 2 R MR 2 R 2 2 MR 5 (sphere) What is the moment of inertia of this M object spun around the red axis? r m 2 2 2 2 I =Σ mo o= mr + mr = 2mr o 2 2 2 2 2 KE R1/2 Iω ( ) =1/2 2(r ω =)mr ω What is the moment of inertia of this M object spun around the red axis? b m 2 2 2 2 I =Σ mo o= mr + mr = 2mr o 2 2 2 2 2 KE R1/2 Iω ( ) =1/2 2(r ω =)mr ω Around the blue axis? I = m r = Mb + Mb = 2Mb 2 Σ o o o 2 2 2 2 2 KE R1/2 Iω ( ) =1/2 2(b ω = )b ω How high do we need to build our rollercoaster hill in order to eProcedure for energy conservation: makes it around the loop if it starts from rest at the top of the hill? •Identify the initial and ﬁnal states of begin the system. end •Deﬁne a zero point for gravitational potential energy •Identify contributions to kinetic and h potential energy in the initial and ﬁnal i R states h f 2R •Identify any losses to friction •Using this info, write down a version 0 m of the energy conservation equation 1 1 1 1 relevant to the system. mgh i kxi+ mvi−W frictionh f Xxf+ mv f •Solve for the variable of interest. 2X 2X X 2 2 begin end • Plug in numbers last mgh = mgh + mv 1 2 i f 2 f Remember: at top of loop: mv 2 ΣF = N + mg = , N = 0 at minimum speed. N mg r R gR = v =minimum speed at top f Now, plug this information into energy conservation statement: 1 2 1 5 mgh =img2R+ m ( )gR hi= 2R+ R = R = 2.5R 2 2 2 mgh = mg2R+ 1 mgR i 2 Indiana Jones and the rolling boulder of doom: mv 2 N + mg = R N mg v = gR h 2R crit 0 m Height of the ramp if the velocity starts from rest: mgh + 1 kx + 1 mv + 1Iω −W = mgh + 1 kx2 + 1mv 2 + 1Iω 2 i 2X i 2X i 2X i Xriction f 2X f 2 f 2 f 1 2 1 2 mgh = mg2R+ mv + f Iω 2 2 2 1 2 1 2 2⎛v f⎞ mgh = mg2R+ mv + ⋅ fmr ⎜ ⎟ 2 2 5 ⎝ r ⎠ gh = 2gR+ v + v 1 2 2 f 5 f gh = 2gR+ gR+ gR 1 2 5 h = 2.7R Indiana Jones and the rolling boulder of doom: mv 2 N + mg = R N mg v = gR h=2r 2R crit 0 m Initial speed of the boulder if the ramp is the same height as the top of the loop: PE + KE = PE + KE grav,i i grav,f f 1 2 1 2 1 2 1 2 mg2R+ mv 2 i 2 Iωi= mg2R+ mv +2 f 2 Iω f 2 2 1 2 1 2 2⎛v i 1 2 1 2 2⎛ vf⎞ mg2R+ mv + ⋅ imR ⎝ ⎠ = mg2R+ mv + ⋅ ⋅fr ⎝ ⎠ 2 2 5 R 2 2 5 r 1 2 1 2 1 1 2gR+ v + vi= 2gRi gR+ gR 2 5 2 5 0.7vi= 0.7gR vi= gR How/why does a yo-yo work? What propels it back up the string? First, we’ll need to know the dimensions and moment of inertia of the yo-yo: mhub0.04 kg hub0.04 m maxle.005 kg axle.005 m 0.01 m 0.08 m side view front view A yo-yo is three cylinders: two hubs and one axle. Since they all rotate on the same axis, the total I for the yo-yo will be the sum of all these parts. I = 1 mr 2 cylinder2 1 2 1 Iyo−yo= 2⋅ m r hub hub+ m axle axle Iyo−yo= I hub+ Ihub+ I axle 2 2 2 1 2 −5 2 Iyo−yo = 0.04kg⋅ 0(04m + ) ⋅0.005kg⋅(0.005m) = 6.4⋅10 kg⋅m 2 What is the linear speed of the yo-yo at the bottom of the 1-m string? 1 2 1 2 mgh =i mv +f Iω f 2 2 2 The yo-yo rolls down the string on the 1 2 1 ⎛ vf ⎞ mgh i mv f I⎜ ⎟ small axle - ω scales with linear 2 2 ⎝ ra ⎠ velocity and the smaller radius. 2mgh Using conservation of energy, we can ﬁnd the vf= translational velocity at the end of the strhub0.04 m I will depend oradiusof the axle, but the r =0.005 m 2 + m moment of the whole object. axle ra mhub0.04 kg 1m maxle0.005 kg 2⋅0.085kg⋅9.8 m ⋅1m 2mgh s2 vf= I = −5 2 = 0.79m /s 2 + m 6.4⋅10 kg⋅m +0.085kg ra (0.005m )2 Compare this velocity to just dropping it: m vf:drop= 2gh = 2⋅9.8 2 ⋅1m = 4.4 m /s s What fraction of the initial potential energy is translational kinetic energy? What fraction is rotational kinetic energy? 1 2 1 2 mgh i mv +f Iω f hub0.04 m 2 2 axle0.005 m 2 PE i mgh = i.085kg⋅9.8m /s ⋅1.0m = 0.83J mhub0.04 kg 1m 1 2 1 2 maxle.005 kg KE trans mv f ⋅0.085kg⋅ 0(79m / s ) = 0.026 J 2 2 2 2 1 2 1 ⎛ vf ⎞ 1 −5 2 ⎛ 0.79m / s ⎞ KE rot= Iω f = I⎜ ⎟ = ⋅6.4⋅10 kg⋅m ⋅ ⎜ ⎟ = 0.80 J 2 2 ⎝ ra ⎠ 2 ⎝ 0.005m ⎠ KE 0.026 J tran= = 0.03= 3% PE i 0.8 J KE rot 0.80J = = 0.96% PE i 0.83J A yo-yo works because of the mis-match between the size of the rolling axle (small) and the moment of the whole object (large). When these are mismatched, most of the gravitational potential energy is still stored as rotational energy at the bottom of the string - friction can then be reintroduced between the string and the axle, and the yo-yo will roll back up - rotational kinetic energy becomes gravitational potential energy again. You are building new a ﬁre safety device for Penn’s high rise dorms as part of a design competition on campus. Your idea is to have a long,heavy cylinder with rope wound around it mounted in the window ledge. One end of the rope is secured to the window ledge and the other end is wound tightly around the cylinder. When a ﬁre is detected,a mechanism will roll the cylinder off the window ledge with its long axis is parallel to the ground. The cylinder will fall straight down without touching the side of the building,while the rope unwinds from a point midway along the cylinder’s length. People will then be able to slide down the rope to safety,with the cylinder weighting down the rope’s end. As part of your design process,ﬁnd the time it takes the cylinder one half its mass times its radius squared.ight of the window from which it is released. The moment of inertia of a cylinder is I=1/2mr 2 Conservation of energy: Now use kinematics to ﬁnd the time to the bottom: 1 2 1 2 2 v2f mgh i mv f + Iω f v f 2ah = a 2 2 2h v 2 4 mgh = 1 mv + ⋅ mr ⋅ 2 ⎛ f⎞ gh h i 2 f 2 2 ⎝ r⎠ 3 2 a = 2h = 3 g 1 1 3 ghi= v f + vf2= v f2 2 4 4 h = 1 at2 2h 2 a = t 4 3gh = v f 2h 3h 0 m t = = 2 g 3 g You are building new a ﬁre safety device for Penn’s high rise dorms as part of a design competition on campus. Your idea is to have a long,heavy cylinder the cylinder. When a ﬁre is detected,a mechanism will roll the cylinder off the window ledge with its long axis is parallel to the ground. The cylinder will fall straight down without touching the side of the building,while the rope unwinds from a point midway along the cylinder’s length. People will then be able to the ground as a function of the height of the window from which it is released. The moment of inertia of a cylinder is one half its mass times its radius squared. I=1/2mr 2 Now using dynamics: Now use kinematics to ﬁnd the time to ΣF = ma Στ = Iα the bottom: 1 2 2h h = at = t T ΣF = T − mg = −ma 2 a Στ = T ⋅r = Iα 2h 3h t = 2 = g h T = mg− ma g ⎛ ⎞ 3 mg Στ = (g− ma r ) I ⎝ ⎠ r 1 2⎛ ⎞ (mg− ma r)= mr ⎜ ⎟ Same answer using 2 ⎝ r energy and dynamics - yay! 1 0 m g−a = 2 a 2 a = g 3 Your new ski-lift design was recently approved! As you’ll recall, the approved design involves a sled that carries people up a mountain with the force exerted by a lightweight cord running over a pulley at the top of the mountain, which is then attached to another weight hanging down a cliff on the far side of the mountain. The new design includes a massive pulley with a moment of inertia of 1.08•10 kg•m and radThe length of the ski slope is 1100 m and makes an angle of 30º with the horizontal; the sled starts from rest at the bottom of the slope. If the total mass of the sled with double the number of passengers is now 750 kg, what must the mass of the weight hanging down the mountain be so that the sled reaches the top of the slope at a now-safe speed of 5 m/s? Find m 2sing dynamics: I=1.08•10 kg•m 2 h=1100 m β=30º m 1750 kg ΣF 1 T −1m gsi1β = m a 1 r=1.0 m v = 5 m/s ΣF = m g−T = m a β 2 2 2 2 Στ = T r2−T r 1 Iα N + T2 h + r 1 T1 T1 I 2 + T1= m a1 m gs1nβ T2 m 1 m 2 T = m g− m a 2 2 2 ⎛ ⎞ Use kinematics to ﬁnd acceleration: (m 2− m a 2 )(m a+ m1gsinβ1r = I ⎝ ⎠ r 2 2 I a + m r(a+ gsinβ) v f v +2ai r 1 m 2 2 r(g−a) v f = a m 1.14⋅10−2 2 2h 1.08⋅10 kg⋅m 2 s + 750kg⋅1m(1.14⋅10−2m +9.8 m sin30º) 1m s2 s2 2 m 2 m −2m ( 5m /s ) m 1m(9.8 2−1.14⋅10 2 ) = a =1.14⋅10 −2 s s 2 2⋅1100m s m 21630 kg Your new ski-lift design was recently approved! As you’ll recall, the approved design involves a sled that carries people up a mountain with the force exerted by a lightweight cord running over a pulley at the top of the mountain, which is then attached to another weight hanging down a cliff on the 6 2 far side of the mountain. The new design includes a massive pulley with a moment of inertia of 1.08•10 kg•The length of them. ski slope is 1100 m and makes an angle of 30º with the horizontal; the sled starts from rest at the bottom of the slope. If the total mass of the sled with double the number of passengers is now 750 kg, what must the mass of the weight hanging down the mountain be so that the sled reaches the top of the slope at a now-safe speed of 5 m/s? Find m 2sing energy: I=1.08•10 kg•m2 h=1100 m β=30º m1=750 kg 1 2 1 2 1 2 m 2h = m g1sinβ + 2 m 1 + 2 m 2 + 2 Iω r=1.0 m β v = 5 m/s 2 1 2 1 2 1 ⎛ ⎞ h m 2h− m 2 = m g1sinβ + m1v + I⎜ ⎟ N + T2 2 2 2 ⎝ r r + 1 T1 T I 2 + 2 1 T2 1 2 1 2 1 ⎛ ⎞ m1g m 2 m 2gh− v )= m 1hsinβ + m1v + I⎜ ⎟ 2 2 2 ⎝ r 2 m ghsinβ + 1 m v + 1 I⎛ ⎞ 1 2 1 2 ⎝ r m2= 1 2 (gh− 2 v ) m2 kg 2 m 1 ⎛ m ⎞ 2 1 6 2⎛5m /s ⎞2 units : s = kg 750kg⋅9.8 2 ⋅1100msin30º+ 750kg⋅ 5 ⎜ ⎟ + 1.08⋅10 kg⋅m ⎜ ⎟ m 2 s 2 ⎝ s⎠ 2 ⎝ 1.0m ⎠ 2 m 2 2 =1630 kg s 9.8m ⋅1.0m− 1 ⎛5 m ⎞ s 2 2 ⎝ s⎠ You have been hired to design a safety system to protect drivers going down hills during an ice storm. The planned system consists of a bumper, which can be considered a stiff spring, at the bottom of the hill. In the scenario you are given, the car starts from rest at the top of a hill which makes an angle q with the horizontal. The distance that the car slides from the top of the hill until it is stopped by the spring is L. For the worst case scenario, assume that there is no frictional force between the car and road due to the ice. If the maximum compression of the spring from its equilibrium position is D, your job is to calculate the required spring constant k in terms of D, L and q. initial N ﬁnal (not to scale) - no friction so this distance doesn’t matter here. L N m b N Fs Fs=-kx 0 m q mbg mbg D 1 2 1 2 1 2 1 2 mgh i kx i xmv iWx= frh x f kx f + mv f Note that this is the same process 2x 2 2 2x with regard to conservation of 1 2 energy: mgLsinq = kD 2 m L 2mgLsinq kg⋅ 2⋅m 2 = k s = N D D m 2 m q 0 m Units check suggests this is correct! Because of your concern that incorrect science is being taught to children when they watch cartoons on TV, you have joined a committee which is reviewing sees him. She grabs a convenient vine and swings towards Tarzan, who has twice her mass, to save him. Luckily, the lowest point of her swing is just wherell tree, Tarzan is standing. When she reaches him, he grabs her and the vine. They both continue to swing to safety over the elephants up to a height which looks to be about 1/2 that of Jane’s original height. To decide if you are going to approve this cartoon, calculate the maximum height Tarzan and Jane can swing as a fraction of her initial height. m =t2m j 1 2 conservation of energy ﬁnds Jane’s velocity m gj = i m vj ji immediately before grabbing Tarzan 2 h i 0 m 2gh =iv ji conservation of momentum ﬁnds their m vj+ji v = mt ti v ( j t) f combined velocity after Jane grabs hi Tarzan - inelastic collision, so energy is - not- conserved. vji vf m ji 2gh =i3m v j f 0 m m j 2gh i = v f 3m j h conservation of energy -after- the collision has i h happened ﬁnds their maximum height after the f 2 collision. 0 m 1 ⎛ 2gh i⎞ ⋅3m j⎜ ⎟ = 3m ghj f 2 ⎝ 3 ⎠ 1/9 the original height (what would actually happen) is pretty different from 1/2 original 1 height (what cartoon shows). Deem this to h = h be inaccuracy in cartoon physics! f 9 i Schedule until Midterm #2: Monday: Angular kinematics, moment of inertia, and torque Wednesday: Quiz on momentum, solutions, lecture on rotational kinetic energy Friday: Conservation of energy with rotating objects Monday: Rolling motion using both torque and energy Wednesday: Review class Thursday: Midterm #2, 5:00 - 5:50 pm, A1 and A6, check Canvas for room assignment. Almost everything we’ve done so far has assumed sliding objects, sometimes with friction, sometimes not - this is obviously not how much of the modern world works... How do deal with turning/torquing/rolling motion? N mg R i

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