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MIT 8 01 - Mechanical Energy and Simple Harmonic Oscillator

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1Mechanical Energy and Simple Harmonic Oscillator8.01Week 09D1Summary: Change in Mechanical EnergyTotal force:Total work:Change in potential energy:Total work done is change in kinetic energy:Mechanical Energy Change:Conclusion:mechanical totalEKUΔ≡Δ+ΔtotalncWKU=Δ +Δtotal total totalcnc=+FFFGGG()final finaltotal total total totalcncinitial initialWd d=⋅= + ⋅∫∫Fr FF rGGGGGtotal totalncWUWK=−Δ + =Δfinaltotal totalcinitialUdΔ=− ⋅∫FrGGModeling the Motion using Force and Energy ConceptsForce and Newton’s Second Law:•Draw all relevant free body force diagrams•Identify non-conservative forces.•Calculate non-conservative workChange in Mechanical Energy:•Choose initial and final states and draw energy diagrams.•Choose zero point P for potential energy for each interaction in which potential energy difference is well-defined.•Identify initial and final mechanical energy.•Apply Energy Law.finalnc ncinitialWd.=⋅∫FrGGtotalncWKU=Δ +ΔMechanical Energy AccountingInitial state: Total initial kinetic energy  Total initial potential energy Total initial mechanical energyFinal state: Total final kinetic energy  Total final potential energy Total final mechanical energy Apply Energy Law: Wnc= Efinalmechanical− Einitialmechanicalinitial 1,initial 2,initialKK K=++⋅⋅⋅initial 1,initial 2,initialUU U= + +⋅⋅⋅Einitialmechanical= Kinitial+Uinitialfinal 1,final 2,finalKK K= + +⋅⋅⋅final 1,final 2,finalUU U=++⋅⋅⋅Efinalmechanical= Kfinal+Ufinal2Example: Energy ChangesA small point like object of mass m rests on top of a sphere of radius R. The object is released from the top of the sphere with a negligible speed and it slowly starts to slide. Find an expression for the angle θwith respect to the vertical at which the object just loses contact with the sphere.Energy Flow diagramsInitial state Final StateExample: Energy ChangesA small point like object of mass m rests on top of a sphere of radius R. The object is released from the top of the sphere with a negligible speed and it slowly starts to slide. Find an expression for the angle θwith respect to the vertical at which the object just loses contact with the sphere.mechanical mechanicalnc final initial0WE E==−⇒initial0K Uinitial=0mechanicalinitial0E Kfinal=12mvf2Ufinal=−mgR(1 − cosθf)mechanical 2final1(1 cos )2ffEmvmgRθ=−−0 = 0 −12mvf2− mgR(1 − cosθf)⎛⎝⎜⎞⎠⎟⇒12mvf2= mgR(1 − cosθf)Recall Modeling the Motion: Newton ‘s Second Law Define system, choose coordinate system. Draw force diagram. Newton’s Second Law for each direction. Example: x-direction Example: Circular motiontotalˆ:.2x2dxFmdt=i2totalˆ:.rvFmR=−rExample (con’t): Free Body Force DiagramNewton’s Second LawConstraint condition:Radial Equation becomesEnergy Condition:Conclusion:2ˆ:cosvNmg mRθ−=−r22ˆ:sindmg mRdtθθ=θ0atfNθθ==2cosffvmg mRθ=⇒12mvf2=R2mgcosθf12mvf2= mgR(1 − cosθf)(1 cos ) cos2ffRmgR mgθθ−=⇒122cos cos33ffθθ−⎛⎞=⇒ =⎜⎟⎝⎠3Table Problem: Loop-the-LoopAn object of mass m is released from rest at a height habove the surface of a table. The object slides along the inside of the loop-the-loop track consisting of a ramp and a circular loop of radius R shown in the figure. Assume that the track is frictionless. When the object is at the top of the track (point a) it pushes against the track with a force equal to three times it’s weight. What height was the object dropped from?Table Problem: Block-Spring System with FrictionA block of mass m slides along a horizontal surface with speed v0. At t=0 it hits a spring with spring constant kand begins to experience a friction force. The coefficient of friction is variable and is given by µk= bx where b is a constant. Find how far the spring has compressed when the block has first come momentarily to rest.Simple Harmonic MotionHooke’s LawDefine system, choose coordinate system.Draw free-body diagram.Hooke’s Lawspringˆkx=−FiG22dxkx mdt−=4Concept QuestionWhich of the following functions x(t) has a second derivative which is proportional to the negative of the function x(t) = Acos2πTt⎛⎝⎜⎞⎠⎟ x(t) = Ae−t /T x(t) = Aet/T x(t) =12at222?dxxdt∝−1.2.3.4.Concept Question 1. 3.2. 4.The first derivative of the sinusoidal functionis:x = Acos2πTt⎛⎝⎜⎞⎠⎟vx(t) = Acos2πTt⎛⎝⎜⎞⎠⎟vx= dx / dtvx(t) =−Asin2πTt⎛⎝⎜⎞⎠⎟ vx(t) =−2πTAsin2πTt⎛⎝⎜⎞⎠⎟ vx(t) =2πTAcos2πTt⎛⎝⎜⎞⎠⎟Simple Harmonic MotionEquation of Motion:Solution: Oscillatory with Period Position:Velocity:Initial Position at t = 0:Initial Velocity at t = 0:General Solution: 22dxkx mdt−= x = Acos2πTt⎛⎝⎜⎞⎠⎟+ Bsin2πTt⎛⎝⎜⎞⎠⎟ vx=dxdt=−2πTAsin2πTt⎛⎝⎜⎞⎠⎟+2πTBcos2πTt⎛⎝⎜⎞⎠⎟ x0≡ x(t = 0) = A vx,0≡ vx(t = 0) =2πTBx = x0cos2πTt⎛⎝⎜⎞⎠⎟+T2πvx,0sin2πTt⎛⎝⎜⎞⎠⎟2mTkπ=Period and Angular FrequencyEquation of Motion:Solution: Oscillatory with Period Tx -component of velocity:x -component of acceleration:Period:Angular frequency x = Acos2πTt⎛⎝⎜⎞⎠⎟+ Bsin2πTt⎛⎝⎜⎞⎠⎟vx≡dxdt=−2πTAsin2πTt⎛⎝⎜⎞⎠⎟+2πTcos2πTt⎛⎝⎜⎞⎠⎟ax≡d2xdt2=−2πT⎛⎝⎜⎞⎠⎟2Acos2πTt⎛⎝⎜⎞⎠⎟−2πT⎛⎝⎜⎞⎠⎟2Bsin2πTt⎛⎝⎜⎞⎠⎟=−2πT⎛⎝⎜⎞⎠⎟2x2222222dx mkx m m x k m Tdt T T kπππ⎛⎞ ⎛⎞−= =− ⇒= ⇒=⎜⎟ ⎜⎟⎝⎠ ⎝⎠ −kx = md2xdt2 ω≡2πT=km5Concept Question: Simple Harmonic Motion1. 2. 3. 4. A block of mass m is attached to a spring with spring constant k is free to slide along a horizontal frictionless surface.At t = 0 the block-spring system is stretched an amount x0 > 0from the equilibrium position and is released from rest. What isthe x -component of the velocity of the block when it first comes back to the equilibrium? vx=−x0T4 vx= x0T4 vx=−kmx0 vx=kmx0Example: Block-Spring System with No FrictionA block of mass m slides along a frictionless horizontal surface with speed vx,0. At t = 0 it hits a spring with spring constant k and begins to slow down. How far is the spring compressed when the block has first come momentarily to rest?Initial and Final ConditionsInitial state: and First comes to rest when Since at timeFinal position vx(t) = vx,0cos2πTt⎛⎝⎜⎞⎠⎟ x(t) =T2πvx,0sin2πTt⎛⎝⎜⎞⎠⎟ 2πTtf=π2


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MIT 8 01 - Mechanical Energy and Simple Harmonic Oscillator

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