Study Guide For Exam Two


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1 Study Gude For Exam Two Physcs 2210 Albretsen Updated: 08/02/2018 All Other Prevous Study Gudes Modules Module 07 Work Work done by a constant force F over a dstance s : Work done by varyng force F over dstance ds : SI Unts are Joules [J] = [N m] Hook's Law for Sprngs Example of a varyng force. Ideal case (frctonless and massless sprng) s a conservatve force. x = Dsplacement from sprng equlbrum. k = Sprng constant. Negatve = Restorng Force. Work done by sprng for arbtrary dsplacement: Knetc Energy W = F s = Fs cos W = F d s F sprng = kx W sprng = 1 2 k x k x 2 f Energy assocated wth a movng object. SI Unts: Joules [J] K = 1 2 m v2 WorkEnergy Theorem: Vald for both varyng and nonvaryng forces. W = Δ K = K f K Power K f K = 1 2 m v 2 f 1 2 m v 2 The rate at whch work s done or the tme rate of energy transfer. Average Power: P= Δ W Δ t
2 Instantaneous Power: P= F v SI Unts: Watts [W] = [J/ s] = [ kg m 2 / s 3 ] Potental Energy Module 08 Descrbes energy due to object's poston n space. Where potental energy s equal to zero s completely arbtrary What matters s the dfference. Examples: Gravtatonal Potental Energy Energy stored n the gravtatonal feld defned as: U gravty mgy Dfference n gravtatonal potental energy when an object changes heght: Δ U =U f U = mgy f mgy Work done by the gravtatonal force on a fallng object: W gravty = mgy mgy f = Δ U gravty Vald as long as y << Earth's radus (can assume g s constant). Sprng Elastc Potental Energy Energy stored n an deal sprng an arbtrary dstance x from equlbrum: U sprng 1 2 k x2 Dfference n sprng elastc potental energy between two ponts: Δ U =U f U = 1 2 k x 2 f 1 2 k x 2 Work done by the sprng force between two ponts: W sprng = 1 2 k x k x 2 f = Δ U sprng Conservatve Forces Occurs when: Work done on a partcle between two ponts s ndependent of path. Work done movng through a closed path s zero. Examples: Gravty Ideal Sprng
3 NonConservatve Forces Causes a change (reducton) n the total mechancal energy. Examples: Frcton Ar Resstance Conservaton of Mechancal Energy Knetc energy and potental energy are types of mechancal energy. The total mechancal energy of an solated system remans constant. If objects n that system nteract only through conservatve forces: Can also rewrte as: E = E f K +U = K f +U f Δ K +Δ U = 0 If there are nonconservatve forces such as frcton nvolved, can express the loss n mechancal energy as Knetc Frcton (ΔK NC ): Knetc Frcton K +U + Δ K NC = K f +U f Δ K +Δ U =Δ K f Δ K NC = f k s Knetc energy lost by constant frcton force f k actng n the opposte drecton of dsplacement s. Lnear Momentum Module 09 Component equatons: p= m v p x =m v x p y = m v y p z = m v z Relaton to Newton's Second Law: F x = dp x SI Unts: [ kg m / s] F= d p F y = dp y F z = dp z
4 Conservaton of Lnear Momentum Assume the system solated from outsde forces. Each mass contrbutes to the total momentum. If the tme dervatve of the total momentum s zero, the momentum must be constant. For example, for a twopartcle system where masses DO NOT stck: p total = p 1 p 2 = constant p = p f p 1 p 2 = p 1f p 2f x components: y components: z components: p 1 x + p 2 x = p 1 fx + p 2 fx m 1 v 1 x v 2 x = m 1 v 1 fx v 2 fx p 1 y + p 2 y = p 1 fy + p 2 fy m 1 v 1 y v 2 y = m 1 v 1 fy v 2 fy p 1 z + p 2 z = p 1 fz + p 2 fz m 1 v 1 z + m 2 v 2 z =m 1 v 1 fz v 2 fz Impulse In collsons, we can use the mpulse approxmaton : Can treat the tmeaveraged force as constant Assume the appled force exerted acts for a short perod of tme. Appled force s much larger than any other forces present. I = Δ p = F Δ t Types of Collsons Perfectly Inelastc Collsons Total lnear momentum s conserved. Total knetc energy s NOT conserved. Masses stck together after collson treat as one mass after collson. Inelastc Collsons Total lnear momentum s conserved. Total knetc energy s NOT conserved. Masses DO NOT stck together after collson. Elastc / Perfectly Elastc Total lnear momentum s conserved. Total knetc energy s conserved. Masses DO NOT stck together after collson.
5 Module 10 Rotatonal Moton Quanttes Arc Length Angular Dsplacement Average Angular Velocty Instantaneous Angular Velocty Average Angular Acceleraton Instantaneous Angular Acceleraton Defntons s = r θ Δ θ θ f θ ω Δ θ Δ t = θ f θ t f t ω lm Δ θ Δ t 0 Δ t = d θ ᾱ Δ ω Δ t = ω f ω t f t α lm Δ ω Δ t 0 Δ t = d ω = d2 θ 2 Angle s measured n radans: Rotatng counter clockwse s postve. Rotatng clockwse s negatve. 2 π radans = 360 degrees =1 revoluton Rotatonal Moton θ= θ o + ω o t α t 2 θ= θ o (ω o + ω) t ω = ω o + α t ω 2 = ω o α (θ θ o ) Lnear Moton x = x o + v o t a t 2 x = x o (v o + v) t v = v o + at v 2 = v o a (x x o ) Relatonshp between angular and lnear quanttes of a partcle n unform crcular moton: v t = r ω a t =r α Rotatonal Knetc Energy Moment of Inerta (I) of rgd object made of n partcles: n I = m r 2 Rotatonal knetc energy of a rotatng rgd object: K R = 1 2 I ω2
6 Module 11 Torque Force F exerted on an object about an axs at a dstance r tends to rotate the object. SI Unts: Newton meter: [N m] (Not joules) Sum each torque that contrbutes to the rotaton and a rgd object to fnd the net torque: τ= I α Instantaneous Power of a rotatng rgd object: WorkEnergy Theorem for a rotatng rgd object: Rollng Moton τ = r F τ= r F sn (θ) P= τ ω W = τ θ=δ K Δ K = 1 2 I ω 2 f 1 2 I ω 2 Must account for both rotatonal moton AND translatonal moton. Translatonal moton: v cm = R ω a cm = R α Total knetc energy: K = 1 2 I ω m v 2 cm Angular Momentum A partcle of mass m and velocty v about an axs at a dstance r has angular momentum about the axs: L= r p= r m v L= mvr sn (ϕ) The torque actng on a partcle s the tme rate of change of the angular momentum: τ = d L Angular momentum of a rgd object rotatng about an axs: L= I ω
7 Conservaton of Angular Momentum If the external net torque actng on a system s zero, then the total angular momentum s constant: Therefore: τ ext = d L = 0 L = constant L = L f I ω = I f ω f Please send comments or correctons to jakealbretsen@weber.edu
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