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AP Physics C Review Mechanics CHSN Review Project This is a review guide designed as preparatory information for the AP1 Physics C Mechanics Exam on May 11, 2009. It may still, however, be useful for other purposes as well. Use at your own risk. I hope you find this resource helpful. Enjoy! This review guide was written by Dara Adib based on inspiration from Shelun Tsai’s review packet. This is a development version of the text that should be considered a work-inprogress. This review guide and other review material are developed by the CHSN Review Project. Copyright © 2009 Dara Adib. This is a freely licensed work, as explained in the Definition of Free Cultural Works (freedomdefined.org). Except as noted under “Graphic Credits” on the following page, the work is licensed under the Creative Commons Attribution-Share Alike 3.0 United States License. To view a copy of this license, visit http://creativecommons.org/licenses/by-sa/3.0/us/ or send a letter to Creative Commons, 171 Second Street, Suite 300, San Francisco, California, 94105, USA. This review guide is provided “as is” without warranty of any kind, either expressed or implied. You should not assume that this review guide is error-free or that it will be suitable for the particular purpose which you have in mind when using it. In no event shall the CHSN Review Project be liable for any special, incidental, indirect or consequential damages of any kind, or any damages whatsoever, including, without limitation, those resulting from loss of use, data or profits, whether or not advised of the possibility of damage, and on any theory of liability, arising out of or in connection with the use or performance of this review guide or other documents which are referenced by or linked to in this review guide. 1 AP is a registered trademark of the College Board, which was not involved in the production of, and does not endorse, this product. 1 “Why do we love ideal worlds? . . . I’ve been doing this for 38 years and school is an ideal world.” — Steven Henning Contents Kinematic Equations 3 Free Body Diagrams 3 Projectile Motion 4 Circular Motion 4 Friction 5 Momentum-Impulse 5 Center of Mass 5 Energy 5 Rotational Motion 7 Simple Harmonic Motion 8 Gravity 9 Graphic Credits • Figure 1 on page 3 is based off a public domain graphic by Concordia College and vectorized by Stannered: http://en.wikipedia.org/wiki/File:Incline.svg. • Figure 2 on page 3 is based off a public domain graphic by Mpfiz: http://en.wikipedia. org/wiki/File:AtwoodMachine.svg. • Figure 5 on page 7 is a public domain graphic by Rsfontenot: http://en.wikipedia.org/ wiki/File:Reference_line.PNG. • Figure 6 on page 7 was drawn by Enoch Lau and vectorized by Stannered: http://en. wikipedia.org/wiki/File:Angularvelocity.svg. It is licensed under the Creative Commons Attribution-Share Alike 2.5 license: http://creativecommons.org/licenses/by-sa/ 2.5/. • Figure 7 on page 8 is based off a public domain graphic by Mazemaster: http://en.wikipedia. org/wiki/File:Simple_Harmonic_Motion_Orbit.gif. • Figure 8 on page 9 is a public domain graphic by Chetvorno: http://en.wikipedia.org/ wiki/File:Simple_gravity_pendulum.svg. 2 Figure 1: Normal Force Kinematic Equations 1 ∆x = at2 + v0 t 2 Figure 2: Atwood’s Machine ∆v = at (v)2 − (v0 )2 = 2a(∆x) Figure 3: Draw a banked curve diagram v0 + v ∆x = ×t 2 Pulled Weights Free Body Diagrams a= F−f Σm N Normal Force f Frictional Force T = ma T Tension mg Weight Elevator Normal force acts upward, weight acts downward. F = ma • Accelerating upward: N = |ma| + |mg| In a particular direction: • Constant velocity: N = |mg| ΣF = (Σm)a • Accelerating downward: N = |mg| − |ma| Atwood’s Machine2 a= 2 Pulley Banked Curve |(m2 − m1 )|g m1 + m2 Friction can act up the ramp (minimum velocity when friction is maximum) or down the ramp (maximum velocity when friction is maximum). videal = and string are assumed to be massless. 3 p rg tan θ Range s vmin = rg(tan θ − µ) µ tan θ + 1 θ represents the smaller angle from the x-axis to the direction of the projectile’s initial motion. Starting from a height of x = 0: s vmax = rg(tan θ + µ) 1 − µ tan θ xmax = Projectile Motion Circular Motion Position Centripetal (radial) (v0 )2 sin 2θ g Centripetal acceleration and force is directed towards the center. It refers to a change in direction. ∆x = vx t 1 ∆y = − gt2 + (vy )0 t 2 ac = v2 r Velocity Fc = mac = θ represents the smaller angle from the x-axis to the direction of the projectile’s initial motion. mv2 r Tangential (vx )0 = v0 cos θ Tangential acceleration is tangent to the object’s motion. It refers to a change in speed. (vy )0 = v0 sin θ at = d|v| dt ∆vx = 0 Combined ∆vy = −gt atotal = q (ac )2 + (at )2 Height θ represents the smaller angle from the x-axis to Vertical loop the direction of the projectile’s initial motion. In a vertical loop, the centripetal acceleration is Starting from a height of x = 0: caused by a normal force and gravity (weight). ymax = (v0 sin θ)2 2g 4 Top Elastic Kinetic energy is conserved. F = ma N + mg = m × N = v2 r m1 v1 + m2 v2 = m1 v10 + m2 v20 mv2 − mg r −(v20 − v10 ) = v2 − v1 Bottom Inelastic F = ma Kinetic energy is not conserved. v2 N − mg = m × r 2 mv + mg N = r m1 v1 + m2 v2 = (m1 + m2 )v0 Center of Mass Friction Friction converts mechanical energy into heat. Static friction (at rest) is generally greater than kinematic friction (in motion). rcm = λ= fmax = µN Σm = p = mv Z dm = λdx (Σm)vCM = Σmv = Σp dp dt Fnet = (Σm)aCM Z I= dm M = dx L Z Momentum-Impulse F= Z Z 1 1 Σmr rdm = xλdx = Σm Σm Σm Energy Fdt = F∆t = ∆p = m∆v Work Collisions Z Total momentum is always conserved when there are no external forces (F = dp dt = 0). 5 W= Fdx = ∆K Power Pavg = Fx W = t t Pinstant = dW = Fv dt Kinetic Energy Linear 1 K = mv2 2 Potential Energy dU F=− dx Z xf ∆U = − FC dx = −WC Z UHooke v= dx dt = ∆x ∆t ω= dθ dt = ∆θ ∆t a= dv dt = ∆v ∆t α= dω dt = ∆ω ∆t ∆x = 12 at2 + v0 t ∆θ = 12 αt2 + ω0 t ∆v = at ∆ω = αt (v)2 − (v0 )2 = 2a(∆x) (ω)2 − (ω0 )2 = 2α(∆ω) ∆x = xi Z 1 = − FHooke dx = − −kxdx = kx2 2 Ug = mgh equilibrium point F = − du dx = 0 (extrema) stable equilibrium U is a minimum v0 + v ×t 2 ∆θ = ω0 + ω ×t 2 F = ma Rx W = x0 Fdx τ = Iα Rθ Wrot = θ0 τdθ W = 12 mv2 − 12 m(v0 )2 Wrot = 21 mω2 − 12 m(ω0 )2 P = Fv Prot = τω p = mv L = Iω F= unstable equilibrium U is a maximum Angular dp dt τ= Figure 4: Rotational Motion Total E = K+U Ei + WNC = Ef WNC represents non-conservative work that converts mechanical energy into other forms of energy. For example, friction converts mechanical energy into heat. 6 dL dt Torque τ = r × F = rF sin θ τ = Iα Moment of Inertia Z 2 I = Σmr = r2 dm Figure 5: Arc Length I = Icm + Mh2 (h represents the distance from the center) Values 1 2 12 ml 1 2 3 ml rod (center) Figure 6: Angular Velocity rod (end) hollow hoop/cylinder mr2 Rotational Motion 1 2 2 mr The same equations for linear motion can be mod- hollow sphere 2 mr2 3 ified for use with rotational motion (Figure 4 on 2 solid sphere 5 mr2 the previous page). solid disk/cylinder Atwood’s Machine Angular Motion θ= s r ω= v r α= at r at = r p a= |(m2 − m1 )|g m1 + m2 + 12 M Angular Momentum L = Iω L = r × p = rp sin θ = rmv sin θ α2 + ω4 ac = ω2 r τ= 1 1 Krolling = Iω2 + mv2 2 2 dL dt Total angular momentum is always conserved when there are no external torques (τ = dL dt = 0). 7 ω = 2πf r A= (x0 )2 + φ = arctan v 2 0 ω −v0 ωx0 1 E = kA2 2 Spring Figure 7: Simple Harmonic Motion Fs = −kx Simple Harmonic Motion r Simple harmonic motion is the projection of uniform circular notion on to a diameter. Likewise, uniform circular motion is the combination of simple harmonic motions along the x-axis and y-axis that differ by a phase of 90◦ . Ts = 2π r ωs = m k k m amplitude (A) maximum magnitude of displacePendulum ment from equilibrium cycle one complete vibration Simple period (T ) time for one cycle s frequency (f) cycles per time T = 2π angular frequency (ω) radians per time r x = Acos(ωt + φ) ω= A cable with a moment of inertia swings back and forth. d represents the distance from the pendulum’s pivot to its center of mass. s I T = 2π mgd a = −ω2 A cos(ωt + φ) = −ω2 x 2π 1 = ω f r f= g L Compound v = −ωA sin(ωt + φ) T= L g 1 ω = T 2π ω= 8 mgd I frictionless pivot Energy amplitude θ U= massless rod E= bob's trajectory massive bob equilibrium position −Gm1 m2 R −GMm 2r v= Figure 8: Simple Pendulum 2πR T r vescape = Torsional 2GM re For orbits around the earth, re represents the raA horizontal mass with a moment of inertia is dius of the earth. suspended from a cable and swings back and forth. r T = 2π ω= I k k I Gravity F= −Gm1 m2 R2 G ≈ 6.67 × 10−11 Nm2 kg2 Kepler’s Laws 1. All orbits are elliptical. 2. Law of Equal Areas. 2 4π 3. T 2 = GM R3 = Ks R3 , where Ks is a uniform constant for all satellites/planets orbiting a specific body 9