Elec and Magnetism Waves and Optics Lab

General Physics I - Mechanics

General Physics II - Electricity and Magnetism

General Physics III - Waves Fluids and Optics

Lecture 07 - Error Analysis  Rejection of Data o First, calculate the mean and the standard deviation o Assuming a Gaussian distribution, this means that 95% of trials should fall within 2 standard deviations o You can calculate the exact value with the t-value calculation:  t sus = | x sus − x | σ o Look up what this corresponds to in the probability t-score table  Probability Outside 2σ = 1 – Probability Within 2σ  Weighted Averages o x wav = ∑ w i x i ∑ w i o σ wav = 1 √ ∑ w i

PHYS | 2CL | Elec and Magnetism Waves and Optics Lab | lecture notes | Lecture 07 - Error Analysis  

Oscillaons F spring =− kx where k = Spring Constant KE spring = − 1 2 kx 2 ω = 2 π T = 2 πf = | v | | r | T = 2 π √ I τ Meter sck pendulum problem, τ = mgh L = rp = rmv ( single particle ) where L = Linear Momentum, r = Posion from Origin, p = momentum L = Iω ( system of particles ) where L = Linear Momentum, I = Inera α = dω dt = τ I = a r F friction = Ia Rotational Energy = 1 2 Iω 2 x = x M cos ( ωt + φ ) v =− x M sin ( ωt + φ ) ω a =− x M cos ( ωt + φ ) ω 2

PHYS | 2A | General Physics I - Mechanics | study guides | Oscillaons F spring =− kx whe

Lecture 01 – Introduction and Errors  Errors o Every measurement we make in lab will look like x = x best ± δx  x is the “measured value” of x  x best is the best estimate we can make of x  δx is the uncertainty OR error in the measurement o When taking measurements, take them to the unit less than the smallest unit, then estimate the uncertainty reasonably  So if our ruler we were using had a smallest increment of 1 cm, then we would measure to 0.1 cm, and write an uncertainty to the same decimal place o If we were to measure 2.3s, 2.4s, 2.5s, and 2.4s, a reasonable value to put would be T = 2.4 ± 0.1 s  Significant Figures o g = 9. 82 ± 0.0283 m s 2 →g = 9.82 ± 0.02 m s 2 o g = 9. 82378 ± 0.02 m s 2 →g = 9.82 ± 0.02 m s 2 o So if we measured v = 6051.78 ± 32 m s  First use sig figs on uncertainty: v = 6051.78 ± 30 m s  Then adjust based estimate based on leading value of uncertainty: v = 605 0 ± 30 m s  Propagation of Uncertainties o If x = x best ± δx and y = y best ±δy and q = f ( x,y ) , then:  δq = √ ( ∂q ∂x δx ) 2 + ( ∂q ∂y δy ) 2 

PHYS | 2CL | Elec and Magnetism Waves and Optics Lab | lecture notes | Lecture 01 – Introduction and 

PHYSICS 2A Prof: Kim Griest (last updated: 12/01/11) Formula sheet Constants and Factors Speed of light: c = 299 , 792 , 458 m/s exactly (about 3 × 10 8 meters/sec) Newton’s constant G =6 . 67 × 10 - 11 m 3 /s 2 kg Earth constants: Acceleration of gravity at surface: g =9 . 8m / s 2 , Mass: M Earth =5 . 97 × 10 24 kg, radius: r Earth =6 . 37 × 10 6 m Mass of proton and neutron about 1.67 × 10 - 27 kg Mass of electron: m e =9 . 11 × 10 - 31 kg Density of air: ρ =1 . 2 kg/m 3 ; Density of water: ρ = 1000 kg/m 3 ; 1 dyne = 10 - 5 Newtons 1 mile = 1609 m; 1 foot = 0.3048 m; 1 foot = 12 inches; 1 mile = 5280 ft 1 pound (lb) = 4.448 Newton, corresponding to the weight from mass of 0.454 kg; 1 ton = 2000 lb Newton = kg m /s 2 ; Joule = Newton m; Watt = Joule/sec Formulas Velocity as a derivative of position: ~v = d~ r/dt Acceleration as a derivative of velocity: ~a = d~v/dt = d 2 ~ r/dt 2 For constant acceleration: ~v = ~v 0 + ~at For constant acceleration: ~ r = ~ r 0 + ~v 0 t + 1 2 ~at 2 For constant acceleration in a straight line: v 2 = v 2 0 +2 a ( x - x 0 ) Frame of reference: ~v 0 = ~v - ~ V ; ~v is velocity w.r.t. frame S , ~v 0 is w.r.t. frame S 0 which moves at ~ V w.r.t. frame S Projectile trajectory: (start at x = 0, y = 0, speed v 0 , angle θ 0 : y = x tan θ 0 - gx 2 / (2 v 2 0 cos 2 θ 0 ) Range of projectile above: x =( v 2 0 /g ) sin 2 θ 0 Circular motion at constant speed: a = v 2 /r , toward center of circle Non-uniform circular motion: radial: a r = v 2 /r , tangential: a t = dv/dt Newton’s force law: ~ F net = d~ p/dt , where momentum is ~ p = m~v ; or if constant mass: ~ F net = m~a Weight: ~ W = m~g Hooke’s law for a spring: F = - kx , where k is the spring constant Friction: Static: F s ≤ μ k N ; Kinetic: F k = μ k N ; N is the Normal Force Drag Force: F D = 1 2 C D ρAv 2 Work ( constant force in 1-D ): W = F x Δ x ; Work ( variable force in 3-D ): W = R r 2 r 1 ~ F · d~ r Kinetic Energy: K = 1 2 mv 2 , Work-Energy theorem: W net =Δ K Power: P = dW/dt ; P = ~ F · ~v Conservative forces: H ~ F · d~ r = 0; Diﬀerence in potental energy is independent of path taken. Potential Energy: Δ U AB = - R B A ~ F · d~ r ; in 1-D: F x = - dU/dx Potential Energy: gravitational near Earth surface: U = mgh ; spring elastic: U = 1 2 kx 2 ; gravita- tional in general U = - GM m/r Newton’s law of Gravity: ~ F = - GMm r 2 ˆ r Orbital period: T 2 =4 π 2 r 3 / ( GM ) Escape velocity: v esc = p 2 GM/r 1 Rotational equations: ω = dθ/dt , α = dω/dt , v t = ωr , a t = αr Torque: ~ τ = ~ r × ~ F = d ~ L/dt = rF sin θ ; direction given by Right Hand Rule Rotational analog of Newton’s law: τ = Iα , where moment of inertia, I = ∑ m i r 2 i for discrete masses and I = R r 2 dm for continuous masses Rotational kinetic energy: K rot = 1 2 Iω 2 ; W rot = τdθ Some moment of inertias : Solid sphere about center: I = 2 5 MR 2 ; Hollow sphere about center: I = 2 3 MR 2 ; Solid cylinder about axis: I = 1 2 MR 2 ; Hollow cylinder about axis: I = MR 2 ; Thin rod about center: I = 1 12 Ml 2 ; Thin rod about end I = 1 3 Ml 2 Angular momentum: ~ L = ~ r × ~ p = I~ ω ; direction given by Right Hand Rule ~ A × ~ B = ˆ i ( A y B z - A z B y ) - ˆ j ( A z B x - A x B z )+ ˆ k ( A x B y - A y B x )= AB sin θ (direction by RHR) Static Equilibrium; ∑ ~ F i = 0 and ∑ ~ τ i =0 Center of mass: ~ r CM =( ∑ m i ~ r i ) /M tot 2

PHYS | 2A | General Physics I - Mechanics | study guides | PHYSICS 2A Prof: Kim Griest (l

Chapter 11 Rolling, Torque, and Angular Momentum http://dx.doi.org/10.1051/0004-6361:200810488 11.2 Rolling as Translation and Rotation Combined Although the center of the rolling disk moves in a straight line parallel to the surface, a point on the rim certainly does not. The point on the rolling disk rim in contact with the surface is momentarily at rest. This motion can be studied by treating it as a combination of translation of the center of mass and rotation of the rest of the object around that center. 11.2 Rolling without slipping ( “smooth rolling” ) 11.4: The Forces of Rolling: Friction and Rolling A wheel rolls horizontally without sliding while accelerating with linear acceleration a com . A static frictional force f s acts on the wheel at P , opposing its tendency to slide. If the wheel does slide when the net force acts on it, the frictional force that acts at P is a kinetic frictional force, f k .The motion then is no longer rolling without slipping (“smooth rolling”), and the above relations does not apply to the motion. (note: no obvious relation between direction of a com ; see later, example of walking yo-yo)  11.3 The Kinetic Energy of Rolling If we view the rolling as pure rotation about an axis through P , then ( ω is the angular speed of the wheel and I P is the rotational inertia of the wheel about the axis through P ). Using the parallel-axis theorem ( I P = I com + Mh 2 ), ( M is the mass of the wheel, I com is its rotational inertia about an axis through its center of mass, and R is the wheel’s radius, at a perpendicular distance h =R ). Using the relation v com = ω R , we get: A rolling object, therefore has two types of kinetic energy: 1.a rotational kinetic energy due to its rotation about its center of mass (= ½ I com ω 2 ), 2.and a translational kinetic energy due to translation of its center of mass (= ½ Mv 2 com ) F net ,x = ma x 11.4: The Forces of Rolling. Example: Rolling Down a Ramp A round uniform body of radius R rolls down a ramp. The forces that act on it are the gravitational force F g , a normal force F N , and a frictional force f s pointing up the ramp. Sample problem: Rolling Down a Ramp A uniform ball, of mass M = 6.00 kg and radius R , rolls smoothly from rest down a ramp at angle θ = 30.0°. a) The ball descends a vertical height h = 1.20m to reach the bottom of the ramp. What is its speed at the bottom? Calculations: Use conservation of energy (no slipping means no dissipation): where I com is the ball’s rotational inertia about an axis through its center of mass, v com is the requested speed at the bottom, and ω is the angular speed there. b) What are the magnitude and direction of the frictional force on the ball as it rolls down the ramp? Calculations: First we need to determine the ball’s acceleration a com, x : We can now solve for the frictional force: K f + U f = K i + U i ( 1 2 I com ω 2 + 1 2 Mv 2 com )+0=0+ Mgh Substituting v com / R for ω , and 2/5 MR 2 for I com ,

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