General Physics I - Mechanics

General Physics II - Electricity and Magnetism

General Physics III - Waves Fluids and Optics

Elec and Magnetism Waves and Optics Lab

a ·  b = a x b x + a y b y a =( a x ,a y )  b =( b x ,b y ) a ·  b = ab cos φ 3.8 Multiplying Vectors B. Multiplying a vector by a vector: Scalar (Dot) Product The scalar product between two vectors is defined as and Here, a and b are the magnitudes of vectors a and b respectively, and ϕ is the angle between the two vectors. The right hand side is a scalar quantity. It is easy to show that (this is not only the definition, it is also the easiest way to compute it) (an alternative expression with nice geometrical meaning; see next slide) a ·  b =0 Geometrical interpretation: Each vector has a component along the other: Corollary: If a and b are perpendicular then

PHYS | 2A | General Physics I - Mechanics | lecture notes | a ·  b = a x b x + a y b y 

13.4: Gravitation Near Earth’s Surface For example, effect of Earth’s rotation: Example, Difference in Accelerations 13.5: Gravitation Inside Earth We stated before that a uniform shell of matter exerts no net gravitational force on a particle located inside it. This is because the attraction from diametrically opposed regions of the shell cancels. Sample Problem Three explorers attempt to travel by capsule through a tunnel directly from the south pole to the north pole. According to the story, as the capsule approaches Earth’s center, the gravitational force on the explorers becomes alarmingly large and then, exactly at the center, it suddenly but only momentarily disappears. Then the capsule travels through the second half of the tunnel, to the north pole. Check this story by finding the gravitational force on the capsule of mass m when it reaches a distance r from Earth’s center. Assume that Earth is a sphere of uniform density ρ (mass per unit volume). Calculations: The force magnitude depends linearly on the capsule’s distance r from Earth’s center. Thus, as r decreases, F also decreases, until it is zero at Earth’s center. 13.6: Gravitational Potential Energy The gravitational potential energy of the two- particle system is: U ( r ) is negative and approaches zero as r approaches infinity (convention on additive constant). If the system contains more than two particles, consider each pair of particles in turn, calculate the gravitational potential energy of that pair with the above relation, as if the other particles were not there, and then algebraically sum the results (bring each one from infinity). That is, 13.6: Gravitational Potential Energy Let us shoot a baseball directly away from Earth along the path in the figure. We want to find the gravitational potential energy U of the ball at point P along its path, at radial distance R from Earth’s center. The work W done on the ball by the gravitational force as the ball travels from point P to a great (infinite) distance from Earth is: where W is the work required to move the ball from point P (at distance R ) to infinity. Work can also be expressed in terms of potential energies as 13.6: Gravitational Potential Energy Path Independence The work done along each circular arc is zero, because the direction of F is perpendicular to the arc at every point. Thus, W includes only contributions along the radial lengths. An arbitrary path can be broken into a sequence of infinitely many infinitesimal radial/arc segments. Hence the gravitational force is a conservative force: the work done by the gravitational force on a particle moving from an initial point i to a final point f is independent of the path taken between the points.

PHYS | 2A | General Physics I - Mechanics | lecture notes | 13.4: Gravitation Near Earth’s

11.5: The Yo-Yo F y = ma y ⇒ T sin θ + N − mg =0 F x = ma x ⇒ T cos θ − f = ma x τ = I α ⇒ fr 2 − Tr 1 = I α = Ia x /r 2 a x = T m + I/r 2 2 (cos θ − r 1 /r 2 ) 11.5: Walk the Yo-Yo: the demo f Eliminating f from the last two: At small angle yo-y walks right, at large angle yo-yo walks left and at critical angle it will slide once the tension is large enough to exceed the force of static friction!!  c =  a ×  b The right-hand rule allows us to find the direction of vector c . It is perpendicular to both vectors a and b, directed as in right had rule. Back to Chapter 3: 3.8: Multiplying Vectors C. Multiplying a vector with a vector: Vector (Cross) Product The result is a new vector c , of magnitude: Here a and b are the magnitudes of vectors a and b respectively, and φ is the smaller of the two angles between a and b vectors. 3.8: Multiplying Vectors: Vector product in unit-vector notation Note that: And, Sample Problem: Vector Product Sample Problem: Vector Product Sample Problem: Vector product, unit vector notation 11.6: Torque Revisited Fig. 11-10 (a) A force F , lying in an x - y plane, acts on a particle at point A . (b) This force produces a torque τ = r x F on the particle with respect to the origin O . By the right-hand rule for vector (cross) products, the torque vector points in the positive direction of z . Its magnitude is given by in (b) and by in (c). • Refers to a particular choice of origin (that is, a choice of coordinate system) • Does not involve a real axis of rotation • Can be used for a single particle • Can be used for a non-rigid collection of particles (not necessarily a rigid body) We will see later how this relates to our previous discussion of torque for a rigid body about a fixed axis of rotation.  τ net =  τ net , 1 +  τ net , 2 + ··· +  τ net ,n = n  i =1  τ net ,i  τ net , 1 +  τ net , 2 =  r 1 × (  F ext , 1 +  F 12 )+  r 2 × (  F ext , 2 +  F 21 ) =  τ ext , 1 +  τ ext , 2 +  r 1 ×  F 12 +  r 2 ×  F 21 =  τ ext , 1 +  τ ext , 2 +  r 1 × ( −  F 21 )+  r 2 ×  F 21 =  τ ext , 1 +  τ ext , 2 +(  r 2 −  r 1 ) ×  F 21 =  τ ext , 1 +  τ ext , 2 For a system (a collection of particles): Do internal forces cancel (as they did for total net force)? Consider pair 12 only: The last line is correct if F 21 is parallel to r 2 − r 1 . This is a vector from 1 to 2, and often (but not always) the force between 1 and 2 is along the line joining them (strong form of Newton’s 3rd Law) Sample problem In the figure three forces, each of magnitude 2.0 N, act on a particle. The particle is in the xz plane at point A given by position vector r , where r = 3.0 m and θ = 30°. Force F 1 is parallel to the x axis, force F 2 is parallel to the z axis and force F 3 is parallel to the y axis. What is the torque, about the origin O , due to each force?

PHYS | 2A | General Physics I - Mechanics | lecture notes | 11.5: The Yo-Yo F y = ma y ⇒ T

Chapter 2 Motion along a straight line 2.2 Motion We find moving objects all around us. The study of motion is called kinematics . Specifically, the description of motion. Examples: • The Earth orbits around the Sun • A roadway moves with Earth’s rotation • Cars move on the roadway • Baseballs fly out to right field Forces cause motion. We will find out, as a result of application of force, if the objects speed up, slow down, or maintain the same rate. The study of what causes motion is called dynamics . The moving object here will be considered as a particle. A particle is an object (a “body”) without spatial extent. That is, in geometry, a point. If we deal with a stiff, extended object, we will assume that all particles on the body move in the same fashion. We will study the motion of a particle, that is, a fixed point in the body, which will represent the entire body. 2.2 Motion This is a mathematical model of reality. There are no objects with zero extent (except, maybe, fundamental particles, like the electron). We will first study motion that takes place in a straight line. 2.3 Position and displacement The location of an object is usually given in terms of a standard reference point, called the origin. The positive direction is taken to be the direction where the coordinates are increasing, and the negative direction as that where the coordinates are decreasing. A change in the coordinates of the position of the body describes the displacement of the body. For example, if the x -coordinate of a body changes from x 1 to x 2 , then the displacement, Δ x = ( x 2 − x 1 ). Displacement is a vector quantity. That is, a quantity that has both magnitude and direction information. An object’s displacement is Δ x = − 4 m means that the object has moved towards decreasing x -axis by 4 m. The direction of motion, here, is toward decreasing x . A common way to describe the motion of an object is to show a graph of the position as a function of time. 2.4 Average Velocity and Average Speed The symbol x ( t ) stands for the displacement x as a function of time t , NOT for the product of x and t !!! Average velocity, or v avg , is defined as the displacement over the time duration. The average velocity has the same sign as the displacement. In the SI system it has units of m/s. But you are more familiar with mph = miles/hour or with kph = km/hr The magnitude of the slope of the x - t graph gives the average velocity Here, the average velocity is:

PHYS | 2A | General Physics I - Mechanics | lecture notes | Chapter 2 Motion along a strai

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

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