PHYS 1114, Lecture 33, April 10 Contents:

PHYS 1114, Lecture 33, April 10 Contents: 1 This class is o cially cancelled, and has been replaced by the common exam Tuesday, April 11, 5:30 PM. A review and Q&A session is scheduled instead during class time. 2 Exam 3 covers chapters 1 10 and 16: (a) All material of previous tests could show up in solving problems related to Chapters 5:3, 8 10 and 16. (b) Chapter 8: Linear momentum, Impulse, Conservation of Momentum, Elastic and Inelastic Collisions, Idea of Rocket Propulsion. (c) Chapter 9: Statics: Extended First Law of Newton: Equilibrium of Forces and Torques. (d) Chapter 10: Angular Acceleration, Kinematics and Dynamics of Rotational Motion, Rotational Inertia, Rotational Work, Rotational Kinetic Energy, Angular Momentum, Conservation of Angular Momentum. (e) Chapter 16: Oscillations and Waves: Hooke s Law, Equations of Motion from Uniform Circular Motion, Simple Pendulum, Damped and Forced Oscillations, Resonance, Propagating Transverse and Longitudinal Waves, Standing Waves, Interference. (f) Chapter 5, Section 3: Elasticity, Stress-Strain. 3 For the examples discussed in the Q&A session see the next seven pages. i

Completely inelastic collision m 1 =1kg v 1 =5m/s m 2 =2kg v 2 =1m/s v m 1 v 1 + m 2 v 2 = (m 1 + m 2 )v 0, (1 kg)(5 m/s) + (2 kg)(1 m/s) = (1 kg + 2 kg)v 0, (5 + 2)6kg m/s = (3 6kg)v 0, v 0 = 7 3 m/s = 2.33 m/s. m 1 =1kg v 1 =5m/s Elastic collision m 2 =2kg v 2 =1m/s 8 m >< 1 v 1 + m 2 v 2 = m 1 v 0 1 + m 2 v 0 2, 1 2 >: m 1v1 2 + 1 2 m 2v2 2 = 1 2 m 0 1v 2 1 + 1 2 m 0 2v 2 2, 0 v 1 v 2 = v 2 v 0 1, (relative velocity changes sign). (1 kg)(5 m/s) + (2 kg)(1 m/s) = (1 kg)v 1 0 + (2 kg)v 2 0, v 1 5 m/s 1 m/s = v 2 0 v 1 0, v 1 0 = v 2 0 4 m/s, v 2 =) 7 m/s = v 0 1 + 2v 0 0 0 0 2 = v 2 4 m/s + 2v 2 =) 11 m/s = 3v 2 8 >< v 0 2 = 11 m/s = 3.67 m/s, =) 3 >: v 0 0 1 = v 2 4 m/s = 1 m/s = 0.33 m/s. 3 ii

Equilibrium: X ~F = 0, X = 0 m=100kg M=1000kg F 1 F 2 mg 30cm 20cm 50cm Mg choose pivot mg F1 = 2 normal force on each left leg F 2 = 2 normal force on each right leg Mg ( F1 + F 2 = mg + Mg, X = F2 100 cm Mg 50 cm mg 30 cm + 0 = 0, F 2 (1.00 6m) = (1000 kg)(9.80 m/s 2 )(0.50 6m) +(100 kg)(9.80 m/s 2 )(0.30 6m), =) F 2 = (500 + 30)(9.80)N = 5194. N = 5.2 kn, F 1 = (1000 + 100)(9.80)N 5194. N = 5586. N = 5.6 kn. iii

Accelerated rotation 2kg 1m 1m 2kg I = 2 (2 kg) (1 m) 2 I = 4 kg m 2 Also, What is, if! goes from 0 to 10 rad/s in 1 min?! =! o + t,! o = 0, 10 rad/s = 0 + (60 s), = 1 6 rad/s2 = 0.167 rad/s 2. How much turned? = o +!t, =) = 0 + = o +! o t + 1 2 t2. 0 + 10 rad/s 2 60 s = 300 rad. = 300 rad 360 2 rad = 17, 000, assuming we have 2 digit accuracy. iv

Rolling on inclined plane m r hoop t? θ d h PE o = mgh = KE f = 1 2 mv2 + 1 2 I!2. For a hoop: I = mr 2 =) v = r!. mgh = 1 2 mv2 + 1 2 mr2 v 2 r = 1 2 mv2 + 1 2 mv2 = mv 2 =) v 2 = gh, v = p gh for a hoop. h d h = sin =) d = sin = vt = p 1 2 gh t s =) t = p 2 h gh sin = 2 h sin g. (This is all done algebraically. Usually, you would have numbers given and you can substitute these early on. More has been done in class.) v

Horizontal spring One example treated was the tranquilizer gun. One main formula in chapter 16 on oscillations is: E = KE + PE el, (A = x max ) E = 1 2 mv2 + 1 2 kx2 = 1 2 ka2 Vertical spring: Bungee cord m v=0 L L v slack x v=0 PE=0 KE=0 E=0 PE=---mgL PE=---mg(L+x)+--kx 1 2 2 KE= --mv 1 2 2 KE=0 E=0 E=0 0 = 1 2 mv2 mgl =) v = p 2gL, 0 = 0 mg(l + x) + 1 2 kx2 =) 1 2 kx2 mgx mgl = 0 q x = mg ± (mg) 2 4( 1 2k)( mgl) 2( 1 2 k) > 0 =) vi

x = mg k r mg + k 2 + 2 mg k L m = 75 kg, g = 9.80 m/s 2, k = 2.0 10 4 N, =) mg = 0.3675 m = 3.7 cm. k L = 5 m =) x max = 0.644 m. (Check this!) x eq? L weight= spring force F sp mg = kx eq m equilibrium x eq x eq = mg k mg Simple pendulum θ L x = L ( in radians) θ T=mgcosθ m mgsinθ m mg mgcosθ θ Restoring force: mg sin mg = kx = kl mg k = mk L,!2 = k m, (since F = kx = ma = m!2 x.) (For circle, radius r, a c = r! 2.) vii

! = r k m = 2 f = 2 T T = 2 s L g L = 1.00 m, g = 9.80 m/s 2 =) T = 2 s 1.00 m 9.80 m/s 2 = 2.0071 s So, if L = 1 m, =) T = 2.01 s 2 s. (More details have been treated in class.) viii

Main Formulae for Third Test: Basic Mathematics: See Lectures 1, 2 and notes for Exam 1 (Lecture 12) for Algebra, Trigonometry, Significant Digits. Reminder of Vectors: Usually best to work with components:5 A = ~ A = (A x, A y ), rather than with magnitude A and angle. ( A 2 = A x 2 + A y 2 tan = A y A x or Ax = A cos A y = A sin A = (Ax, A y ) B = (B x, B y ) A ± B = (Ax ± B x, A y ± B y ) A = ( A x, A y ) (Rescaling) A A -A 3 2 A B B A B B A B A A+B 1

Coordinates: 8 < r = r o + r = : o + t = t o + t 8 < : r = r r o = Displacement = o = Angular Displacement t = t t o = Elapsed Time On Line : 8 < : r = x, v = v, a = a, r o = x o, v o = v o, t = t, (t o = 0 usually). The vector character now means that the x, v, and a can be positive, zero, or negative. In Plane : r = (x, y) = Position Vector, v = (v x, v y ) = Velocity, a = (a x, a y ) = Acceleration q Speed = Magnitude of velocity = v = vx 2 + vy 2 On Circle : = Position Angle, r = Radius,! = Angular Velocity, = Angular Acceleration Zero Acceleration = Constant Velocity: a = 0 () v v v o, = 0 ()!!! o r = r o + v t,! =! o + t 2

Translational & Rotational Kinematics Translational Motion Rotational Motion r = r r o = o SI Unit: m SI Unit: rad v = r t = r r o! = t t o t = o t t o SI Unit: m/s SI Unit: rad/s ā = v t = v v o =! t t o t =!! o t t o SI Unit: m/s 2 SI Unit: rad/s 2 v = v o + a t! =! o + t v = 1 2 (v + v o)! = 1 2 (! +! o) r r o = 1 2 (v + v o) t o = 1 2 (! +! o) t r r o = v o t + 1 2 a t2 o =! o t + 1 2 t2 2a x x = v 2 x v 2 ox 2a y y = v 2 y v 2 oy 2 =! 2! 2 o In the above table, a and are assumed to be constant. Rotational linear quantity is radius times angular quantity: s = r, v = r!, a t = r, a r = r! 2 a c 1 rev = 360 = 2 rad =)! = 2 f Area Circle = r 2 Perimeter Circle = 2 r 3

Translational & Rotational Dynamics: Translational Motion Rotational Motion Force F Torque = F? r = F r? = F r sin, with = 6 (F, r) F net = X F = ma SI Unit: N = kg m/s 2 net = X = I SI Unit: N m Total Mass Moment of Inertia M = X m I = X m r 2 r is distance of m to axis Linear Momentum Angular Momentum p = m v L = I! SI Unit: kg m/s SI Unit: kg m 2 /s F net = X F = p t net = X = L t 6 Translational and Rotational Work and Energy: Translational Motion Rotational Motion W = F k d = F d cos W rot = with = (F, d) = angular displacement SI Unit: J SI Unit: J KE trans = 1 2 mv2 KE rot = 1 2 I!2 SI Unit: J SI Unit: J 4

Extended Newton s Laws of Motion First Law of Newton: An object continues in a state of rest or in a state of motion at a constant speed along a straight line, unless compelled to change that state by a net force; also, it is in a state of constant zero or nonzero angular velocity, unless compelled to change that state by a net torque. Second Law of Newton: X X Fx = ma x, Fy = ma y, X = I NB First Law follows from second as no acceleration takes place if there is no net force or torque. Third Law of Newton: Action = Reaction Fourth Law of Newton: Universal Gravity: F = G m 1m 2 r 2 where G is Newton s constant =) F = mg with g = G M E R E 2 = 9.80 m/s2 Examples of Forces: Gravity or Weight Normal Force Static and Kinetic Friction Tension 5

Free-Body Diagram for a Body on an Inclined Plane: Friction always opposes motion. If at rest or not accelerated: F N = mg cos, F fr = mg sin. q F N mg sinq Static and Kinetic Friction Forces: Static: mg q F fr mg cosq F fr apple F max fr = µ s F N Kinetic: F fr = µ k F N Work-Energy Theorem: (KE = Kinetic Energy) W net = KE = KE f KE o, Work done = Change in KE Conservative Forces: Work W C is independent of path. Nonconservative Forces: Work W NC depends on the path. In formula: W net =W C +W NC, W C = PE = (PE f PE o ). Work-Energy Theorem: (Rewritten) W NC = KE + PE = E. Conservation of Mechanical Energy: If W NC = 0 then KE + PE = KE o + PE o = constant E In general: Total energy (mechanical and non-mechanical, like heat, electrical energy, etc.) is conserved. 6

Gravitational Potential Energy Elastic Potential Energy PE grav = mgy PE spring = 1 2 kx2 F P = kx, F S = kx Here F P is force on spring, F S is force by spring. Total PE = Sum of all PE contributions. Total KE = Sum of all KE contributions. Average Power: P = work done time lapsed = W t = F d t = F v sometimes Unit: joule/second = J/s = W = watt Note: W = work, W = watt, d= distance, d = deci. Impulse and (Linear) Momentum: Impulse = F net t = p momentum = p = mv Unit: newton second = N s = kg m/s Second Law of Newton said di erently: Impulse = Change in Momentum 7

Principle of Conservation of Linear Momentum: If no net external force is applied to a system of bodies, then (by Third Law of Newton all internal forces cancel): If P all F = 0, then P all mv = constant Completely Inelastic Collision: (Bodies merge afterward) m 1 v 1 + m 2 v 2 = m 1 v 0 1 + m 2 v 0 2 = (m 1 + m 2 )v 0 v 0 1 = v 0 2 v 0 Elastic Collision: Both total momentum vector and total kinetic energy are conserved (3 equations for 4 unknowns): m 1 v 1 + m 2 v 2 = m 1 v1 0 + m 2 v2 0 1 2 m 1v1 2 + 1 2 m 2v2 2 = 1 2 m 1v1 0 2 + 1 2 m 2v2 0 2 One-dimensional Elastic Collision: m 1 v 1 + m 2 v 2 = m 1 v1 0 + m 2 v2 0 1 2 m 1v1 2 + 1 2 m 2v2 2 = 1 2 m 1v1 0 2 + 1 2 m 2v2 0 2 Here the quadratic (second) equation can be replaced by v 0 2 v 0 1 = (v 2 v 1 ) = v 1 v 2 8

Center of Mass: Position vector of CM is weighted average r CM = P mr P m = m 1r 1 + m 2 r 2 + m 1 + m 2 + Tangential and Radial Variables in Circular Motion: v = v t, a = a t + a r, F = F t + F r Here a t and F t are directed along (the tangent to) the circle and a r and F r are directed toward the center, with magnitudes a r = v t 2 r = v2 r = r!2 = a c, F r = ma r = F c (a c is Centripetal Acceleration and F c is Centripetal Force). Example: Loop-the-loop with normal and gravity forces. If no friction present, can calculate velocity versus height by energy conservation 1 2 mv2 + mgy = constant. Then normal force plus normal component of weight must equal F c. Uniform circular motion: v = 2 r T with period = T = 1 f = 1 frequency 9

Equilibrium Conditions for a Rigid Body: X Fx = 0, X Fy = 0, X = 0 (Here the axis for the torques can be chosen at will.) Hooke s Law and Elasticity in Extended Objects: Stress = constant Strain Stress = Force per Area, Strain = Relative Deformation [The following three formulae will be provided if needed: Stretching and Compression: F A = Y L L 0, Y is Elastic or Young s Modulus Shear Deformation: F A = S L L 0, S is Shear Modulus Volume Deformation (under compression): P = B V V 0, B is Bulk Modulus, P is Pressure However, you are responsible for their correct interpretation and application.] 10

Ideal Spring: Restoring Force: F = where k is the spring constant and x is the displacement from unstrained length. Elastic Potential Energy: kx PE el = 1 2 kx2 Harmonic Motion Circular Motion ( Take x-component ) on Reference Circle x = A cos = A cos!t =!t, r = A v = A! sin = A! sin!t v t = r! = A! a = A! 2 cos = A! 2 cos!t a r = r! 2 = A! 2 =) F = ma = m! 2 x =) k = m! 2 or! = 2 T = 2 f = r k m, f = 1 T Pendulum: Mass m hanging on string of length L making angle with the vertical, then x = L, F = mg sin mg, for small, or k = F x = mg L so that! = 2 f = 2 T = r g L 11

Wave Motion: For a transverse wave the disturbance is perpendicular to the direction of propagation of the wave. For a longitudinal wave it is parallel. Wave velocity: v = T = f where = wavelength, which passes by in period T. Wave Intensity: Power per Cross Area For spherical waves: I = P A = P, where r is radius. 4 r2 Linear Superposition and Interference: If two or more waves are present simultaneously at the same place, the resultant wave is the sum of the individual waves. If two waves at some place have the same direction, it is called constructive interference. If the direction is opposite, it is called destructive interference. Standing Wave: A wave that does not seem to move. The simplest case shows alternatingly nodes (N) and antinodes (A). Beats: Waves with two nearly equal frequencies produce beats with the di erence frequency, (i.e. the beat frequency is the di erence of the two frequencies). 12