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problem_pdf5

Course Number: PHYSICS 160, Spring 2008

College/University: Kentucky

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chaney (glc568) Torque, Angular Momentum, and Rotational Equilibrium murthy (21118) 1 This print-out should have 51 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. 001 (part 1 of 2) 10.0 points A small, solid sphere of mass 0.8 kg and radius 21 cm rolls without slipping along the track consisting of slope and loop-the-loop with radius 2.95 m at the...

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(glc568) chaney Torque, Angular Momentum, and Rotational Equilibrium murthy (21118) 1 This print-out should have 51 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. 001 (part 1 of 2) 10.0 points A small, solid sphere of mass 0.8 kg and radius 21 cm rolls without slipping along the track consisting of slope and loop-the-loop with radius 2.95 m at the end of the slope. It starts from rest near the top of the track at a height h, where h is large compared to 21 cm. The acceleration of gravity is 9.8 m/s2 . Hint: The moment of inertia for a solid 2 sphere is m r 2 . 5 0.8 kg horizontal direction. The ball lands in the catching cup of the cart because both the cart and ball have the same horizontal component of velocity. Now let the ballistics cart move at an angle with the horizontal as shown in the gure below. The cart (including wheels) has a mass M and the moment of inertia of each of the 1 four wheels is m R2 . 2 y x 21 cm x Using conservation of energy (assuming no friction between cart and axle) and assuming pure rolling motion (no slipping), nd the acceleration of the cart along the incline. h 1. ax = 2. ax = 3. ax = What is the minimum value of h (in terms of the radius of the loop R) such that the sphere completes the loop? Answer in units of m. 002 (part 2 of 2) 10.0 points What are the force component in the horizontal direction on the sphere at the point P, which has coordinates (R, 0) if we take the center of the loop as origin, and if h = 3 R ? Answer in units of N. 003 (part 1 of 3) 10.0 points In a demonstration known as ballistics cart, a ball is projected vertically upward from a cart moving with constant velocity along the 4. ax = 5. ax = 6. ax = 7. ax = 8. ax = 9. ax = 10. ax = 2m g cos M + 2m M g sin M +m m g sin M + 2m M g cos M + 2m 2M g sin M + 2m m g cos M + 2m 2m g sin M + 2m M g sin M + 2m 2M g cos M + 2m M g cos M +m 63 2.95 m P chaney (glc568) Torque, Angular Momentum, and Rotational Equilibrium murthy (21118) 2 004 (part 2 of 3) 10.0 points Find the distance db that the ball travels measured along the incline. Assume the cart is initially at rest. 1. db = 2. db = 3. db = 4. db = 5. db = 6. db = 7. db = 8. db = 9. db = 10. db = 2 3 vy0 cos2 g sin 2 vy0 sin g cos2 2 2 vy0 cos2 g sin 2 2 vy0 sin g cos2 2 3 vy0 cos g sin2 2 vy0 cos g sin2 2 2 vy0 cos g sin2 2 2 vy0 sin2 g cos 2 vy0 sin2 g cos 2 vy0 cos2 g sin 5. x = 6. x = 7. x = 8. x = 9. x = 10. x = 4m M +2m 2m M +2m 2m M +2m 4m M +2m 3m M +2m 4m M +2m sin2 cos cos2 sin cos sin2 cos2 sin tan sin2 sin cos2 2 vy0 g 2 vy0 g 2 vy0 g 2 vy0 g 2 vy0 g 2 vy0 g 006 10.0 points A bowling ball has a mass of 3.9 kg, a moment of inertia of 0.022464 kg m2 , and a radius of 0.12 m. It rolls along the lane without slipping at a linear speed of 2.7 m/s. What is the kinetic energy of the rolling ball? Answer in units of J. 007 10.0 points Consider a wheel(solid disk) of radius 0.267 m, 1 mass 12 kg and moment of inertia I = M R2 2 The wheel rolls without slipping in a straight line in an uphill direction 29 above the horizontal. The wheel starts at angular speed 34.6 rad/s but the rotation slows down as the wheel rolls uphill, and eventually the wheel comes to a stop and rolls back downhill. How far does the wheel roll in the uphill direction before it stops? The acceleration of gravity is 9.8 m/s2 . Answer in units of m. 008 (part 1 of 3) 10.0 points 2 A uniform solid sphere I = m r 2 of mass 5 2.3 kg and radius r = 0.435 m, is placed on the inside surface of a hemispherical bowl of radius R = 2.25 m. The sphere is released from rest at an angle = 50.6 from the vertical and rolls without slipping (see the gure). 005 (part 3 of 3) 10.0 points The x component of the acceleration of the ball released by the cart is g sin , so the x component of the carts acceleration is smaller M than that of the ball by the factor . M + 2m Find x, where x is the amount by which the ball overshoots the car. 1. x = 2. x = 3. x = 4. x = 2m M + 2m 2m M + 2m 3m M + 2m 4m M + 2m 2 vy0 g 2 sin vy0 cos2 g 2 tan2 vy0 sin g 2 cos vy0 sin2 g sin2 cos chaney (glc568) Torque, Angular Momentum, and Rotational Equilibrium murthy (21118) 3 R 3 m R2 and it rolls without slipping. 4 F M R r The acceleration of gravity is 9.8 m/s . How much potential energy has the sphere lost when it reaches the bottom of the bowl? Answer in units of J. 009 (part 2 of 3) 10.0 points What is the translational velocity of the sphere when it reaches the bottom of the bowl? Answer in units of m/s. 010 (part 3 of 3) 10.0 points What is the angular velocity of the sphere when it reaches the bottom of the bowl? Answer in units of rad/s. 011 10.0 points A solid sphere has a radius of 0.93 m and a mass of 65 kg. How much work is required to get the sphere rolling with an angular speed of 32 rad/s on a horizontal surface? Assume the sphere starts from rest and rolls without slipping. Answer in units of J. 012 10.0 points A large, cylindrical roll of tissue paper of initial radius 6.5 cm lies on a long, horizontal surface with the open end of the paper nailed to the surface. The roll is given a slight shove (v0 0) and commences to unroll. The acceleration of gravity is 9.8 m/s2 . Determine the speed of the center of mass of the roll when its radius has diminished to 0.28 cm. Answer in units of m/s. 013 (part 1 of 3) 10.0 points A spool of mass 15.9 kg and radius 3.73 m is unwound by a constant force 21.5 N pulling on the massless rope wrapped around it. Assume the moment of inertia of the spool about O is 2 When the center of the spool has moved a distance d, nd the total kinetic energy. Think about how far the force has traveled. The acceleration of gravity is 9.8 m/s2 . 1. 5F d 3 2. F d 3. 2F d 3 4. 2 F d 5. 6. 7. 8. 9. 10. 4F d 3 7F d 3 Fd 2 Fd 3 3F d 2 8F d 3 014 (part 2 of 3) 10.0 points Find the acceleration. Answer in units of m/s2 . 015 (part 3 of 3) 10.0 points Find the critical value beyond which the spool starts to slip. Assume the coecient of static friction between the spool and the surface is 0.4. Answer in units of N. 016 (part 1 of 3) 10.0 points chaney (glc568) Torque, Angular Momentum, and Rotational Equilibrium murthy (21118) 4 A string is wound around a uniform disc whose mass is 2.9 kg and radius is 0.24 m , see gure below. The disc is released from rest with the string vertical and its top end tied to a xed support. The acceleration of gravity is 9.8 m/s2 . 2 h 0.24 m 2.9 kg As the disc descends, calculate the tension in the string. Answer in units of N. 017 (part 2 of 3) 10.0 points Calculate the magnitude of the acceleration of its center of mass. 1. acm = 5.94533 m/s2 2. acm = 6.53333 m/s2 3. acm = 5.684 m/s2 4. acm = 7.252 m/s2 018 (part 3 of 3) 10.0 points After starting from rest, calculate the speed of its center of mass when the center of mass has fallen h = 1.4 m. Answer in units of m/s. 019 (part 1 of 4) 10.0 points Assume: Drag (friction) is negligible. Given: g = 9.81 m/s2 . The density of this large Yo-Yo like solid is uniform throughout. The Yo-Yo like solid has a mass of 2.7 kg . Front View A cord is wrapped around the stem of the Yo-Yo like solid and attached to the ceiling. The radius of the stem is 4 m and the radius of the disk is 5 m. 5m 2.7 kg 4m Cross sectional Side View Calculate the moment of inertia about the center of mass (axis of rotation). Answer in units of kg m2 . 020 (part 2 of 4) 10.0 points What is the vertical acceleration of the center of mass of the Yo-Yo? Answer in units of m/s2 . 021 (part 3 of 4) 10.0 points What is magnitude of the torque the cord exerts on the center of mass of the Yo-Yo? Answer in units of N m. 022 (part 4 of 4) 10.0 points The Yo-Yo is released from rest at height h. Find the velocity v of the center of mass of the disk at the height h = 16 m . Answer in units of m/s. 023 10.0 points A force Fx = (1.7 N) + (2.7 N) is applied to an object that is pivoted about a xed axis aligned along the z coordinate axis. The force is applied at the point x = (4.5 m)+(4.3 m) . Find the z-component of the net torque. h chaney (glc568) Torque, Angular Momentum, and Rotational Equilibrium murthy (21118) 5 Answer in units of N m. 024 (part 1 of 3) 10.0 points A force F = Fx + Fy + Fz k acts on i j a particle located at X = (x, y, z). Given Fx = 24.1 N, Fy = 1.15 N, Fz = 1.39 N, x = 1.13 m, y = 4.2 m and z = 3.38 m, calculate the three components of the torque vector = x + y + z k . i j First, calculate the x component. Answer in units of N m. 025 (part 2 of 3) 10.0 points Second, calculate the y component. Answer in units of N m. F 9. Horizontally Eastward 10. Vertically Down 028 (part 2 of 2) 10.0 points Calculate the magnitude of the torque. Answer in units of N m. 029 10.0 points You are using a wrench to tighten a nut. F rod 026 (part 3 of 3) 10.0 points Finally, calculate the z component. Answer in units of N m. 027 (part 1 of 2) 10.0 points A 2.82 m long rod of negligible weight is attached on one end to a ball joint which allows the rod to rotate in all directions. The free end of the rod points 15 above the Eastward horizontal direction. A 29.3 N force directed vertically Down is applied at the rods free end. What is the direction of the torque due to this force (relative to the pivot end of the rod)? 1. Horizontally Northward 2. Horizontally Westward 3. Up and Southwest 4. Insucient data to determine the direction. 5. Vertically 6. Up Down and Eastward 7. Horizontally Southward 8. The torque vanishes wrench wrench nut 1 nut 2 rod F wrench F wrench nut 3 nut 4 Which of the following correctly lists the arrangements (above) in descending order of torque magnitude? 1. 3 4 2 1 2. 2 1 3 4 3. 1 3 2 4 4. 3 2 4 1 5. 3 1 2 4 6. 1 2 3 4 7. 2 1 4 3 chaney (glc568) Torque, Angular Momentum, and Rotational Equilibrium murthy (21118) 6 8. 2 4 1 3 9. 4 2 3 1 10. 2 4 3 1 030 (part 1 of 2) 10.0 points In the Bohrs model of the hydrogen atom, the electron moves in a circular orbit of radius 4.005 1011 m around the proton. Assume that the orbital angular momentum of the electron is equal to h. Calculate the orbital speed of the electron. Answer in units of m/s. 031 (part 2 of 2) 10.0 points Calculate the angular frequency of the electrons motion Answer in units of s1 . 032 (part 1 of 2) 10.0 points A bowling ball is given an initial speed v0 on an alley such that it initially slides without rolling. The coecient of friction between ball and alley is . Find the speed of the balls center of mass vCM at the time pure rolling motion occurs. 1. vCM = 2. vCM = 3. vCM = 4. vCM = 5. vCM = 1 5 3 5 5 7 3 7 1 3 v0 v0 v0 v0 v0 22 2 2 v0 7g 2 3 v0 4. x = 14 g 2 12 v0 5. x = 25 g 3. x = 034 10.0 points A small metallic bob is suspended from the ceiling by a thread of negligible mass. The ball is then set in motion in a horizontal circle so that the thread describes a cone. The acceleration of gravity is 9.8 m/s2 . 9.8 m/s2 Calculate the magnitude of the angular momentum of the bob about the supporting point. Answer in units of kg m2 /s. 035 10.0 points A cue stick strikes a cue ball and delivers a horizontal impulse in such a way that the ball rolls without slipping as it starts to move. At what height above the balls center (if the radius of the ball is 1.3 cm) was the blow struck? Answer in units of cm. 036 10.0 points Hint: The moment of inertia for a uniform 1 disk is I = m r 2 . 2 Two uniform disks with the same mass are connected by a light inextensible string supported by a massless pulley, on a frictionless axis. The string is attached to a point on the 2. 7 m v 9 kg 033 (part 2 of 2) 10.0 points Since the frictional force provides the deceleration, from Newtons second law it follows that aCM = g. Find the distance x it has traveled. When pure rolling motion occurs, vCM = R . 2 5 v0 1. x = 81 g 2 12 v0 2. x = 49 g chaney (glc568) Torque, Angular Momentum, and Rotational Equilibrium murthy (21118) 7 circumference of disk A (on the left). The string is wound around disk B (on the right) so that the disk will rotate like a yo-yo when dropped. 1. a = 1 4 3 2. a = 2 1 3. a = 2 2 4. a = 3 g g g g 5. a = 2 g A B 6. a = g 038 (part 1 of 2) 10.0 points A uniform solid disk is set into rotation with an angular speed 0 about an axis through its center. While still rotating at this speed, the disk is placed into contact with a horizontal surface and released. Describe the outcome when the disks are simultaneously released from rest at the same height above the oor. 1. Disk B (on the right) will reach the oor rst. 2. Both disks will reach the oor at the same time. 3. Disk A (on the left) will reach the oor rst. 4. Both disks will remain stationary. 037 10.0 points A cord is wrapped around a ywheel of inertia 1 m R2 and radius R . A mass m is I= 2 suspended by the cord. The system is released from rest and the suspended mass falls. What is the angular speed of the disk once pure rolling takes place? 1. = 2. = 3. = 1 0 2 2 0 3 1 0 3 1 0 5 1 0 7 R g I T 4. = 5. = m What is the acceleration of the suspended mass m? 039 (part 2 of 2) 10.0 points Hint: Consider torques about the center of mass. Find the fractional loss in kinetic energy from the time the disk is released until pure rolling occurs. 1. 3 E = E 7 chaney (glc568) Torque, Angular Momentum, and Rotational Equilibrium murthy (21118) 8 2. E E E 3. E E 4. E E 5. E 1 2 2 = 3 2 = 5 1 = 3 = 3. F cos + W = f 4. F sin = f 5. F sin = W 042 (part 3 of 3) 10.0 points Find the smallest coecient of friction needed for the wall to keep the sphere from slipping. 1 sin 1 2. = tan 1 3. = cos 1. = 4. = cos P 5. = tan 6. = sin W To what does the torque equation i 040 (part 1 of 3) 10.0 points A solid sphere of radius R and mass M is held against a wall by a string being pulled at an angle . f is the magnitude of the frictional force and W = M g . F R i = 0 about point O (the center of the sphere) lead? 1. W = f 2. F sin cos = f 3. F = f 4. F + W = f 5. F sin = f 6. F cos2 = f 041 (part 2 of 3) 10.0 points To what does the vertical component of the force equation lead? 1. F sin + f = W 2. F sin = f + W 043 (part 1 of 2) 10.0 points A solid sphere of radius R and mass M is placed in a wedge as shown in the gure. The inner surfaces of the wedge are frictionless. R M B A Determine the force exerted by the wedge on the sphere at the left contact point. 1. FA = 2 M g sin sin( + ) cos 2. FB = M g cos( + ) chaney (glc568) Torque, Angular Momentum, and Rotational Equilibrium murthy (21118) 9 3. FA = M g sin sin( + ) cos 4. FA = M g sin( + ) cos 5. FA = M g cos( + ) sin 6. FA = M g cos( + ) 4. m1 = m2 5. a2 m1 = b2 m2 046 10.0 points A uniform 150 g rod with length 46 cm rotates in a horizontal plane about a xed, vertical, frictionless pin through its center. Two small 44 g beads are mounted on the rod such that they are able to slide without friction along its length. Initially the beads are held by catches at positions 12 cm on each sides of the center, at which time the system rotates at an angular speed of 25 rad/s . Suddenly, the catches are released and the small beads slide outward along the rod. Find the angular speed of the system at the instant the beads reach the ends of the rod. Answer in units of rad/s. 047 10.0 points A bullet of mass 3m moving with velocity v strikes tangentially the edge of a spoked wheel of radius R, and the bullet sticks to the edge of the wheel. The spoked wheel has a mass 7m which is concentrated on its rim (neglect the mass of the spokes). The wheel, initially at rest, begins to rotate about its center, which remains xed on a frictionless axle. What is the angular velocity of the spoked wheel after the collision? 1. 0.3 v R 044 (part 2 of 2) 10.0 points Determine the force exerted by the wedge on the sphere at the right contact point. 1. FB = M g 2. FB 3. FB 4. FB 5. FB 6. FB sin cos( + ) cos =Mg cos( + ) sin =Mg sin( + ) sin =Mg cos( + ) sin =Mg sin( + ) cos =Mg sin( + ) 045 10.0 points Consider the wheel-and-axle system shown below. b a m1 m2 Which of the following expresses the condition required for the system to be in static equilibrium? 1. b2 m1 = a2 m2 2. a m1 = b m2 3. a m2 = b m1 v R v 3. 0.547723 R v 4. 3.33333 R v 5. 0.428571 R v 6. 0.15 R v 7. 0.176471 R v 8. 2.33333 R 2. 0.230769 chaney (glc568) Torque, Angular Momentum, and Rotational Equilibrium murthy (21118) 10 9. 1.42857 10. 0.7 v R v R i f 048 10.0 points A gure skater on ice spins on one foot. She pulls in her arms and her angular speed increases. Choose the best statement below. 1. Her angular speed increases because her angular momentum increases. 2. Her angular speed increases because air friction is reduced as her arms come in. 3. Her angular speed increases because she is undergoing uniformly accelerated angular motion. 4. Her angular speed is unrelated to her arms. She pulls them in at the same time as she speeds up her spin because it looks better this way. 5. Her angular speed increases because her angular momentum remains the same but her moment of inertia decreases. 6. Her angular speed increases because by pulling in her arms she creates a net torque in the direction of rotation. 7. Her angular speed increases because her potential energy increases as her arms come in. 049 10.0 points A student sits on a rotating stool holding two 2 kg objects. When his arms are extended horizontally, the objects are 0.7 m from the axis of rotation, and he rotates with angular speed of 0.72 rad/sec. The moment of inertia of the student plus the stool is 8 kg.m2 and is assumed to be constant. The student then pulls the objects horizontally to a radius 0.37 m from the rotation axis. (a) (b) Calculate the change in kinetic energy of the system. Ignore the dierence in the moment of inertia contributed by the two dierent arm shapes. Answer in units of J. 050 (part 1 of 2) 10.0 points A 3 kg bicycle wheel rotating at a 2472 rev/min angular velocity has its shaft supported on one side, as shown in the gure. When viewing from the left (from the positive x-axes), one sees that the wheel is rotating in a clockwise manner. The distance from the center of the wheel to the pivot point is 0.4 m. The wheel is a hoop of radius 0.4 m, and its shaft is horizontal. Assume all of the mass of the system is located at the rim of the bicycle wheel. The acceleration of gravity is 9.8 m/s2 . 2 re v /min 0. 4 m z 247 y x 0.4 m radius mg Find the change in the precession angle after a 1.2 s time interval. Answer in units of . 051 (part 2 of 2) 10.0 points The direction of precession as viewed from the top is 1. opposite to the direction of rotation of the chaney (glc568) Torque, Angular Momentum, and Rotational Equilibrium murthy (21118) 11 wheel. 2. static, since angular momentum is conserved. 3. along the direction of rotation of the wheel. 4. clockwise. 5. counter-clockwise.

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EE 200 Fall 2008 (Weber) Homework 1At the upper right corner on page 1 of all homeworks, show: Last name, First name Date EE 200, Homework # Show intermediate steps whenever possible. 1. Write the algebraic expression(s) that dene the function plott
USC - EE - 200
EE 200 Fall 2008 (Weber) Homework 2At the upper right corner on page 1 of all homeworks, show: Last name, First name Date EE 200, Homework # Show intermediate steps whenever possible. 1. Problem 3 in chapter 3 of the text. 2. Design a state machine
USC - EE - 200
EE 200 Fall 2008 (Weber) Homework 3At the upper right corner on page 1 of all homeworks, show: Last name, First name Date EE 200, Homework # Show intermediate steps whenever possible. 1. Problem 6 in chapter 5. The answer for each section should sho
USC - EE - 200
EE 200 Fall 2008 (Weber) Homework 4At the upper right corner on page 1 of all homeworks, show: Last name, First name Date EE 200, Homework # Show intermediate steps whenever possible. 1. Problem 8 in chapter 8. 2. Problem 10 in chapter 8. You dont h
USC - EE - 200
EE 200 Fall 2008 (Weber) Homework 5At the upper right corner on page 1 of all homeworks, show: Last name, First name Date EE 200, Homework # Show intermediate steps whenever possible. 1. A discrete-time system has an impulse response of7h(n) =k=
USC - EE - 200
EE 200 Fall 2008 (Weber) Homework 5 Solutions1. a.h(n).. nb. y(n) = k=h(k)x(n k)7=k= m=0 7(k m) cos (k m) cos (n k) 4 (n k) 4=7m=0 k==m=0cos 0 (n m) 4= c.H() = m=h(m)eim7=7m= k=0(m k)eim
USC - EE - 200
EE 200 Fall 2008 (Weber) Homework 1 Solutions1. The function denition is divided into four parts: x 2 x y= x + 2 0 2. (f1 f2 )(x) = 0.5signum(2 sin(x)if 2 x 1 if 1 x 1 if 1 x 2 otherwise(f2 f1 )(x) = 2 sin( 0.5signum(x)(f2 f
USC - EE - 200
EE 200 Fall 2008 (Weber) Homework 2 Solutions1.{1}/0 {0}/0 {1}/0 {1}/10{0}/0 {0}/0111Sequence found10{0}/0 {0}/0 {1}/0 {0}/0 {1}/1 {1}/0 {0}/0Sequence found{1}/101112. The following states are needed to make the state machi
USC - EE - 200
EE 200 Fall 2008 (Weber) Homework 3 Solutions1. a. s(0) = s(1) = 1 0 1 0 1 1 1 0 = 1 0The state does not change when the input is zero so for n Integers, s(n) = b. s(0) = s(1) = s(2) = s(3) = 0 1 1 0 1 0 1 0 1 1 1 1 1 1 0 1 1 1 2 1 = = = 1 1 2 1
USC - EE - 200
EE 200 Fall 2008 (Weber) Homework 4 Solutions1. Problem 8 x(t) = 1 + cos(t) + cos(2t) = ei0t + 0.5eit + 0.5eit + 0.5ei2t + 0.5ei2t y(t) = H(0)ei0t + 0.5H()eit + 0.5H()eit + 0.5H(2)ei2t + 0.5H(2)ei2t = (1)ei0t + 0.5(1)eit + 0.5(1)eit + 0.5(0)ei2t +
USC - BME - 423
BME 423 Lecture Notes - Dec. 1, 2008I. Clinical Studies A. Experimental Studies B. Observational Studies II. What do the Data Show? A. Randomize and Control B. Problems with the Population III. Ethical Issues and Clinical Trials A. Declaration of H
USC - BME - 423
USC - BME - 423
USC - BME - 423
UCSD - PHYS - phys 2d
1-5This is a case of dilation. T = " T# in this problem with the proper time T " = T 0) # v& 2 , T = +1 " % ( . + . * $ c' -"1 22 v ) #T & , T0 / = +1 " % 0 ( . c * $ T' + .12;) # L 2 &2 + + 0 in this case T = 2T0 , v = *1" % ( . + $ L0
UCSD - PHYS - phys 2d
2-2(a)Scalar equations can be considered in this case because relativistic and classical velocities are in the same direction.p = " mv = 1.90mv = = 0.85c(b) 2-3mv $1 # ( v c )2 ' & ) % (12*1 $1 # (v c) 2 ' & ) % (12$ + 1 .2' = 1.90 *
UCSD - PHYS - phys 2d
3The Quantum Theory of Light2 dB E2 r = r dt r dB E = 2 dt 3-1(a)(b) r dB r dB If r remains constant, then: E = Eq = e so that Fdt = dt = m e dv , or 2 dt 2 dt re dv = dB 2m e v+ vB E er B dv = 2