SOL30
11 Pages

SOL30

Course Number: PHY phy 303k, Spring 2010

College/University: University of Texas

Word Count: 3340

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Thevalingam, Donald Homework 30 Due: Apr 30 2007, midnight Inst: Eslami This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points The center of mass of a pitched baseball or radius 5.36 cm moves at 18.3 m/s. The ball spins about an axis through its center of mass with an...

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Donald Thevalingam, Homework 30 Due: Apr 30 2007, midnight Inst: Eslami This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points The center of mass of a pitched baseball or radius 5.36 cm moves at 18.3 m/s. The ball spins about an axis through its center of mass with an angular speed of 50.2 rad/s. Calculate the ratio of the rotational energy to the translational kinetic energy. Treat the ball as a uniform sphere. Correct answer: 0.00864758 . Explanation: The wanted ratio is g I T 1 R m Find the total kinetic energy of the system, when the weight is moving at a speed v . 1. K = 1 m v2 2 3 2. K = m v 2 correct 4 3. K = m v 2 4 m v2 5 1 5. K = m v 2 3 5 6. K = m v 2 2 3 7. K = m v 2 2 2 8. K = m v 2 3 5 9. K = m v 2 4 Explanation: The total kinetic energy is the sum of the kinetic energy of the hanging weight, Km = 1 m v 2 , plus the kinetic energy of rotation of 2 1 1 the disk, Kd = I 2 = m v 2 . Therefore, 2 4 K = Kg + Kd 1 1 = m v2 + m v2 2 4 3 = m v2 . 4 4. K = I 2 ratio = 2 2 mv 2 2mr 2 2 5 = mv 2 22 2r = 5v 2 2(0.0536 m/cm)2 (50.2 rad/s)2 = 5(18.3 m/s)2 = 0.00864758 keywords: 002 (part 1 of 1) 10 points A circular disk with a mass m and radius R is mounted at its center, about which it can rotate freely. The disk has moment of inertia 1 I = m R2 . 2 A light cord wrapped around it supports a weight m g . Thevalingam, Donald Homework 30 Due: Apr 30 2007, midnight Inst: Eslami keywords: 003 (part 1 of 3) 10 points For any given rotational axis, the radius of gyration, K , of a rigid body is dened by the I , where M is the total expression K 2 = M mass of the body and I is the moment of inertia about the given axis. In other words, the radius of gyration is the distance between an imaginary point mass M , and the axis of rotation with I for the point mass about that axis is the same as for the rigid body. Find the radius of gyration of a solid disk of radius 3.93 m rotating about a central axis. Correct answer: 2.77893 m. Explanation: In all part the axis is the central axis, as stated in the problem. The moment of inertia of a solid disk with radius R and mass M is M R2 Id = . 2 Hence, M R2 Kd = =R 2 2M = (3.93 m) 2 = 2.77893 m . 004 (part 2 of 3) 10 points Find the radius of gyration of a uniform rod of length 7.73 m rotating about a central axis. Correct answer: 2.23146 m. Explanation: The moment of inertia of a uniform rod of length L and mass M is M L2 Ir = 12 Hence, Kr = L M L2 = 3 12 M 6 (7.73 m) 3 = 6 = 2.23146 m . keywords: 2 005 (part 3 of 3) 10 points Find the radius of gyration of a solid sphere of radius 3.93 m rotating about a central axis. Correct answer: 2.48555 m. Explanation: The moment of inertia of a solid sphere with radius R and mass M is Is = Hence, Ks = 2 M r2 =r 5M 2 = (3.93 m) 5 = 2.48555 m . 2 5 2 M r2 . 5 006 (part 1 of 1) 10 points A pendulum is made of a rod whose moment 1 M L2 , of inertia about its center of mass is 12 where its mass is 1.1 kg and and its length is 2.3 m, and a thin cylindrical disk, whose moment of inertia about its center of mass 1 is M R2 , where its mass is 2.9 kg and its 2 radius is 1.3 m. 2.3 m 1.1 kg 1.3 m 2.9 kg What is the moment of inertia of the pendulum about the pivot point? Correct answer: 19.7312 kg m2 . Explanation: Thevalingam, Donald Homework 30 Due: Apr 30 2007, midnight Inst: Eslami keywords: Let : 1 M L2 , 12 1 Idisk = M R2 , 2 L = 2. 3 m , Mr = 1.1 kg , R = 1.3 m , and Md = 2.9 kg , Irod = 3 007 (part 1 of 1) 10 points Consider three objects of equal masses but dierent shapes: a solid disk, a thin ring, and a thin hollow square. The sizes of the three objects are related according to the following picture square ring disk Basic Concept: The parallel axis theorem Ipp = Icm + M L2 , where Icm is the moment of inertia about the center of mass, M is the mass, and L is the distance from the center of mass to the pivot point. Solution: For the rod using the parallel axis theorem for the moment of inertia Irpp (about the pivot point) Irpp = 1 L Mr L 2 + M r 12 2 1 (1.1 kg) (2.3 m)2 = 12 2. 3 m + (1.1 kg) 2 2 = 1.93967 kg m . 2 R R 2R Note: The ring and the square are hollow and their perimeters carry all the mass, but the disk is solid and has uniform mass density over its whole area. Compare the three objects moments of inertia when rotated around their respective centers of mass. Which of the following conditions is correct? cm cm cm 1. Iring > Idisk > Isquare 2 cm cm cm 2. Isquare > Iring > Idisk correct cm cm cm 3. Isquare > Idisk > Iring cm cm cm 4. Idisk > Isquare > Iring cm cm cm 5. Idisk > Iring > Isquare cm cm cm 6. Iring > Isquare > Idisk For the cylindrical disk using the parallel axis theorem for the moment of inertia Icpp (about the pivot point) Icpp = 1 Md R 2 + M d L 2 2 1 = (2.9 kg) (1.3 m)2 2 + (2.9 kg) (2.3 m)2 = 17.7915 kg m2 . Explanation: The moment of inertia I = though as I = M r2 , where M is the objects net mass and r 2 is the average distance2 of massive points making up the object from the rotation axis. For a hoop or a thin ring, all massive point are at the same distance R from the axis, hence cm r2 = R2 = Iring = M R2 . r2 dm can be The total moment of inertia is I = Irpp + Icpp = (1.93967 kg m2 ) + (17.7915 kg m2 ) = 19.7312 kg m2 . Thevalingam, Donald Homework 30 Due: Apr 30 2007, midnight Inst: Eslami For the solid disk, the distance ranges from zero in the center to R at the perimeter, hence the average distance is less than R , (m) cm r2 < R2 = Idisk < M R2 . 4 5 4 3 2 1 01 23 4567 x (m) cm r2 > R2 = Isquare < M R2 . Therefore, without any calculations we can say that cm cm cm Isquare > Iring > Idisk . Calculate the x coordinate of the center of mass of the metal plate. Correct answer: 2 m. Explanation: Basic Concept: The center of mass xcoordinate is x dm x where m m , If you want actual values of the three moments of inertia, you can nd two of them in the textbook, namely cm Iring = M R2 and cm Idisk = 1 M R2 . 2 For the square, we need to calculate: A hollow square is a sum of four thin rods, each having M mass and length 2 R , and the center of 4 each rod is located at the distance R from the squares center. Hence, using the parallel axis theorem cm Irod = Mrod R2 + Irod M2 1M = R+ (2 R)2 4 12 4 1 = M R2 , 3 dm , and dm = y dx , where mass of the plate. is the areal density area Solution: Let (x1 , y1 ) = (0 m, 0 m) (x2 , y2 ) = (6 m, 0 m) (x3 , y3 ) = (0 m, 3 m) . The equation for the hypotenuse is y y2 y3 y 2 = . x x2 x3 x 2 The slope of the hypotenuse is s= y3 y 2 x3 x 2 3m0m 1 = = . 0m6m 2 and therefore cm Isquare = 4 Irod 4 = M R2 . 3 And this conrms than indeed cm cm cm Isquare > Iring > Idisk . Rewriting the equation, we have y = s (x x 2 ) + y 2 1 (x 6 m) + 0 m = 2 = s (x x2 ) , where y2 = 0 . keywords: 008 (part 1 of 2) 10 points A uniform at plate of metal is situated in the reference frame shown in the gure below. On the other hand, on the perimeter of the hollow square, the distance from the center ranges from R to R 2, so the average distance is greater than R , y 0 8 9 10 (1) (2) Thevalingam, Donald Homework 30 Due: Apr 30 2007, midnight Inst: Eslami The x-coordinate of the center of mass is x2 5 of the plate is = m x2 xcm = x2 x1 x1 x y dx x2 y dx y dx = s 0 x1 x2 m (x x2 ) dx m x = x1 x s (x x2 ) dx x2 x1 x2 s (x x2 ) dx x (x x2 ) dx (x x2 ) dx x2 = 0 x2 0 13 1 x (x 2 ) x 2 2 =3 12 x (x 2 ) x 2 13 1 x 2 (x 2 ) x 2 2 2 =3 12 x (x 2 ) x 2 22 x3 = 22 3 x2 1 = x2 3 1 = (6 m) 3 = 2 m. 0 2 12 x (x2 ) x s 2 0 m = y3 12 x (x 2 ) x 2 x2 22 m m = = 1 y3 2 1 x2 x 2 y3 2 x2 2 2m = x 2 y3 2 (1 kg) = (6 m) (3 m) = 0.111111 kg/m2 , = (4) (3) Vertical Center-of-Mass: Using Eq. 3, the y -coordinate of the center of mass of the metal plate is y= 1 y3 3 1 = (3 m) 3 = 1 m. y3 from Eq. 1. The area of a x2 1 1 triangle is A = (base) (height) = x2 y3 , 2 2 which agrees with Eq. 4. Conventional Solution: The moment of x inertia Iy =xcm of the triangle about a y -axis through its center of mass x-coordinate can be accomplished in two steps. where s = x First: The moment of inertia Iy =xcm about its center of mass (xcm , ycm ) is determined. Second: The parallel axis theorem is used. The moment of inertia about the center of x mass Iy =xcm is x Iy =xcm x2 = x1 (x xcm )2 dm x2 009 (part 2 of 2) 10 points Given: The mass of the plate is m = 1 kg . Calculate the moment of inertia of the triangle with the y -axis as the axis of rotation. Correct answer: 6 kg m2 . Explanation: m Areal Density: The areal density = A = x1 (x xcm )2 y dx (x xcm )2 (x x2 ) dx x3 [2 xcm + x2 ] x2 +[x2 + 2 xcm x2 ] x cm x2 x2 = s 0 = s 0 Thevalingam, Donald Homework 30 Due: Apr 30 2007, midnight Inst: Eslami [x2 x2 ] dx cm = s 14 1 x [2 xcm + x2 ] x3 4 3 1 + [x2 + 2 xcm x2 ] x 2 cm [x 2 x 2 ] x cm x2 0 x2 6 = x1 x2 y dx x2 (x x2 ) dx x x2 = s 0 14 1 x2 [2 xcm + x2 ] x3 s = 2 4 3 1 + x2 + 2 xcm x2 x2 2 2 cm x 2 x 2 cm 2 14 y3 x x2 12 2 1 1 + xcm x3 x2 x2 2 3 2 cm 2 12 2 x xcm x2 + x2 =m cm 62 3 12 22 12 =m x x+ x 62 92 92 1 = (5) m x2 2 18 1 (1 kg) (6 m)2 = 18 = 2 kg m2 . = using Eq. 1 for s , Eq. 4 for , and Eq. 3 for xcm . Using the parallel axis theorem, we have x x Iy =0 = Iy =xcm + m x2 cm 1 m x2 + m x 2 = 2 cm 18 12 12 x+ x =m 18 2 9 2 1 = m x2 2 6 1 = (1 kg) (6 m)2 6 2m x 2 y3 2 14 1 x (x2 ) x3 4 3 0 14 14 = s x x 42 32 2m y3 1 x4 = x 2 y3 x2 12 2 1 = m x2 2 6 1 = (1 kg) (6 m)2 6 = s = 6 kg m2 , using Eq. 1 for s and Eq. 4 for . Note: This problem has a dierent triangle for most students. keywords: 010 (part 1 of 1) 10 points Calculate the moment of inertia for 4m L L 3m L Axis L m L 2m Rods of length L are massless. 1. 52mL2 2. 41mL2 3. 32mL2 correct 4. 30mL2 5. 16mL2 Explanation: = 6 kg m2 . Standard Solution: The moment of inerx tia Iy =0 of the triangle about the x-coordinate is x Iy =0 x2 x1 x2 dm Thevalingam, Donald Homework 30 Due: Apr 30 2007, midnight Inst: Eslami m L2 + 2 m (2 L)2 + 3 m ( (2 L)2 + L2 ) + 4 m (L 2 + L 2 ) = m L2 + 8 m L2 + 3 m (5 L2 ) + 4 m (2 L2 ) = (1 + 8 + 15 + 8) m L2 = 32 m L2 keywords: 011 (part 1 of 1) 10 points Three identical thin rods of length 20 m and mass 51.7 kg are placed perpendicular to each other with their centers intersecting each other, as shown below. This setup is rotated about an axis parallel to the z -axis that passes through the end of one rod along the y -axis and is perpendicular to another along the x-axis. x 51.7 kg per rod z m = L 2 2 7 1 m L2 . 4 For the rod along the y -axis, from the table 1 m L2 . 3 In the rod along the x-axis, the bit of material m d x and between x and x + d x has mass L 2 L 2+ is at a distance r = x from the 2 axis of rotation, so the total moment of inertia for this rod is Iy = L/2 Ix = 2 0 x2 + L2 4 m dx L L/2 of xis a ion tat ro m x3 L 2 x + =2 L 3 4 0 3 2m L L3 = + L 24 8 1 = m L2 . 3 Hence, the overall moment of inertia of the three-rod system is I = I z + Iy + Ix 111 = ++ m L2 433 11 = m L2 12 11 (51.7 kg) (20 m)2 = 12 = 18956.7 kg m2 . 20 m y Determine the moment of inertia of this arrangement. Correct answer: 18956.7 kg m2 . Explanation: Let : m = 51.7 kg L = 20 m . and (1) Since r is much smaller than L, we can use the thin rod approximation. Let the junction be at the center of our coordinate system. The axis of rotation is parallel to the z -axis and the y -axis is along the rod touching the axis of rotation. For the rod along the z -axis the parallel axis theorem gives: m r2 +m Iz = 2 L 2 2 Alternate Solution: The moment of intertia of a thin rod about its center-of-mass is 1 Icm = mL2 . Therefore the center-of-mass 12 moment-of-inertia of the three perpendicular rods about an axis concentric to one of the rods is Icm = Iz + Iy + Ix 1 1 =0+ m L2 + m L2 12 12 1 2 = mL . 6 Thevalingam, Donald Homework 30 Due: Apr 30 2007, midnight Inst: Eslami Using the parallel axis theorm, we have I = Icm + 3 m = L 2 2 8 1 3 m L2 + m L2 6 4 11 2 = mL , 12 which is the same as Eq. 1. keywords: 012 (part 1 of 1) 10 points A rod can pivot at one end and is free to rotate without friction about a vertical axis, as shown. A force F is applied at the other end, at an angle to the rod. L m 013 (part 1 of 1) 10 points A system of two wheels xed to each other is free to rotate about a frictionless axis through the common center of the wheels and perpendicular to the page. Four forces are exerted tangentially to the rims of the wheels, as shown below. 2F 3R 2R F F F What is the magnitude of the net torque on the system about the axis? 1. = 0 F If F were to be applied perpendicular to the rod, at what distance d from the axis of rotation should it be applied in order to produce the same torque ? 1. d = 2 L 2. d = L tan 3. d = L 4. d = L sin correct 5. d = L cos Explanation: The torque the force generates is = F L sin . Thus the distance in question should be L sin . keywords: 2. = 2 F R correct 3. = 14 F R 4. = F R 5. = 5 F R Explanation: The three forces F give counter-clockwise torques while the other force 2 F gives a clockwise torque. So the total torque is = F i Ri = 2F R . = 2 F 3 R + F 3 R + F 3 R + F 2 R keywords: 014 (part 1 of 1) 10 points Given: A circular shaped object with an inner radius of 14 cm and an outer radius of 26 cm. There are three forces (acting perpendicular to the axis of rotation) whose magnitudes are 10 N, 26 N, and 14 N acting on the object, as Thevalingam, Donald Homework 30 Due: Apr 30 2007, midnight Inst: Eslami shown in the gure. The force of magnitude 26 N is 23 below horizontal. 10 N 26 cm 9 23 26 N as shown. The moment of inertia of the rod about an axis perpendicular to the rod and through the center of mass is given by I = m2 . 12 The acceleration of gravity is 9.8 m/s2 . F pivot 1.2 kg 2.1 m 14 N 6.5 kg Find the magnitude of the net torque on the wheel about the axle through the center of the object. Correct answer: 2.6 N m. Explanation: Let : a = 14 cm , b = 26 cm , F1 = 10 N , F2 = 26 N , F3 = 14 N , and = 23 . b F1 14 cm If a 5 N force at an angle of 106 to the horizontal acts on the rod as shown, what is the magnitude of the resulting angular acceleration about the pivot point? Correct answer: 1.2782 rad/s2 . Explanation: Let : = 2.1 m , m = 1.2 kg , = 106 , and F = 5 N. Basic Concepts: =rF =I Solution: By the parallel axis theorem, the moment of inertia of a stick pivoted at the end is I = Icm + m d2 1 m 2+m = 12 1 = m 2. 3 2 F2 a The total torque is M F3 = a F 2 b F1 b F3 = (14 cm) (26 N) = 2. 6 N m = 2. 6 N m . keywords: 015 (part 1 of 1) 10 points A uniform horizontal rod of mass 1.2 kg and length 2.1 m is free to pivot about one end (26 cm) (10 N) + (14 N) 2 The sum of the torques (counterclockwise rotation dened as positive) is = F sin m g Hence F sin m g 1 m 3 2 2 = I . = 2 10 6 Thevalingam, Donald Homework 30 Due: Apr 30 2007, midnight Inst: Eslami = = 6 F sin 3 m g 2m 6 (5 N) sin(106 ) 2 (1.2 kg) (2.1 m) 3 (1.2 kg) (9.8 m/s2 ) 2 (1.2 kg) (2.1 m) Ft M g 2 = 0, = 1.2782 rad/s2 . keywords: 016 (part 1 of 5) 10 points A long, 5 kg uniform rod and 0.6 m length is supported at the left end by a horizontal axis into the page and perpendicular to the rod, as shown. The acceleration of gravity is 9.8 m/s2 . The right end is connected to the ceiling by a thin vertical thread so that the rod is horizontal. The moment of inertia of the rod M2 about the axis at the end of the rod is . 3 Thread 0. 6 m 5 kg 10 Mg and is directed upward So Fa = 2 Alternate Solution: For the last two points, sum the torques about any other axis and also use F = 0. For example, summing torques about left end of rod where Ft is the force exerted by the thread Mg 2 (5 kg) (9.8 m/s2 ) = = 24.5 N . 2 Fa = And F = 0 gives Ft + Fa M g = 0, so Fa and is directed upward. 017 (part 2 of 5) 10 points The thread is then burned by a match. Find the angular acceleration of the rod about the axis immediately after the thread breaks. Correct answer: 24.5 rad/s2 . Explanation: Using r = I and calculating the torque about the axis end of the rod 2 Horizontal Axis Determine the magnitude and direction of the force exerted on the rod by the axis. Correct answer: 24.5 N. Explanation: Let : = 0.6 m , M = 5 kg , and g = 9. 8 m /s 2 . = rF = 0 (or CW = CCW ) . Summing torques about the right end of the rod Fa M g exerted by axis 2 = 0, where Fa is force Mg Solving for = 2 =M 3 . 3g 2 3 9.8 m/s2 = 2 0.6 m = 24.5 rad/s2 . 018 (part 3 of 5) 10 points Find the translational acceleration of the center of mass of the rod immediately after the thread breaks. Correct answer: 7.35 m/s2 . Explanation: Thevalingam, Donald Homework 30 Due: Apr 30 2007, midnight Inst: Eslami Using the relation between translational and angular acceleration, ac = r. Substituting r = and from previous 2 part ac = 3 g 4 3 = (9.8 m/s2 ) 4 = 7.35 m/s2 . 019 (part 4 of 5) 10 points Find the force exerted on the end of the rod by the axis immediately after the thread breaks. Correct answer: 12.25 N. Explanation: Using Newtons second law: F = Ma But F = M g Fr , so M g Fr = M a . 3 Substituting a = g and solving for Fr 4 Fr = 1 Mg 4 1 = (5 kg) (9.8 m/s2 ) 4 = 12.25 N . Substituting I = = = = M 3 2 11 and solving for Mg sin I 3g sin 3 (9.8 m/s2 ) sin 20 0 .6 m = 4.09377 rad/s . Alternate Solution: Work done by gravitational force Wg equals the increase in kinetic energy of rotation Krot Wg = M g cos d = 2 Krot = Mg sin 2 1 I 2 2 1 I 2 = M g sin . 2 2 Substituting I = = 3g M 3 2 and solving for sin = 4.09377 rad/s . 020 (part 5 of 5) 10 points The rod rotates about the axis and swings down from the horizontal position. When the rod is at an angle = 20 with the horizontal, nd the angular velocity of the rod. Correct answer: 4.09377 rad/s. Explanation: Let : = 20 . Using conservation of energy the increase in kinetic energy of rotation Krot is equal to the decrease in potential energy U : 1 I 2 = m g h 2 keywords: Krot = U = 1 I 2 = M g sin . 2 2

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University of Texas - PHY - phy 303k
Thevalingam, Donald Homework 31 Due: May 24 2007, midnight Inst: Eslami This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. The due time is Central time. 001 (part 1 o
University of Texas - PHY - phy 303k
Thevalingam, Donald Homework 32 Due: May 24 2007, midnight Inst: Eslami This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. The due time is Central time. 001 (part 1 o
University of Texas - PHY - phy 303k
Version 040 Quiz 2 coker (58245) This print-out should have 24 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. Quiz 2 covers only topics covered in class lectures over Chs. 6, 7, 8 and 9 in Oha
University of Texas - PHY - phy 303k
homework 07 GADHIA, TEJAS Due: Mar 7 2007, 4:00 am So we obtain Question 1 part 1 of 1 10 points Two identical stars, a xed distance D apart, revolve in a circle about their mutual center of mass, as shown below. Each star has mass M and speed v . G is th
University of Texas - PHY - phy 303k
homework 05 GADHIA, TEJAS Due: Feb 21 2007, 4:00 am Question 1 part 1 of 1 10 points Given: g = 9.8 m/s2 . To test the performance of its tires, a car travels along a circular track of radius R = 379 m. The track is perfectly at (no banking). The car incr
University of Texas - PHY - phy 303k
homework 06 GADHIA, TEJAS Due: Feb 28 2007, 4:00 am Question 1 part 1 of 2 10 points While skiing in Jackson, Wyoming, your friend Ben (75 kg) started his descent down the bunny run, 11 m above the bottom of the run. If he started at rest and converted al
University of Texas - PHY - phy 303k
Patel (ppp285) HW05 TSOI (58160) This print-out should have 25 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 1.7864 kg raindrop falls vertically at constant sp
University of Texas - PHY - phy 303k
Patel (ppp285) HW04 TSOI (58160) This print-out should have 20 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 As shown in the gure, a block is pushed up against t
University of Texas - PHY - phy 303k
Patel (ppp285) HW06 TSOI (58160) This print-out should have 24 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 Three 6 kg masses are located at points in the xy pl
University of Texas - PHY - phy 303k
midterm 02 GADHIA, TEJAS Due: Mar 7 2007, 11:00 pm1Mechanics - Basic Physical ConceptsMath: Circle: 2 r, r2 ; Sphere: 4 r2 , (4/ r3 3) b2 Quadratic Eq.: a x2 + b x + c = 0, x = b 2 a4 a cCartesian and polar coordinates: y x = r cos , y = r sin , r2 =
University of Texas - PHY 303K - 303K
bhakta (vsb255) HW 10 coker (58245) This print-out should have 24 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. 001 10.0 points A 2 kg steel ball strikes a wall with a speed of 14.9 m/s at an
USC - CHEM - 322AL
USC - CHEM - 322AL
USC - CHEM - 322AL
USC - CHEM - 322AL
USC - CHEM - 322AL
USC - CHEM - 322AL
USC - CHEM - 322AL
USC - CHEM - 322AL
USC - CHEM - 322AL
USC - CHEM - 322AL
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USC - CHEM - 322AL
USC - CHEM - 322AL
Quiz 8.05 Predict the products of the following addition reactions. Indicate important stereochemical details in the products. CH3CH2CH2CH=CH2 + HICH3CH2CH2CHICH3as a racemic formCH3+ HICH3 IachiralCH3 CH3+ HICH3 H+ CH3 A I BCH3 H+ I CH3CH3 H C
USC - CHEM - 322AL
Acid-Base ReactionspKa 4.82 2.86Ka 1.5 x 10-5 139 x 10-5G (kcal/M) 6.7ClCO2H CO 2+H+Cl CO2H CO 24.0+H+4.05 4.528.9 x 10-5 3.0 x 10-55.7ClCO2H Cl CO2HCO 2+H+6.3ClClCO 2+H+Inductive effect (through bond) falls off with distance (bo
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Bromination Reactions StereochemistryBr a Br top H H3C CH3 H bottom H H3C Br H H3C a c b d CH3 H c d H H3C CH3 H b H H3C Br H H3C CH3 H Br Br CH3 H Br d = Br c = Br Br = Br Br H H3C CH3 H Br Br CH3 H Br b = Br Br = Br Br a = Br Br = Br Br Br = Br BrHow
USC - CHEM - 322AL
USC - CHEM - 322AL
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USC - CHEM - 322AL
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USC - CHEM - 322AL
Bromination of Higher Alkanes: SelectivityBrominations of alkanes proceed much more slowly than chlorinations, but with much greater selectivity.CH3 Br2 CH3-C-CH3 h!, 127o C H isobutane CH3 CH3 CH3-C-CH3 + CH3-C-CH2Br H Br tert-butyl bromide isobutyl br
USC - CHEM - 322AL
A Word on ThermochemistryH HO H H+H !+Br !HO+BrDriving Force: G = - RT ln(Keq) G = - 2.3 RT log(Keq) = 2.3 * 0.0020 kcal/mol-K * T * pKeq or at room temp (298 K) G = 1.4 pKeq THERMODYNAMICS: A 1.4 kcal/mol difference between starting materials an
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Alcohols and EthersAlcohol: Any kind of R3C-OH functional group.(a secondary alcohol) OH H H HO (a phenol) estradiol H HO HO HO OOH OHoctanolOHwater!-D-glucose (a simple sugar)1-octanol (a primary alcohol)Kow = [drug]octanol / [drug]water logP = l
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Quiz 12.02 Provide the organic products of the two reactions below. Assign formal oxidation states to the reacting carbon centers in the starting materials and the products. (A) O CH3CH (i) LiAlH4/ether (ii) H2OO (B) CH3COCH2CH3How many equivalents of h
USC - CHEM - 322AL
The Story of the HydrideMolecular Structure (refresher) 2!#CspHs!#Csp2Csp2 &quot;#CpCp &quot;CpCp!#Csp2Hs2 H's !Csp2Hs MOCsp2p!Csp2Csp2 MOCsp2p !Csp2Hs MO2 H's&quot;CpCp!Csp2HsH!Csp2Csp2H!Csp2HsHHMolecular Structure (refresher) 2!#CspHs!#Csp2Osp2&quot;
USC - CHEM - 322AL
USC - CHEM - 322AL
USC - CHEM - 322AL
USC - CHEM - 322AL
USC - CHEM - 322AL
USC CHEM 322aL/325aL F 2009JUNG/WILLIAMS/ELLERNWater-Referenced pKa sH2SO4-9, 2 H2NNH2 NH213H Cl-7H OH15.7H OH2-1.7HO OCH316H3PO42, 7, 12H20 CH3O H 4.7 O CH3 H CH 25H2CO36, 10H NH238HO10HCH244HN H2CH311HCH350
USC - CHEM - 322AL
Practice Exam 1 (9/2/09)(15 pts) 1. Eight constitutional isomers of C5H8O containing a double bond and an aldehyde.(10 pts) 2. Draw (a) resonance structure(s) of the following species.O(10 pts) 3. Label the following sets of structures as geometrical
USC - CHEM - 322AL
Practice Exam 1 (9/2/09)(15 pts) 1. Eight constitutional isomers of C5H8O containing a double bond and an aldehyde. There are eight isomers. CHO CHO CHOCHO CHO CHOCHO CHO (10 pts) 2. Draw (a) resonance structure(s) of the following species.OO(10 pts
USC - CHEM - 322AL
Practice Exam 2 (9/9/09)1. Predict the direction in which the reaction proceeds by placing a reaction arrow between the reactant and the product. Show the curved (mechanism) arrow for the reaction in the provided space. O OH O O Na+NaOH+H2O2. In the
USC - CHEM - 322AL
Practice Exam 3 (9/16/09)1. Using systematic nomenclature, assign a name for the following compound. Br2. Given the following compound name, draw a valid structure.5-chloro-4-cyclopentyl-6-hepten-3-ol3. Draw the two possible chair conformations with a
USC - CHEM - 322AL
Practice Exam 4 (9/23/09)I. Provide the stereochemisty by either R or S configuration on the designated carbon. OH H Me OMe Br H Carbon in the back bearing H, Br, and OMe group CHO H OH HO Me HO H Me Carbon in the middle bearing HO and MeII. Given the f
USC - CHEM - 322AL
Practice Exam 5 (9/30/09)I. Predict the major product(s) for the following reactions and show the stereochemistry if any. (10 pts) Br CH3COO room temperature (10 pts) Me ClMeOH 25 C(10 pts) (R)-2-chlorobutane NaOH 25 CII. In elimination reactions, t-b
USC - CHEM - 322AL
Practice Exam 6 (10/07/09)I. Give the systematic name for the following structure. ClBr II. Give the structure for the following compound.(E)-1-chloro-2-fluoro-1-iodo-1-penteneIII. Predict the major product. Me OH H heatIV. For the following eliminat
USC - CHEM - 322AL
Practice Exam 7 (10/12/09)I. Complete the following syntheses showing isolable intermediates and conditions. Ia.fromO PhtoPhPhIb.HOfrom toHint: A double bond between two rings is too strained. The species just below is not a viable intermediate.
USC - CHEM - 322AL
USC - CHEM - 322AL
USC - CHEM - 322AL
USC - CHEM - 322AL
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USC - CHEM - 322AL
USC - CHEM - 322AL
HKU - MEEM - IS6007
ISyE 3103Supply Chain Modeling: Transportation and LogisticsSpring 2006 Extra Regression Forecasting ProblemThis is an extra problem1 provided for those of you who would like some additional practice in the mechanics of specifying regression models for
HKU - ISE - meem 6015
ISyE 3103Supply Chain Modeling: Transportation and LogisticsSpring 2006 Winters Method Forecasting ExampleA scatterplot of the following actual demand data suggests the existence of seasonal (quarterly) variation coupled with an overall increasing tren
National Taiwan University - MGT - 7625
t 3Steven P. Robbins Mary Coulter * @ + D @ + D 7 7 @ @Copyright 2005 Prentice Hall, Inc. All rights reserved.32 * * * * D * * @7 @ + @@ + D * D * *Copyright 2005 Prentice Hall, Inc. All rights reserved.33Copyright 2005 Prentice Hall, In
National Taiwan University - MGT - 7625
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National Taiwan University - MGT - 7625
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National Taiwan University - MGT - 7625
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National Taiwan University - MGT - 7625
Steven P. Robbins Mary Coultert 7x +` x *H Copyright 2005 Prentice Hall, Inc. All rights reserved.721. 2. 3. 4.* x+ * h G u x + @ , * G u x + Hx +xCopyright 2005 Prentice Hall, Inc. All rights reserved.73Copyright 2005 Prentice Hall, Inc. A
National Taiwan University - MGT - 7625
Steven P. Robbins Mary Coultert 8- - -j * *D *~ v - ` J *1 a~ J 2 7t @ 5` 2 @ g@ ^ TM @ [ D f@ [ D -7@Copyright 2005 Prentice Hall, Inc. All rights reserved.82 2 * * P f @ [ D * D * * * ^ 7 @Copyright 2005 Prentice Hall, Inc. All righ
National Taiwan University - MGT - 7625
Steven P. Robbins Mary Coultert 9- - + 0 6 * Copyright 2005 Prentice Hall, Inc. All rights reserved.92 + + + + Copyright 2005 Prentice Hall, Inc. All rights reserved.93@ k 8 A + S - p +7@Copyright 2005 Prentice Hall, Inc. All rights reserv