This preview has intentionally blurred parts. Sign up to view the full document

View Full Document

Unformatted Document Excerpt

10 homework FRENNEA, KYLE Due: Apr 6 2007, 4:00 am Question 1 part 1 of 1 10 points A one-piece cylinder is shaped as in the figure below, with a core section protruding from the larger drum. The cylinder is free to rotate around the central axis shown in the drawing. A rope wrapped around the drum, of radius 1.29 m, exerts a force F1 to the right on the cylinder. A rope wrapped around the core, of radius 0.413 m, exerts a force F2 downward on the cylinder. y F1 R1 1 weight of mass 2 m is attached to the end of this rope. R m 2m At some moment, the weight is falling with a speed v . The total kinetic energy K of the system at this moment is given by x R2 O 1. K = [1 + k] m v 2. K = 3. K = 4. K = 5. K = 6. K = 1+ 2k 2 1+k 2 1+k 2 k 1+ 2 1+ 2k 2 m v2 m v2 mv m v 2 correct mv F2 Let F1 = 4.59 N and F2 = 8.59 N What is the net torque acting on the cylinder about the rotation axis (which is the z axis in the figure)? Correct answer: -2.37343 N m (tolerance 1 %). Explanation: The torque due to F1 is -R1 F1 and is negative because it tends to produce a clockwise rotation. The torque due to F2 is +R2 F2 and is positive since it tends to produce a counterclockwise rotation. Therefore, the net torque about the rotation axis is net = R2 F2 - R1 F1 = -2.37343 N m . 7. K = [1 + k] m v 2 8. K = [1 + 2 k] m v 2 9. K = 1 + k mv 2 10. K = [1 + 2 k] m v Explanation: Basic Concepts: Rotational kinetic energy is 1 Erot = I 2 2 Ktot = Kwheel + Kmass At the moment the weight is descending with a speed v, the tangential speed of the wheel is Question 2 part 1 of 2 10 points Consider a circular wheel with a mass m, and a radius R. The moment of inertia about the center of the wheel is I = k m R2 , where k is a constant in the range between 0.5 k 1.0 . A rope wraps around the wheel. A homework 10 FRENNEA, KYLE Due: Apr 6 2007, 4:00 am also v. Then the angular velocity of the wheel v is = . So the total kinetic energy is R 1 1 K = 2 m v2 + I 2 2 2 1 1 = 2 m v 2 + k m (R )2 2 2 1 1 = 2 m v2 + k m v2 2 2 k = 1+ m v2 . 2 Question 3 part 2 of 2 10 points 1 Assume: k = . 2 If the system is released from rest, find the speed v at the moment when the weight has descended a vertical distance h. 1. v = 2. v = 3. v = 4. v = 5. v = 6. v = 7. v = 8. v = 9. v = 10. v = 4 gh 5 8 gh 7 8 gh 11 2gh 8 gh 9 16 gh 11 8 g h correct 5 gh 4 gh 3 = Then we have v= 8 gh. 5 5 m v2 . 4 2 Question 4 part 1 of 4 10 points A 2900 kg block is lifted at a constant speed of 6 cm/s by a steel cable that passes over a massless pulley to a motor-driven winch. The radius of the winch drum is 25 cm and the winch drum has a mass of 3000 kg. The acceleration due to gravity is 9.81 m/s2 . 25 cm 3000 kg 2900 kg What force must be exerted by the cable? Correct answer: 28.449 kN (tolerance 1 %). Explanation: Let : m = 2900 kg , v = 6 cm/s = 0.06 m/s , r = 25 cm = 0.25 m . 6 cm/s and kN N Note: There is no acceleration in this system and the mass of the winch drum is not required. Because the block is lifted at constant speed, the cable tension is T = mg = (2900 kg) (9.81 m/s2 ) = 28.449 kN . Question 5 part 2 of 4 10 points 2 gh 3 Explanation: The mechanical energy is conserved. So 2mgh = 1 + k 2 m v2 homework 10 FRENNEA, KYLE Due: Apr 6 2007, 4:00 am What torque does the cable exert on the winch drum? Correct answer: 7.11225 kN m (tolerance 1 %). Explanation: The torque is =Tr = (28.449 kN) (0.25 m) = 7.11225 kN m . Question 6 part 3 of 4 10 points What is the angular velocity of the winch drum? Correct answer: 0.24 rad/s (tolerance 1 %). Explanation: v = r, so the angular velocity is = v r 0.06 m/s = 0.25 m = 0.24 rad/s . 3 A pulley (in the form of a uniform disk) with mass 51 kg and a radius 27 cm is attached to the ceiling, in a uniform gravitational field, and rotates with no friction about its pivot. The acceleration of gravity is 9.8 m/s2 . These masses are connected by a massless inextensible cord. T1 , T2 , and T3 are magnitudes of the tensions. T3 27 cm T2 39 kg 4.1 m 27 kg Determine the acceleration of the mass 27 kg. Correct answer: 1.28525 m/s2 (tolerance 1 %). 51 kg T1 Explanation: Question 7 part 4 of 4 10 points What power must be developed by the motor to drive the winch drum? Correct answer: 1.70694 kW (tolerance 1 %). Explanation: The power is P =Tv = (28.449 kN) (0.06 m/s) = 1.70694 kW . Question 8 part 1 of 6 10 points Assume: The positive y direction is up. Consider the free body diagrams kW 1000 W Let : a1 = acceleration of the 27 kg mass M2 = 39 kg , m1 = 27 kg , Mp = 51 kg , = 4.1 m , s = R, v = R, a = R, 1 I = M R2 , and 2 1 1 Kdisk = I 2 = M v 2 . 2 4 homework 10 FRENNEA, KYLE Due: Apr 6 2007, 4:00 am Question 9 part 2 of 6 10 points 4 T2 T1 M2 M2 g m1 m1 g Determine the acceleration of the center of mass acm of M2 + m1 . Correct answer: -0.233681 m/s2 (tolerance 1 %). Explanation: The larger mass accelerates down and the smaller mass accelerates up, so the center of mass of the system accelerates down. The acceleration of the center of mass is acm M2 aM 2 + m1 am1 . M2 + m1 a The equations of motion using free body diagrams are Fy : Fy : : T1 - m1 g = m1 a1 T2 - M2 g = M2 a2 = -M2 a1 I = (T2 - T1 ) r . a1 r a Thus, since Using "up" as the positive direction, then aM 2 = a2 = -a1 and am1 = +a1 . From Eq. (3), and using Eqs. (4), (5), and (6), we have acm = -2 M2 (M2 - m1 ) - m1 [M2 - m1 ] g (M2 + m1 ) {Mp + 2 (M2 + m1 )} [M2 - m1 ]2 g = -2 [M2 + m1 ] {Mp + 2 (M2 + m1 )} [12 kg]2 = -2 (9.8 m/s2 ) [66 kg] {183 kg} = -0.233681 m/s2 . T1 = m1 (g + a1 ) (1) T2 = M2 (g - a1 ) (2) I = [M2 (g - a1 ) - m1 (g + a1 )] r a1 = [(M2 - m1 ) g - (M2 + m1 ) a1 ] r . I r for a disk, I = 1 Mp r 2 , so 2 [Mp + 2 (M2 + m1 )] a1 = 2 (M2 - m1 ) g . Solving for a1 , we have a1 = 2 [M2 - m1 ] g {Mp + 2 (M2 + m1 )} 2 [(12 kg)] = (9.8 m/s2 ) {183 kg} = 1.28525 m/s2 . (3) The negative sign indicates the center of mass acceleration is down. Question 10 part 3 of 6 10 points Determine the magnitude of the tension T1 . Correct answer: 299.302 N (tolerance 1 %). Explanation: Substituting a1 from Eq. (3) into Eq. (1) gives T1 = m1 (g + a1 ) m1 [Mp + 4 M2] g = {Mp + 2 (M2 + m1 )} (27 kg) [207 kg] (9.8 m/s2 ) = {183 kg} = 299.302 N , (7) where [M2 + m1 ] = (39 kg) + (27 kg) = 66 kg [M2 - m1 ] = (39 kg) + (27 kg) = 12 kg {Mp + 2 (M2 + m1 )} = (51 kg) + 2 [66 kg] = 183 kg . (4) (5) (6) homework 10 FRENNEA, KYLE Due: Apr 6 2007, 4:00 am using Eqs. (4), (5), & (6) and where [Mp + 4 M2] = (51 kg) + 4 (39 kg) = 207 kg . Question 11 part 4 of 6 10 points Determine the magnitude of the difference in the tensions T = T2 - T1 . Correct answer: 32.7738 N (tolerance 1 %). Explanation: Substituting a1 from Eq. (3) into Eq. (2) gives T2 = M2 (g - a1 ) M2 [Mp + 4 m1 ] g = {Mp + 2 (M2 + m1 )} (39 kg) [159 kg] (9.8 m/s2 ) = {183 kg} = 332.075 N , using Eqs. (4), (5), & (6) and where [Mp + 4 m1 ] = (51 kg) + 4 (27 kg) = 159 kg . Therefore T2 > T1 by the amount T = T2 - T1 = (332.075 N) - (299.302 N) = 32.7738 N . Question 12 part 5 of 6 10 points Determine the magnitude of the tension T3 . Correct answer: 1131.18 N (tolerance 1 %). Explanation: Using Eqs. (7) & (8), we have T3 = T1 + T2 + Mp g = (299.302 N) + (332.075 N) + (51 kg) (9.8 m/s2 ) = 1131.18 N . (9) (8) Question 13 part 6 of 6 10 points 5 Determine the magnitude of the difference in the force of gravity on the sum of the three masses and the tension in the cord holding up the pulley Fg = (Mp + M2 + m1 ) g - T3 . Correct answer: 15.423 N (tolerance 1 %). Explanation: F = [Mp + M2 + m1 ] g - T3 = [(51 kg) + (39 kg) + (27 kg)] (9.8 m/s2 ) - (1131.18 N) = (1146.6 N) - (1131.18 N) = 15.423 N . Using Eqs. (7) & (8), we have F = (M2 + m1 + Mp ) g - (T1 + T2 + Mp ) g = [M2 - m1 ] a1 = [(39 kg) - (27 kg)] (1.28525 m/s2 ) = 15.423 N . Since M2 > m1 the center of mass of the blocks is accelerating down and the tension T3 is less that the force of gravity of the sum of the masses. However, if M2 = m1 then the difference would be zero. Question 14 part 1 of 2 10 points Assume: When the disk lands on the surface it does not bounce. The disk has mass m and outer radius R with a radial mass distribution (which may not be uniform) so that its moment of inertia 8 is m R2 . 9 A disk is given a hard kick (impulse) along a horizontal surface at time t0 . The kicking force acts along a horizontal line through the homework 10 FRENNEA, KYLE Due: Apr 6 2007, 4:00 am disk's center, so the disk acquires a linear velocity v0 but no initial angular velocity. The coefficient of friction between the disk and the surface is . The kinetic friction force between the surface and the disk slows down its linear motion while at the same time making the disk spin on its axis at an accelerating rate. Eventually, the disk's rotation catches up with its linear motion, and the disk begins to roll at time trolling without slipping on the surface. The acceleration of gravity is g . m R , radius v0 8 I= m R2 9 How long t = trolling - t0 does it take for the ball to roll without slipping? 1. t = 2. t = 3. t = 4. t = 5. t = 6. t = 7. t = 8. t = 9. t = 10. t = 3 v0 10 g 4 v0 9 g 5 v0 14 g 2 v0 5 g 7 v0 15 g 6 v0 13 g 1 v0 5 g 8 v0 correct 17 g 5 v0 13 g 9 v0 19 g 6 Using the frictional force f , we can determine the acceleration f = mg, and Fsurf ace = m a , or m a = m g , so a = g. Since vsurf ace = v0 - a t , we have = v0 - g t . (2) (1) After pure rolling begins at trolling there is no longer any frictional force and consequently no acceleration. From the perspective of the center of the disk, let the tangential velocity of the rim of the disk be vdisk and the angular velocity be ; the angular acceleration is = I , so = I mgR = 8 m R2 9 9 g . = 8 R The time dependence of is (3) = t 9 g = t , so 8 R vdisk R 9 = gt. (4) 8 When the disk reaches pure rolling, the velocity from the perspective of the surface will be the same as the velocity from the perspective of the center of the disk; that is, there will be no slipping. Setting the velocity vdisk from Eq. 4 equal to vsurf ace from Eq. 2 gives vdisk = vsurf ace 9 g t = v0 - g t , 8 17 g t = v0 , so 8 8 v0 . t= 17 g or Explanation: From the perspective of the surface, let the speed of the center of the disk be vsurf ace . (5) homework 10 FRENNEA, KYLE Due: Apr 6 2007, 4:00 am Question 15 part 2 of 2 10 points Once the disk rolls without slipping, what is its linear speed? 9 v0 13 8 2. vrolling = v0 13 4 3. vrolling = v0 7 5 4. vrolling = v0 9 10 5. vrolling = v0 19 9 v0 6. vrolling = 14 5 7. vrolling = v0 7 9 8. vrolling = v0 correct 17 7 9. vrolling = v0 13 8 v0 10. vrolling = 11 Explanation: Using Eqs. 1 and 5, we have 1. vrolling = vsurf ace = v0 - a t = v0 - g t = v0 - g = v0 = 8 v0 17 g 8 1- 17 (6) Question 16 part 1 of 1 10 points 7 A 3.23 kg hollow cylinder with inner radius 0.28 m and outer radius 0.58 m rolls without slipping when it is pulled by a horizontal string with a force of 26.3 N, as shown in the diagram below. Its moment of inertia about the center of 1 2 2 mass is m Rout + Rin . 2 3.23 kg 26.3 N 0.58 m 0.28 m What is the acceleration of the cylinder's center of mass? Correct answer: 10.074 m/s2 (tolerance 1 %). Explanation: Let : Rin = 0.28 m , Rout = 0.58 m , F = 26.3 N , m = 3.23 kg , and f = force of friction . F 0.28 m f Using a free-body diagram, we have F + f = ma (1) a 0.58 m 9 v0 , 17 or using Eqs. 4 and 5, we have vdisk = 9 gt 8 8 v0 9 = g 8 17 g 9 v0 . 17 (6) = From the torque equation (using the center of the cylinder as the axis of rotation), we have a (F - f ) Rout = I = I Rout a (2) F -f =I 2 Rout homework 10 FRENNEA, KYLE Due: Apr 6 2007, 4:00 am Adding equation (1) and (2) together, m 2F =a m+ I 2 Rout 2 m Rout + I , so =a 2 Rout 2 2 F Rout a= 2 m Rout + I 2 F 4 Rout = 2 2 m 3 Rout + Rin (26.3 N) 4 (0.58 m)2 = (3.23 kg) 3 (0.58 m)2 + (0.28 m)2 = 10.074 m/s2 . 0.81 m m x 0.81 m y 8 The Moment of Inertia: The mass of the hollow cylinder is Note: Figure is not drawn to scale What is the new angular speed if the spokes are shortened to 0.63 m? (An effect similar to this ocurred in the early stages of the formation of our galaxy. As the massive cloud of gas and dust contracted, an initially small rotation increased with time.) Correct answer: 2.8102 rev/s (tolerance 1 %). Explanation: Basic Concepts: Rout m= Rin Rout dm 2 L r dr Rin L = I = M r 2 L2 = L1 and Given: r1 = 0.81 m r2 = 0.63 m 1 = 1.7 rev/s The total mass of the system is the sum of all point masses: M = 4m Solution: Angular momentum of the system is conserved, so 2 2 M r 2 2 = M r 1 1 2 2 r 2 2 = r 1 1 r 2 1 2 = 1 2 r2 (0.81 m)2 (1.7 rev/s) = (0.63 m)2 = 2.8102 rev/s = 2 2 = L Rout - Rin , Rout Iz = Rin Rout r 2 dm r 2 2 L r dr Rin = 1 4 4 = L Rout - Rin 2 1 2 2 L Rout - Rin = 2 1 2 2 = m Rout + Rin . 2 2 2 Rout + Rin Question 17 part 1 of 1 10 points The figure shows a system of point masses that rotates at an angular speed of 1.7 rev/s. The masses are connected by light, flexible spokes that can be lengthened or shortened. homework 10 FRENNEA, KYLE Due: Apr 6 2007, 4:00 am Solution: Question 18 part 1 of 4 10 points Assume: This collision takes place in outer space where there is no gravitational field and no friction. small A piece of putty of mass 47 g and negligible size has a speed of 1.5 m/s. It makes a collision with a rod of length 5 cm and mass 86 g (initially at rest) such that the putty hits the very end of the rod. The putty sticks to the end of the rod and spins around after the collision. Note: The rod has a moment of inertia 1 equal to m r 2 about its center-of-mass. 12 The piece of putty is a point mass. 1.5 m/s 47 g + cm 9 p = m1 v = (m1 + m2 ) V Therefore V = m1 v m1 + m2 (0.047 kg) = (1.5 m/s) (0.047 kg) + (0.086 kg) = 0.530075 m/s . Question 19 part 2 of 4 10 points What is the angular momentum of the system relative to the center-of-mass after the collision? Correct answer: 0.00113966 kg m2 /s (tolerance 1 %). Explanation: V + 0 (a) 86 g (b) (c) 1 + cm Figure: The piece of putty and rod: (a) before they collide, (b) at the time of the collision, and (c) after they collide. After the collisions the center-ofmass has a linear velocity V and an angular velocity about the centerof-mass "+ cm". What is the velocity V of the center-of-mass of the system after the collision? Correct answer: 0.530075 m/s (tolerance 1 %). Explanation: Let : m1 = 47 g , m2 = 86 g , = 5 cm , and v = 1.5 m/s . Basic Concepts: Conservation of linear momentum and angular momentum. 2 2 0 (b) Basic Concepts: Conservation of linear momentum and angular momentum. When the piece of putty touches the rod, the distance from the center-of-mass to the center of the rod of mass m2 is 2 m2 = 1 m1 - 2 m1 2 m2 = 2 2 (m1 + m2 ) = m1 , so 2 homework 10 FRENNEA, KYLE Due: Apr 6 2007, 4:00 am 2 2 = m1 + m2 (0.047 kg) (0.05 m) = 2 (0.047 kg) + (0.086 kg) m1 = 0.00883459 m and the distance from the center-of-mass to the smaller puck m1 is 1 = - 2 = 0.0161654 m . 2 = 0.00113966 kg m2 /s 3.6911 10-5 kg m2 10 = 30.8759 rad/s . Question 21 part 4 of 4 10 points Ef % of the enEi ergy of the system after the collision to the energy of the system before the collision? Correct answer: 68.6131 % (tolerance 1 %). What is the percent ratio Explanation: 1 m1 v 2 2 1 = (0.047 kg) (1.5 m/s)2 2 = 0.052875 N . The angular momentum of the system about the center-of-mass is L = r p = 1 m1 v = (0.0161654 m) (0.047 kg) (1.5 m/s) = 0.00113966 kg m2 /s . Question 20 part 3 of 4 10 points What is the system's angular speed about the center-of-mass after the collision? Correct answer: 30.8759 rad/s (tolerance 1 %). Explanation: The moment of inertia of the system about its center-of-mass using the parallel axis theorem is cm I = Iputty + Irod + m2 2 2 1 2 = m1 2 + m2 + 2 2 1 12 = (0.047 kg) (0.0161654 m)2 + (0.086 kg) 1 (0.05 m)2 + (0.00883459 m)2 12 = (1.22821 10-5 kg m2 ) +(2.4629 10-5 kg m2 ) = 3.6911 10-5 kg m2 . Ei = Ef = 1 1 [m1 + m2 ] V 2 + I 2 2 2 = (0.0186852 N) + (0.017594 N) = 0.0362792 N . Ef (100) Ei (0.0362792 N) = (100) (0.052875 N) R= = 68.6131 % . Question 22 part 1 of 3 10 points A mass m is attached to a cord passing through a small hole in a frictionless horizontal surface. The mass is initially orbiting with speed v0 in a circle of radius r0 . The cord is then slowly pulled from below, decreasing the radius of the circle to r. What is the speed of the mass when the radius is r? From the angular momentum definition it follows that L I , so = L I homework 10 FRENNEA, KYLE Due: Apr 6 2007, 4:00 am r r0 r0 + r 2. v = v0 r0 r0 correct 3. v = v0 r 1. v = v0 4. None of these r0 - r r0 r0 6. v = 2 v0 r r0 7. v = v0 r0 - r r0 - r 8. v = v0 r0 + r r0 + r 9. v = v0 r0 - r r0 10. v = v0 r0 + r Explanation: Basic Concepts 5. v = v0 =rF d dt Solution = rF = r F sin(0) = 0 5. W = 6. W = 7. W = 8. W = L = ext m (r0 v0 )2 (r + r0 )3 m (r v0 )2 3. T = 3 r0 11 2. T = 4. None of these m (r0 v0 )2 5. T = correct r3 2 m (r0 v0 )2 6. T = r3 Explanation: T = m v2 m (r0 v0 )2 = r r3 Question 24 part 3 of 3 10 points How much work W is done in moving m from r0 to r? Note: The tension depends on r. 1. W = 2 2 m v0 3 1 2 2. W = m v0 2 3. None of these 4. W = 1 2 m v0 2 1 2 m v0 2 1 2 m v0 3 1 2 m v0 3 2 2 m v0 3 r2 2 r0 2 r0 r2 2 r0 r2 r2 2 r0 r2 2 r0 -1 -1 -1 -1 -2 correct 2 r0 -2 r2 (r0 + r)2 -1 r2 Angular moment is conserved, so Lf = Li m r v = m r0 v0 Therefore, v= r0 v0 r Question 23 part 2 of 3 10 points Find the tension in the cord as a function of r. m (r0 v0 )2 1. T = 2 r3 Explanation: The work is done by the centripetal force in the negative direction. Method 1: W = =- F d T dr homework 10 FRENNEA, KYLE Due: Apr 6 2007, 4:00 am m (r0 v0 )2 dr (r )3 r0 m (r0 v0 )2 1 1 = - 2 2 2 r r0 2 1 2 r0 = m v0 -1 2 r2 =- Method 2: W = K = 1 1 2 m v 2 - m v0 2 2 2 r0 -1 r2 r 12 Find the angular momentum of the planet that arises from its orbital motion about the Sun. Correct answer: 2.65135 1040 kg m2 /s (tolerance 1 %). Explanation: Given : r = 1.493 1011 m , = 1 rev/365 days . L = I = M r 2 L = 5.97 1024 kg 1.493 1011 m 1 rev 2 rad 365 day 1 rev 1h 1 day 24 h 3600 s = 2.65135 1040 kg m2 /s . and Question 27 part 1 of 1 10 points The figure shows the rear view of a space capsule that is rotating about its longitudinal axis at 1 rev/min. The occupants want to stop this rotation. They have small jets mounted tangentially at a distance 2.2 m from the axis, as indicated, and can eject 9.5 g/s of gas from each jet with a nozzle velocity of 516 m/s. The moment of inertia of the ship about its axis (assumed to be constant) is 2290 kg m2 . 9.5 g/s at 516 m/s 2 and The angular momentum is 1 2 = m v0 2 Question 25 part 1 of 2 10 points Assume: You are on a planet similar to Earth where : R = 6.34 106 m , M = 5.97 1024 kg , = 1 rev/day . Find the angular momentum of the planet that arises from its spinning motion on its axis. Correct answer: 6.98038 1033 kg m2 /s (tolerance 1 %). Explanation: The angular momentum is L=I= 2 M R2 5 L= 2 (5.97 1024 kg) 5 (6.34 106 m)2 2 rad 1 rev 1 day 24 h 1 rev/min 1 rev 1 day 1h 3600 s 2.2 m 2290 kg m2 = 6.98038 1033 kg m2 /s . 9.5 g/s at 516 m/s Figure: Rear view of spacecraft. Question 26 part 2 of 2 10 points homework 10 FRENNEA, KYLE Due: Apr 6 2007, 4:00 am For how long must they turn on these jets to stop the rotation? Correct answer: 11.1183 s (tolerance 1 %). Explanation: Let : R = 2.2 m , v = 516 m/s , m = 9.5 g/s , t = 1 rev/min , I = 2290 kg m2 . The angular momentum is L = I , and the force exerted by the gas on the spaceship is F = ma = m m v = v . t t 13 1 m L2 . The moment of inertia of the sphere 3 2 about its center-of-mass is Isphere = m r 2 . 5 The acceleration of gravity g = 9.8 m/s2 . and 8 kg Cm 40 m a ss 20 radius 9.3 m 8 kg Determine the moment of inertia of the rod plus mass system with respect to the pivot point. Correct answer: 23987.4 kg m2 (tolerance 1 %). Explanation: Basic Concepts: The moment of inertia, the center of mass and the energy conservation. The rotational kinetic energy is Krot = 1 I 2 . 2 Let : m = 8 kg , L = 40 m , r = 9.3 m , = 20 . The magnitude of the net torque exerted by the jet is L t I 2 m vR = . t t net = 2 F R = Thus t = I m vR 2 t (2290 kg m2 )(1 rev/min) = 2 (0.0095 kg/s) (516 m/s) (2.2 m) 2 rad 1 min rev 60 s = 11.1183 s . Question 28 part 1 of 3 10 points Consider a thin 40 m rod pivoted at one end whose mass is 8 kg . A uniform density 8 kg spherical object (whose radius is 9.3 m) is attached to the end of the rod. The moment of inertia of the rod about an end is Irod = and Solution: The moment of inertia of the rod, Irod , with respect to the pivot point is Irod = 1 m L2 , 3 and the moment of inertia Isphere of the mass m with respect to the pivot point is Isphere = 2 m r 2 + m (L + r)2 , 5 Then, the moment of inertia of the system I is I = Irod + Im homework 10 FRENNEA, KYLE Due: Apr 6 2007, 4:00 am 1 2 m L2 + m (L + r)2 + m r 2 3 5 1 2 = (8 kg) (40 m) 3 + (8 kg) [(40 m) + (9.3 m)]2 2 + (8 kg) (9.3 m)2 5 = = 23987.4 kg m2 . Question 29 part 2 of 3 10 points Note: The length Cmass in the figure represents the location of the center-of-mass of the rod plus mass system. Determine the position of the center of mass from the pivot point; i.e., find Cmass . Correct answer: 34.65 m (tolerance 1 %). 1 The center-of-mass of the rod is L. The 2 center-of-mass of the sphere is at L + r from the pivot. Then, the center of mass of the rod plus mass system is L + m [L + r] 2 m+m L [L + r] + 4 2 3L r + 4 2 3 (40 m) (9.3 m) + 4 2 34.65 m . m Explanation: The torque equation is = r F = r F = I . 14 The torque exerted by gravity on the rod is = (m + m) g Cmass cos . Therefore, the angular acceleration of the rod is = I Cmass (m + m) g cos = I (34.65 m) (16 kg) (9.8 m/s2 ) cos(20 ) = (23987.4 kg m2 ) = 0.21284 rad/s2 . Question 31 part 1 of 2 10 points A hollow cylinder has length a 1.5 m, mass 1 kg, and radius 0.1 m. The cylinder is free to rotate about a vertical axis that passes through its center and is perpendicular to the cylinder's axis. Inside the cylinder are two masses of 0.3 kg each, attached to springs of spring constant k and unstretched lengths 0.6 m. The inside walls of the cylinder are frictionless. 1.5 m 0.3 kg 0.3 kg Cmass = = = = = Question 30 part 3 of 3 10 points Note: The rod is initially at rest at 20 below the horizontal. What is the angular acceleration of the rod immediately after it is released? Correct answer: 0.21284 rad/s2 (tolerance 1 %). Explanation: 1.2 m Determine the value of the spring constant k if the masses are located 1 m from the center of the cylinder when the cylinder rotates at 32 rad/s. Correct answer: 512 N/m (tolerance 1 %). Explanation: Let : m = 1 kg , x = 1 m , and = 32 rad/s . 1 kg homework 10 FRENNEA, KYLE Due: Apr 6 2007, 4:00 am Applying Newton's second law, the net inward force acting on each of the 0.3 kg masses is Fradial = k x = m (x + x) 2 . Thus k= m (x + x) 2 x (1 kg) (1 m) (32 rad/s)2 = 0.6 m = 512 N/m . Thus the work needed is W = Krot + Uspring 1 1 = I 2 + k (x)2 2 2 1 = (0.794 N m2 ) (32 rad/s)2 2 1 + (512 N/m) (0.6 m)2 2 = 498.688 J . 15 Question 33 part 1 of 3 10 points A uniform plank of length 2.25 m and mass 33.2 kg is supported by three ropes, as indicated in the figure. The acceleration of gravity is 9.8 m/s2 . Question 32 part 2 of 2 10 points How much work was needed to bring the system from rest to 32 rad/s? Correct answer: 498.688 J (tolerance 1 %). Explanation: The moment of inertia of the 1 kg cylinder is 1 1 IM = M r 2 + M L2 2 12 The moment of inertia of the 0.3 kg disk is Im = 1 m r2 4 T2 T1 T3 0.514 m 42 2.25 m Applying the parallel-axis theorem, the moment of inertia of the 0.3 kg disk with respect to the center of the 1 kg cylinder is Im = 1 m r 2 + m x2 , 4 Find the tension T1 when a(n) 714 N person is 0.514 m from the left end. Correct answer: 486.885 N (tolerance 1 %). Explanation: Given : L = 2.25 m , m = 33.2 kg , = 42 , W = 714 N , and s = 0.514 m . T 1 in s L @ so the total moment of inertia of the system is I = IM + 2 Im 1 1 1 = M r2 + M L2 + 2 m r 2 + m x2 2 12 4 1 1 (1 kg) (1.5 m)2 = (1 kg) (0.1 m)2 + 2 12 1 + 2 (0.3 kg) (0.1 m)2 + (1 m)2 4 = 0.794 N m2 . W @s 1 mg@ L 2 homework 10 FRENNEA, KYLE Due: Apr 6 2007, 4:00 am Applying the second (rotational) condition of equilibrium (with axis of rotation at the left end), = ccw - cw = 0 16 Applying translational equilibrium horizontally, Fx = Flef t - Fright = 0 T1 cos - T3 = 0 mgL T1 L sin - W s - =0 2 2 T1 L sin = 2 W s + m g L Ws mg T1 = + L sin 2 sin (366.996 N m) = (2.25 m) sin(0.733038 rad) (325.36 N) + 2 sin(0.733038 rad) = 243.763 N + 243.121 N = 486.885 N . Question 34 part 2 of 3 10 points Find the tension T2 . Correct answer: 713.571 N (tolerance 1 %). Explanation: Applying translational equilibrium vertically, T3 = T1 cos = (486.885 N) cos(0.733038 rad) = 361.826 N . Question 36 part 1 of 1 10 points A 23.3 kg person climbs up a uniform ladder with negligible mass. The upper end of the ladder rests on a frictionless wall. The bottom of the ladder rests on a floor with a rough surface where the coefficient of static friction is 0.1 . The angle between the horizontal and the ladder is 66 . The acceleration of gravity is 9.8 m/s2 . 4.3 m =0 66 d 23.3 kg = 0.1 Note: Figure is not to scale. When the person attempts to climb the ladder, how far up the ladder d will the person reach before the ladder slips (kaboom)? Correct answer: 0.965796 m (tolerance 1 %). Explanation: Let : = 66 , L = 4.3 m , W = m g = 228.34 N , = 0.1 . and Fy = Fup - Fdown = 0 T2 + T1 sin - W - m g = 0 T2 = W + m g - T1 sin = 714 N + 325.36 N - (486.885 N) sin(0.733038 rad) = 713.571 N . Question 35 part 3 of 3 10 points Find the tension T3 . Correct answer: 361.826 N (tolerance 1 %). Explanation: homework 10 FRENNEA, KYLE Due: Apr 6 2007, 4:00 am 17 Nw f P ivot mg Nf Using the above equilibrium conditions, we have (taking the sum of the torques at the bottom of the ladder) Fx : Fy : : f - Nw = 0 , Nf - W = 0 , and (1) (2) W d cos - Nw L sin = 0 , (3) where d is the distance of the person from the bottom of the ladder. Using Eq. 2, Eq. 3 becomes f L sin = W d cos , Wd . so f= L tan (4) The ladder may slip when f = fmax = Nw , from Eq. 4 fmax W Wd . L tan Thus, solving for d , we have d L tan (0.1) (4.3 m) tan 66 0.965796 m . (5) ... View Full Document

End of Preview

Sign up now to access the rest of the document