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09 homework FRENNEA, KYLE Due: Mar 28 2007, 4:00 am Solution: Horizontally Question 1 part 1 of 1 10 points Two ice skaters approach each other at right angles. Skater A has a mass of 72.5 kg and travels in the +x direction at 2.88 m/s. Skater B has a mass of 62.5 kg and is moving in the +y direction at 0.982 m/s. They collide and cling together. Find the final speed of the couple. Correct answer: 1.6121 m/s (tolerance 1 %). Explanation: From conservation of momentum p = 0 mA vA^ + mB vB^ = (mA + mB ) vf i Therefore vf = (mA vA )2 + (mB vB )2 mA + mB (208.8 kg m/s)2 + (61.375 kg m/s)2 = 72.5 kg + 62.5 kg = 1.6121 m/s Question 2 part 1 of 2 10 points before m1 v1 m1 after v3 v4 v2 m2 m2 1 m1 v1 + m2 (0) = m1 v3 cos + m2 v4 cos , so that m2 v4 cos = m1 v1 - m1 v3 cos (1) = (0.6 kg) (8.2 m/s) - (0.6 kg) (6.4 m/s) cos 67 = 3.41959 kg m/s . Vertically m1 (0) + m2 v2 = m1 v3 sin + m2 v4 sin , so that m2 v4 sin = m2 v2 - m1 v3 sin (2) = (1.6 kg) (5.6 m/s) - (0.6 kg) (6.4 m/s) sin 67 = 5.42526 kg m/s . Thus tan = m2 v4 sin m2 v4 cos (5.42526 kg m/s) = (3.41959 kg m/s) = 1.58652 , and = arctan(1.58652) = 57.7764 . Question 3 part 2 of 2 10 points With what speed does the can move immediately after the collision? Correct answer: 4.00815 m/s (tolerance 1 %). Explanation: Using equation (1) above, v4 = m1 v1 - m1 v3 cos m2 cos (0.6 kg) (8.2 m/s) = (1.6 kg) cos 57.7764 (0.6 kg) (6.4 m/s) cos(67 ) - (1.6 kg) cos(57.7764 ) = 4.00815 m/s An m2 = 1.6 kg can of soup is thrown upward with a velocity of v2 = 5.6 m/s. It is immediately struck from the side by an m1 = 0.6 kg rock traveling at v1 = 8.2 m/s. The rock ricochets off at an angle of = 67 with a velocity of v3 = 6.4 m/s. What is the angle of the can's motion after the collision? Correct answer: 57.7764 (tolerance 1 %). Explanation: Basic Concepts: Conservation of Momentum pbef ore = paf ter . homework 09 FRENNEA, KYLE Due: Mar 28 2007, 4:00 am or using equation (2) above, m2 v2 - m1 v3 sin m2 sin (5.6 m/s) = sin(57.7764 ) (0.6 kg) (6.4 m/s) sin(67 ) - (1.6 kg) sin(57.7764 ) = 4.00815 m/s . p2 N m2 v2 p1 m1 v1 pf = mf vf 2 v4 = Basic Concepts: Momentum Conserva1 p = mv. tion, K = mv 2 , 2 During the collision, the total momentum of the two car system will be conserved pf = p1 + p2 = m1 v1 + m2 v2 pf x ^ + pf y ^ = m1 v1 ^ + m2 v2 ^ . i i Looking at the x and y components of momentum, pf x = m2 v2 = (3651.6 kg) (180 km/h) 103 m 1h km 3600 s = 182580 kg m/s , and pf y = m1 v1 = (2040 kg) (219.6 km/h) 1h 103 m km 3600 s = 124440 kg m/s . Since we are asked to find the angle from the y-axis instead of the x, tan = pf x pf y 182580 kg m/s = 124440 kg m/s = 1.46721 182580 kg m/s = arctan 124440 kg m/s = 55.7231 . Question 5 part 2 of 2 10 points Question 4 part 1 of 2 10 points A 2040 kg car skidding due north on a level frictionless icy road at 219.6 km/h collides with a 3651.6 kg car skidding due east at 180 km/h in such a way that the two cars stick together. 180 km/h N 3651.6 kg 219.6 km/h 2040 kg vf At what angle (-180 +180 ) East of North do the two coupled cars skid off at? Correct answer: 55.7231 (tolerance 1 %). Explanation: Let : m1 v1 m2 v2 = 2040 kg , = 219.6 km/h , = 3651.6 kg , and = 180 km/h . homework 09 FRENNEA, KYLE Due: Mar 28 2007, 4:00 am How much kinetic energy is lost in the collision? Correct answer: 4.07108 106 J (tolerance 1 %). Explanation: To find the energy lost in the collision, we need to know the total kinetic energy of the system before and after the collision. (There is no change in potential energy since the cars remain on level ground.) The initial and final energies are given by Ei = Ki = = 1 1 2 2 m1 v1 + m2 v2 2 2 Question 6 part 1 of 2 10 points 3 A ball moving at 18 m/s makes an offcenter elastic collision with another ball of equal mass that is initially at rest. The incoming ball is deflected at an angle of 43 from its original direction of motion. Find the speed of the first ball after the collision. Correct answer: 13.1644 m/s (tolerance 1 %). Explanation: 1 (2040 kg) (61 m/s)2 2 1 + (3651.6 kg) (50 m/s)2 2 = 8.35992 106 J , and Ef = Kf 1 2 = (m1 + m2 ) vf . 2 To find vf , we need to go back to the momentum equation; pf , m1 + m2 or in component form, pf x vf x = , m1 + m2 pf y . vf y = m1 + m2 vf = Let : v = 18 m/s = 43 . and Because of the elastic collision, and the fact that the masses are equal, the angle of deflection for the recoiling ball is 47 . Applying conservation of momentum parallel to the original line of motion and pxi = pxf m v = m v1 cos 1 + m v2 cos 2 v2 cos 2 = v - v1 cos 1 . (1) Now that we know the components of vf , we can find its magnitude 2 2 2 vf = vf x + vf y . Applying conservation of momentum perpendicular to the original line of motion, pyi = pyf 0 = m v1 sin 1 - m v2 sin 2 v2 sin 2 = v1 sin 1 . Dividing Eq. 2 by Eq. 1, v2 sin 2 v1 sin 1 = v2 cos 2 v - v1 cos 1 tan 2 (v - v1 cos 1 ) = v1 sin 1 v1 (sin 1 + tan 2 cos 1 ) = v tan 2 We can substitute this into the equation for the final kinetic energy to get Ef = 1 2 m1 + m2 = 4.28884 106 J . p2 x f + p2 y f (2) The energy lost is then Elost = -E = Ei - Ef = (8.35992 106 J) - (4.28884 106 J) = 4.07108 106 J . homework 09 FRENNEA, KYLE Due: Mar 28 2007, 4:00 am 4 v1 = v tan 2 sin 1 + tan 2 cos 1 (18 m/s) tan 47 = sin 43 + tan 47 cos 43 If the car is moving at a constant speed, there is no tangential acceleration, thus the acceleration is purely radial, ar = v2 . R = 13.1644 m/s . Question 7 part 2 of 2 10 points Find the speed of the second ball. Correct answer: 12.276 m/s (tolerance 1 %). Explanation: The magnitude of the velocity is v2 = v1 sin 1 sin 2 (13.1644 m/s) sin 43 = sin 47 = 12.276 m/s . Question 10 part 1 of 3 10 points The figure below represents, at a given instant, the total acceleration of a particle moving clockwise in a circle of radius 1.63 m. The magnitude of the total acceleration of the particle at the given instant is 10.5 m/s2 ; the angle between the vector of total acceleration and the radius-vector of the particle is 30 . v 1. 63 m Question 8 part 1 of 2 10 points A racing car travels on a circular track of radius 163 m. The car moves with a constant linear speed of 20.9 m/s. Find its angular speed. Correct answer: 0.128221 rad/s (tolerance 1 %). Explanation: The linear speed v and the angular speed are related by, v = R v = . R Question 9 part 2 of 2 10 points Find the magnitude of its acceleration. Correct answer: 2.67982 m/s2 (tolerance 1 %). Explanation: 30 At this instant of time, find the centripetal acceleration of the particle. Correct answer: 9.09327 m/s2 (tolerance 1 %). Explanation: The radial (i.e., centripetal) acceleration ar of the particle is ar = a cos = (10.5 m/s2 ) cos (30 ) = 9.09327 m/s2 . Question 11 part 2 of 3 10 points Find the speed of the particle at the same moment. Correct answer: 3.84994 m/s (tolerance 1 %). Explanation: 10.5 m/s2 homework 09 FRENNEA, KYLE Due: Mar 28 2007, 4:00 am From the formula for the radial (i.e., centripetal) acceleration v2 ar = r we obtain that the speed of the particle is v = r ar = (1.63 m) (9.09327 m/s2 ) Question 14 part 1 of 1 10 points = 3.84994 m/s . Question 12 part 3 of 3 10 points Find particle's tangential acceleration at the same moment. Correct answer: 5.25 m/s2 (tolerance 1 %). Explanation: From the formula Solution: t = avg 3.1 rad = 2.2 rad/s = 1.40909 s 5 A woman passes through a revolving door with a tangential speed of 1.2 m/s. If she is 0.62 m from the center of the door, what is the door's angular speed? Correct answer: 1.93548 rad/s (tolerance 1 %). Explanation: Basic Concept: vt = r a2 = a2 + a2 t r we obtain at = = a2 - a2 r (10.5 m/s2 )2 - (9.09327 m/s2 )2 Given: vt = 1.2 m/s r = 0.62 m Solution: = vt r 1.2 m/s = 0.62 m = 1.93548 rad/s = 5.25 m/s2 . Question 13 part 1 of 1 10 points A girl ties a toy airplane to the end of a string and swings it around her head. The plane's average angular speed is 2.2 rad/s. In what time interval will the plane move through an angular displacement of 3.1 rad? Correct answer: 1.40909 s (tolerance 1 %). Explanation: Basic Concept: Question 15 part 1 of 2 10 points t You want to design a large, permanent space station so that no artificial gravity is necessary. You decide to shape it like a large coffee can of radius 164 m and rotate it about its central axis. The acceleration of gravity is 9.8 m/s2 . avg = Given: avg = 2.2 rad/s = 3.1 rad homework 09 FRENNEA, KYLE Due: Mar 28 2007, 4:00 am Thus the effective gravity the jogger feels is 6 radius 164 m ag = 2 vef f r [r + vm ]2 = r [(164 m) (0.244451 rad/s) + (2.7 m/s)]2 = (164 m) = 11.1645 m/s2 . Question 17 part 1 of 1 10 points What rotational speed would be required to simulate gravity? Correct answer: 0.244451 rad/s (tolerance 1 %). Explanation: Basic Concepts The centripetal acceleration is the simulated gravity. ac = g = so that = = g r (9.8 m/s2 ) (164 m) v2 (r )2 = = r 2 r r A girl throws a long stick with a length of 0.46 m and a mass of 0.2 kg into the air in such a way that the center of mass rises vertically. At the moment it leaves her hand, the stick is horizontal and the speed of the end of the stick nearest to her is zero. When the center of mass of the stick reaches its highest point, the stick is horizontal, and it has made 18 complete revolutions. Assume: The stick's cross sectional area and mass is uniform. The acceleration of gravity is 9.8 m/s2 . vcm0 0.2 kg 0.23 m 0.46 m vend0 = 0.244451 rad/s . Question 16 part 2 of 2 10 points If an astronaut jogged in the direction of the rotation at 2.7 m/s , what is the simulated gravitational acceleration the astronaut would feel? Correct answer: 11.1645 m/s2 (tolerance 1 %). Explanation: The linear speed of the rim is vrim = r , so that his effective speed is vef f = vrim + vm = r + vm . How long did it take for the center of mass to reach its highest point? Correct answer: 1.62921 s (tolerance 1 %). Explanation: From kinematics v0 = g t From definition of angular velocity v0 = r = From kinematics =t = . t . 2 (1) (2) (3) homework 09 FRENNEA, KYLE Due: Mar 28 2007, 4:00 am Substituting from Eq. 3 into Eq. 2 gives v0 = . 2 t = gt. 2 t Since, = n (2 ) , we have t= = n g (6) (4) Question 20 part 1 of 2 10 points 7 Substituting v0 from Eq. 1 into Eq. 4 gives (5) (18) (0.46 m) (9.8 m/s2 ) = 1.62921 s . A small turtle, appropriately named "Dizzy", is placed on a horizontal, rotating turntable at a distance of 26.9 cm from its center. Dizzy's mass is 71.6 g, and the coefficient of static friction between his feet and the turntable is 0.42. The acceleration of gravity is 9.8 m/s2 . Find the maximum number of revolutions per second the turntable can have if Dizzy is to remain stationary relative to the turntable. Correct answer: 0.622561 rev/s (tolerance 1 %). Explanation: The force of static friction is what keeps Dizzy from flying off the turntable by providing the centripetal force which keeps him going in a circular path of radius 26.9 cm . m v2 centripetal The force should not exceed r the maximum force of static friction. This gives max 2 Ff = s m g = m r max Question 18 part 1 of 2 10 points A flywheel with a very low friction bearing takes 1.4 hours to stop after the motor power is turned off. The flywheel was originally rotating at 32 rpm. What was the initial rotation rate in radians per second? Correct answer: 3.35103 rad/s (tolerance 1 %). Explanation: = 2f = 3.35103 rad/s. Question 19 part 2 of 2 10 points How many revolutions does the flywheel make before it stops? Correct answer: 1344 revs (tolerance 1 %). Hence, the maximum speed Dizzy can have without flying off the turntable follows from 2 max = max s g r s g = r (0.42) (9.8 m/s2 ) (0.269 m) 1 rev 2 rad = Explanation: Let us first find the angular acceleration of the flywheel. It is = -0 = -0.000664887 rad/s2 . t = (3.91166 rad/s) = 0.622561 rev/s . The angle rotated is then 1 = 0 t + t2 2 = 8444.6 rad = 1344 revs. Question 21 part 2 of 2 10 points The turntable starts from rest at t = 0, and has a uniform acceleration of 1.17 rad/s2 . Find the time at which Dizzy begins to slip. homework 09 FRENNEA, KYLE Due: Mar 28 2007, 4:00 am Correct answer: 3.3433 s (tolerance 1 %). max - 0 (3.91166 rad/s) - 0 = (1.17 rad/s2 ) = 3.3433 s . 8 Explanation: t= 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 Solution: The moment of inertia of the rod, Irod , with respect to the pivot point is Irod = 1 m L2 , 3 Question 22 part 1 of 4 10 points Consider a rod of length L and mass m which is pivoted at one end. An object with mass m is attached to the free end of the rod. The acceleration of gravity g = 9.8 m/s2 . Note: Contray to the diagram shown below, consider the mass at the end of the rod to be a point particle. and the moment of inertia Im of the mass m with respect to the pivot point is Imass = m L2 . Then, the moment of inertia of the system I is I = Irod + Im 1 = m L2 + m L2 3 4 = m L2 . 3 Question 23 part 2 of 4 10 points Note: The length C in the figure represents the location of the center-of-mass of the rod plus mass system (but is not drawn to scale). Determine the position of the center of mass from the pivot point; i.e., find C . 1. C = L 2. C = 5 L 8 7 3. C = L 8 4. None of these. 3 L correct 4 1 6. C = L 2 5. C = Explanation: 1 The center of mass of the rod is L. Then, 2 the center of mass of the rod plus mass system m C L 23 m Determine the moment of inertia, I, of the system with respect to the pivot point. 1. I = 4 m L2 correct 3 13 2. I = m L2 12 3. I = L2 4. I = 5 m L2 3 5 5. I = m L2 4 3 6. I = m L2 2 7. None of these. homework 09 FRENNEA, KYLE Due: Mar 28 2007, 4:00 am is 1 L+L m 2 C= m+m 1 1 = L+ L 4 2 3 = L. 4 Question 24 part 3 of 4 10 points The unit is released from rest in the horizontal position. What is the kinetic energy of the unit when the rod momentarily has a vertical orientation? 1. K = 1 mgL 2 Question 25 part 4 of 4 10 points 9 Given: The kinetic energy is KV at the vertical position and the moment of inertia of the system is I. Find the angular velocity of the system in terms of KV and I. 1. = 2 KV correct I 2. None of these. 3. = KV I 2 KV 4. = I KV 2I 5. = 6. = 2. K = 2 m g L 3. K = m g L 5 mgL 2 3 5. K = m g L correct 2 4. K = 6. None of these. Explanation: The potential energy, U , released can be computed in two ways. Method 1: U = U = mg cm of rod KV I Explanation: The rotational kinetic energy of the system is given by 1 KV = I 2 , 2 where is the angular velocity of the system. Then, 2 KV = . I Question 26 part 1 of 2 10 points A uniform flat plate of metal is situated in the reference frame shown in the figure below. 5 4 3 2 1 0 0 1 2 3 4 5 6 7 8 x (m) 9 10 (m) + U mass 3 = mgL. 2 Method 2: L + mgL 2 U |rod+m = U |cm of = (2 m) g = 3 mgL. 2 rod+m 3 L 4 y Calculate the x coordinate of the center of homework 09 FRENNEA, KYLE Due: Mar 28 2007, 4:00 am mass of the metal plate. Correct answer: 1 m (tolerance 1 %). Explanation: Basic Concept: The center of mass xcoordinate is x dm x where m m , x2 10 = 0 x (x - x2 ) dx x2 0 (x - x2 ) dx x2 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 ) = (3 m, 0 m) (x3 , y3 ) = (0 m, 4 m) . The equation for the hypotenuse is y3 - y2 y - y2 = . x - x2 x3 - x 2 The slope of the hypotenuse is s= y3 - y2 x3 - x2 4m-0m 4 = =- . 0m-3m 3 1 3 1 x - (x2 ) x2 2 = 3 1 2 x - (x2 ) x 2 1 3 1 x2 - (x2 ) x2 2 2 = 3 1 2 x - (x2 ) x2 2 2 x3 = 22 3 x2 1 = x2 3 1 = (3 m) 3 = 1 m. 0 (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 = (4 m) 3 = 1.33333 m . (1) Rewriting the equation, we have y = s (x - x2 ) + y2 4 (x - 3 m) + 0 m = - 3 = s (x - x2 ) , Question 27 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: 1.5 kg m2 (tolerance 1 %). Explanation: m Areal Density: The areal density = A of the plate is = m x2 (2) where y2 = 0 . The x-coordinate of the center of mass is x2 xcm = x1 x2 x1 x y dx x2 y dx x s (x - x2 ) dx x2 y dx x1 = x1 = s 0 m x2 x1 s (x - x2 ) dx (x - x2 ) dx homework 09 FRENNEA, KYLE Due: Mar 28 2007, 4:00 am 2 1 2 s x - (x2 ) x 2 0 m = y3 1 2 - x - (x2 ) x2 x2 2 2 m m = = 1 y3 2 1 x x2 y 3 2 x2 2 2 2m (4) = x2 y 3 2 (1 kg) = (3 m) (4 m) = 0.166667 kg/m2 , y3 from Eq. 1. The area of a where s = - x2 1 1 triangle is A = (base) (height) = x2 y3 , 2 2 which agrees with Eq. 4. Conventional Solution: The moment of x=x inertia Iy cm of the triangle about a y-axis through its center of mass x-coordinate can be accomplished in two steps. 11 = m x = s 1 4 1 x2 - [2 xcm + x2 ] x3 2 4 3 1 + x2 + 2 xcm x2 x2 2 2 cm -x2 x2 cm 2 - y3 1 4 - x x2 12 2 1 1 + xcm x3 - x2 x2 2 3 2 cm 2 1 2 2 =m x - xcm x2 + x2 cm 6 2 3 1 2 2 2 1 2 x - x + x =m 6 2 9 2 9 2 1 m x2 (5) = 2 18 1 = (1 kg) (3 m)2 18 = 0.5 kg m2 . = using Eq. 1 for s , Eq. 4 for , and Eq. 3 for xcm . Using the parallel axis theorem, we have x=0 x=x Iy = Iy cm + m x2 cm 1 m x2 + m x2 = 2 cm 18 1 2 1 2 =m x + x 18 2 9 2 1 = m x2 2 6 1 = (1 kg) (3 m)2 6 2m x2 y 3 x=x First: The moment of inertia Iy cm 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=x mass Iy cm is x=x Iy cm = x2 x1 (x - xcm )2 dm x2 x1 = = s 0 (x - xcm )2 y dx x2 (x - xcm ) (x - x2 ) dx x2 2 = 1.5 kg m2 . Standard Solution: The moment of inerx=0 tia Iy of the triangle about the x-coordinate is x=0 Iy x2 = s 0 x3 - [2 xcm + x2 ] x2 +[x2 + 2 xcm x2 ] x cm -[x2 x2 ] cm dx x2 dm x1 x2 = s 1 4 1 x - [2 xcm + x2 ] x3 4 3 1 + [x2 + 2 xcm x2 ] x 2 cm x2 0 = x1 x2 y dx x2 = s 0 x2 (x - x2 ) dx 1 4 1 x - (x2 ) x3 4 3 x2 0 -[x2 x2 ] x cm = s homework 09 FRENNEA, KYLE Due: Mar 28 2007, 4:00 am 1 4 1 4 x - x 4 2 3 2 y3 1 2m - - x4 = x2 y 3 x2 12 2 1 = m x2 2 6 1 = (1 kg) (3 m)2 6 = s = 1.5 kg m2 , using Eq. 1 for s and Eq. 4 for . Note: This problem has a different triangle for most students. Question 28 part 1 of 2 10 points The figure below shows a rigid 3-mass system which can rotate about an axis perpendicular to the system. The mass of each connecting rod is negligible. Treat the masses as particles. The x-axis is along the horizontal direction with the origin at the left-most mass 2 kg. 2 kg 5 kg 3 kg (3 kg)(2 (3 m) (2 kg) + (5 kg) + (3 kg) 11 kg = (3 m) 10 kg = 3.3 m . + Question 29 part 2 of 2 10 points 12 Note: The position x0 about which the rigid body rotates is not necessarily to scale. x0 2 kg 5 kg 3 kg L x L 3m x 3m The masses are separated by rods of length 3 m, so that the entire length is 2 (3 m). Determine the x-coordinate of the center of mass for the three-mass system with respect to the origin. Correct answer: 3.3 m (tolerance 1 %). Explanation: First find the center of mass. Define the origin to coincide with the far left mass M . XCM mi xi = mi (2 kg)(0 ) = (2 kg) + (5 kg) + (3 kg) (5 kg)(1 (3 m) + (2 kg) + (5 kg) + (3 kg) Explanation: The moment of inertia of a system of point 2 particles is given by I = mi ri . Label the three moments of inertia as Ileft , Imiddle , and Iright . Remembering that the distances ri are with respect to the axis of rotation, we have Ileft Imiddle 1 = (2 kg) (3 m) 3 2 = (5 kg) (3 m) 3 5 (3 m) 3 2 Find the moment of inertia of the 3-mass system about a rotation axis perpendicular to the x-axis and passing through the point (3 m) x0 = , with respect to the origin at the 3 left-most mass 2 kg. Correct answer: 97 kg m2 (tolerance 1 %). 2 2 Iright = (3 kg) Hence . I = Ileft + Imiddle + Iright 1 (3 m)2 = (2 kg) 9 4 (3 m)2 + (5 kg) 9 homework 09 FRENNEA, KYLE Due: Mar 28 2007, 4:00 am + (3 kg) = 25 9 (3 m)2 13 97 kg (3 m)2 9 = 97 kg m2 . Question 30 part 1 of 1 10 points negligible mass) a distance 13 m from the center of mass of the rod. Initially the rod makes an angle of 44 with the horizontal. The rod is released from rest at an angle of 44 with the horizontal, as shown in the figure below The acceleration of gravity is 9.8 m/s2 . Hint: The moment of inertia of the rod 1 about its center-of-mass is Icm = m 2 . 12 A spherical object (with non-uniform density) of mass 14 kg and radius 0.26 m rolls along a horizontal surface at a constant linear speed without slipping. The moment of inertia of the object about a diameter is Icm = (0.33) M R2 . The object's rotational kinetic energy Krot about its own center is what fraction Ktotal of the object's total kinetic energy? Correct answer: 0.24812 (tolerance 1 %). Explanation: We begin by using 44 1.1 kg O 13 m What is the angular speed of the rod at the instant the rod is in a horizontal position? Correct answer: 0.983243 rad/s (tolerance 1 %). Explanation: Let : = 13 m , d = = 13 m , = 44 , and m = 1.1 kg . Basic Concepts: 1 I 2 2 =rF =I KR = 13 m v = R, where R is the radius of the sphere. Then, Krot = 1 Icm 2 2 1 = (0.33) M R2 2 2 = (0.165) M v 2 The total kinetic energy is given by Ktotal = Krot + Kcm 1 1 = Icm M v 2 + M v 2 2 2 2 = (0.165) M v + (0.5) M v 2 = (0.665) M v 2 Krot (0.165) M v 2 = Ktotal (0.665) M v 2 = 0.24812 . Question 31 part 1 of 1 10 points A uniform rod of mass 1.1 kg is 13 m long. The rod is pivoted about a horizontal, frictionless pin at the end of a thin extension (of Ki + Ui = Kf + Uf . Solution: I = Icm + m d2 1 = m 2 + m 2 12 13 m 2 = 12 13 (1.1 kg) (13 m)2 = 12 = 201.392 kg m2 . (1) homework 09 FRENNEA, KYLE Due: Mar 28 2007, 4:00 am Since the rod is uniform, its center of mass is located a distance from the pivot. The vertical height of the center of mass above horizontal is sin . Using conservation of energy and substituting I from Eq. 1, we have Kf = Ui 1 I 2 = m g sin 2 14 13 m 2 2 = m g sin 24 24 g sin 2 = 13 24 g sin = 13 = 24 (9.8 m/s2 ) sin(44 ) 13 (13 m) (2) = 0.983243 rad/s . ... View Full Document