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- Physics Homework #9 (SOLUTIONS)

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4/4/09 1:53 AMMasteringPhysics Page 1 of 13http://session.masteringphysics.com/myct [ Assignment View ] Course PHY121(TTH) HW #9 Due at 11:59pm on Sunday, April 5, 2009 View Grading Details Center of Mass and External Forces Learning Goal: Understand that, for many purposes, a system can be treated as a point-like particle with its mass concentrated at the center of mass. A complex system of objects, both point-like and extended ones, can often be treated as a point particle, located at the system's center of mass. Such an approach can greatly simplify problem solving. Before you use the center of mass approach, you should first understand the following terms: System: Any collection of objects that are of interest to you in a particular situation. In many problems, you have a certain freedom in choosing your system. Making a wise choice for the system is often the first step in solving the problem efficiently. Center of mass: The point that represents the "average" position of the entire mass of a system. The postion of the center of mass can be expressed in terms of the position vectors of the particles as . The x coordinate of the center of mass can be expressed in terms of the x coordinates of the particles as . Similarly, the y coordinate of the center of mass can be expressed. Internal force: Any force that results from an interaction between the objects inside your system. As we will show, the internal forces do not affect the motion of the system's center of mass. External force: Any force acting on an object inside your system that results from an interaction with an object outside your system. Consider a system of two blocks that have masses and . Assume that the blocks are point-like particles and are located along the x axis at the coordinates and as shown . In this problem, the blocks can only move along the x axis. Part A Find the x coordinate of the center of mass of the system. Express your answer in terms of , , , and . ANSWER: = Part B If , then the center of mass is located: ANSWER: to the left of at a distance much greater than to the left of at a distance much less than to the right of at a distance much less than to the right of at a distance much greater than to the right of at a distance much less than to the left of at a distance much less than 4/4/09 1:53 AMMasteringPhysics Page 2 of 13http://session.masteringphysics.com/myct Part C If , then the center of mass is located: ANSWER: at at half-way between and the answer depends on and Part D Recall that the blocks can only move along the x axis. The x components of their velocities at a certain moment are and . Find the component of the velocity of the center of mass at that moment. Keep in mind that, in general: . Express your answer in terms of , , , and . ANSWER: = Because and are the x components of the velocities of and their values can be positive or negative or equal to zero. Part E Suppose that and have equal magnitudes. Also, is directed to the right and is directed to the left. The velocity of the center of mass is then: ANSWER: directed to the left directed to the right zero the answer depends on the ratio Part F Assume that the x components of the blocks' momenta at a certain moment are and . Find the x component of the velocity of the center of mass at that moment. Express your answer in terms of , , , and . ANSWER: = Part G Suppose that . Which of the following must be true? ANSWER: none of the above Part H Assume that the blocks are accelerating, and the x components of their accelerations at a certain moment are and . Find the x component of the acceleration of the center of mass at that moment. Keep in mind that, in general, . Express your answer in terms of , , , and . ANSWER: = Because and are the x components of the velocities of and their values can be positive or negative or equal to zero. We will now consider the effect of external and internal forces on the acceleration of the center of mass. Part I Consider the same system of two blocks. An external force is now acting on block . No forces are applied to block as shown . Find the x component of the acceleration of the center 4/4/09 1:53 AMMasteringPhysics Page 3 of 13http://session.masteringphysics.com/myct of mass of the system. Hint I.1 Using Newton's laws Hint not displayed Express your answer in terms of the x component of the force, ,and . ANSWER: = Part J Consider the same system of two blocks. Now, there are two forces involved. An external force is acting on block and another external force is acting on block . Find the x component of the acceleration of the center of mass of the system. Express your answer in terms of the x components and of the forces, and . ANSWER: = Note that, in both cases, the acceleration of mass can be found as where is the net external force applied to the system, and is the total mass of the system. Even though each force is only applied to one object, it affects the acceleration of the center of mass of the entire system. This result is especially useful since it can be applied to a general case, involving any number of objects moving in all directions and being acted upon by any number of external forces. Part K Consider the previous situation. Under what condition would the acceleration of the center of mass be zero? Keep in mind that and represent the components, of the corresponding forces. ANSWER: Part L Consider the same system of two blocks. Now, there are two internal forces involved. An internal force is applied to block by block and another internal force is applied to block by block . Find the x component of the acceleration of the center of mass of the system. Express your answer in terms of the x components and of the forces, and . 4/4/09 1:53 AMMasteringPhysics Page 4 of 13http://session.masteringphysics.com/myct ANSWER: = Newton's 3rd law tells you that . From your answers above, you can conclude that .The internal forces do not change the velocity of the center of mass of the system. In the absence of any external forces, and is constant. You just demonstrated this to be the case for the two-body situation moving along the x axis; however, it is true in more general cases as well. Exercise 8.20 Block in the figure has mass 1.00 , and block has mass 3.00 . The blocks are forced together, compressing a spring between them; then the system is released from rest on a level, frictionless surface. The spring, which has negligible mass, is not fastened to either block and drops to the surface after it has expanded. Block acquires a speed of 1.60 . Part A What is the final speed of block ? ANSWER: = 4.80 Part B How much potential energy was stored in the compressed spring? ANSWER: = 15.4 Exercise 8.43 A 10.0- marble slides to the left with a velocity of magnitude 0.400 on the frictionless, horizontal surface of an icy New York sidewalk and has a head-on, elastic collision with a larger 30.0- marble sliding to the right with a velocity of magnitude 0.200 . Let be to the right. (Since the collision is head-on, all the motion is along a line.) Part A 4/4/09 1:53 AMMasteringPhysics Page 5 of 13http://session.masteringphysics.com/myct Find the magnitude of the velocity of 30.0- marble after the collision. ANSWER: = 0.100 Part B Find the magnitude of the velocity of 10.0- marble after the collision. ANSWER: = 0.500 Part C Find the direction of the velocity of each marble after the collision. ANSWER: The 10.0 g marble is moving to the left and the 30.0 g marble is moving to the right. The 30.0 g marble is moving to the left and the 10.0 g marble is moving to the right. Both marbles are moving to the right. Both marbles are moving to the left. Part D Calculate the change in momentum (that is, the momentum after the collision minus the momentum before the collision) for 30.0- marble. ANSWER: = ?9.00×10 ?3 Part E Calculate the change in momentum for 10.0- marble. ANSWER: = 9.00×10 ?3 Part F Compare the values you get for each marble. ANSWER: The changes in momentum have the same magnitude and the same sign. The changes in momentum have different magnitudes and opposite sign. The changes in momentum have the same magnitude and opposite sign. The changes in momentum have different magnitudes and the same sign. Part G Calculate the change in kinetic energy (that is, the kinetic energy after the collision minus the kinetic energy before the collision) for 30.0- marble. ANSWER: = ?4.50×10 ?4 Part H Calculate the change in kinetic energy for 10.0- marble. ANSWER: = 4.50×10 ?4 Part I Compare the values you get for each marble. ANSWER: The changes in kinetic energy have the same magnitude and the same sign. The changes in kinetic energy have different magnitudes and opposite sign. The changes in kinetic energy have the same magnitude and opposite sign. The changes in kinetic energy have different magnitudes and the same sign. Angular Motion with Constant Acceleration Learning Goal: To understand the meaning of the variables that appear in the equations for rotational kinematics with constant angular acceleration. Rotational motion with a constant nonzero acceleration is not uncommon in the world around us. For instance, many machines have spinning parts. When the machine is turned on or off, the spinning parts tend to change the rate of their rotation with virtually constant angular acceleration. Many introductory problems in rotational kinematics involve motion of a particle with 4/4/09 1:53 AMMasteringPhysics Page 6 of 13http://session.masteringphysics.com/myct constant, nonzero angular acceleration. The kinematic equations for such motion can be written as and . Here, the symbols are defined as follows: is the angular position of the particle at time . is the initial angular position of the particle. is the angular velocity of the particle at time . is the initial angular velocity of the particle. is the angular acceleration of the particle. is the time that has elapsed since the particle was located at its initial position. In answering the following questions, assume that the angular acceleration is constant and nonzero: . Part A True or false: The quantity represented by is a function of time (i.e., is not constant). ANSWER: true false Part B True or false: The quantity represented by is a function of time (i.e., is not constant). ANSWER: true false Keep in mind that represents an initial value, not a variable. It refers to the angular position of an object at some initial moment. Part C True or false: The quantity represented by is a function of time (i.e., is not constant). ANSWER: true false Part D True or false: The quantity represented by is a function of time (i.e., is not constant). ANSWER: true false The angular velocity always varies with time when the angular acceleration is nonzero. Part E Which of the following equations is not an explicit function of time ? Keep in mind that an equation that is an explicit function of time involves as a variable. ANSWER: An equation that is not an explicit function of time is useful when you do not know or do not need the time. Part F In the equation , what does the time variable represent? Choose the answer that is always true. Several of the statements may be true in a particular problem, but only one is always true. ANSWER: the moment in time at which the angular velocity equals 4/4/09 1:53 AMMasteringPhysics Page 7 of 13http://session.masteringphysics.com/myct the moment in time at which the angular velocity equals the time elapsed from when the angular velocity equals until the angular velocity equals Consider two particles A and B. The angular position of particle A, with constant angular acceleration, depends on time according to . At time , particle B, which also undergoes constant angular acceleration, has twice the angular acceleration, half the angular velocity, and the same angular position that particle A had at time . Part G Which of the following equations describes the angular position of particle B? Hint G.1How to approach the problem Hint not displayed ANSWER: Note that particle B has a smaller initial angular velocity but greater angular acceleration. Also, it has been in motion for less time than particle A. Part H How long after the time does the angular velocity of particle B equal that of particle A? Hint H.1How to approach the problem Hint not displayed ANSWER: The two particles never have the same angular velocity. Circular Motion Tutorial Learning Goal: Understand how to find the equation of motion of a particle undergoing uniform circular motion. Consider a particle--the small red block in the figure--that is constrained to move in a circle of radius . We can specify its position solely by , the angle that the vector from the origin to the block makes with our chosen reference axis at time . Following the standard conventions we measure in the counterclockwise direction from the positive x axis. Part A What is the position vector as a function of angle . For later remember that is itself a function of time. Hint A.1x coordinate 4/4/09 1:53 AMMasteringPhysics Page 8 of 13http://session.masteringphysics.com/myct Hint not displayed Hint A.2y coordinate Hint not displayed Give your answer in terms of , , and unit vectors and corresponding to the coordinate system in the figure. ANSWER: = Answer not displayed Uniform Circular Motion A frequently encountered kind of circular motion is uniform circular motion, where changes at a constant rate . In other words, . Usually, . Part B For uniform circular motion, find at an arbitrary time . Give your answer in terms of and . ANSWER: = Answer not displayed Part C Part not displayed Part D Find , a position vector at time . Hint D.1Finding Hint not displayed Give your answer in terms of and unit vectors and/or . ANSWER: = Answer not displayed Part E Determine an expression for the position vector of a particle that starts on the positive y axis at (i.e., at , ) and subsequently moves with constant . Hint E.1Adding a phase Hint not displayed Hint E.2Finding a phase Hint not displayed Express your answer in terms of , , , and unit vectors and . ANSWER: = Answer not displayed Exercise 9.4 A fan blade rotates with angular velocity given by . Part A Calculate the angular acceleration as a function of time. ANSWER: = For the next two parts assume that the constants and are defined as follows: 4/4/09 1:53 AMMasteringPhysics Page 9 of 13http://session.masteringphysics.com/myct = 4.65 = 0.755 Part B Calculate the instantaneous angular acceleration at = 2.9 . ANSWER: = -4.38 Part C Calculate the average angular acceleration for the time interval to = 2.9 . ANSWER: = -2.19 Exercise 9.10 An electric fan is turned off, and its angular velocity decreases uniformly from 490 to 170 in a time interval of length 3.60 . Part A Find the angular acceleration in . ANSWER: = -1.48 Part B Find the number of revolutions made by the motor in the time interval of length 3.60 . ANSWER: = 19.8 Part C How many more seconds are required for the fan to come to rest if the angular acceleration remains constant at the value calculated in part A? ANSWER: = 1.91 Exercise 9.16 A computer disk drive is turned on starting from rest and has constant angular acceleration. Part A If it took 0.890 for the drive to make its second complete revolution, how long did it take to make the first complete revolution? ANSWER: = 2.15 Part B What is its angular acceleration, in ? ANSWER: = 2.72 Introduction to Moments of Inertia Learning Goal: To understand the definition and the meaning of moment of inertia; to be able to calculate the moments of inertia for a group of particles and for a continuous mass distribution with a high degree of symmetry. By now, you may be familiar with a set of equations describing rotational kinematics. One thing that you may have noticed was the similarity between translational and rotational formulas. Such similarity also exists in dynamics and in the work-energy domain. For a particle of mass moving at a constant speed , the kinetic energy is given by the formula . If we consider instead a rigid object of mass rotating at a constant angular speed , the kinetic energy of such an object cannot be found by 4/4/09 1:53 AMMasteringPhysics Page 10 of 13http://session.masteringphysics.com/myct using the formula directly: different parts of the object have different linear speeds. However, they all have the same angular speed. It would be desirable to obtain a formula for kinetic energy of rotational motion that is similar to the one for translational motion; such a formula would include the term instead of . Such a formula can, indeed, be written: for rotational motion of a system of small particles or for a rigid object with continuous mass distribution, the kinetic energy can be written as . Here, is called the moment of inertia of the object (or of the system of particles). It is the quantity representing the inertia with respect to rotational motion. It can be shown that for a discrete system, say of particles, the moment of inertia (also known as rotational inertia) is given by . In this formula, is the mass of the ith particle and is the distance of that particle from the axis of rotation. For a rigid object, consisting of infinitely many particles, the analogue of such summation is integration over the entire object: . In this problem, you will answer several questions that will help you better understand the moment of inertia, its properties, and its applicability. It is recommended that you read the corresponding sections in your textbook before attempting these questions. Part A On which of the following does the moment of inertia of an object depend? Check all that apply. ANSWER: linear speed linear acceleration angular speed angular acceleration total mass shape and density of the object location of the axis of rotation Unlike mass, the moment of inertia depends not only on the amount of matter in an object but also on the distribution of mass in space. The moment of inertia is also dependent on the axis of rotation. The same object, rotating with the same angular speed, may have different kinetic energy depending on the axis of rotation. Consider the system of two particles, a and b, shown in the figure . Particle a has mass , and particle b has mass . Part B What is the moment of inertia of particle a? ANSWER: undefined: an axis of rotation has not been specified. Part C Find the moment of inertia of particle a with respect to the x axis (that is, if the x axis is the axis of rotation), the moment of inertia of particle a with respect to the y axis, and the moment of inertia of particle a with respect to the z axis (the axis that passes through the origin perpendicular to both the x and y axes). 4/4/09 1:53 AMMasteringPhysics Page 11 of 13http://session.masteringphysics.com/myct Express your answers in terms of and separated by commas. ANSWER: , , = Part D Find the total moment of inertia of the system of two particles shown in the diagram with respect to the y axis. Express your answer in terms of and . ANSWER: = For parts E through G, suppose that both particles rotate with the same angular speed about the y axis while maintaining their distances from the y axis. Part E Using the total moment of inertia of the system found in Part D, find the total kinetic energy of the system. Express your answer in terms of , , and . ANSWER: = It is useful to see how the formula for rotational kinetic energy agrees with the formula for the kinetic energy of an object that is not rotating. To see the connection, let us find the kinetic energy of each particle. Part F Using the formula for kinetic energy of a moving particle , find the kinetic energy of particle a and the kinetic energy of particle b. Hint F.1 Find the linear speed Hint not displayed Express your answers in terms of , , and separated by a comma. ANSWER: , = Part G Using the results for the kinetic energy of each particle, find the total kinetic energy of the system of particles. Express your answer in terms of , , and . ANSWER: = Not surprisingly, the formulas and give the same result. They should, of course, since the rotational kinetic energy of a system of particles is simply the sum of the kinetic energies of the individual particles making up the system. Exercise 9.34 Four small spheres, each of which you can regard as a point of mass 0.200 , are arranged in a square 0.400 on a side and connected by light rods . 4/4/09 1:53 AMMasteringPhysics Page 12 of 13http://session.masteringphysics.com/myct Part A Find the moment of inertia of the system about an axis through the center of the square, perpendicular to its plane (an axis through point O in the figure). ANSWER: = 6.40×10 ?2 Part B Find the moment of inertia of the system about an axis bisecting two opposite sides of the square (an axis along the line AB in the figure). ANSWER: = 3.20×10 ?2 Part C Find the moment of inertia of the system about an axis that passes through the centers of the upper left and lower right spheres and through point O. ANSWER: = 3.20×10 ?2 Kinetic Energy and Moment of Inertia Consider a particle of mass that is revolving with angular speed around an axis. The perpendicular distance from the particle to the axis is . Part A Find the kinetic energy of the rotating particle. Hint A.1Find the speed of the particle Hint not displayed Express your answer in terms of , , and . ANSWER: = Part B The kinetic energy of a rotating body is generally written as , where is the moment of inertia. Find the moment of inertia of the particle described in the problem introduction with respect to the axis about which it is rotating. Hint B.1How to approach the problem Hint not displayed Give your answer in terms of and , not . ANSWER: = Consider a system of several point masses all rotating about the same axis. The total kinetic energy of the system is the sum of the kinetic energies of all of the masses. For point masses, . Comparing this to the expression for the kinetic energy given in Part C, we see that the moment of inertia is 4/4/09 1:53 AMMasteringPhysics Page 13 of 13http://session.masteringphysics.com/myct . In other words, the total moment of inertia is the sum of the moments of inertia of each mass. Part C Find the moment of inertia of a hoop of radius and mass with respect to an axis perpendicular to the hoop and passing through its center. Hint C.1Find the moment of inertia of a segment of the hoop Hint not displayed Express your answer in terms of and . ANSWER: = Summary10 of 11 items complete (90.4% avg. score) 10.95 of 12 points clockwork Homework #9 Solutions

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Author: Anonymous

Textbook: University Physics (12th Edition)

University Physics Vol 1 (Chapters 1-20) (12th Edition) (Chapters 1-20 v. 1)

Created: 2009-04-04

Updated: 2009-04-04

File Size: 13 page(s)

Views: 19179

Textbook: University Physics (12th Edition)

University Physics Vol 1 (Chapters 1-20) (12th Edition) (Chapters 1-20 v. 1)

Created: 2009-04-04

Updated: 2009-04-04

File Size: 13 page(s)

Views: 19179

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