Homework 6-1

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3/1/24, 11:42 PM Homework 6-1 1/26 Homework 6-1 Due: 11:59pm on Tuesday, February 20, 2024 To understand how points are awarded, read the Grading Policy for this assignment. Problem 10.73 The spring in the figure has a spring constant of 1300 . It is compressed 14.0 , then launches a 200 block. The horizontal surface is frictionless, but the block's coefficient of kinetic friction on the incline is 0.200. Part A What distance does the block sail through the air? Express your answer with the appropriate units. ANSWER: Correct ± The Power of One Learning Goal: To learn the definition of power and how power, force, and velocity are related. The definition of work done by a constant force does not include time. For practical purposes, however, it is often important to know how fast work is being done. The rate at which work is being done is called power . The average power can be calculated as , where is the amount of work done during the time interval . The power created by a force may be a constant; that is, work is being done at constant rate. However, this is not always the case. If the rate of performing work is changing, it makes sense to talk about the instantaneous power , defined as https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID= 0 8.20

3/1/24, 11:42 PM Homework 6-1 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID= 0 2/26 . The SI unit of power is the watt (W). One watt is defined as the power created when one joule of work is done each second. In equation form, one writes . A commonly used unit of work is the kilowatt-hour (kW-hour). One kilowatt-hour is the amount of work done in one hour when the power is one kilowatt. In equation form, this is - . In this problem, you will answer several questions that will help familiarize you with power and enable you to derive a formula relating power, force, and velocity. A sled of mass is being pulled horizontally by a constant horizontal force of magnitude . The coefficient of kinetic friction is . During time interval , the sled moves a distance , starting from rest. Part A Find the average power created by the force . Express your answer in terms of the given quantities and, if necessary, appropriate constants. You may or may not use all of the given quantities. Hint 1. Find the work done by the force . Find the work done by the force . Express your answer in terms of the given quantities and, if necessary, appropriate constants. You may or may not use all of the given quantities. ANSWER: ANSWER: Correct Part B Find the average velocity of the sled during that time interval. Express your answer in terms of the given quantities and, if necessary, appropriate constants. You may or may not use all of the given quantities. ANSWER: = =

3/1/24, 11:42 PM Homework 6-1 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID= 0 3/26 Correct Part C Find the average power created by the force in terms of the average speed of the sled. Express your answer in terms of and . ANSWER: Correct You just obtained a very useful formula for the average power: . If an object is moving at a constant speed, and the force is also constant, this formula can be used to find the average power. If is changing, the formula can be used to find the instantaneous power at any given moment (with the quantity in this case meaning the instantaneous velocity, of course). Part D A sled of mass is being pulled horizontally by a constant upward force of magnitude that makes an angle with the direction of motion. The coefficient of kinetic friction is . The average velocity of the sled is . Find the average power created by force . Express your answer in terms of , , and . Hint 1. Find the work done When the directions of the force and the velocity vectors are the same, . When the two vectors make an angle , which of the following formulas can be applied? ANSWER: ANSWER: = = =

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3/1/24, 11:42 PM Homework 6-1 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID= 0 4/26 Correct Another way to express this formula is this , where is the component of force parallel to the velocity of the object. Let us now consider several questions that include numeric data. A sled is being pulled along a horizontal surface by a horizontal force of magnitude 600 N. Starting from rest, the sled speeds up with acceleration 0.08 for 1 minute. Part E Find the average power created by force . Express your answer in watts to three significant figures. Hint 1. Find the work done by force . Find the work done by force . Express your answer in joules to three significant figures. Hint 1. Find the displacement of the sled Which formula is useful in finding the displacement of the sled? ANSWER: ANSWER: ANSWER: = 8.64×10 4 = 1440

3/1/24, 11:42 PM Homework 6-1 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID= 0 5/26 Correct Part F Find the instantaneous power created by force at . Express your answer in watts to three significant figures. Hint 1. Find the speed Find the speed of the sled after 10 s.. Express your answer in meters per second. ANSWER: ANSWER: Correct Part G Find the instantaneous power created by the normal force at . The magnitude of the acceleration due to gravity is . Express your answer in watts to three significant figures. Hint 1. Work done by the normal force Recall that the normal force is directed upward, while the sled is moving horizontally. Given these directions, what is the angle between the force and the direction of motion? What does this tell you about how much work the normal force does? ANSWER: Correct When vectors and are perpendicular, the power created by force is zero. Problem 9.37 - Enhanced - with Hints and Feedback = 0.8 = 480 = 0

3/1/24, 11:42 PM Homework 6-1 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID= 0 6/26 Part A How much work must you do to push a 11.0 block of steel across a steel table ( = 0.60) at a steady speed of 1.30 for 4.70 ? Express your answer with the appropriate units. Hint 1. How to approach the problem Recall the expression for work in terms of force and displacement. Write the equilibrium equation for the block to find the force you have to apply to keep the block moving at a steady speed. ANSWER: Correct Part B What is your power output while doing so? Express your answer with the appropriate units. Hint 1. How to approach the problem Use your answer from the previous part to calculate power using its definition. Alternatively, you can use the expression for power in terms of force and velocity. ANSWER: Correct Problem 9.41 A 64 human sprinter can accelerate from rest to 10 in 3.1 . During the same time interval, a 26 greyhound can go from rest to 20 . Part A What is the average power output of the human? Average power over a time interval is . Express your answer to two significant figures and include the appropriate units. ANSWER: = 395 = 84.1

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3/1/24, 11:42 PM Homework 6-1 7/26 Correct Part B What is the average power output of the greyhound? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct 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 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 i th 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: . = 1.0 = 1.7 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID= 0

3/1/24, 11:42 PM Homework 6-1 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID= 0 8/26 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: Correct 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: linear speed linear acceleration angular speed angular acceleration total mass shape and density of the object location of the axis of rotation

3/1/24, 11:42 PM Homework 6-1 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID= 0 9/26 Correct 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). Express your answers in terms of and separated by commas. ANSWER: Correct 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: Correct 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. Remember that both particles rotate about the y axis. Express your answer in terms of , , and . ANSWER: undefined: an axis of rotation has not been specified. , , = = =

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3/1/24, 11:42 PM Homework 6-1 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID= 0 10/26 Correct Next we will prove explicitly that the rotational kinetic energy formula ( ) agrees with the linear kinetic energy formula ( ). To see the connection, let us find the kinetic energy of particle a and b. 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. Remember that both particles rotate about the y axis. Express your answers in terms of , , and separated by a comma. Hint 1. Find the linear speed Using the formula , find the linear speed of particle a. Express your answer in terms of and . ANSWER: ANSWER: Correct Part G Now, using the results of Part F, find the total kinetic energy of the system. Remember that both particles rotate about the y axis. Express your answer in terms of , , and . ANSWER: Correct 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. Moment of Inertia and Center of Mass for Point Particles = , = =

3/1/24, 11:42 PM Homework 6-1 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID= 0 11/26 Ball a, of mass , is connected to ball b, of mass , by a massless rod of length . The two vertical dashed lines in the figure, one through each ball, represent two different axes of rotation, axes a and b . These axes are parallel to each other and perpendicular to the rod. The moment of inertia of the two-mass system about axis a is , and the moment of inertia of the system about axis b is . It is observed that the ratio of to is equal to 3: Assume that both balls are pointlike; that is, neither has any moment of inertia about its own center of mass. Part A Find the ratio of the masses of the two balls. Express your answer numerically. Hint 1. How to approach the problem Find an expression for and for in terms of the masses and . Substitute these expressions into the formula given in the problem introduction and then solve for the ratio of the masses. Hint 2. Find Find , the moment of inertia of the system about axis a . Express your answer in terms of any or all of the following quantities: , , and . Hint 1. Formula for the moment of inertia The formula for the moment of inertia of an object consisting of particles with masses , at distances from the rotation axis is . ANSWER: ANSWER: = = 0.333

3/1/24, 11:42 PM Homework 6-1 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID= 0 12/26 Correct Part B Find , the distance from ball a to the system's center of mass. Express your answer in terms of , the length of the rod. Hint 1. How to approach the problem To find , compute the position of the center of mass of the system, using a coordinate system in which ball a is at the origin. In these coordinates, ball a is at distance zero from the origin, ball b is at distance from the origin, and the center of mass is at distance from the origin. Using these values in the equation for center of mass, you obtain . Use the result from Part A to eliminate and from this equation and obtain in terms of . Hint 2. Find in terms of What is in terms of ? ANSWER: ANSWER: Correct Scaling of Moments of Inertia Learning Goal: To understand the concept of moment of inertia and how it depends on mass, radius, and mass distribution. In rigid-body rotational dynamics, the role analogous to the mass of a body (when one is considering translational motion) is played by the body's moment of inertia. For this reason, conceptual understanding of the motion of a rigid body requires some understanding of moments of inertia. This problem should help you develop such an understanding. The moment of inertia of a body about some specified axis is , where is a dimensionless constant, is the mass of the body, and is the perpendicular distance from the axis of rotation. Therefore, if you have two similarly shaped objects of the same size but with one twice as massive as the other, the more massive object should have a moment of inertia twice that of the less massive one. Furthermore, if you have two similarly shaped objects of the same mass, but one has twice the size of the other, the larger object should have a moment of inertia that is four times that of the smaller one. = =

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3/1/24, 11:42 PM Homework 6-1 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID= 0 13/26 Part A Two spherical shells have their mass uniformly distrubuted over the spherical surface. One of the shells has a diameter of 2 meters and a mass of 1 kilogram. The other shell has a diameter of 1 meter. What must the mass of the 1-meter shell be for both shells to have the same moment of inertia about their centers of mass? ANSWER: Correct It is important to understand how the dimensionless constant in the moment of inertia formula given in the problem introduction is determined. Consider a disk and a thin ring, both having the same outer radius and mass . The moment of inertia of the disk is , while the moment of inertia of the ring is . (For each object, the axis is perpendicular to the plane of the object and passes through the object's center of mass.) The factor of for the disk gives an indication of how the mass is distributed in that object. If both the disk and ring are spinning with the same angular velocity , the ring should have a greater kinetic energy, because all of the mass of the ring has linear speed , whereas the linear speeds of the different parts of the disk vary, depending on how far the part is from the center, and these speeds vary from zero to . In general, the value of reﬂects the distribution of mass within the object. A number close to indicates that most of the mass is located at a distance from the center of mass close to , while a number much less than indicates that most of the mass is located near the center of mass. Part B Consider the moment of inertia of a solid uniform disk, versus that of a solid sphere, about their respective centers of mass. Assume that they both have the same mass and outer radius, that they have uniform mass distributions, and that the disk is rotated about an axis perpendicular to its face. What is the relation between the moment of inertia of the disk and that of the sphere ? Hint 1. How to approach the problem Draw a figure of each object and the axis. Consider two "slices" parallel to the axis at different radii and the ratio of masses inside these slices. Which object has a greater percentage of the mass closer to the axis? ANSWER: Correct The Parallel-Axis Theorem = 4 kg

3/1/24, 11:42 PM Homework 6-1 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID= 0 14/26 Learning Goal: To understand the parallel-axis theorem and its applications To solve many problems about rotational motion, it is important to know the moment of inertia of each object involved. Calculating the moments of inertia of various objects, even highly symmetrical ones, may be a lengthy and tedious process. While it is important to be able to calculate moments of inertia from the definition ( ), in most cases it is useful simply to recall the moment of inertia of a particular type of object. The moments of inertia of frequently occurring shapes (such as a uniform rod, a uniform or a hollow cylinder, a uniform or a hollow sphere) are well known and readily available from any mechanics text, including your textbook. However, one must take into account that an object has not one but an infinite number of moments of inertia. One of the distinctions between the moment of inertia and mass (the latter being the measure of tranlsational inertia) is that the moment of inertia of a body depends on the axis of rotation. The moments of inertia that you can find in the textbooks are usually calculated with respect to an axis passing through the center of mass of the object. However, in many problems the axis of rotation does not pass through the center of mass. Does that mean that one has to go through the lengthy process of finding the moment of inertia from scratch? It turns out that in many cases, calculating the moment of inertia can be done rather easily if one uses the parallel-axis theorem. Mathematically, it can be expressed as , where is the moment of inertia about an axis passing through the center of mass, is the total mass of the object, and is the moment of inertia about another axis, parallel to the one for which is calculated and located a distance from the center of mass. In this problem you will show that the theorem does indeed work for at least one object: a dumbbell of length made of two small spheres of mass each connected by a light rod (see the figure). NOTE: Unless otherwise noted, all axes considered are perpendicular to the plane of the page. Part A Using the definition of moment of inertia, calculate , the moment of inertia about the center of mass, for this object. Express your answer in terms of and . Hint 1. Location of the center of mass The center of mass is halfway between the two spheres, at point A (see the figure). Hint 2. Finding the moment of inertia for each sphere. Since each sphere is located at distance from the center of mass, one sphere's individual moment of inertia is . ANSWER: =

3/1/24, 11:42 PM Homework 6-1 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID= 0 15/26 Correct Part B Using the definition of moment of inertia, calculate , the moment of inertia about an axis through point B, for this object. Point B coincides with (the center of) one of the spheres (see the figure). Express your answer in terms of and . Hint 1. Finding the contribution of each sphere Now the axis of rotation passes through the left sphere, so the moment of inertia due to that sphere is trivially obtained; Also, keep in mind that the right sphere is now located at distance from the axis of rotation. ANSWER: Correct Part C Now express for this object using the parallel-axis theorem. Express your answer in terms of and . ANSWER: Correct Part D Using the definition of moment of inertia, calculate , the moment of inertia about an axis through point C, for this object. Point C is located a distance from the center of mass (see the figure). Express your answer in terms of and . Hint 1. Finding the contribution of each sphere Each sphere is now located a distance from the axis of rotation. ANSWER: = =

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3/1/24, 11:42 PM Homework 6-1 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID= 0 16/26 Correct Part E Now express for this object using the parallel-axis theorem. Express your answer in terms of and . ANSWER: Correct Not surprisingly, the parallel-axis theorem yields the correct result each time! Let us now apply that theorem to a more general case. Consider an irregular object of mass . Its moment of inertia measured with respect to axis A (parallel to the plane of the page), which passes through the center of mass (see the second diagram), is given by . Axes B, C, D, and E are parallel to axis A; their separations from axis A are shown in the diagram. In the subsequent questions, the subscript indicates the axis with respect to which the moment of inertia is measured: for instance, is the moment of inertia about axis C. Part F Which moment of inertia is the smallest? ANSWER:

3/1/24, 11:42 PM Homework 6-1 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID= 0 17/26 Correct One of the important results obtained from the parallel-axis theorem is that for any object, is always the smallest among the family of moments of inertia with respect to various parallel axes. Part G Which moment of inertia is the largest? ANSWER: Correct Part H Which moments of inertia are equal? ANSWER: Correct Part I Which moment of inertia equals ? ANSWER: and and and No two moments of inertia are equal.

3/1/24, 11:42 PM Homework 6-1 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID= 0 18/26 Correct Part J Axis X, not shown in the diagram, is parallel to the axes shown. It is known that . Which of the following is a possible location for axis X? ANSWER: Correct 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 . between axes A and C between axes C and D between axes D and E to the right of axis E

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3/1/24, 11:42 PM Homework 6-1 19/26 Part A Find the kinetic energy of the rotating particle. Express your answer in terms of , , and . Hint 1. Find the speed of the particle What is the magnitude of the velocity of the particle? Express your answer in terms of and . ANSWER: ANSWER: Correct 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. Give your answer in terms of and , not . Hint 1. How to approach the problem Set the kinetic energy that you found in Part B equal to , and then solve for . ANSWER: Correct 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 . In other words, the total moment of inertia is the sum of the moments of inertia of each mass. Part C https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID= 0 = = =

3/1/24, 11:42 PM Homework 6-1 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID= 0 20/26 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. Express your answer in terms of and . Hint 1. Find the moment of inertia of a segment of the hoop Imagine that this hoop is cut into identical segments, as shown in the figure. Each of these segments can be considered a separate particle of mass located at a distance from the axis. What is the moment of inertia of one such segment? Express your answer in terms of , , and . ANSWER: ANSWER: Correct Kinetic Energy of a Dumbbell = =

3/1/24, 11:42 PM Homework 6-1 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID= 0 21/26 This problem illustrates the two contributions to the kinetic energy of an extended object: rotational kinetic energy and translational kinetic energy. You are to find the total kinetic energy of a dumbbell of mass when it is rotating with angular speed and its center of mass is moving translationally with speed . Denote the dumbbell's moment of inertia about its center of mass by . Note that if you approximate the spheres as point masses of mass each located a distance from the center and ignore the moment of inertia of the connecting rod, then the moment of inertia of the dumbbell is given by , but this fact will not be necessary for this problem. Part A Find the total kinetic energy of the dumbbell. Express your answer in terms of , , , and . Hint 1. How to approach the problem Compute separately the rotational and translational kinetic energies of the dumbbell. Then add the two to find the total kinetic energy. Hint 2. Find the rotational kinetic energy What is the dumbbell's rotational kinetic energy ? Give your answer in terms of some or all of the variables , , and . Hint 1. Formula for rotational kinetic energy The formula for the rotational kinetic energy of a body with moment of inertia about some axis, rotating with angular velocity about it, is . ANSWER: Hint 3. Find the translational kinetic energy What is the dumbbell's translational kinetic energy ? Give your answer in terms of some or all of the variables , , and . ANSWER: =

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3/1/24, 11:42 PM Homework 6-1 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID= 0 22/26 ANSWER: Correct Part B The rotational kinetic energy term is often called the kinetic energy in the center of mass, while the translational kinetic energy term is called the kinetic energy of the center of mass. You found that the total kinetic energy is the sum of the kinetic energy in the center of mass plus the kinetic energy of the center of mass. A similar decomposition exists for angular and linear momentum. There are also related decompositions that work for systems of masses, not just rigid bodies like a dumbbell. It is important to understand the applicability of the formula . Which of the following conditions are necessary for the formula to be valid? Check all that apply. ANSWER: Correct Constrained Rotation and Translation Learning Goal: To understand that contact between rolling objects and what they roll against imposes constraints on the change in position (velocity) and angle (angular velocity). The way in which a body makes contact with the world often imposes a constraint relationship between its possible rotation and translational motion. A ball rolling on a road, a yo-yo unwinding as it falls, and a baseball leaving the pitcher's hand are all examples of constrained rotation and translation. In a similar manner, the rotation of one body and the translation of another may be constrained, as happens when a fireman unrolls a hose from its storage drum. Situations like these can be modeled by constraint equations , relating the coupled angular and linear motions. Although these equations fundamentally involve position (the angle of the wheel at a particular distance down the road), it is usually the = = The velocity vector must be perpendicular to the axis of rotation. The velocity vector must be perpendicular or parallel to the axis of rotation. The moment of inertia must be taken about an axis through the center of mass.

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3/1/24, 11:42 PM Homework 6-1 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID= 0 23/26 relationship of velocities and accelerations that are relevant in solving a problem involving such constraints. The velocities are needed in the conservation equations for momentum and angular momentum, and the accelerations are needed for the dynamical equations. It is important to use the standard sign conventions: positive for counterclockwise rotation and positive for motion toward the right. Otherwise, your dynamical equations will have to be modified. Unfortunately, a frequent result will be the appearance of negative signs in the constraint equations. Consider a measuring tape unwinding from a drum of radius . The center of the drum is not moving; the tape unwinds as its free end is pulled away from the drum. Neglect the thickness of the tape, so that the radius of the drum can be assumed not to change as the tape unwinds. In this case, the standard conventions for the angular velocity and for the (translational) velocity of the end of the tape result in a constraint equation with a positive sign (e.g., if , that is, the tape is unwinding, then also). Part A Assume that the function represents the length of tape that has unwound as a function of time. Find , the angle (in radians) through which the drum will have rotated, as a function of time. Hint 1. Find the amount of tape that unrolls in one complete revolution of the drum If the measuring tape unwinds one complete revolution ( ), how much tape, , will have unwound? ANSWER: ANSWER: =

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3/1/24, 11:42 PM Homework 6-1 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID= 0 24/26 Correct Part B The tape is now wound back into the drum at angular rate . With what velocity will the end of the tape move? (Note that our drawing specifies that a positive derivative of implies motion away from the drum. Be careful with your signs! The fact that the tape is being wound back into the drum implies that , and for the end of the tape to move closer to the drum, it must be the case that . Hint 1. How to approach the probelm The function is given by the derivative of with respect to time. Compute this derivative using the expression for found in Part A and the fact that . Express your answer in terms of and . ANSWER: ANSWER: Correct Part C Since is a positive quanitity, the answer you just obtained implies that will always have the same sign as . If the tape is unwinding, both quanitites will be positive. If the tape is being wound back up, both quantities will be negative. Now find , the linear acceleration of the end of the tape, in terms of , the angular acceleration of the drum, where . ANSWER: =

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3/1/24, 11:42 PM Homework 6-1 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID= 0 25/26 Correct Part D Perhaps the trickiest aspect of working with constraint equations for rotational motion is determining the correct sign for the kinematic quantities. Consider a tire of radius rolling to the right, without slipping, with constant x velocity . Find , the (constant) angular velocity of the tire. Be careful of the signs in your answer; recall that positive angular velocity corresponds to rotation in the counterclockwise direction. ANSWER:

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3/1/24, 11:42 PM Homework 6-1 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID= 0 26/26 Correct This is an example of the appearance of negative signs in constraint equations--a tire rolling in the positive direction translationally exhibits negative angular velocity, since rotation is clockwise. Part E Assume now that the angular velocity of the tire, which continues to roll without slipping, is not constant, but rather that the tire accelerates with constant angular acceleration . Find , the linear acceleration of the tire. ANSWER: Correct Score Summary: Your score on this assignment is 100%. You received 11 out of a possible total of 11 points.

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Homework 6-1

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