# mastering physics assignment 8

## Physics 140 with Uher at University of Michigan - Ann Arbor *

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Assignment 8 Due: 11:59pm on Tuesday, March 9, 2010 Note: To understand how points are awarded, read your instructor's Grading Policy. [Return to Standard Assignment View] Ladybugs on a Rotating Disk Two ladybugs sit on a rotating disk (the ladybugs are at rest with respect to the surface of the disk and do not slip). Ladybug 1 is halfway between ladybug 2 and the axis of rotation. Part A What is the angular speed of ladybug 1? ANSWER: one-half the angular speed of ladybug 2 the same as the angular speed of ladybug 2 two times the angular speed of ladybug 2 one-quarter the angular speed of ladybug 2 Correct Part B What is the ratio of the linear speed of ladybug 2 to that of ladybug 1? Hint B.1 Relation between linear and angular speeds Hint not displayed Answer numerically. ANSWER: 2 Correct Part C At the instant shown in the figure, what is the direction of the the radial component of the acceleration of ladybug 2? Hint C.1 Radial acceleration of an object moving on a circle The radial acceleration of an object moving on a circle is the centripetal acceleration, where is the tangential velocity of the object, is the distance from the axis of rotation and is the unit position vector of the object w.r.t. the origin. ANSWER: Correct Part D What is the ratio of the radial acceleration of ladybug 2 to that of ladybug 1? Hint D.1 Radial acceleration of an object moving on a circle Hint not displayed Answer numerically. ANSWER: 2 Correct MasteringPhysics: Assignment Print View http://session.masteringphysics.com/myct/assignmentPrint?ass... 1 of 21 4/17/10 5:30 PM Although the circular trajectory of ladybug 2 has twice the radius as that of ladybug 1, ladybug 2 also has twice the linear velocity of ladybug 1. Thus, according to the formula , where is centripital acceleration, ladybug 2 has twice the centripetal acceleration of ladybug 1. Part E What is the direction of the vector representing the angular velocity of ladybug 2? Hint E.1 Direction of the angular velocity vector The direction of the angular velocity vector is given by the right-hand rule. Curl the fingers of your right hand along the direction of rotation, and your thumb will point along the direction of the angular velocity vector. ANSWER: Correct Part F In general, velocity is represented by a vector. Let represent the velocity of ladybug 1. Angular velocity is also often represented by a vector, so let that of ladybug 1 be given by . Take to be the vector from the axis of rotation to ladybug 1. Which of the following equations correctly describes the relationship between the velocity , angular velocity , and position of ladybug 1? ANSWER: Correct Part G Now assume that at the moment pictured in the figure, the disk is rotating but slowing down. What is the direction of the tangential component of the acceleration (i.e. acceleration tangent to the motion) of ladybug 2? ANSWER: Correct An Exhausted Bicyclist An exhausted bicyclist pedals somewhat erratically when exercising on a static bicycle. The angular velocity of the wheels follows the equation , where represents time (measured in seconds). Part A There is a spot of paint on the front wheel of the bicycle. Take the position of the spot at time to be at angle radians with respect to an axis parallel to the ground (and perpendicular to the axis of rotation of the tire) and measure positive angles in the direction of the wheel's rotation. What angular displacement has the spot of paint undergone between time 0 and 2 seconds? Hint A.1 How to approach the problem Hint not displayed Hint A.2 Find the angular displacement Hint not displayed Express your answer in radians. ANSWER: = 0.793 Correct Part B MasteringPhysics: Assignment Print View http://session.masteringphysics.com/myct/assignmentPrint?ass... 2 of 21 4/17/10 5:30 PM Express the angular displacement undergone by the spot of paint at seconds in degrees. Hint B.1 Radians and degrees Hint not displayed ANSWER: = 45.5 Correct Part C What distance has the spot of paint moved in 2 seconds if the radius of the wheel is 50 centimeters? Hint C.1 Distance traveled by a particle that rotates about a certain axis Consider a particle that rotates about the z axis in the xy plane. The angular diplacement traveled by the particle is given by , where is the distance traveled by the particle in the xy plane and is the distance of the particle from the z axis. Express your answer in centimeters. ANSWER: = 39.7 Correct Part D Which one of the following statements describes the motion of the spot of paint at seconds? Hint D.1 How to approach the problem As the bicyclist pedals, the wheels rotate in the positive direction as we have defined it. This makes the angular velocity of the wheel always nonnegative, and consequently the angular velocity of the spot of paint is always nonnegative. In addition, the angular velocity of the spot of paint increases if its angular acceleration is positive, and it decreases if its angular acceleration is negative. Hint D.2 Find the angular acceleration at seconds What is the angular acceleration of the spot of paint at seconds? Hint D.2.1 Find the angular acceleration of the spot of paint Hint not displayed Express your answer in radians per second per second. ANSWER: = 0.827 Correct ANSWER: The angular acceleration of the spot of paint is constant and the magnitude of the angular speed is decreasing. The angular acceleration of the spot of paint is constant and the magnitude of the angular speed is increasing. The angular acceleration of the spot of paint is positive and the magnitude of the angular speed is decreasing. The angular acceleration of the spot of paint is positive and the magnitude of the angular speed is increasing. The angular acceleration of the spot of paint is negative and the magnitude of the angular speed is decreasing. The angular acceleration of the spot of paint is negative and the magnitude of the angular speed is increasing. Correct Moment of Inertia and Center of Mass for Point Particles 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. Hint A.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 A.2 Find MasteringPhysics: Assignment Print View http://session.masteringphysics.com/myct/assignmentPrint?ass... 3 of 21 4/17/10 5:30 PM Find , the moment of inertia of the system about axis a. Hint A.2.1 Formula for the moment of inertia Hint not displayed Express your answer in terms of any or all of the following quantities: , , and . ANSWER: = Correct Express your answer numerically. ANSWER: = 0.333 Correct Part B Find , the distance from ball A to the system's center of mass. Hint B.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 B.2 Find in terms of What is in terms of ? ANSWER: = Correct Now substitute in your formula for . Express your answer in terms of , the length of the rod. ANSWER: = Correct Flywheel Kinematics A heavy flywheel is accelerated (rotationally) by a motor that provides constant torque and therefore a constant angular acceleration . The flywheel is assumed to be stationary at time in Parts A, B, and C of this problem. Part A Find the time it takes to accelerate the flywheel to 10.0 rps (revolutions per second) if is 5.00 radians/s . Hint A.1 A linear analogy Hint not displayed Hint A.2 Angular velocity as a function of time Hint not displayed Hint A.3 A reminder about unit conversion. Hint not displayed Give a numerical answer, in seconds, to one decimal place. ANSWER: = 12.6 Correct s Part B Find the time to accelerate the flywheel from rest up to angular velocity . Hint B.1 Angular velocity as a function of time Hint not displayed Express your answer in terms of and . MasteringPhysics: Assignment Print View http://session.masteringphysics.com/myct/assignmentPrint?ass... 4 of 21 4/17/10 5:30 PM ANSWER: = Correct Part C Find the angle through which the flywheel will have turned during the time it takes for it to accelerate from rest up to angular velocity . Hint C.1 A linear analogy The relationship between velocity and displacement for an object undergoing constant linear accleration is given by . An analogous formula for angular velocity applies when the angular acceleration is constant. Hint C.2 Angular velocity as a function of displacement For an object undergoing constant angular acceleration , the relationship between angular velocity and angular displacement is , where is the angular velocity at the initial time , and the initial angular displacement. Express your answer in terms of any of the following: , , and/or . ANSWER: = Correct Part D Assume that the motor has accelerated the wheel up to an angular velocity with angular acceleration in time . At this point, the motor is turned off, and a brake is applied that decelerates the wheel with a constant angular acceleration of . Find , the time it will take the wheel to stop after the brake is applied (that is, the time for the wheel to reach zero angular velocity). Hint D.1 How to approach the problem Solve this part using the same technique that you used for Part B. Just keep in mind that the initial conditions have changed (in this case, the wheel is initially spinning) and that the angular acceleration is different. Express your answer in terms of any of the following: , , and/or . 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. 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: = 4 Correct kg 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. MasteringPhysics: Assignment Print View http://session.masteringphysics.com/myct/assignmentPrint?ass... 5 of 21 4/17/10 5:30 PM 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 B.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 An Electric Ceiling Fan An electric ceiling fan is rotating about a fixed axis with an initial angular velocity of 0.230 . The angular acceleration is 0.905 . Its blades form a circle of diameter 0.770 . Part A Compute the angular velocity of the fan after time 0.210 has passed. Hint A.1 Angular velocity and acceleration Hint not displayed Express your answer numerically in revolutions per second. ANSWER: 0.420 Correct Part B Through how many revolutions has the blade turned in the time interval 0.210 from Part A? Hint B.1 Angle and angular velocity Hint not displayed Express the number of revolutions numerically. ANSWER: 6.83×10 −2 Correct Part C What is the tangential speed of a point on the tip of the blade at time = 0.210 ? Hint C.1 Relating angular and linear speed Hint not displayed Hint C.2 Converting revolutions to radians Hint not displayed Express your answer numerically in meters per second. ANSWER: = 1.02 Correct Part D What is the magnitude of the resultant acceleration of a point on the tip of the blade at time = 0.210 ? Hint D.1 How to approach the problem Hint not displayed Hint D.2 Find the centripetal acceleration Hint not displayed Hint D.3 Find the tangential acceleration Hint not displayed Hint D.4 Calculating the vector sum Hint not displayed Express the acceleration numerically in meters per second squared. ANSWER: = 3.46 Correct MasteringPhysics: Assignment Print View http://session.masteringphysics.com/myct/assignmentPrint?ass... 6 of 21 4/17/10 5:30 PM Two Worlds on a String Two balls, A and B, with masses and are connected by a taut, massless string, and are moving along a horizontal frictionless plane. The distance between the centers of the two balls is . At a certain instant, the velocity of ball B has magnitude and is directed perpendicular to the string and parallel to the horizontal plane, and the velocity of ball A is zero. Part A Find , the tension in the string. Hint A.1 Descibe the nature of the motion Describe the ongoing (not the instantaneous) motion of the system qualitatively. ANSWER: Ball B moves in a circle around ball A while ball A remains at rest. Both balls move in a circle about the midpoint of the string while sliding along the plane translationally. Both balls move in a circle about the center of mass of the system while sliding along the plane translationally. Correct Hint A.2 The key idea For a ball to move in a circle, it must be subject to a centripetal force. It is the tension in the string that provides this force. Therefore, , where is the linear speed of the rotational motion (relative to the point about which the ball rotates) and is the radius of the motion. As the answer to the first hint suggests, and are not the same as and . Hint A.3 Find the velocity of the center of mass Find , the translational speed of the system's center of mass. Hint A.3.1 How to compute the velocity of the center of mass You can calculate the velocity of the center of mass of a system by computing the total momentum of the system and dividing by the total mass of the system. Express your answer in terms of , , and . ANSWER: = Answer Requested Hint A.4 Find the rotational speed As the system slides across the horizontal plane, it will rotate about its center of mass. Find , the linear speed of ball B's rotational motion relative to the center of mass. Hint A.4.1 How to compute ball B's rotational speed To find the speed with which ball B rotates about the system's center of mass, you must subtract the translational speed of the center of mass from the ball's total speed : . Express your answer in terms of , , and . ANSWER: = Correct Hint A.5 Find the radius of rotation Find , the radius of rotational motion of ball B. Hint A.5.1 How to approach the question Since ball B rotates about the center of mass of the system, will be the distance from the ball to the system's center of mass. In other words, to find , calculate the distance between the center of mass and ball B. Hint A.5.2 Position of the center of mass Recall that the position of the center of mass of a system is equal to the weighted average position of all the individual objects constituting the system (in this case, the two balls). The weighting factor for each object is its mass. Express your answer in terms of , , and . MasteringPhysics: Assignment Print View http://session.masteringphysics.com/myct/assignmentPrint?ass... 7 of 21 4/17/10 5:30 PM ANSWER: = Answer Requested Hint A.6 Acceleration of ball B Once you know ball B's radius of rotation and rotational speed , you can compute its acceleration using the centripetal acceleration formula: . Find , the magnitude of the acceleration of ball B. Express your answer in terms of , , , and . ANSWER: = Answer Requested The the only force acting on the ball of mass is the tension in the string. Now that you know the acceleration, you can compute the tension in the string using Newton's second law. Express in terms of , , , and . ANSWER: = Correct Note that your answer is "symmetric" between the parameters and . This is as it should be: The tension should be the same regardless of whether or initially moves. Only their relative velocity matters. The Parallel-Axis Theorem 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. Hint A.1 Location of the center of mass The center of mass is halfway between the two spheres, at point A (see the figure). Hint A.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 . Express your answer in terms of and . ANSWER: = Correct Part B MasteringPhysics: Assignment Print View http://session.masteringphysics.com/myct/assignmentPrint?ass... 8 of 21 4/17/10 5:30 PM 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). Hint B.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. Express your answer in terms of and . ANSWER: = Correct Part C Now calculate 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). Hint D.1 Finding the contribution of each sphere Hint not displayed Express your answer in terms of and . ANSWER: = Correct Part E Now calculate 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: 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 MasteringPhysics: Assignment Print View http://session.masteringphysics.com/myct/assignmentPrint?ass... 9 of 21 4/17/10 5:30 PM Which moment of inertia is the largest? ANSWER: Correct Part H Which moments of inertia are equal? ANSWER: and and and No two moments of inertia are equal. Correct Part I Which moment of inertia equals ? ANSWER: 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: between axes A and C between axes C and D between axes D and E to the right of axis E Correct Kinetic Energy of a Dumbbell 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. Hint A.1 How to approach the problem Hint not displayed Hint A.2 Find the rotational kinetic energy Hint not displayed Hint A.3 Find the translational kinetic energy Hint not displayed Express your answer in terms of , , , and . MasteringPhysics: Assignment Print View http://session.masteringphysics.com/myct/assignmentPrint?ass... 10 of 21 4/17/10 5:30 PM 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: 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. Correct Problem 9.86 The pulley in the figure has radius 0.160 and a moment of inertia 0.480 . The rope does not slip on the pulley rim. Part A Use energy methods to calculate the speed of the 4.00-kg block just before it strikes the floor. ANSWER: = 2.81 Correct Problem 9.96 A thin, uniform rod is bent into a square of side length . Part A If the total mass is , find the moment of inertia about an axis through the center and perpendicular to the plane of the square. Use the parallel-axis theorem. Express your answer in terms of the variables and . ANSWER: = Correct Problem 9.89 Two metal disks, one with radius = 2.37 and mass = 0.850 and the other with radius = 5.04 and mass = 1.61 , are welded together and mounted on a frictionless axis through their common center. . MasteringPhysics: Assignment Print View http://session.masteringphysics.com/myct/assignmentPrint?ass... 11 of 21 4/17/10 5:30 PM Part A What is the total moment of inertia of the two disks? ANSWER: = 2.28×10 −3 Correct Part B A light string is wrapped around the edge of the smaller disk, and a 1.50-kg block, suspended from the free end of the string. If the block is released from rest at a distance of 2.05 above the floor, what is its speed just before it strikes the floor? ANSWER: = 3.29 Answer Requested Part C Repeat the calculation of part B, this time with the string wrapped around the edge of the larger disk. ANSWER: = 5.01 Correct Problem 9.97 A cylinder with radius and mass has density that increases linearly with distance from the cylinder axis, , where is a positive constant. Part A Calculate the moment of inertia of the cylinder about a longitudinal axis through its center in terms of and . ANSWER: = Correct Part B Is your answer greater or smaller than the moment of inertia of a cylinder of the same mass and radius but of uniform density? ANSWER: Greater Smaller Correct Finding Torque A force of magnitude making an angle with the x axis is applied to a particle located along axis of rotation A, at Cartesian coordinates in the figure. The vector lies in the xy plane, and the four axes of rotation A, B, C, and D all lie perpendicular to the xy plane. A particle is located at a vector position with respect to an axis of rotation (thus points from the axis to the point at which the particle is located). The torque about this axis due to a force acting on the particle is given by , where is the angle between and , is the magnitude of , is the magnitude of , the component of that is perpendicualr to is the moment arm, and is the component of the force that is perpendicular to . Sign convention: If you use the first of the two equalities above, you will obtain the correct sign for torque when you do the calculation. If you use either of the other two expressions for the torque, you will need to determine the sign by analyzing the direction of the rotation that the torque would tend to produce. Recall that negative torque about an axis corresponds to clockwise rotation. In this problem, you must express the angle in the above equation in terms of , , and/or when entering your answers. Keep in mind that and . Part A MasteringPhysics: Assignment Print View http://session.masteringphysics.com/myct/assignmentPrint?ass... 12 of 21 4/17/10 5:30 PM What is the torque about axis A due to the force ? Hint A.1 When force is applied at the pivot point Hint not displayed Express the torque about axis A at Cartesian coordinates . ANSWER: = Correct Part B What is the torque about axis B due to the force ? (B is the point at Cartesian coordinates , located a distance from the origin along the y axis.) Hint B.1 Finding with respect to an axis Hint not displayed Hint B.2 A helpful figure Hint not displayed Express the torque about axis B in terms of , , , , and/or other given coordinate data. ANSWER: = Correct Part C What is the torque about axis C due to ? (C is the point at Cartesian coordinates , a distance along the x axis.) Hint C.1 A helpful figure Hint not displayed Hint C.2 Clockwise or counterclockwise? Hint not displayed Express the torque about axis C in terms of , , , , and/or other given coordinate data. ANSWER: = Correct Part D What is the torque about axis D due to ? (D is the point located at a distance from the origin and making an angle with the x axis.) Express the torque about axis D in terms of , , , , and/or other given coordinate data. ANSWER: = Correct You could have found by using any of the three equations listed at the top of the page, , and making the following associations: and . Calculating Torques Using Two Standard Methods Learning Goal: To understand the two most common procedures for finding torques when the forces and displacements are all in one plane: the moment arm method and the tangential force method. The purpose of this problem is to give you further practice finding torques in two-dimensional situations. In this case it is overkill to use the full cross product definition of the torque because the only nonzero component of the torque is the component perpendicular to the plane containing the problem. There are two common methods for finding torque in a two-dimensional problem: the tangential force method and the moment arm method. Both of these methods will be illustrated in this problem. Throughout the problem, torques that would cause counterclockwise rotation are considered to be positive. Consider a uniform pole of length , attached at its base (via a pivot) to a wall. The other end of the pole is attached to a cable, so that the pole makes an angle with respect to the wall, and the cable is horizontal. The tension in the cable is . The pole is attached to the wall at point A. MasteringPhysics: Assignment Print View http://session.masteringphysics.com/myct/assignmentPrint?ass... 13 of 21 4/17/10 5:30 PM Tangential force method The tangential force method involves finding the component of the applied force that is perpendicular to the displacement from the pivot point to where the force is applied. This perpendicular component of the force is called the tangential force. Part A What is , the magnitude of the tangential force that acts on the pole due to the tension in the rope? Express your answer in terms of and . ANSWER: = Correct When using the tangential force method, you calculate the torque using the equation , where is the distance from the pivot to the point where the force is applied. The sign of the torque can be determined by checking which direction the tangential force would tend to cause the pole to rotate (where counterclockwise rotation implies positive torque). Part B What is the magnitude of the torque on the pole, about point A, due to the tension in the rope? Express your answer in terms of , , and . ANSWER: = Correct Moment arm method The moment arm method involves finding the effective moment arm of the force. To do this, imagine a line parallel to the force, running through the point at which the force is applied, and extending off to infinity in either direction. You may shift the force vector anywhere you like along this line without changing the torque, provided you do not change the direction of the force vector as you shift it. It is generally most convenient to shift the force vector to a point where the displacement from it to the desired pivot point is perpendicular to its direction. This displacement is called the moment arm. For example, consider the force due to tension acting on the pole. Shift the force vector to the left, so that it acts at a point directly above the point A in the figure. The moment arm of the force is the distance between the pivot and the tail of the shifted force vector. The magnitude of the torque about the pivot is the product of the moment arm and force, and the sign of the torque is again determined by the sense of the rotation of the pole it would cause. Part C Find , the length of the moment arm of the force. Express your answer in terms of and . ANSWER: = Correct To calculate the torque using the moment arm method, use the equation , where is the moment arm perpendicular to the applied force. MasteringPhysics: Assignment Print View http://session.masteringphysics.com/myct/assignmentPrint?ass... 14 of 21 4/17/10 5:30 PM Part D Find the magnitude of the torque on the pole, about point A, due to the tension in the rope. Express your answer in terms of , , and . ANSWER: = Correct For this problem, the two methods of finding torque involved nearly the same of amount of algebra, and either method could be used. Of course, both methods lead to the same final result. Now consider a woman standing on the ball of her foot as shown . A normal force of magnitude acts upward on the ball of her foot. The Achilles' tendon is attached to the back of the foot. The tendon pulls on the small bone in the rear of the foot with a force . This small bone has a length , and the angle between this bone and the Achilles' tendon is . The horizontal displacement between the ball of the foot and the point P is . Part E Suppose you were asked to find the torque about point P due to the normal force in terms of given quantities. Which method of finding the torque would be the easiest to use? ANSWER: tangential force method moment arm method Correct Part F Find , the torque about point P due to the normal force. Express your answer in terms of and any of the other quantities given in the figure. ANSWER: = Correct Part G Suppose you were asked to find the torque about point P due to the force of magnitude in the Achilles' tendon. Which of the following statements is correct? ANSWER: The tangential force method must be used. The moment arm method must be used. Either method may be used. Neither method can be used. Correct Part H Find , the torque about point P due to the force applied by the Achilles' tendon. Express your answer in terms of , , and . ANSWER: = Correct Torque and Angular Acceleration Learning Goal: To understand and apply the formula to rigid objects rotating about a fixed axis. To find the acceleration of a particle of mass , we use Newton's second law: , where is the net force acting on the particle. To find the angular acceleration of a rigid object rotating about a fixed axis, we can use a similar formula: , where is the net torque acting on the object and is its moment of inertia. In this problem, you will practice applying this formula to several situations involving angular acceleration. In all of these situations, two objects of masses and are attached to a seesaw. The seesaw is made of a bar that has length and is pivoted so that it is free to rotate in the vertical plane without friction. You are to find the angular acceleration of the seesaw when it is set in motion from the horizontal position. In all cases, assume that . MasteringPhysics: Assignment Print View http://session.masteringphysics.com/myct/assignmentPrint?ass... 15 of 21 4/17/10 5:30 PM Part A The seesaw is pivoted in the middle, and the mass of the swing bar is negligible. Find the angular acceleration of the seesaw. Hint A.1 Find the moment of inertia Hint not displayed Hint A.2 Find the net torque Hint not displayed Express your answer in terms of some or all of the quantities , , , as well as the acceleration due to gravity . ANSWER: = Correct Part B In what direction will the seesaw rotate, and what will the sign of the angular acceleration be? ANSWER: The rotation is in the clockwise direction and the angular acceleration is positive. The rotation is in the clockwise direction and the angular acceleration is negative. The rotation is in the counterclockwise direction and the angular acceleration is positive. The rotation is in the counterclockwise direction and the angular acceleration is negative. Correct Part C Now consider a similar situation, except that now the swing bar itself has mass . Find the angular acceleration of the seesaw. Hint C.1 What has changed? Compared to the previous situation, what quantities have changed? ANSWER: net torque only moment of inertia only both torque and moment of inertia Correct In calculating the torque due to gravity, the weight of the bar is applied at the center of mass of the bar. The distance between the center of mass and the pivot is used in the calculation of the torque. In this case, the center of mass of the bar coincides with the pivot point. Therefore, the corresponding torque is zero. Hint C.2 Find the moment of inertia What is the new moment of inertia ? Recall that the moment of inertia for a uniform thin rod of mass and length , pivoted at its center, is given by . Express your answer in terms of some or all of the quantities , , , and . MasteringPhysics: Assignment Print View http://session.masteringphysics.com/myct/assignmentPrint?ass... 16 of 21 4/17/10 5:30 PM ANSWER: = Correct Express your answer in terms of some or all of the quantities , , , , as well as the acceleration due to gravity . ANSWER: = Correct Part D In what direction will the seesaw rotate and what will the sign of the angular acceleration be? ANSWER: The rotation is in the clockwise direction and the angular acceleration is positive. The rotation is in the clockwise direction and the angular acceleration is negative. The rotation is in the counterclockwise direction and the angular acceleration is positive. The rotation is in the counterclockwise direction and the angular acceleration is negative. Correct Part E This time, the swing bar of mass is pivoted at a different point, as shown in the figure. Find the magnitude of the angular acceleration of the swing bar. Be sure to use the absolute value function in your answer, since no comparison of , , and has been made. Hint E.1 What has changed? Compared to the previous situation, what quantities have changed? ANSWER: net torque only moment of inertia only both torque and moment of inertia Correct This time, the torque due to the weight of the bar is not zero. Additionally, the torques due to the weights of the blocks are different, since their lever arms have changed. Hint E.2 Find the moment of inertia Find the total moment of inertia of the bar and the two masses. Hint E.2.1 Find the moment of inertia of the bar alone What is the moment of inertia of the bar alone? Hint E.2.1.1 Use the parallel-axis theorem The moment of inertia of a uniform bar of mass and length about the axis passing through the midpoint (center of mass) of the bar and perpendicular to the it (as in the previous situation) is . In the current situation, the axis passes at distance from the center of mass. The value of the moment of inertia with respect to this new axis can be determined using the parallel-axis theorem: , where is the moment of inertia of a body of mass about an axis through its center of mass, and is the moment of inertia of that object through an axis parallel to the original axis through its center of mass, but displaced from it by a distance . Express your answer in terms of and . ANSWER: = Correct Express your answer in terms of some or all of the quantities , , , and . MasteringPhysics: Assignment Print View http://session.masteringphysics.com/myct/assignmentPrint?ass... 17 of 21 4/17/10 5:30 PM ANSWER: = Correct Hint E.3 Find the net torque Find the magnitude of the net torque acting on the system. Be sure to use the absolute value function in your answer, since no comparison of , , and has been given. Express your answer in terms of some or all of the quantities , , , , as well as the acceleration due to gravity . Enter the absolute value function abs(). For instance, enter abs(x*y) for . ANSWER: = Answer Requested Since we cannot numerically compare the quantities , , and , we cannot determine the direction of the net torque. Express your answer in terms of some or all of the quantities , , , , as well as the acceleration due to gravity . Enter the absolute value function as abs(). For instance, enter abs(x*y) for . ANSWER: = Correct Since we cannot numerically compare the quantities , , and , we cannot determine the direction of rotation or the sign of the angular acceleration. Part F If is 24 kilograms, is 12 kilograms, and is 10 kilograms, what is the direction of rotation and the sign of the angular acceleration? ANSWER: The rotation is in the clockwise direction and the angular acceleration is positive. The rotation is in the clockwise direction and the angular acceleration is negative. The rotation is in the counterclockwise direction and the angular acceleration is positive. The rotation is in the counterclockwise direction and the angular acceleration is negative. Correct Pulling a String to Accelerate a Wheel A bicycle wheel is mounted on a fixed, frictionless axle, as shown . A massless string is wound around the wheel's rim, and a constant horizontal force of magnitude starts pulling the string from the top of the wheel starting at time when the wheel is not rotating. Suppose that at some later time the string has been pulled through a distance . The wheel has moment of inertia , where is a dimensionless number less than 1, is the wheel's mass, and is its radius. Assume that the string does not slip on the wheel. Part A Find , the angular acceleration of the wheel, which results from pulling the string to the left. Use the standard convention that counterclockwise angular accelerations are positive. Hint A.1 Relate torque about the axle to force applied to the wheel Hint not displayed Hint A.2 Relate torque on wheel to angular acceleration Hint not displayed Express the angular acceleration, , in terms of , , , and (but not ). ANSWER: = Correct Part B MasteringPhysics: Assignment Print View http://session.masteringphysics.com/myct/assignmentPrint?ass... 18 of 21 4/17/10 5:30 PM The force pulling the string is constant; therefore the magnitude of the angular acceleration of the wheel is constant for this configuration. Find the magnitude of the angular velocity of the wheel when the string has been pulled a distance . Note that there are two ways to find an expression for ; these expressions look very different but are equivalent. Hint B.1 What the no-slip case means The assumption that the string does not slip on the wheel tells you that the translational velocity of the string equals the velocity of a particle on the rim ( ) of the wheel. Hint B.2 Review of translational motion with constant acceleration In kinematics, you learned that for translational motion with constant acceleration the velocity is given by . The wheel is stationary at , so the displacement of the string, , will be proportional to . Hint B.3 When has the string been pulled a distance ? Hint not displayed Hint B.4 Relating translational acceleration and angular acceleration Hint not displayed Express the angular velocity of the wheel in terms of the displacement , the magnitude of the applied force, and the moment of inertia of the wheel , if you've found such a solution. Otherwise, following the hints for this part should lead you to express the angular velocity of the wheel in terms of the displacement , the wheel's radius , and . ANSWER: = Correct This solution can be obtained from the equations of rotational motion and the equations of motion with constant acceleration. An alternate approach is to calculate the work done over the displacement by the force and equate this work to the increase in rotational kinetic energy of rotation of the wheel Part C Find , the speed of the string after it has been pulled by over a distance . Hint C.1 Relating the speed of the string to the angular velocity Find the speed of the string in terms of . (This relationship will be valid at any time .) ANSWER: = Answer Requested Express the speed of the string in terms of , , , and ; do not include , , or in your answer. ANSWER: = Correct Note that this is the speed that an object of mass (which is less than ) would attain if pulled a distance by a force with constant magnitude . Problem 10.75: Rolling Stones A solid, uniform spherical boulder starts from rest and rolls down a 50.0-m high hill, as shown in the figure . The top half of the hill is rough enough to cause the boulder to roll without slipping, but the lower half is covered with ice and there is no friction. Part A What is the translational speed of the boulder when it reaches the bottom of the hill? MasteringPhysics: Assignment Print View http://session.masteringphysics.com/myct/assignmentPrint?ass... 19 of 21 4/17/10 5:30 PM ANSWER: = 29.0 Correct Problem 10.87 A uniform rod of length rests on a frictionless horizontal surface. The rod pivots about a fixed frictionless axis at one end. The rod is initially at rest. A bullet traveling parallel to the horizontal surface and perpendicular to the rod with speed strikes the rod at its center and becomes embedded in it. The mass of the bullet is one-fourth the mass of the rod. Part A What is the final angular speed of the rod? Express your answer in terms of the variables and . ANSWER: = Correct Part B What is the ratio of the kinetic energy of the system after the collision to the kinetic energy of the bullet before the collision? ANSWER: = 0.158 Correct Problem 10.85 A 4.00 ball is dropped from a height of 11.5 above one end of a uniform bar that pivots at its center. The bar has mass 9.50 and is 6.20 in length. At the other end of the bar sits another 4.30 ball, unattached to the bar. The dropped ball sticks to the bar after the collision. Part A How high will the other ball go after the collision? ANSWER: = 1.40 Correct Problem 10.70 A thin-walled, hollow spherical shell of mass and radius starts from rest and rolls without slipping down the track shown in the figure . Points and are on a circular part of the track having radius . The diameter of the shell is very small compared to and , and rolling friction is negligible. Part A What is the minimum height for which this shell will make a complete loop-the-loop on the circular part of the track? Express your answer in terms of the variables , , and appropriate constants. ANSWER: = Correct Part B How hard does the track push on the shell at point , which is at the same level as the center of the circle? Express your answer in terms of the variables , , and appropriate constants. ANSWER: = Correct Part C MasteringPhysics: Assignment Print View http://session.masteringphysics.com/myct/assignmentPrint?ass... 20 of 21 4/17/10 5:30 PM Suppose that the track had no friction and the shell was released from the same height you found in part (a). Would it make a complete loop-the-loop? ANSWER: yes no Correct Part D In part (c), how hard does the track push on the shell at point , the top of the circle? Express your answer in terms of the variables , , and appropriate constants. ANSWER: = All attempts used; correct answer displayed Part E How hard did the track push on the shell at point in part (a)? Express your answer in terms of the variables , , and appropriate constants. ANSWER: = 0.00×10 0 Correct Score Summary: Your score on this assignment is 96.9%. You received 203.4 out of a possible total of 210 points. MasteringPhysics: Assignment Print View http://session.masteringphysics.com/myct/assignmentPrint?ass... 21 of 21 4/17/10 5:30 PM MasteringPhysics: Assignment Print View

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