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MasteringPhysics: Assignment Print View Page 1 of 12 Assignment Display Mode: View Printable Answers IUPhysicsP201F2009 Assignment 4a Due at 11:00pm on Tuesday, September 23, 2008 View Grading Details A Mass on a Turntable: Conceptual Description: An object is sitting on a rotating turntable. The student is asked for the direction of velocity, acceleration, and net force. Based on Mechanics Baseline Test. A small metal cylinder rests on a circular turntable that is rotating at a constant rate, as illustrated in the diagram. Part A Which of the following sets of vectors best describes the velocity, acceleration, and net force acting on the cylinder at the point indicated in the diagram? http://session.masteringphysics.com/myct/assignmentPrint?assignmentID=1144566 9/17/2008 MasteringPhysics: Assignment Print View Page 2 of 12 Hint A.1 The direction of acceleration can be determined from Newton's second law According to Newton's second law, the acceleration of an object has the same direction as the net force acting on that object. ANSWER: a b c d e Part B Let be the distance between the cylinder and the center of the turntable. Now assume that the cylinder is moved to a new location from the center of the turntable. Which of the following statements accurately describe the motion of the cylinder at the new location? Part B.1 Find the speed of the cylinder Find the speed of the cylinder at the new location. Assume that the cylinder makes one complete turn in a period of time . Express your answer in terms of and . ANSWER: = Now compare your result with the speed of the cylinder before it is moved. Part B.2 Find the acceleration of the cylinder Find the magnitude of the acceleration of the cylinder at the new location. Assume that the cylinder makes one complete turn in a period of time . Hint B.2.a Centripetal acceleration Recall that the acceleration of an object that moves in a circular path of radius with constant speed has magnitude given by . Note that both the velocity and radius of the trajectory change when the cylinder is moved. Express your answer in terms of and . ANSWER: http://session.masteringphysics.com/myct/assignmentPrint?assignmentID=1144566 9/17/2008 MasteringPhysics: Assignment Print View Page 3 of 12 = Now compare your result with the acceleration of the cylinder before it is moved. Check all that apply. ANSWER: The speed of the cylinder has decreased. The speed of the cylinder has increased. The magnitude of the acceleration of the cylinder has decreased. The magnitude of the acceleration of the cylinder has increased. The speed and the acceleration of the cylinder have not changed. Accelerating along a Racetrack Description: Determine the direction of acceleration of cars at various points along a race track A road race is taking place along the track shown in the figure . All of the cars are moving at constant speeds. The car at point F is traveling along a straight section of the track, whereas all the other cars are moving along curved segments of the track. Part A Let be the velocity of the car at point A. What can you say about the acceleration of the car at that point? Hint A.1 Acceleration along a curved path Since acceleration is a vector quantity, an object moving at constant speed along a curved path has nonzero acceleration because the direction of its velocity is changing, even though the magnitude of its velocity (the speed) is constant. Moreover, if the speed is constant, the object's acceleration is always perpendicular to the velocity vector at each point along the curved path and is directed toward the center of curvature of the path. ANSWER: The acceleration is parallel to . http://session.masteringphysics.com/myct/assignmentPrint?assignmentID=1144566 9/17/2008 MasteringPhysics: Assignment Print View Page 4 of 12 The acceleration is perpendicular to and directed toward the inside of the track. The acceleration is perpendicular to and directed toward the outside of the track. The acceleration is neither parallel nor perpendicular to . The acceleration is zero. Part B Let be the velocity of the car at point C. What can you say about the acceleration of the car at that point? Hint B.1 Acceleration along a curved path Since acceleration is a vector quantity, an object moving at constant speed along a curved path has nonzero acceleration because the direction of its velocity is changing, even though the magnitude of its velocity (the speed) is constant. Moreover, if the speed is constant, the object's acceleration is always perpendicular to the velocity vector at each point along the curved path and is directed toward the center of curvature of the path. ANSWER: The acceleration is parallel to . The acceleration is perpendicular to and pointed toward the inside of the track. The acceleration is perpendicular to and pointed toward the outside of the track. The acceleration is neither parallel nor perpendicular to . The acceleration is zero. Part C Let be the velocity of the car at point D. What can you say about the acceleration of the car at that point? Hint C.1 Acceleration along a curved path Since acceleration is a vector quantity, an object moving at constant speed along a curved path has nonzero acceleration because the direction of its velocity is changing, even though the magnitude of its velocity (the speed) is constant. Moreover, if the speed is constant, the object's acceleration is always perpendicular to the velocity vector at each point along the curved path and is directed toward the center of curvature of the path. ANSWER: The acceleration is parallel to . The acceleration is perpendicular to and pointed toward the inside of the track. The acceleration is perpendicular to and pointed toward the outside of the track. The acceleration is neither parallel nor perpendicular to . The acceleration is zero. Part D Let be the velocity of the car at point F. What can you say about the acceleration of the car at that point? Hint D.1 Acceleration along a straight path http://session.masteringphysics.com/myct/assignmentPrint?assignmentID=1144566 9/17/2008 MasteringPhysics: Assignment Print View Page 5 of 12 The velocity of an object that moves along a straight path is always parallel to the direction of the path, and the object has a nonzero acceleration only if the magnitude of its velocity changes in time. ANSWER: The acceleration is parallel to . The acceleration is perpendicular to and pointed toward the inside of the track. The acceleration is perpendicular to and pointed toward the outside of the track. The acceleration is neither parallel nor perpendicular to . The acceleration is zero. Part E Assuming that all cars have equal speeds, which car has the acceleration of the greatest magnitude, and which one has the acceleration of the least magnitude? Hint E.1 How to approach the problem Recall that the magnitude of the acceleration of an object that moves at constant speed along a curved path is inversely proportional to the radius of curvature of the path. Use A for the car at point A, B for the car at point B, and so on. Express your answer as the name the car that has the greatest magnitude of acceleration followed by the car with the least magnitude of accelation, and separate your answers with a comma. ANSWER: D F Part F Assume that the car at point A and the one at point E are traveling along circular paths that have the same radius. If the car at point A now moves twice as fast as the car at point E, how is the magnitude of its acceleration related to that of car E. Part F.1 Find the acceleration of the car at point E Let be the radius of the two curves along which the cars at points A and E are traveling. What is the magnitude of the acceleration of the car at point E? Hint F.1.a Uniform circular motion The magnitude of the acceleration of an object that moves with constant speed along a circular path of radius is given by . Express your answer in terms of the radius of curvature and the speed of car E. ANSWER: = Part F.2 Find the acceleration of the car at point A http://session.masteringphysics.com/myct/assignmentPrint?assignmentID=1144566 9/17/2008 MasteringPhysics: Assignment Print View Page 6 of 12 If , what is the acceleration of the car at point A? Let be the radius of the two curves along which the cars at points A and E are traveling. Hint F.2.a Uniform circular motion The magnitude of the acceleration of an object that moves with constant speed along a circular path of radius is given by . Express your answer in terms of the speed of the car at E and the radius . ANSWER: = Since the magnitude of the acceleration of an object that moves with constant speed along a circular path is proportional to the square of the speed, the acceleration of the car at point A is proportional to . ANSWER: The magnitude of the acceleration of the car at point A is twice that of the car at point E. The magnitude of the acceleration of the car at point A is the same as that of the car at point E. The magnitude of the acceleration of the car at point A is half that of the car at point E. The magnitude of the acceleration of the car at point A is four times that of the car at point E. Velocity of a Roller Coaster Ranking Task Description: Conceptual question on the velocity of different roller-coaster carts going over a semi-circular hill. (ranking task) Six roller-coaster carts pass over the same semicircular "bump." The mass of each cart (including passenger) and the normal force of the track on the cart at the top of each bump are given in the figures. http://session.masteringphysics.com/myct/assignmentPrint?assignmentID=1144566 9/17/2008 MasteringPhysics: Assignment Print View Page 7 of 12 Part A Rank the speeds of the different carts as each passes over the top of the bump. Hint A.1 Newton's 2nd law Newton's 2nd law is valid for any motion, whether the motion is along a straight line or a circular path. The net force acting on the cart must be equal to the product of its mass and acceleration, , and since the cart is moving along a circular path, its acceleration (toward the center of the circular path) has magnitude . Part A.2 Determine the net force on the cart Defining the down direction (toward the center of the circular path) as positive, which of the following expressions describes the component of the net force acting on the cart in the vertical direction? ANSWER: Taking the expressions for the net force on the cart and the centripetal acceleration of the cart and substituting into Newton's 2nd law, , results in . Since the radius is the same for every cart, we can ignore and write . But is simply so , http://session.masteringphysics.com/myct/assignmentPrint?assignmentID=1144566 9/17/2008 MasteringPhysics: Assignment Print View Page 8 of 12 or . Rank from largest to smallest. To rank items as equivalent, overlap them. ANSWER: View Taking the expressions for the net force on the cart and the centripetal acceleration of the cart and substituting into Newton's 2nd law, , results in . Since the radius is the same for every cart, we can ignore and write . But is simply so , or . Therefore, larger implies smaller . http://session.masteringphysics.com/myct/assignmentPrint?assignmentID=1144566 9/17/2008 MasteringPhysics: Assignment Print View Page 9 of 12 Orbital Speed of a Satellite Description: Short conceptual problem comparing the orbital motion (speed, radius, and period) of two different satellites. This problem is based on Young/Geller Conceptual Analysis 6.6 Part A Two identical satellites orbit the earth in stable orbits. One satellite orbits with a speed at a distance from the center of the earth. The second satellite travels at a speed that is less than . At what distance from the center of the earth does the second satellite orbit? Hint A.1 How to approach the problem Satellites orbiting any large object, like the earth or the moon, are in free fall. The only force that acts on a satellite that maintains a stable orbit is the gravitational attraction of the object on the satellite. This force is given by Newton's law of gravitation , where is the gravitational constant, is the mass of the large object (e.g., in this case the earth), is the mass of the satellite, and is the distance from the center of the object to the satellite. According to Newton's 2nd law the force on the satellite is also given by , where for objects undergoing uniform circular motion with speed . Use this information to find a relationship between a satellite's speed and its orbital distance. Hint A.2 Working with the equations Because the gravitational attraction of the earth on the satellite is the only force acting on the satellite, you know that . You also know that the satellite undergoes uniform circular motion, meaning that . Combining these expressions yields . http://session.masteringphysics.com/myct/assignmentPrint?assignmentID=1144566 9/17/2008 MasteringPhysics: Assignment Print View Page 10 of 12 Simplify this expression by getting all the distance terms on one side of the equation. ANSWER: At a distance that is less than . At a distance equal to . At a distance greater than . The speed of a satellite of mass orbiting a distance from the center of a larger object of mass is given by the relationship , where is the gravitational constant. We can’t choose the radius of the orbit and the speed of the satellite independently; choosing a value of automatically determines . Also note that the satellite’s motion does not depend on its mass . Part B Now assume that a satellite of mass is orbiting the earth at a distance from the center of the earth with speed . An identical satellite is orbiting the moon at the same distance with a speed . How does the time it takes the satellite circling the moon to make one revolution compare to the time it takes the satellite orbiting the earth to make one revolution? Hint B.1 How to approach this problem To compare the time it takes the satellite circling the moon to make one revolution to the time it takes the satellite orbiting the earth to make one revolution you should first compare the speeds and of the satellite. Then you can relate the speeds to the time it takes the satellites to make one revolution. Part B.2 Compare the speeds of the satellites How does the speed of the satellite orbiting the moon, , compare to the speed of the satellite orbiting the earth, ? ANSWER: is greater than . is equal to . is less than . Hint B.3 Relationship between speed and the time for one revolution The time it takes a satellite to make one revolution is called the period . The period depends on the speed of the satellite and its radius and is given by . http://session.masteringphysics.com/myct/assignmentPrint?assignmentID=1144566 9/17/2008 MasteringPhysics: Assignment Print View Page 11 of 12 ANSWER: is less than . is equal to . is greater than . The period of a satellite orbiting an object of mass at a distance from its center is given by . This relationship is also known as Kepler's 3rd law. Weight on a Neutron Star Description: Given your weight on earth, calculate your weight on the surface of a neutron star. Neutron stars, such as the one at the center of the Crab Nebula, have about the same mass as our sun but a much smaller diameter. Part A If you weigh 650 on the earth, what would be your weight on the surface of a neutron star that has the same mass as our sun and a diameter of 25.0 ? Take the mass of the sun to be = 1.99×1030 , the gravitational constant to be = 6.67×10−11 , and the acceleration due to gravity at the earth's surface to be = 9.810 . Hint A.1 How to approach the problem "Weight" is the same as "the force of gravitational attraction." Use Newton's law of gravitation to calculate the gravitational force that would be exerted on you if you were standing on the surface of the star. Be careful when determining the values needed for the equation. Hint A.2 Law of universal gravitation The gravitational force exerted on a mass by a second mass is , where is the distance between the two masses, and 6.67×10−11 is the universal gravitational constant. Part A.3 Calculate your mass Calculate your mass if you weigh 650 on earth. http://session.masteringphysics.com/myct/assignmentPrint?assignmentID=1144566 9/17/2008 MasteringPhysics: Assignment Print View Page 12 of 12 Express your answer in kilograms. ANSWER: = Part A.4 Calculate the distance between you and the star Calculate your distance from the center of the star if you are standing on its surface. Express your answer in meters. ANSWER: = Express your weight in newtons. ANSWER: = This is over times your weight on earth! You probably shouldn't venture there.... Summary 0 of 5 items complete (0% avg. score) 0 of 5 points http://session.masteringphysics.com/myct/assignmentPrint?assignmentID=1144566 9/17/2008

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