Document Sample

MasteringPhysics: Assignment Print View Page 1 of 13 Assignment Display Mode: View Printable Answers IUPhysicsP201F2009 Assignment09a Due at 11:00pm on Tuesday, October 28, 2008 View Grading Details Torques on a Seesaw: A Tutorial Description: Balancing seesaw: intuition, how it relates to torque, and finally balancing using horizontal force. (version for algebra-based courses) Learning Goal: To make the connection between intuitive understanding of a seesaw and the standard formalism for torque. This problem deals with the concept of torque, the "twist" that an off-center force applies to a body that tends to make it rotate. Use your intuition to try to answer the following question. If your intuition fails, work the rest of the problem and return here when you feel that you are more comfortable with torques. Part A Marcel is helping his two children, Jacques and Gilles, to balance on a seesaw so that they will be able to make it tilt back and forth without the heavier child, Jacques, simply sinking to the ground. Given that Jacques, whose weight is , is sitting at distance to the left of the pivot, at what distance should Marcel place Gilles, whose weight is , to the right of the pivot to balance the seesaw? Hint A.1 How to approach the problem Consider whether increases or decreases as each of the variables that it depends on, , , and , is made http://session.masteringphysics.com/myct/assignmentPrint?assignmentID=1142221 9/17/2008 MasteringPhysics: Assignment Print View Page 2 of 13 larger or smaller. Express your answer in terms of , , and . ANSWER: = Now consider this problem as a more formal introduction to torque. The torque of each child about the pivot point is the product of the child's weight and the distance of the child (strictly speaking, the child's center of mass) from the pivot. The sign of the torque is taken to be positive by convention if it would cause a counterclockwise rotation of the seesaw. The distance is measured perpendicular to the line of force and is called the moment arm. Part B Find the torque about the pivot due to the weight of Gilles on the seesaw. Express your answer in terms of and . ANSWER: = Marcel wants the seesaw to balance, which means that there can be no angular acceleration about the pivot. For the angular acceleration to be zero, the sum of the torques about the pivot must equal zero: . Part C Determine , the sum of the torques on the seesaw. Consider only the torques exerted by the children. Hint C.1 Torque from the weight of the seesaw The seesaw is symmetric about the pivot, and so the gravitational force on the seesaw produces no net torque. More generally, when determining torques, the gravitational force on an object in a uniform gravitational field can be taken to act at the object's center of mass. Here, the center of mass is directly above the pivot, so the weight of the seesaw has zero moment arm and produces no torque about the pivot. Express your answer in terms of , , , and . ANSWER: If you did not solve for the distance required to balance the seesaw in Part A, do so now. The equation applies to any body that is not rotationally accelerating. Combining this equation with (which applies to any body that is not accelerating linearly) gives a pair of equations that are sufficient to form the basis of statics. The art of applying these equations to large or complicated structures http://session.masteringphysics.com/myct/assignmentPrint?assignmentID=1142221 9/17/2008 MasteringPhysics: Assignment Print View Page 3 of 13 constitutes a significant part of mechanical and civil engineering. Gilles has an identical twin, Jean, also of weight . The two twins now sit on the same side of the seesaw, with Gilles at distance from the pivot and Jean at distance . Part D Where should Marcel position Jacques to balance the seesaw? Part D.1 Find the sum of the torques when the seesaw is balanced For the seesaw to balance, the sum of the torques must be zero. What is the sum of the torques due to the three children? Express your answer in terms of , , , , and . ANSWER: Now solve your torque equation for . Express your answer in terms of , , , and . ANSWER: = Bad news! When Marcel finds the distance from the previous part, it turns out to be greater than , the distance from the pivot to the end of the seesaw. Hence, even with Jacques at the very end of the seesaw, the twins Gilles and Jean exert more torque than Jacques does. Marcel now elects to balance the seesaw by pushing sideways on an ornament (shown in red) that is at height above the pivot. http://session.masteringphysics.com/myct/assignmentPrint?assignmentID=1142221 9/17/2008 MasteringPhysics: Assignment Print View Page 4 of 13 Part E With what force in the rightward direction, , should Marcel push? If you think the force should be toward the left, your answer should be negative. Hint E.1 Sign conventions It is easy to make sign errors in torque problems, and experience shows that it is better to use standard conventions (here, that goes to the right, the direction of positive x) than to change the direction of positive torque or displacement to suit your convenience. (You are likely to forget your unconventional choice at a later point in the problem.) In this case, your intuition correctly expects that Marcel must push to the left to make things balance; the equations will confirm this by giving a negative result for . (A positive result would mean that the force would be directed to the right in the figure.) Part E.2 Find the torque due to Marcel's push The sum of all four torques (due to each of the three children, plus Marcel's) must be zero. What is the torque due to Marcel's push? Keep in mind that a positive torque will cause counterclockwise rotation of the seesaw. Express your answer in terms of the unknown force and the height at which it is applied. ANSWER: = Express your answer in terms of , , , , , and . ANSWER: = This answer will necessarily be negative, because you were told that to balance the seesaw with the twins on the right, Jacques had to be "beyond the end" of the seesaw. Therefore, when indicates Jacque's position, will be less than zero. Hence Marcel must push to the left, as you would expect. Young's Modulus Description: Contains several questions, conceptual and quantitative, leading to better understanding of Young's modulus. (version for algebra-based courses) Learning Goal: To understand the meaning of Young's modulus and to perform some real-life calculations related to the stretching of steel. Hooke's law states that for springs and other "elastic" objects http://session.masteringphysics.com/myct/assignmentPrint?assignmentID=1142221 9/17/2008 MasteringPhysics: Assignment Print View Page 5 of 13 , where is the magnitude of the stretching force, is the corresponding elongation of the spring from equilibrium, and is a constant that depends on the geometry and the material of the spring. If the deformations are small enough, most materials, in fact, behave like springs: Their deformation is directly proportional to the external force. Therefore, it may be useful to operate with an expression that is similar to Hooke's law but describes the properties of various materials, as opposed to objects such as springs. Consider, for instance, a bar of initial length and cross-sectional area stressed by a force of magnitude . As a result, the bar stretches by . Let us define two new terms: Tensile stress is the ratio of the stretching force to the cross-sectional area: . Tensile strain is the ratio of the elongation of the rod to the initial length of the bar: . It turns out that the ratio of the tensile stress to the tensile strain is a constant as long as the tensile stress is not too large. That constant, which is an inherent property of a material, is called Young's modulus and is given by Part A What is the SI unit of Young's modulus? Hint A.1 Look at the dimensions If you look at the dimensions of Young's modulus, you will see that they are equivalent to the dimension of pressure. Use the SI unit of pressure. ANSWER: http://session.masteringphysics.com/myct/assignmentPrint?assignmentID=1142221 9/17/2008 MasteringPhysics: Assignment Print View Page 6 of 13 Part B Consider a metal bar of initial length and cross-sectional area . The Young's modulus of the material of the bar is . Find the "spring constant" of such a bar for low values of tensile strain. Hint B.1 How to approach the problem Consider the equation defining . Then isolate and compare the result with Hooke's law, . Express your answer in terms of , , and . ANSWER: = Part C Ten identical steel wires have equal lengths and equal "spring constants" . The wires are connected end to end, so that the resultant wire has length . What is the "spring constant" of the resulting wire? Hint C.1 How to determine the new spring constant Use the expression for the spring constant determined in Part B. From the expression derived in Part B, you can determine what happens to the spring constant when the length of the spring increases. ANSWER: Part D Ten identical steel wires have equal lengths and equal "spring constants" . The wires are slightly twisted together, so that the resultant wire has length and is 10 times as thick as each individual wire. What is the "spring constant" of the resulting wire? Hint D.1 How to determine the new spring constant Use the expression for the spring constant determined in Part B. From the expression derived in Part B, you can determine what happens to the spring constant when the area of the spring increases. ANSWER: http://session.masteringphysics.com/myct/assignmentPrint?assignmentID=1142221 9/17/2008 MasteringPhysics: Assignment Print View Page 7 of 13 Part E Consider a steel guitar string of initial length and cross-sectional area . The Young's modulus of the steel is . How far ( ) would such a string stretch under a tension of 1500 ? Express your answer in millimeters using two significant figures. ANSWER: = A Bar Suspended by Two Wires Description: A nonuniform bar is suspended by two strings at different angles. Use balance of forces and torques to find the center of mass of the bar. (version for algebra-based courses) A nonuniform horizontal bar of mass is supported by two massless wires against gravity. The left wire makes an angle with the horizontal, and the right wire makes an angle . The bar has length . Part A What is the position of the center of mass of the bar, measured as distance from the bar's left end? Hint A.1 The nature of the problem The bar is at rest. Therefore, both the net force and the net torque acting on the bar have zero magnitude. Part A.2 Find the sum of the x components of the forces Assume that the tensions in the left and right wires are and , respectively. Which expression represents accurately the sum of the x components of the forces? Because the bar is at rest, these forces should sum to zero. http://session.masteringphysics.com/myct/assignmentPrint?assignmentID=1142221 9/17/2008 MasteringPhysics: Assignment Print View Page 8 of 13 Use the sign convention indicated in the figure. ANSWER: = Part A.3 Find the sum of the y components of the forces Assuming that the tensions in the left and right wires are and , respectively, which expression accurately represents the sum of the y components of the forces? Because the bar is at rest, these components should sum to zero. Use the sign convention indicated in the figure. ANSWER: = Part A.4 Find the sum of the torques about the left end of the bar The net torque is zero about any axis you select. Here we ask you to find the net torque of the system about the left end of the bar (you may, of course, try any other point if you like and check that it gives the same final answer for the center of mass ). Assume that the tensions in the left and right wires are and , respectively. The weight of the bar is . Using the sign convention shown in the figure, which expression accurately gives the sum of the torques about the left end of the bar. ANSWER: = Part A.5 Eliminate weight from your equations You should have three equations by now. It is possible to eliminate two variables and solve for in terms of the others. As an intermediate step, solve your torque equation for in terms of , , , etc. Then solve your y- component force equation for and substitute back into your expression for . In other words, find an expression for using the torque and equations. http://session.masteringphysics.com/myct/assignmentPrint?assignmentID=1142221 9/17/2008 MasteringPhysics: Assignment Print View Page 9 of 13 Express your answer in terms of , , , , and . ANSWER: = You now have an expression for that contains only and . Now you must eliminate and from your expression. Hint: Use what you know about the x components of the forces acting on the bar. Express your answer in terms of , , and . ANSWER: = An Amusement Park Airplane Ride Description: Determine the strain on a rod that is attached to an amusement park airplane of known weight. When the system rotates about the center, students must determine if the temsion in the rod increases or decreases. (version for algebra-based courses) An amusement park ride consists of airplane-shaped cars suspended by steel rods. Each rod has length 15 and cross-sectional area 8.00 . Take Young's modulus for steel to be . Part A How much is the rod stretched (change in length of ) when the ride is at rest? Assume that each airplane with two riders has a total weight of 1900 and that the rods are vertical when the ride is at rest. Neglect the stretch due to the weight of the rod itself. http://session.masteringphysics.com/myct/assignmentPrint?assignmentID=1142221 9/17/2008 MasteringPhysics: Assignment Print View Page 10 of 13 Part A.1 Calculate the tensile stress Calculate the tensile stress on the rod from the car hanging below it. Hint A.1.a Definition of tensile stress Tensile stress on a rod is given by the equation where is the force directed along the length of the rod and is the cross-sectional area of the rod. Express your answer in pascals. ANSWER: = Part A.2 Calculate the tensile strain Calculate the tensile strain given the tensile stress and Young's modulus. Hint A.2.a Definition of tensile strain The tensile strain on the rod is defined as , where is the original (unstressed) length of the rod and is the elongation, that is, the length by which it stretches Hint A.2.b Definition of Young's modulus Young's modulus for a material is given by the equation . Use this and the results for the tensile stress on the rod to determine the strain and thus the extension of the rod. ANSWER: = Now use the definition of tensile strain and solve for the elongation . Express your answer in meters. ANSWER: = http://session.masteringphysics.com/myct/assignmentPrint?assignmentID=1142221 9/17/2008 MasteringPhysics: Assignment Print View Page 11 of 13 Part B When the ride starts operating as shown in the diagram, does the tension in the rods increase or decrease? ANSWER: It increases. It decreases. Note that the tensile force equals the tension force in the rod. When the ride is spinning, the rod is no longer vertical. Applying Newton's 2nd law in the vertical direction, we see that a component of the tension must support the weight of the car. This means that the tension is greater than the car's weight. Also, we have a confession to make: The mass of the rod in this case is comparable with the mass of the airplane so it does add to the total stretch. However, the contribution of the weight of the rod to the total stretch is not that easy to find because the mass is distributed over the length of the rod. In this case, we had to settle for a simplified model of the situation—something that physicists often have to do. Moments around a Rod Description: Find the rotational acceleration about a bent rod that is fixed at a corner and has various forces applied at different points and angles. Problem is numerical with moment of inertia and forces specified. A rod is bent into an L shape and attached at one point to a pivot. The rod sits on a frictionless table and the diagram is a view from above. This means that gravity can be ignored for this problem. There are three forces that are applied to the rod at different points and angles: , , and . Note that the dimensions of the bent rod are in centimeters in the figure, although the answers are requested in SI units (kilograms, meters, seconds). Part A If and , what does the magnitude of have to be for there to be rotational equilibrium? Part A.1 Finding torque about pivot from What is the magnitude of the torque | | provided by around the pivot point? http://session.masteringphysics.com/myct/assignmentPrint?assignmentID=1142221 9/17/2008 MasteringPhysics: Assignment Print View Page 12 of 13 Give your answer numerically in newton-meters to two significant figures. ANSWER: | |= Answer numerically in newtons to two significant figures. ANSWER: = N Part B If the L-shaped rod has a moment of inertia , , , and again , how long a time would it take for the object to move through ( /4 radians)? Assume that as the object starts to move, each force moves with the object so as to retain its initial angle relative to the object. Part B.1 Find the net torque about the pivot What is the magnitude of the total torque | | around the pivot point? Answer numerically in newton-meters to two significant figures. ANSWER: | |= Part B.2 Calculate Given the total torque around the pivot point, what is , the magnitude of the angular acceleration? Hint B.2.a Equation for If you know the magnitude of the total torque ( ) and the rotational inertia ( ), you can then find the rotational acceleration ( ) from Express your answer numerically in radians per second squared to two significant figures. ANSWER: = Hint B.3 Description of angular kinematics Now that you know the angular acceleration, this is a problem in rotational kinematics; find the time needed to go through a given angle . For constant acceleration ( ) and starting with (where is angular speed) the relation is given by http://session.masteringphysics.com/myct/assignmentPrint?assignmentID=1142221 9/17/2008 MasteringPhysics: Assignment Print View Page 13 of 13 which is analogous to the expression for linear displacement ( ) with constant acceleration ( ) starting from rest, . Express the time in seconds to two significant figures. ANSWER: = s Part C Now consider the situation in which and , but now a force with nonzero magnitude is acting on the rod. What does have to be to obtain equilibrium? Part C.1 Find the required component of Only the tangential (perpendicular) component of (call it ) provides a torque. What is ? Answer in terms of . You will need to evaluate any trigonometric functions. ANSWER: = Give a numerical answer, without trigonometric functions, in newtons, to two significant figures. ANSWER: = N Summary 0 of 5 items complete (0% avg. score) 0 of 5 points http://session.masteringphysics.com/myct/assignmentPrint?assignmentID=1142221 9/17/2008

DOCUMENT INFO

Shared By:

Categories:

Tags:
Horizontal

Stats:

views: | 5002 |

posted: | 10/16/2011 |

language: | English |

pages: | 13 |

Description:
Horizontal bar originated in Germany. 18th century Western European countries appear to seize wire acrobatics big loop action, inspired by this 1811 German gymnast Yang in Berlin on the outskirts of Hudson Hyde gymnastics field instead of using a Mugang wire acrobatics show, the first installation the world's first deputy horizontal bar. 1812 will Mugang to iron, later changed to steel, the bar increases flexibility and endurance. 1920s as an independent event. Bar length 2.4 m, diameter 2.8 cm, 2.55 ~ 2.75 m high. Bars at both ends were fixed on the pillar. 1896 is listed as an Olympic sport.

OTHER DOCS BY bestt571

How are you planning on using Docstoc?
BUSINESS
PERSONAL

By registering with docstoc.com you agree to our
privacy policy and
terms of service, and to receive content and offer notifications.

Docstoc is the premier online destination to start and grow small businesses. It hosts the best quality and widest selection of professional documents (over 20 million) and resources including expert videos, articles and productivity tools to make every small business better.

Search or Browse for any specific document or resource you need for your business. Or explore our curated resources for Starting a Business, Growing a Business or for Professional Development.

Feel free to Contact Us with any questions you might have.