## Assignment 10 Due: 11:59pm on Wednesday, April 23, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Conceptual Question 12.3 Part A The figure shows three rotating disks, all of equal mass. Rank in order, from largest to smallest, their rotational kinetic energies to . Rank from largest to smallest. To rank items as equivalent, overlap them. ANSWER: Ka Kc Correct Conceptual Question 12.6 You have two steel solid spheres. Sphere 2 has twice the radius of sphere 1. Part A By what factor does the moment of inertia of sphere 2 exceed the moment of inertia of sphere 1? ANSWER: I2 I1 Correct Problem 12.2 A high-speed drill reaches 2500 in 0.59 . Part A What is the drill’s angular acceleration? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B Through how many revolutions does it turn during this first 0.59 ? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct I2/I1 = 32 rpm s = 440 rad s2 s = 12 rev Constant Angular Acceleration in the Kitchen Dario, a prep cook at an Italian restaurant, spins a salad spinner and observes that it rotates 20.0 times in 5.00 seconds and then stops spinning it. The salad spinner rotates 6.00 more times before it comes to rest. Assume that the spinner slows down with constant angular acceleration. Part A What is the angular acceleration of the salad spinner as it slows down? Express your answer numerically in degrees per second per second. Hint 1. How to approach the problem Recall from your study of kinematics the three equations of motion derived for systems undergoing constant linear acceleration. You are now studying systems undergoing constant angular acceleration and will need to work with the three analogous equations of motion. Collect your known quantities and then determine which of the angular kinematic equations is appropriate to find the angular acceleration . Hint 2. Find the angular velocity of the salad spinner while Dario is spinning it What is the angular velocity of the salad spinner as Dario is spinning it? Express your answer numerically in degrees per second. Hint 1. Converting rotations to degrees When the salad spinner spins through one revolution, it turns through 360 degrees. ANSWER: Hint 3. Find the angular distance the salad spinner travels as it comes to rest Through how many degrees does the salad spinner rotate as it comes to rest? Express your answer numerically in degrees. Hint 1. Converting rotations to degrees 0 = 1440 degrees/s = − 0 One revolution is equivalent to 360 degrees. ANSWER: Hint 4. Determine which equation to use You know the initial and final velocities of the system and the angular distance through which the spinner rotates as it comes to a stop. Which equation should be used to solve for the unknown constant angular acceleration ? ANSWER: ANSWER: Correct Part B How long does it take for the salad spinner to come to rest? Express your answer numerically in seconds. = 2160 degrees = 0 + 0t+ 1 2 t2 = 0 + t = + 2( − ) 2 20 0 = -480 degrees/s2 Hint 1. How to approach the problem Again, you will need the equations of rotational kinematics that apply to situations of constant angular acceleration. Collect your known quantities and then determine which of the angular kinematic equations is appropriate to find . Hint 2. Determine which equation to use You have the initial and final velocities of the system and the angular acceleration, which you found in the previous part. Which is the best equation to use to solve for the unknown time ? ANSWER: ANSWER: Correct ± A Spinning Electric Fan An electric fan is turned off, and its angular velocity decreases uniformly from 540 to 250 in a time interval of length 4.40 . Part A Find the angular acceleration in revolutions per second per second. Hint 1. Average acceleration Recall that if the angular velocity decreases uniformly, the angular acceleration will remain constant. Therefore, the angular acceleration is just the total change in angular velocity divided by t t = 0 + 0t+ 1 2 t2 = 0 + t = + 2( − ) 2 20 0 t = 3.00 s rev/min rev/min s the total change in time. Be careful of the sign of the angular acceleration. ANSWER: Correct Part B Find the number of revolutions made by the fan blades during the time that they are slowing down in Part A. Hint 1. Determine the correct kinematic equation Which of the following kinematic equations is best suited to this problem? Here and are the initial and final angular velocities, is the elapsed time, is the constant angular acceleration, and and are the initial and final angular displacements. Hint 1. How to chose the right equation Notice that you were given in the problem introduction the initial and final speeds, as well as the length of time between them. In this problem, you are asked to find the number of revolutions (which here is the change in angular displacement, ). If you already found the angular acceleration in Part A, you could use that as well, but you would end up using a more complex equation. Also, in general, it is somewhat favorable to use given quantities instead of quantities that you have calculated. ANSWER: = -1.10 rev/s2 0 t 0 − 0 = 0 + t = 0 + t+ 1 2 t2 = + 2( − ) 2 20 0 − 0 = (+ )t 1 2 0 ANSWER: Correct Part C How many more seconds are required for the fan to come to rest if the angular acceleration remains constant at the value calculated in Part A? Hint 1. Finding the total time for spin down To find the total time for spin down, just calculate when the velocity will equal zero. This is accomplished by setting the initial velocity plus the acceleration multipled by the time equal to zero and then solving for the time. One can then just subtract the time it took to reach 250 from the total time. Be careful of your signs when you set up the equation. ANSWER: Correct Problem 12.8 A 100 ball and a 230 ball are connected by a 34- -long, massless, rigid rod. The balls rotate about their center of mass at 130 . Part A What is the speed of the 100 ball? Express your answer to two significant figures and include the appropriate units. ANSWER: 29.0 rev rev/min 3.79 s g g cm rpm g Correct Problem 12.10 A thin, 60.0 disk with a diameter of 9.00 rotates about an axis through its center with 0.200 of kinetic energy. Part A What is the speed of a point on the rim? Express your answer with the appropriate units. ANSWER: Correct Problem 12.12 A drum major twirls a 95- -long, 470 baton about its center of mass at 150 . Part A What is the baton’s rotational kinetic energy? Express your answer to two significant figures and include the appropriate units. ANSWER: v = 3.2 ms g cm J 3.65 ms cm g rpm K = 4.4 J Correct Net Torque on a Pulley The figure below shows two blocks suspended by a cord over a pulley. The mass of block B is twice the mass of block A, while the mass of the pulley is equal to the mass of block A. The blocks are let free to move and the cord moves on the pulley without slipping or stretching. There is no friction in the pulley axle, and the cord’s weight can be ignored. Part A Which of the following statements correctly describes the system shown in the figure? Check all that apply. Hint 1. Conditions for equilibrium If the blocks had the same mass, the system would be in equilibrium. The blocks would have zero acceleration and the tension in each part of the cord would equal the weight of each block. Both parts of the cord would then pull with equal force on the pulley, resulting in a zero net torque and no rotation of the pulley. Is this still the case in the current situation where block B has twice the mass of block A? Hint 2. Rotational analogue of Newton’s second law The net torque of all the forces acting on a rigid body is proportional to the angular acceleration of the body net and is given by , where is the moment of inertia of the body. Hint 3. Relation between linear and angular acceleration A particle that rotates with angular acceleration has linear acceleration equal to , where is the distance of the particle from the axis of rotation. In the present case, where there is no slipping or stretching of the cord, the cord and the pulley must move together at the same speed. Therefore, if the cord moves with linear acceleration , the pulley must rotate with angular acceleration , where is the radius of the pulley. ANSWER: Correct Part B What happens when block B moves downward? Hint 1. How to approach the problem To determine whether the tensions in both parts of the cord are equal, it is convenient to write a mathematical expression for the net torque on the pulley. This will allow you to relate the tensions in the cord to the pulley’s angular acceleration. Hint 2. Find the net torque on the pulley Let’s assume that the tensions in both parts of the cord are different. Let be the tension in the right cord and the tension in the left cord. If is the radius of the pulley, what is the net torque acting on the pulley? Take the positive sense of rotation to be counterclockwise. Express your answer in terms of , , and . net = I I a a = R R a = a R R The acceleration of the blocks is zero. The net torque on the pulley is zero. The angular acceleration of the pulley is nonzero. T1 T2 R net T1 T2 R Hint 1. Torque The torque of a force with respect to a point is defined as the product of the magnitude times the perpendicular distance between the line of action of and the point . In other words, . ANSWER: ANSWER: Correct Note that if the pulley were stationary (as in many systems where only linear motion is studied), then the tensions in both parts of the cord would be equal. However, if the pulley rotates with a certain angular acceleration, as in the present situation, the tensions must be different. If they were equal, the pulley could not have an angular acceleration. Problem 12.18 Part A In the figure , what is the magnitude of net torque about the axle? Express your answer to two significant figures and include the appropriate units. F O F l F O = Fl net = R(T2 − T1 ) The left cord pulls on the pulley with greater force than the right cord. The left and right cord pull with equal force on the pulley. The right cord pulls on the pulley with greater force than the left cord. ANSWER: Correct Part B What is the direction of net torque about the axle? ANSWER: Correct Problem 12.22 An athlete at the gym holds a 3.5 steel ball in his hand. His arm is 78 long and has a mass of 3.6 . Assume the center of mass of the arm is at the geometrical center of the arm. Part A What is the magnitude of the torque about his shoulder if he holds his arm straight out to his side, parallel to the floor? Express your answer to two significant figures and include the appropriate units. = 0.20 Nm Clockwise Counterclockwise kg cm kg ANSWER: Correct Part B What is the magnitude of the torque about his shoulder if he holds his arm straight, but below horizontal? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Parallel Axis Theorem The parallel axis theorem relates , the moment of inertia of an object about an axis passing through its center of mass, to , the moment of inertia of the same object about a parallel axis passing through point p. The mathematical statement of the theorem is , where is the perpendicular distance from the center of mass to the axis that passes through point p, and is the mass of the object. Part A Suppose a uniform slender rod has length and mass . The moment of inertia of the rod about about an axis that is perpendicular to the rod and that passes through its center of mass is given by . Find , the moment of inertia of the rod with respect to a parallel axis through one end of the rod. Express in terms of and . Use fractions rather than decimal numbers in your answer. Hint 1. Find the distance from the axis to the center of mass Find the distance appropriate to this problem. That is, find the perpendicular distance from the center of mass of the rod to the axis passing through one end of the rod. = 41 Nm 45 = 29 Nm Icm Ip Ip = Icm + Md2 d M L m Icm = m 1 12 L2 Iend Iend m L d ANSWER: ANSWER: Correct Part B Now consider a cube of mass with edges of length . The moment of inertia of the cube about an axis through its center of mass and perpendicular to one of its faces is given by . Find , the moment of inertia about an axis p through one of the edges of the cube Express in terms of and . Use fractions rather than decimal numbers in your answer. Hint 1. Find the distance from the axis to the axis Find the perpendicular distance from the center of mass axis to the new edge axis (axis labeled p in the figure). ANSWER: d = L 2 Iend = mL2 3 m a Icm Icm = m 1 6 a2 Iedge Iedge m a o p d ANSWER: Correct Problem 12.26 Starting from rest, a 12- -diameter compact disk takes 2.9 to reach its operating angular velocity of 2000 . Assume that the angular acceleration is constant. The disk’s moment of inertia is . Part A How much torque is applied to the disk? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B How many revolutions does it make before reaching full speed? Express your answer using two significant figures. ANSWER: d = a 2 Iedge = 2ma2 3 cm s rpm 2.5 × 10−5 kg m2 = 1.8×10−3 Nm Correct Problem 12.23 An object’s moment of inertia is 2.20 . Its angular velocity is increasing at the rate of 3.70 . Part A What is the total torque on the object? ANSWER: Correct Problem 12.31 A 5.1 cat and a 2.5 bowl of tuna fish are at opposite ends of the 4.0- -long seesaw. N = 48 rev kgm2 rad/s2 8.14 N m kg kg m Part A How far to the left of the pivot must a 3.8 cat stand to keep the seesaw balanced? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Static Equilibrium of the Arm You are able to hold out your arm in an outstretched horizontal position because of the action of the deltoid muscle. Assume the humerus bone has a mass , length and its center of mass is a distance from the scapula. (For this problem ignore the rest of the arm.) The deltoid muscle attaches to the humerus a distance from the scapula. The deltoid muscle makes an angle of with the horizontal, as shown. Use throughout the problem. Part A kg d = 1.4 m M1 = 3.6 kg L = 0.66 m L1 = 0.33 m L2 = 0.15 m = 17 g = 9.8 m/s2 Find the tension in the deltoid muscle. Express the tension in newtons, to the nearest integer. Hint 1. Nature of the problem Remember that this is a statics problem, so all forces and torques are balanced (their sums equal zero). Hint 2. Origin of torque Calculate the torque about the point at which the arm attaches to the rest of the body. This allows one to balance the torques without having to worry about the undefined forces at this point. Hint 3. Adding up the torques Add up the torques about the point in which the humerus attaches to the body. Answer in terms of , , , , , and . Remember that counterclockwise torque is positive. ANSWER: ANSWER: Correct Part B Using the conditions for static equilibrium, find the magnitude of the vertical component of the force exerted by the scapula on the humerus (where the humerus attaches to the rest of the body). Express your answer in newtons, to the nearest integer. T L1 L2 M1 g T total = 0 = L1M1g − Tsin()L2 T = 265 N Fy Hint 1. Total forces involved Recall that there are three vertical forces in this problem: the force of gravity acting on the bone, the force from the vertical component of the muscle tension, and the force exerted by the scapula on the humerus (where it attaches to the rest of the body). ANSWER: Correct Part C Now find the magnitude of the horizontal component of the force exerted by the scapula on the humerus. Express your answer in newtons, to the nearest integer. ANSWER: Correct ± Moments around a Rod 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). |Fy| = 42 N Fx |Fx| = 254 N F 1 F 2 F 3 Part A If and , what does the magnitude of have to be for there to be rotational equilibrium? Answer numerically in newtons to two significant figures. Hint 1. Finding torque about pivot from What is the magnitude of the torque | | provided by around the pivot point? Give your answer numerically in newton-meters to two significant figures. ANSWER: ANSWER: Correct 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. Express the time in seconds to two significant figures. F3 = 0 F1 = 12 N F 2 F 1 1 F 1 | 1 | = 0.36 N m F2 = 4.5 N I = 9 kg m2 F1 = 12 N F2 = 27 N F3 = 0 t 45 Hint 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: Hint 2. Calculate Given the total torque around the pivot point, what is , the magnitude of the angular acceleration? Express your answer numerically in radians per second squared to two significant figures. Hint 1. Equation for If you know the magnitude of the total torque ( ) and the rotational inertia ( ), you can then find the rotational acceleration ( ) from ANSWER: Hint 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 which is analogous to the expression for linear displacement ( ) with constant acceleration ( ) starting from rest, | p ivot| | p ivot| = 1.8 N m vot Ivot pivot = Ipivot. = 0.20 radians/s2 = 0 = 1 , 2 t2 x a . ANSWER: Correct 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? Give a numerical answer, without trigonometric functions, in newtons, to two significant figures. Hint 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: ANSWER: Correct x = 1 a 2 t2 t = 2.8 s F1 = 12 N F2 = 0 F3 F3 F 3 F 3 F3t F3t F3 F3t = 1 2 F3 F3 = 9.0 N Problem 12.32 A car tire is 55.0 in diameter. The car is traveling at a speed of 24.0 . Part A What is the tire’s rotation frequency, in rpm? Express your answer to three significant figures and include the appropriate units. ANSWER: Correct Part B What is the speed of a point at the top edge of the tire? Express your answer to three significant figures and include the appropriate units. ANSWER: Correct Part C What is the speed of a point at the bottom edge of the tire? Express your answer as an integer and include the appropriate units. ANSWER: cm m/s 833 rpm 48.0 ms 0 ms Correct Problem 12.33 A 460 , 8.00-cm-diameter solid cylinder rolls across the floor at 1.30 . Part A What is the can’s kinetic energy? Express your answer with the appropriate units. ANSWER: Correct Problem 12.45 Part A What is the magnitude of the angular momentum of the 780 rotating bar in the figure ? g m/s 0.583 J g ANSWER: Correct Part B What is the direction of the angular momentum of the bar ? ANSWER: Correct Problem 12.46 Part A What is the magnitude of the angular momentum of the 2.20 , 4.60-cm-diameter rotating disk in the figure ? 3.27 kgm2/s into the page out of the page kg ANSWER: Correct Part B What is its direction? ANSWER: Correct Problem 12.60 A 3.0- -long ladder, as shown in the following figure, leans against a frictionless wall. The coefficient of static friction between the ladder and the floor is 0.46. 3.66×10−2 kgm /s 2 x direction -x direction y direction -y direction z direction -z direction m Part A What is the minimum angle the ladder can make with the floor without slipping? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 12.61 The 3.0- -long, 90 rigid beam in the following figure is supported at each end. An 70 student stands 2.0 from support 1. = 47 m kg kg m Part A How much upward force does the support 1 exert on the beam? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B How much upward force does the support 2 exert on the beam? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Enhanced EOC: Problem 12.63 A 44 , 5.5- -long beam is supported, but not attached to, the two posts in the figure . A 22 boy starts walking along the beam. You may want to review ( pages 330 – 334) . For help with math skills, you may want to review: F1 = 670 N F2 = 900 N kg m kg The Vector Cross Product Part A How close can he get to the right end of the beam without it falling over? Express your answer to two significant figures and include the appropriate units. Hint 1. How to approach the problem Draw a picture of the four forces acting on the beam, indicating both their direction and the place on the beam that the forces are acting. Choose a coordinate system with a direction for the axis along the beam, and indicate the position of the boy. What is the net force on the beam if it is stationary? Just before the beam tips, the force of the left support on the beam is zero. Using the zero net force condition, what is the force due to the right support just before the beam tips? For the beam to remain stationary, what must be zero besides the net force on the beam? Choose a point on the beam, and compute the net torque on the beam about that point. Be sure to choose a positive direction for the rotation axis and therefore the torques. Using the zero torque condition, what is the position of the boy on the beam just prior to tipping? How far is this position from the right edge of the beam? ANSWER: Correct d = 2.0 m Problem 12.68 Flywheels are large, massive wheels used to store energy. They can be spun up slowly, then the wheel’s energy can be released quickly to accomplish a task that demands high power. An industrial flywheel has a 1.6 diameter and a mass of 270 . Its maximum angular velocity is 1500 . Part A A motor spins up the flywheel with a constant torque of 54 . How long does it take the flywheel to reach top speed? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B How much energy is stored in the flywheel? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part C The flywheel is disconnected from the motor and connected to a machine to which it will deliver energy. Half the energy stored in the flywheel is delivered in 2.2 . What is the average power delivered to the machine? Express your answer to two significant figures and include the appropriate units. ANSWER: m kg rpm N m t = 250 s = 1.1×106 E J s Correct Part D How much torque does the flywheel exert on the machine? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 12.71 The 3.30 , 40.0-cm-diameter disk in the figure is spinning at 350 . Part A How much friction force must the brake apply to the rim to bring the disk to a halt in 2.10 ? P = 2.4×105 W = 1800 Nm kg rpm s Express your answer with the appropriate units. ANSWER: Correct Problem 12.74 A 5.0 , 60- -diameter cylinder rotates on an axle passing through one edge. The axle is parallel to the floor. The cylinder is held with the center of mass at the same height as the axle, then released. Part A What is the magnitude of the cylinder’s initial angular acceleration? Express your answer to two significant figures and include the appropriate units. ANSWER: 5.76 N kg cm = 22 rad s2 Correct Part B What is the magnitude of the cylinder’s angular velocity when it is directly below the axle? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 12.82 A 45 figure skater is spinning on the toes of her skates at 0.90 . Her arms are outstretched as far as they will go. In this orientation, the skater can be modeled as a cylindrical torso (40 , 20 average diameter, 160 tall) plus two rod-like arms (2.5 each, 67 long) attached to the outside of the torso. The skater then raises her arms straight above her head, where she appears to be a 45 , 20- -diameter, 200- -tall cylinder. Part A What is her new rotation frequency, in revolutions per second? Express your answer to two significant figures and include the appropriate units. ANSWER: Incorrect; Try Again Score Summary: = 6.6 rad s kg rev/s kg cm cm kg cm kg cm cm 2 = Your score on this assignment is 95.7%. You received 189.42 out of a possible total of 198 points.

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