4/24/2014Chapter 9 Homework Chapter 9 Homework Due: 10:00pm on Wednesday, April 9, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Exercise 9.1 Part A What angle in radians is subtended by an arc of 1.56m in length on the circumference of a circle of radius 2.56m ? ANSWER: θ = 0.609 rad Correct Part B What is this angle in degrees? ANSWER: θ = 34.9 ∘ Correct Part C An arc of length 14.2cm on the circumference of a circle subtends an angle of 124∘ . What is the radius of the circle? ANSWER: r = 6.56 cm Correct Part D The angle between two radii of a circle with radius 1.47m is 0.660rad . What length of arc is intercepted on the circumference of the circle by the two radii? ANSWER: http://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2832426 1/27 4/24/2014 Chapter 9 Homework L = 0.970 m Correct Circular Motion Tutorial Learning Goal: Understand how to find the equation of motion of a particle undergoing uniform circular motion. Consider a particle--the small red block in the figure--that is constrained to move in a circle of radius R. We can specify its position solely by θ(t), the angle that the vector from the origin to the block makes with our chosen reference axis at time t. Following the standard conventions we measure θ(t) in the counterclockwise direction from the positive x axis. Part A ⃗ What is the position vector r (t) as a function of angle θ(t). For later remember that Give your answer in terms of R, , and unit vectors ^ i and θ(t) ^ j θ(t) is itself a function of time. corresponding to the coordinate system in the figure. Hint 1. x coordinate What is the x coordinate of the particle? Your answer should be in terms of R and . θ(t) ANSWER: x = Rcos(θ(t)) Hint 2. y coordinate http://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2832426 2/27 4/24/2014 Chapter 9 Homework What is the y coordinate of the particle? Your answer should be in terms of R and . θ(t) ANSWER: y = Rsin(θ(t)) ANSWER: r ⃗ (t) = ^ ^ Rcos(θ(t)) i + Rsin(θ(t))j Correct Uniform Circular Motion A frequently encountered kind of circular motion is uniform circular motion, where θ(t) changes at a constant rate ω. In other words, ω= Usually, dθ(t) dt . . θ(t = 0) = 0 Part B For uniform circular motion, find θ(t) at an arbitrary time t. Give your answer in terms of ω and t. ANSWER: θ(t) = ωt Correct Part C What does ⃗ r (t) become now? Express your answer in terms of R, ω ^ , t, and unit vectors ^ i and j . ANSWER: r ⃗ (t) = Rcos(ωt)^ i + Rsin(ωt)^ j http://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2832426 3/27 4/24/2014 Chapter 9 Homework Correct Part D Find r ,⃗ a position vector at time t . = 0 Give your answer in terms of R and unit vectors ^ i and/or ^ j. Hint 1. Finding Simply plug t r⃗ = 0 into your expression for the components of r⃗ (t). ANSWER: r⃗ = ^ Ri Correct Part E Determine an expression for the position vector of a particle that starts on the positive y axis at , (x0 , y 0 ) = (0, R) ) and subsequently moves with constant ω. Express your answer in terms of R, ω t = 0 (i.e., at t = 0 ^ , t, and unit vectors ^ i and j . Hint 1. Adding a phase You can think of changing the initial position as adding a phase angle ϕ to the equation for θ(t). That is, θ(t) = ωt + ϕ . Hint 2. Finding a phase From previous parts you found that x = R cos(θ(t)) and y = R sin(θ(t)) . What should the angle θ be for x and y to be equal to 0 and R respectively? Express your answer as a fraction of the number π, for example (3/4)π or . (1/4)π ANSWER: θ = 1.57 ANSWER: http://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2832426 4/27 4/24/2014 Chapter 9 Homework ryaxis (t) = ^ ^ Rcos(ωt + 1.57) i + Rsin(ωt + 1.57)j Correct From this excersice you have learned that even though the motion takes place in the plane there is only one degree of freedom, angle θ , and that changing the initial coordinates introduces a phase angle in the equation. Pushing a Merry-Go-Round A child is pushing a playground merry-go-round. The angle through which the merry-go-round has turned varies with time according to θ(t) = γt + βt3 , where γ = 0.400 rad/s and β = 0.0120 rad/s3 . Part A Calculate the angular velocity of the merry-go-round as a function of time. Express your answer in radians per second in terms of γ , β , and t. Hint 1. Position versus velocity Recall that the angular velocity of an object is just the time derivative of its angular position. ANSWER: ω(t) = γ + 3βt 2 rad/sec Correct Part B What is the initial value ω0 of the angular velocity? Express your answer in radians per second. Hint 1. Position versus velocity Recall that the angular velocity of an object is just the time derivative of its angular position. The initial value is just the value at t = 0 s. ANSWER: ω0 = 0.4 rad/s http://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2832426 5/27 4/24/2014 Chapter 9 Homework Correct Part C Calculate the instantaneous value of the angular velocity ω(t) at time t = 5.00 s . Express your answer in radians per second. ANSWER: ω(5.00) = 1.3 rad/s Correct Part D Calculate the average angular velocity ωav for the time interval t = 0 to t = 5.00 seconds. Express your answer in radians per second. Hint 1. How to approach the problem In order to find the average angular velocity, just take the total angular displacement and divide by the total time. You can find the total angular displacement from the formula in the introduction for angular displacement θ(t). ANSWER: ωav = 0.7 rad/s Correct 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 magnitude of the angular acceleration of the salad spinner as it slows down? Express your answer numerically in radians per second per second. http://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2832426 6/27 4/24/2014 Chapter 9 Homework 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 α. 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 radians per second. Hint 1. Converting rotations to radians When the salad spinner spins through one revolution, it turns through 2π radians. ANSWER: ω0 = 25.1 radians/s Hint 3. Find the angular distance the salad spinner travels as it comes to rest Through how many radians Δθ = θ − θ 0 does the salad spinner rotate as it comes to rest? Express your answer numerically in radians. Hint 1. Converting rotations to radians One revolution is equivalent to 2π radians. ANSWER: Δθ = 37.7 radians 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 α? ANSWER: θ = θ0 + ω 0 t + 1 2 αt 2 ω = ω0 + αt ω 2 = ω 2 0 + 2α(θ − θ 0 ) http://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2832426 7/27 4/24/2014 Chapter 9 Homework ANSWER: α = 8.38 2 radians/s Correct Part B How long does it take for the salad spinner to come to rest? Express your answer numerically in seconds. 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 t. 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 t? ANSWER: θ = θ0 + ω 0 t + 1 2 αt 2 ω = ω0 + αt ω 2 2 = ω0 + 2α(θ − θ 0 ) ANSWER: t = 3.00 s Correct Marching Band A marching band consists of rows of musicians walking in straight, even lines. When a marching band performs in an event, such as a parade, and must round a curve in the road, the musician on the outside of the curve must walk around the curve in the same amount of time as the musician on the inside of the curve. This motion can be approximated by a disk rotating at a constant rate about an axis perpendicular to its plane. In this case, the axis of rotation is at the inside of the curve. http://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2832426 8/27 4/24/2014 Chapter 9 Homework Consider two musicians, Alf and Beth. Beth is four times the distance from the inside of the curve as Alf. Part A If Beth travels a distance s during time Δt, how far does Alf travel during the same amount of time? Hint 1. Find the angle through which Alf rotates If Beth rotates through an angle of θ during time Δt, through what angle does Alf rotate during the same amount of time? Hint 1. Angular velocity At any given instant, every part of a rigid body has the same angular velocity the relationship ω= Δθ Δt ω , where ω is given by . ANSWER: 4θ 2θ θ 1 2 1 4 θ θ Hint 2. Arc length If an angle θ (measured in radians) is subtended by an arc of length s on a circle of radius r, as shown in the figure, then s = rθ http://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2832426 . 9/27 4/24/2014 Chapter 9 Homework Use this formula to compare the lengths of the arcs that Alf and Beth trace out during equal time intervals. ANSWER: 4s 2s 1 2 1 4 s s s Correct The musician on the outside of the curve must travel farther than the musician on the inside of the curve in order to maintain the marching band's straight, even rows. Part B If Alf moves with speed v, what is Beth's speed? Speed in this case means the magnitude of the linear velocity, not the magnitude of the angular velocity. ANSWER: 4v v 1 4 v Correct The musician on the outside of the curve must travel faster than the musician on the inside of the curve. This is why most of the musicians on the outside of a curve appear to be jogging while their colleagues on the inside of the curve march in place. Constrained Rotation and Translation Learning Goal: To understand that contact between rolling objects and what they roll against imposes constraints on the change in position(velocity) and angle (angular velocity). The way in which a body makes contact with the world often imposes a constraint relationship between its possible rotation and translational motion. A ball rolling on a road, a yo-yo unwinding as it falls, and a baseball leaving the pitcher's hand are all examples of constrained rotation and translation. In a similar manner, the rotation of one body and the translation of another may be constrained, as happens when a fireman unrolls a hose from its storage drum. Situations like these can be modeled by constraint equations, relating the coupled angular and linear motions. Although http://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2832426 10/27 4/24/2014 Chapter 9 Homework these equations fundamentally involve position (the angle of the wheel at a particular distance down the road), it is usually the relationship of velocities and accelerations that are relevant in solving a problem involving such constraints. The velocities are needed in the conservation equations for momentum and angular momentum, and the accelerations are needed for the dynamical equations. It is important to use the standard sign conventions: positive for counterclockwise rotation and positive for motion toward the right. Otherwise, your dynamical equations will have to be modified. Unfortunately, a frequent result will be the appearance of negative signs in the constraint equations. Consider a measuring tape unwinding from a drum of radius r. The center of the drum is not moving; the tape unwinds as its free end is pulled away from the drum. Neglect the thickness of the tape, so that the radius of the drum can be assumed not to change as the tape unwinds. In this case, the standard conventions for the angular velocity ω and for the (translational) velocity v of the end of the tape result in a constraint equation with a positive sign (e.g., if v > 0, that is, the tape is unwinding, then ω > 0 also). Part A Assume that the function x(t) represents the length of tape that has unwound as a function of time. Find θ(t), the angle through which the drum will have rotated, as a function of time. Express your answer (in radians) in terms of x(t) and any other given quantities. Hint 1. Find the amount of tape that unrolls in one complete revolution of the drum If the measuring tape unwinds one complete revolution (θ ), how much tape, = 2π x2π , will have unwound? ANSWER: x2π = 2πr ANSWER: θ(t) = x(t) radians r Correct Part B http://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2832426ω(t) 11/27 4/24/2014 Chapter 9 Homework The tape is now wound back into the drum at angular rate ω(t). With what velocity will the end of the tape move? (Note that our drawing specifies that a positive derivative of x(t) implies motion away from the drum. Be careful with your signs! The fact that the tape is being wound back into the drum implies that tape to move closer to the drum, it must be the case that ω(t) < 0 , and for the end of the . v(t) < 0 Answer in terms of ω(t) and other given quantities from the problem introduction. Hint 1. How to approach the probelm The function ω(t) is given by the derivative of θ(t) with respect to time. Compute this derivative using the expression for θ(t) found in Part A and the fact that dx(t) dt = v(t). Express your answer in terms of v(t) and r. ANSWER: ω(t) = v(t) r ANSWER: v(t) = rω(t) Correct Part C Since r is a positive quanitity, the answer you just obtained implies that v(t) will always have the same sign as . If the tape is unwinding, both quanitites will be positive. If the tape is being wound back up, both quantities ω(t) will be negative. Now find a(t), the linear acceleration of the end of the tape. Express your answer in terms of α(t), the angular acceleration of the drum: α(t) = dω(t) . dt ANSWER: a(t) = rα(t) Correct Part D Perhaps the trickiest aspect of working with constraint equations for rotational motion is determining the correct sign for the kinematic quantities. Consider a tire of radius r rolling to the right, without slipping, with constant x http://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2832426 12/27 4/24/2014 Chapter 9 Homework velocity vx . Find ω, the (constant) angular velocity of the tire. Be careful of the signs in your answer; recall that positive angular velocity corresponds to rotation in the counterclockwise direction. Express your answer in terms of vx and r. ANSWER: ω = −vx r Correct This is an example of the appearance of negative signs in constraint equations--a tire rolling in the positive direction translationally exhibits negative angular velocity, since rotation is clockwise. Part E Assume now that the angular velocity of the tire, which continues to roll without slipping, is not constant, but rather that the tire accelerates with constant angular acceleration α. Find ax , the linear acceleration of the tire. Express your answer in terms of α and r. ANSWER: ax = −rα Correct Linear and Rotational Quantities Conceptual Question A merry-go-round is rotating at constant angular speed. Two children are riding the merry-go-round: Ana is riding at point A and Bobby is riding at point B. http://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2832426 13/27 4/24/2014 Chapter 9 Homework Part A Which child moves with greater magnitude of velocity? Hint 1. Distinguishing between velocity and angular velocity Ana’s (or Bobby’s) velocity is determined by the actual distance traveled (typically in meters) in a given time interval. The angular velocity is determined by the angle through which he rotates (typically in radians) in a given time interval. ANSWER: Ana has the greater magnitude of velocity. Bobby has the greater magnitude of velocity. Both Ana and Bobby have the same magnitude of velocity. Correct Part B Who moves with greater magnitude of angular velocity? Hint 1. Distinguishing between velocity and angular velocity Ana’s (or Bobby’s) velocity is determined by the actual distance he travels (typically in meters) in a given time interval. His angular velocity is determined by the angle through which he rotates (typically in radians) in a given time interval. ANSWER: http://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2832426 14/27 4/24/2014 Chapter 9 Homework Ana has the greater magnitude of angular velocity. Bobby has the greater magnitude of angular velocity. Both Ana and Bobby have the same magnitude of angular velocity. Correct Part C Who moves with greater magnitude of tangential acceleration? Hint 1. Distinguishing tangential, centripetal, and angular acceleration Ana’s tangential and centripetal acceleration are components of his acceleration vector. During circular motion, if Ana’s speed is changing (meaning the merry-go-round is speeding up or slowing down) he will have a nonzero tangential acceleration. However, even if the merry-go-round is turning at constant angular speed, he will experience a centripetal acceleration, because the direction of his velocity vector is changing (you can’t move along a circular path unless your direction of travel is changing!). Both tangential and centripetal accelerations have units of m/s2 , since they are the two-dimensional components of linear acceleration. Angular acceleration, on the other hand, is a measure of the change in Ana’s angular velocity. If his rate of rotation is changing, he will have a nonzero angular acceleration. Thus, angular acceleration has units of rad/s2 . ANSWER: Ana has the greater magnitude of tangential acceleration. Bobby has the greater magnitude of tangential acceleration. Both Ana and Bobby have the same magnitude of tangential acceleration. Correct Both Ana and Bobby are maintaining a constant speed, so they both have a tangential acceleration of zero (thus they are equal)! Part D Who has the greater magnitude of centripetal acceleration? Hint 1. Distinguishing tangential, centripetal, and angular acceleration Ana’s tangential and centripetal acceleration are components of his acceleration vector. For circular motion, if Ana’s speed is changing (meaning the merry-go-round is speeding up or slowing down) he will have a nonzero tangential acceleration. However, even if the merry-go-round is turning at constant angular speed, he will experience a centripetal acceleration, because the direction of his velocity vector is changing (you can’t move along a circular path unless your direction of travel is changing!). http://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2832426 m/ 2 15/27 4/24/2014 Chapter 9 Homework Both tangential and centripetal accelerations have units of m/s2 , since they are the two-dimensional components of linear acceleration. Angular acceleration, on the other hand, is a measure of the change in Ana’s angular velocity. If his rate of rotation is changing, he will have a nonzero angular acceleration. Thus, angular acceleration has units of rad/s2 . ANSWER: Ana has the greater magnitude of centripetal acceleration. Bobby has the greater magnitude of centripetal acceleration. Both Ana and Bobby have the same magnitude of centripetal acceleration. Correct Part E Who moves with greater magnitude of angular acceleration? Hint 1. Distinguishing tangential, centripetal, and angular acceleration Ana’s tangential and centripetal acceleration are components of his acceleration vector. For circular motion, if Ana’s speed is changing (meaning the merry-go-round is speeding up or slowing down) he will have a nonzero tangential acceleration. However, even if the merry-go-round is turning at constant angular speed, he will experience a centripetal acceleration, because the direction of his velocity vector is changing (you can’t move along a circular path unless your direction of travel is changing!). Both tangential and centripetal accelerations have units of m/s2 , since they are the two-dimensional components of linear acceleration. Angular acceleration, on the other hand, is a measure of the change in Ana’s angular velocity. If his rate of rotation is changing, he will have a nonzero angular acceleration. Thus, angular acceleration has units of rad/s2 . ANSWER: Ana has the greater magnitude of angular acceleration. Bobby has the greater magnitude of angular acceleration. Both Ana and Bobby have the same magnitude of angular acceleration. Correct Both Ana and Bobby are maintaining a constant angular velocity, so they both have an angular acceleration of zero (thus they are equal)! Exercise 9.20 http://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2832426 −7 m 16/27 4/24/2014 Chapter 9 Homework A compact disc (CD) stores music in a coded pattern of tiny pits 10−7 m deep. The pits are arranged in a track that spirals outward toward the rim of the disc; the inner and outer radii of this spiral are 25.0 mm and 58.0 mm, respectively. As the disc spins inside a CD player, the track is scanned at a constant linear speed of 1.25 m/s. Part A What is the angular speed of the CD when scanning the innermost part of the track? ANSWER: ω = 50.0 rad/s Correct Part B What is the angular speed of the CD when scanning the outermost part of the track? ANSWER: ω = 21.6 rad/s Correct Part C The maximum playing time of a CD is 74.0 min. What would be the length of the track on such a maximumduration CD if it were stretched out in a straight line? ANSWER: L = 5.55 km Correct Part D What is the average angular acceleration of a maximum-duration CD during its 74.0-min playing time? Take the direction of rotation of the disc to be positive. ANSWER: αav = −6.41×10−3 2 rad/s Correct http://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2832426 17/27 4/24/2014 Chapter 9 Homework Exercise 9.26 Part A Derive an equation for the radial acceleration that includes v and ω, but not r. ANSWER: arad = vω Correct Part B You are designing a merry-go-round for which a point on the rim will have a radial acceleration of 0.500 m/s2 when the tangential velocity of that point has magnitude 2.00 m/s. What angular velocity is required to achieve these values? ANSWER: ω = 0.250 rad/s Correct Weight and Wheel Consider a bicycle wheel that initially is not rotating. A block of mass m is attached to the wheel and is allowed to fall a distance h. Assume that the wheel has a moment of inertia I about its rotation axis. Part A Consider the case that the string tied to the block is attached to the outside of the wheel, at a radius rA . Find ωA , the angular speed of the wheel after the block has fallen a distance h, for this case. Express ωA in terms of m, g, , h rA , and I . http://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2832426 18/27 4/24/2014 Chapter 9 Homework Hint 1. How to approach this problem The most straighforward way to solve this problem is to use conservation of mechanical energy. The total initial energy of the system is equal to the total final energy of the system (where the system consists of the wheel and the block). In other words, E i = Ebf + Ewf . Where Ei is the initial energy of the system, Ebf is the final energy of the block and Ewf is the final energy of the wheel. Hint 2. Initial energy of the system Initially, the wheel is not rotating. The initial energy of the system consists of the gravitational potential energy stored in the block, since it is not moving either. Supposing that the gravitiational potential energy of the block is zero at "ground level," find the initial energy of the system. ANSWER: Ei = mgh Hint 3. Final energy of block Find the final energy of the block. Express the final energy of the block in terms of given quantities (excluding final angular velocity of the wheel, ωA . ) and the unknown h Hint 1. Final velocity of the block Find vf , the magnitude of the final velocity of the block. Express the velocity in terms of rA and the final angular velocity of the wheel, ωA . ANSWER: vf = rA ωA http://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2832426 19/27 4/24/2014 Chapter 9 Homework ANSWER: E bf = 1 2 m ( r A ωA ) 2 Hint 4. Final energy of wheel Find the final kinetic energy of the wheel. Express your answer in terms of I (the wheel's moment of inertia) and ωA . ANSWER: E wf = 1 2 I ωA 2 ANSWER: − − − − − − − ωA = 2mgh √ mrA 2 +I Correct Part B Now consider the case that the string tied to the block is wrapped around a smaller inside axle of the wheel of radius rB . Find ωB , the angular speed of the wheel after the block has fallen a distance h, for this case. Express ωB in terms of m, g, , h rB , and I . Hint 1. Similarity to previous part The derivation of ωB is exactly the same as the derivation for ωA , using rB instead of rA . http://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2832426 20/27 4/24/2014 Chapter 9 Homework ANSWER: − − − − − − − ωB = √ 2mgh mrB 2 +I Correct Part C Which of the following describes the relationship between ωA and ωB ? Hint 1. How to approach this question To figure out which angular velocity is greater ( ωA or ωB ), you only need to consider the radius dependence of the expression for ω. Ignoring all of the other parameters, you should have found that ω goes as 1/radius (where "radius" refers to where the string is attached, which is not necessarily the outer radius of the wheel). The problem then reduces to figuring out which is greater, 1/rA or 1/rB . ANSWER: ωA > ωB ωB > ωA ωA = ωB Correct This is related to why gears are found on the inside rather than the outside of a wheel. Exercise 9.30 Four small spheres, each of which you can regard as a point of mass 0.200 kg, are arranged in a square 0.400 m on a side and connected by light rods . http://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2832426 21/27 4/24/2014 Chapter 9 Homework Part A Find the moment of inertia of the system about an axis through the center of the square, perpendicular to its plane (an axis through point O in the figure). ANSWER: I = 6.40×10−2 2 kg ⋅ m Correct Part B Find the moment of inertia of the system about an axis bisecting two opposite sides of the square (an axis along the line AB in the figure). ANSWER: I = 3.20×10−2 2 kg ⋅ m Correct Part C Find the moment of inertia of the system about an axis that passes through the centers of the upper left and lower right spheres and through point O. ANSWER: I = 3.20×10−2 2 kg ⋅ m http://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2832426 22/27 4/24/2014 Chapter 9 Homework Correct Exercise 9.31 Calculate the moment of inertia of each of the following uniform objects about the axes indicated. Consult Table Moments of Inertia of Various Bodies in the Textbook as needed. Part A A thin 2.50-kg rod of length 90.0cm , about an axis perpendicular to it and passing through one end. ANSWER: I = 0.675 2 kg ⋅ m Correct Part B A thin 2.50-kg rod of length 90.0cm , about an axis perpendicular to it and passing through its center. ANSWER: I = 0.169 2 kg ⋅ m Correct Part C A thin 2.50-kg rod of length 90.0cm , about an axis parallel to the rod and passing through it. ANSWER: I = 0 2 kg ⋅ m Correct Part D A 4.50-kg sphere 30.0cm in diameter, about an axis through its center, if the sphere is solid. ANSWER: http://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2832426 23/27 4/24/2014 Chapter 9 Homework I = 4.05×10−2 2 kg ⋅ m Correct Part E A 4.50-kg sphere 30.0cm in diameter, about an axis through its center, if the sphere is a thin-walled hollow shell. ANSWER: I = 6.75×10−2 2 kg ⋅ m Correct Part F An 7.50-kg cylinder, of length 14.0cm and diameter 10.0cm , about the central axis of the cylinder, if the cylinder is thin-walled and hollow. ANSWER: I = 1.88×10−2 2 kg ⋅ m Correct Part G An 7.50-kg cylinder, of length 14.0cm and diameter 10.0cm , about the central axis of the cylinder, if the cylinder is solid. ANSWER: I = 9.38×10−3 2 kg ⋅ m Correct Exercise 9.35 A wagon wheel is constructed as shown in the figure . The radius of the wheel is 0.300 m, and the rim has mass 1.45 kg . Each of the eight spokes, that lie along a diameter and are 0.300 m long, has mass 0.200kg . http://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2832426 24/27 4/24/2014 Chapter 9 Homework Part A What is the moment of inertia of the wheel about an axis through its center and perpendicular to the plane of the wheel? ANSWER: I = 0.179 2 kg ⋅ m Correct Exercise 9.44 A light, flexible rope is wrapped several times around a hollow cylinder with a weight of 40.0 N and a radius of 0.25 m, that rotates without friction about a fixed horizontal axis. The cylinder is attached to the axle by spokes of a negligible moment of inertia. The cylinder is initially at rest. The free end of the rope is pulled with a constant force P for a distance of 5.00 m, at which point the end of the rope is moving at 6.00 m/s. Part A If the rope does not slip on the cylinder, what is the value of P? ANSWER: P = 14.7 N Correct Parallel Axis Theorem Icm http://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2832426 25/27 4/24/2014 Chapter 9 Homework The parallel axis theorem relates Icm , the moment of inertia of an object about an axis passing through its center of mass, to Ip , the moment of inertia of the same object about a parallel axis passing through point p. The mathematical 2 statement of the theorem is Ip = Icm + M d , where d is the perpendicular distance from the center of mass to the axis that passes through point p, and M is the mass of the object. Part A Suppose a uniform slender rod has length L and mass . The moment of inertia of the rod about about an axis m that is perpendicular to the rod and that passes through its center of mass is given by Iend I cm = 1 12 mL 2 . Find , the moment of inertia of the rod with respect to a parallel axis through one end of the rod. Express Iend in terms of m and L . 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 d 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. ANSWER: d = L 2 ANSWER: Iend = mL 2 3 Correct Part B Now consider a cube of mass m with edges of length a. The moment of inertia Icm of the cube about an axis through its center of mass and perpendicular to one of its faces is given by I cm = 1 6 2 ma . Find Iedge , the moment of inertia about an axis p through one of the edges of the cube Express Iedge in terms of m and a. Use fractions rather than decimal numbers in your answer. http://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2832426 26/27 4/24/2014 Chapter 9 Homework Hint 1. Find the distance from the o axis to the p axis Find the perpendicular distance d from the center of mass axis to the new edge axis (axis labeled p in the figure). ANSWER: d = a √2 ANSWER: Iedge = 2ma 2 3 Correct Score Summary: Your score on this assignment is 99.8%. You received 14.97 out of a possible total of 15 points. http://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2832426 27/27