Version 00 olling & Angular Momentum ramadoss (7 This print-out should have 76 questions Multiple-choice questions may continue on the next column or page find all choices before answering One version of this assignment with keys is on my website AP B 998 MC 6 00 00 points A kg object moves in a circle of radius 4 m at a constant speed of 3 m/s A net force of 45 N acts on the object What is the magnitude of the angular momentum L of the object with respect to an axis perpendicular to the circle and through its center? L 4 kg m correct s L 9 N m kg 3 L m s 4 L 35 kg m s 5 L 8 N m kg The angular momentum is L mvr ( kg(3 m/s(4 m 4 kg m /s AP M 998 MC 3 33 00 (part of 00 points A wheel with rotational inertia I is mounted on a fixed, frictionless axle The angular speed ω of the wheel is increased from zero to ω f in a time interval T What is the average net torque τ on the wheel during this time interval? τ Iω f T τ Iω f T 3 τ ω f T 4 τ Iω f T correct 5 τ ω f T The change of angular momentum of the wheel is L Iω f, so the average net torque on the wheel during this time interval is τ L T Iω f T 003 (part of 00 points What is the average power input to the wheel during this time interval? P I ω f T P I ω f T 3 P Iω f T 4 P Iω f T correct 5 P Iω f T The change in the energy of the wheel is Iω f in the time interval T, so the average power input to the wheel is Iω f T Iω f T AP M 998 MC 6 004 00 points A wheel of mass M and radius rolls on a level surface without slipping If the angular velocity of the wheel about its center is ω, what is its linear momentum relative to the surface? p M ω correct
Version 00 olling & Angular Momentum ramadoss (7 p M ω 3 p M ω 4 p M ω 5 p 0 First, we note that the wheel is rotating about its center at an angular velocity of ω, so the velocity difference between the center ofthewheel andthelowestpoint(thepointof the wheel that touches the surface is ω in magnitude in the horizontal direction Since the wheel rolls without slipping (which means the velocity of the lowest point is 0 the velocity v of the center of the wheel is ω Obviously, the center of the wheel is also the center ofthe mass, so the linear momentum is p M ω T y < M b g and T x < M a g 3 T y < M b g and T x M a g 4 T y < M b g and T x > M a g correct 5 T y M b g and T x < M a g 6 T y > M b g and T x > M a g 7 T y > M b g and T x M a g 8 T y M b g and T x M a g 9 T y M b g and T x > M a g The magnitude of the tension in the string is the same at all points along the string: T T y T x Consider the free body diagrams Atwood Machine 07 005 (part of 3 00 points Consider a massless, frictionless pulley attached to the ceiling A massless, inextensible string is attached to the masses M b and M a, wherem b > M a ThetensionsT y,t z,t x,and the gravitational constant g are magnitudes a Ty Mbg M b M a Tx Mag a T y T z ω Since the larger mass will move down and the smaller mass up, we can take motion downward as positive for M b and motion upward as positive for M a Applying Newton s second law to M a and M b, respectively, F Tx M a g M a a ( M b l T x M a T x M a (g +a > M a g, and for mass M b, F Mb g T y M b a ( What is trueabout the tensions T y and T x? T y > M b g and T x < M a g T y M b (g a < M b g 006 (part of 3 00 points Which relationship about T z is true?
Version 00 olling & Angular Momentum ramadoss (7 3 T z < M b g +M a g correct T z M b g +M a g 3 T z > M b g +M a g The forces are in equilibrium for the pulley, so T z T y +T x M b (g a+m a (g +a M b g +M a g (M b M a a < M b g +M a g 007 (part 3 of 3 00 points What is the magnitude of the acceleration of the center of mass of the system? M b +M a g M b M a ( Mb M a g correct M b +M a 3 M b +M a M a M b g 4 M a M b g M b +M a ( Mb +M a 5 g M b M a 6 M b M a M b +M a g Since T y T x, adding equations and yields M b g M a g M a a+m b a a M b M a M a +M b g Assume that the CM moves downward(and that a CM > 0: (M b +M a a CM M b a M a a a CM M b M a a M b +M a ( Mb M a g > 0, M b +M a so the assumption was correct keywords: Bohr Model of Hydrogen 0 008 (part of 00 points In the Bohr s model of the hydrogen atom, theelectronmovesinacircularorbitofradius 4544 0 m around the proton Assume that the orbital angular momentum of the electron is equal to h Calculate the orbital speed of the electron Correct answer: 60077 0 7 m/s The angular momentum of the electron in the ground state of the hydrogen atom (this the case herein the Bohr s model is h, therefore: and solving for v, L mvr h v h mr (66607 0 34 Js (90939 0 3 kg(4544 0 m 60077 0 7 m/s 009 (part of 00 points Calculate the angular frequency of the electron s motion Correct answer: 358 0 7 s The angular frequency is given by ω v r (60077 07 m/s (4544 0 m 358 0 7 s Bullet otates a od 0 00 (part of 00 points
Version 00 olling & Angular Momentum ramadoss (7 4 AwoodenblockofmassM hangsfromarigid rod of length l having negligible mass The rod is pivoted at its upper end A bullet of mass m traveling horizontally and normal to the rod with speed v hits the block and gets embedded in it v m M What is the angular momentum L of the block-bullet system, with respect to the pivot point immediately after the collision? ( M m L vl M +m L (M mvl 3 L mvl correct 4 L (m+mvl 5 L M vl If τ ext 0, then L 0 The net angular momentum of the system conserves, and L i L f L mvl 0 (part of 00 points What is the fraction K f K i (the final kinetic energy compared to the initial kinetic energy in the collision? K f M K i M m K f M K i M +m 3 K f K i m M m 4 K f K i m m+m l 5 K f m K i m+m correct By conservation of the angular momentum L i L f L mvl (m+mv f l ( m v f v m+m K i mv and K f I ω f where I (M +m l and ω f v f l, so K f (M +mv f and K f K i m M +m v mv m M +m Bullet Hits a Cube 0 0 00 points Assume: A bullet of mass m and cube of mass M undergo an inelastic collision, where m M Note: The moment of inertia of this cube (with edges of length a and mass M about 8M a an axis along one of its edges is 3 A solid cube is resting on a horizontal surface The cube is constrained to rotate about an axis at its bottom right edge (due to a small obstacle on the table A bullet with speed v min is shot at the left-hand face at a height of 4 a The bullet gets embedded in 3 the cube ω a 4 3 a m v min M M g Find the minimum value of v min required to tip the cube so that it falls its right-hand face
Version 00 olling & Angular Momentum ramadoss (7 5 v min m ( 3ga 5 M v min M ( 3ga correct m 3 v min m ( 3ga 3 M 4 v min m ( ga M 5 v min m ( 5ga M Basic Concepts: L const U + K 0 For the cube to tip over the center of mass (CM must rise so that it is over the axis of rotation AB To do(this the CM must be raised a distance of a From conservation of energy ( M ga I cubeω ( From conservation of angular momentum ( 4a 8M a 3 mv ω 3 ( mv ω ( M a Thus, substituting Eq into, we have ( 8M a ( m v ( 3 4M a M ga Solving for v yields v min M m 3ga ( Child on a Merrygoround 03 00 points A child is standing on the edge of a merry-goround that is rotating with frequency f The child then walks towards the center of the merry-go-round For the system consisting of the child plus the merry-go-round, what remains constant as the child walks towards the center? (neglect friction in the bearing neither mechanical energy nor angular momentum mechanical energy and angular momentum 3 only mechanical energy 4 only angular momentum correct The forces external to the system are not exerting any torque, so the angular momentumisconserved Ontheotherhand, thefrictionforceactingonthechildisdoingwork,because she is moving towards the center (which is the direction of that force Circular Trajectory 04 00 points A particle of mass m moves in a circle of radius at a constant speed v Assume: The motion begins from the point Q, which has coordinates (, 0 Determine the angular momentum of the particle about point P, which has coordinates (,0 as a function of time [ ( ] L m v vt sin +π + ˆk [ ( L vt mv cos +π + ˆk 3 None of these [ ( ] 4 L vt mv sin + ˆk [ ( ] 5 L m v vt cos + ˆk [ ( ] 6 L vt mv sin +π + ˆk [ ( ] 7 L m v vt cos +π + ˆk
Version 00 olling & Angular Momentum ramadoss (7 6 [ 8 L mv cos 9 L m v [ sin Basic Concept: ( ] vt + ] + ( vt ˆk ˆk correct axis Aparticleofmassm 36kgandinitial velocity v 0 33 m/s (perpendicular to the cylinder s axis flies too close to the cylinder s edge, collides with the cylinder and sticks to it L r m v Solution: The angular displacement of the particle around the circle is θ ωt vt The vector from the center of the circle to the mass is then cosθî+ sinθĵ The vector from the point P to the point of mass is r î+ cosθî+ sinθĵ {[ ( ] ( vt vt r +cos î+sin The velocity is So v d r dt v sin ( vt î+v cos ( vt ĵ L r m v mv{[+cos(ωt]î+sin(ωtĵ} [ sin(ωtî+cos(ωtĵ] mv {[+cos(ωt][cos(ωt] [sin(ωt][ sin(ωt]} [ ( ] ˆk vt mv cos + ˆk } ĵ Collision With a Cylinder 05 00 points A solid cylinder of mass M 45 kg, radius 044mand uniform density ispivotedon a frictionless axle coaxial with its symmetry Before the collision, the cylinder was not rotating What is the magnitude of its angular velocity after the collision? Correct answer: 03448 rad/s Basic Concept: Conservation of Angular Momentum, L particle z + L cylinder z const The axle allows the cylinder to rotate without friction around a fixed axis but it keeps this axisfixed Letthez coordinateaxisrunalong this axis of rotation; then the axle may exert arbitrary torques in x and y directions but τ z 0 Consequently, the z componenent of the angular momentum must be conserved, L z const, hence when the particle collides with the cylinder L before z,part + Lbefore z,cyl L z,net L after z,part + Lafter z,cyl Before the collision, the cylinder did not rotate hence L before z,cyl 0 while the particle had angular momentum L before part r P 0 r m v 0 Both the radius-vector r and the velocity v 0 of the particle lie in the xy plane ( to the z axis, and according to the picture, at the moment of collision the radius vector has magnitude r equal to the cylinder s radius
Version 00 olling & Angular Momentum ramadoss (7 7 and direction perpendicular to the particle s velocity Hence, its angular momentum is parallel to the z axis and has magnitude L before part L before z,part mv 0 Altogether, before the collision L before z,net mv 0 and therefore, after the collision we should also have L after z,net mv 0 ( After the collision, the cylinder and the particle rotate as a single rigid body of net moment of inertia I net I cyl + I part M + m ( relatve to cylinder s axis For a rigid rotation like this, the angular momentum points in z direction and its magnitude is L z,net ωi net (3 Combining eqs (, ( and (3 together, we immeditaly obtain ω after Lafter z,net I net v 0 mv 0 M +m m 03448 rad/s M +m Component of Net Torque 06 00 points A force F (3 Nî + (6 Nĵ is applied to an object that is pivoted about a fixed axis aligned along the z coordinate axis The force is applied at the point (38 mî+ (54 mĵ Find the z-component of the net torque In this case, the net torque only has a z- component Component of Net Torque 03 07 00 points A force F (4 Nî+(5 Nĵ is applied to an object that is pivoted about a fixed axis aligned along the z coordinate axis The force isappliedatthepoint r (43mî+(5mĵ Find the z-component of the net torque Correct answer: 49 N m Basic Concept: τ r F Solution: From the definition of torque τ r F [xî+yĵ] [F x î+f y ĵ] [(43 mî+(5 mĵ] [(4 Nî+(5 Nĵ] [(43 m(5 N (5 m(4 N]ˆk ( 49 Nmˆk In this case, the net torque only has a z- component Conical Pendulum 04 08 00 points A small metallic bob is suspended from the ceiling by a thread of negligible mass The ballisthen set inmotioninahorizontal circle so that the thread describes a cone Correct answer: 54 N m Basic Concept: τ r F Solution: From the definition of torque τ F [ x î+ y ĵ] [F x î+f y ĵ] [(38 mî+(54 mĵ] [(3 Nî+(6 Nĵ] [(38 m(6 N (54 m(3 N]ˆk ( 54 Nmˆk 98 m/s 5 v m kg
Version 00 olling & Angular Momentum ramadoss (7 8 Calculate the magnitude of the angular momentum of the bob about a vertical axis through the supporting point The acceleration of gravity is 98 m/s Correct answer: 9649 kg m /s Let : l m, θ 5, g 98 m/s, and m kg Consider the free body diagram ( kg sin 5 (98 m/s ( m 3 cos5 9649 kg m /s Cylinder rolls 09 00 points The coefficient of static friction between a certain cylinder and a horizontal floor is 04 If the rotational inertia of the cylinder about its symmetry axis is given by I (/M, then the maximum acceleration the cylinder can have without slipping is: T θ 04 g 0 g 3 g 4 0 g 5 08 g correct mg Newton s second law in the vertical and horizontal projections, respectively, gives T cosθ mg 0 T cosθ mg and T sinθ mω lsinθ 0 T mω l, where the radius of the orbit is lsinθ Dividing, cosθ mg mω l g ω lcosθ r and v are perpendicular, where r l sinθ,sotheangularmomentuml m r v will be L mω(lsinθ gl m sin 3 θ cosθ If the cylinder is not slipping, we know that the angular acceleration is related to the linear acceleration by a α Let us consider the relevant force to be applied at the center of the cylinder Then, the only horizontal forces applied to the cylinder are the accelerating force and the frictional force Taking the center of the cylinder to be the axis of rotation, we realize that the applied force exerts no torque (since r 0, and the rotational analogue of Newton s second law reads: Iα τ ( net M α f s M( a µ s Mg a µ s g (04(g 08 g
Version 00 olling & Angular Momentum ramadoss (7 9 Decelerated Grinding Wheel 00 (part of 00 points The motor driving a grinding wheel with a rotational inertia of 08 kgm is switched off when the wheel has a rotational speed of 39 rad/s After 66 s, the wheel has slowed down to 3 rad/s What is the absolute value of the constant torque exerted by friction to slow the wheel down? Correct answer: 0945455 N m We have so that τ I ω ω 0 t τ t L (Iω, (08 kgm (39 rad/s 3 rad/s 66 s 0945455 Nm 0 (part of 00 points If this torque remains constant, how long after the motor is switched off will the wheel come to rest? Correct answer: 33 s When the wheel comes to rest, its angular speed is ω 0; hence t Iω 0 τ I(ω 0 ω τ (08 kgm (39 rad/s (0945455 N m 33 s Disk and Hoop ace 0 (part of 00 points A uniform solid disk and a uniform hoop are placed side by side at the top of an incline of height h If they are released from rest and roll without slipping, determine their speeds when they reach the bottom (d disk, h hoop 4 v d 3 hg, v h hg correct v d hg, v h hg 3 v d hg, v h 3 hg 4 v d 3 hg, v h 3hg 5 v d hg, v h hg Because they roll without slipping, v rω and the total kinetic energy is K mv + Iω, so for the disk, since I mr, K mv d + 4 mv d 3 4 mv d 3 4 mv d mhg, 4 v d 3 hg Similarly, for the hoop, since I mr conservation of energy gives us v h hg 03 (part of 00 points Which object reaches the bottom first? the hoop both at the same time 3 the disk correct
Version 00 olling & Angular Momentum ramadoss (7 0 From the previous result one can see that for any fallen height h, the disk has a velocity greater than the velocity of the hoop, so the disk will reach the bottom first (its velocity has been greater than hoop s all the time keywords: Disk and Hoop ace 0 04 (part of 00 points A uniform solid disk and a uniform hoop are placed side by side at the top of an incline of height h If they are released from rest and roll without slipping, determine their speeds when they reach the bottom (d disk, h hoop v d 3 hg, v h 3hg v d 3hg, v h hg 3 v d 3hg, v h 3 hg 4 v d hg, v h hg 5 v d hg, v h hg 4 6 v d 3 hg, v h hg correct 7 v d 3 hg, v h hg 8 v d hg, v h hg 9 v d hg, v h 3 hg 0 v d 3hg, v h hg Because they roll without slipping, v rω and the total kinetic energy is K mv + Iω, so for the disk, since I mr, K mv d + 4 mv d 3 4 mv d 3 4 mv d mhg, 4 v d 3 hg Similarly, for the hoop, since I mr, conservation of energy gives us v h hg 05 (part of 00 points Whatistheratiooftheiraccelerationsasthey roll down the incline? a disk a hoop 4 3 correct a disk a hoop 3 a disk a hoop 3 4 a disk a hoop 3 5 a disk a hoop 3 6 a disk a hoop 7 a disk a hoop 8 a disk a hoop 9 a disk a hoop 0 a disk a hoop 3 3 4 3 Both disk and hoop started from rest, so v as When the two roll down the same height h on the incline, they have covered the same distance s, so a disk a hoop a disk s a hoop s (v disk (v hoop 4 3
Version 00 olling & Angular Momentum ramadoss (7 keywords: Door Handle Torque 06 00 points A door is opened in the usual way, in the direction indicated D C In which direction is the net torque vector τ due to the force applied to the door handle? The force is toward you, away from the door B D A B 3 Insufficient information is given 4 C 5 A correct r τ τ r F F r points from the hinge to the spot where F acts, so by the righthand rule, r F points in the direction A, along the hinge line Figure Skater 07 (part of 00 points Afigureskaterrotatingononespotwithboth arms and one leg extended has moment of inertia I i She then pulls in her arms and the extended leg, reducing her moment of inertia to 075I i What is the ratio of her final to initial kinetic energy? 3/4 3 / 4 6/9 5 4/3 correct 6 3/8 7 8/3 8 9/6 9 The angular momentum was conserved, so L i L f Since L Iω and KE r Iω, So, KE f KE i KE r L I L f I f L i I i I i I f I i 075I i 4 3 08 (part of 00 points Consider the following statements for the figure skater: I Angular momentum was conserved II Mechanical energy was conserved III The kinetic energy changed because of energy dissipation due to friction IV Her rotation rate changed in response toatorqueexertedbypullinginher armsand leg Which is the correct combination of statements? I, II, IV I and II
Version 00 olling & Angular Momentum ramadoss (7 3 I, II, III 4 I correct 5 II (I The angular momentum was conserved since there was no external torque acting on the skater (II From part, we found that the kinetic energy increased and the potential energy didn t change, so the total mechanical energy increased (II is wrong (III We found that the total kinetic energy increased, and therefore, we can infer that there was no energy dissipation (IVTheforcepullingherarmsinwouldbe perpendicular to her rotation, and therefore, exert no torque Ergo, the only correct statement is (I Figure Skater Spin 0 09 00 points A figure skater on ice spins on one foot She pulls in her arms and her rotational speed increases Choose the best statement below: Her angular speed increases because she is undergoing uniformly accelerated angular motion Her angular speed increases because her potential energy increases as her arms come in 3 Her angular speed increases because her angular momentum is the same but her moment of inertia decreases correct 4 Her angular speed increases because by pulling inher armsshe creates a net torque in the direction of rotation 5 Her angular speed increases because air friction is reduced as her arms come in 6 Her angular speed increases because her angular momentum increases The initial angular momentum of the figure skater isi i ω i After she pulls in her arms, the angular momentum of her is I f ω f Note that I f < I i because her arms now rotate closer to the rotation axis and reduce the moment of inertia Since the net external torque is zero, angular momentum remains unchanged, and so I i ω i I f ω f L Therefore, ω f > ω i Grindstone Energy 030 00 points A constant torque of 7 N m is applied to a grindstone whose moment of inertia is 065 kg m Using energy principles, and neglecting friction, find the angular speed after the grindstone has made 54 rev Correct answer: 403 rad/s Given : τ 7 N m, I 065 kg m, and N 54 rev W net τ θ I w f Iw i Σ τ ext d L dt If the torque is constant, then we can write for the angle of rotation of the grindstone: τ Iα, where α is the angular acceleration Therefore, α τ I d θ dt, and θ N π, where N is the number of revolutions Thus, ω αθ,
Version 00 olling & Angular Momentum ramadoss (7 3 ω αθ τ I πn 4π(7 N m(54 rev 065 kg m 403 rad/s Horizontal otation 03 (part of 3 00 points A mass m is attached to a cord passing through a small hole in a frictionless horizontal surface The mass is initially orbiting withspeedv 0 inacircleofradiusr 0 Thecord is then slowly pulled from below, decreasing the radius of the circle to r What is the speed of the mass when the radius is r? v v 0 r 0 r correct v v 0 r r 0 3 v v 0 r 0 +r r 0 r 4 v v 0 r 0 r r 0 5 None of these 6 v v 0 r 0 +r r 0 7 v v 0 r 0 r 8 v v 0 r 0 r 0 +r 9 v v 0 r 0 r r 0 +r 0 v v 0 r 0 r 0 r Basic Concepts Solution τ r F d L dt τ ext τ r F r F sin(0 0 Angular moment is conserved, so Therefore, L f L i mrv mr 0 v 0 v r 0v 0 r 03 (part of 3 00 points Find the tension in the cord as a function of r T m(r 0v 0 r 3 T m(r 0v 0 (r+r 0 3 3 None of these 4 T m(rv 0 r 3 0 5 T m(r 0v 0 r 3 6 T m(r 0v 0 r 3 T mv r correct m(r 0v 0 r 3 033 (part 3 of 3 00 points How much work W is done in moving m from r 0 to r? Note: The tension depends on r W ( r 3 mv 0 r0 W ( r 0 mv 0 r correct 3 None of these 4 W ( r 0 3 mv 0 r
Version 00 olling & Angular Momentum ramadoss (7 4 5 W ( r mv 0 r0 6 W ( r 0 3 mv 0 r 7 W ( r 3 mv 0 r0 8 W ( (r0 +r mv 0 r Thework isdoneby thecentripetal force in the negative direction Method : W F d l T dr Method : r m(r 0 v 0 r 0 (r 3 dr m(r 0v 0 ( r r0 ( r 0 mv 0 r W K mv mv 0 r 0 0( mv r Horizontal Circle 0 034 (part of 00 points A ball is rotating in a horizontal circle at the end of a string of length 3 m at an angular velocity of rad/s The string is gradually shortened to 5 m without any force being exerted in the direction of the ball s motion Find the new angular velocity of the ball Correct answer: 78 rad/s Since there is no force exerted in the direction of the ball s motion, the total angular momentum of the ball is conserved We have I ω I 0 ω 0 ω ml ω ml 0ω 0 ( l0 l ω 0 ( 3 m ( rad/s 5 m 78 rad/s 035 (part of 00 points Find its new linear speed Correct answer: 43 m/s The new linear speed is v l ω (5 m(78 rad/s 43 m/s Ice Skater pulls arms in 036 00 points AniceskaterwithrotationalinertiaI 0 isspinning with angular velocity ω 0 She pulls her arms in, decreasing her rotational inertia to I 0 /3 Her angular velocity becomes: 3ω 0 ω 0 / 3 3 3ω 0 correct 4 ω 0 /3 5 ω 0 Angular momentum is given by L Iω, and is conserved on ice, so: I 0 ω 0 Iω ( 3 I 0 ω ω 3ω 0
Version 00 olling & Angular Momentum ramadoss (7 5 Impulse on a Cue Ball 037 00 points A cue stick strikes a cue ball and delivers a horizontal impulse in such a way that the ball rolls without slipping as it starts to move At what height above the ball s center (if the radius of the ball is083 cm was the blow struck? Correct answer: 033 cm Basic Concepts: L Iω p mv ph I c ω 0 ( p M v 0 ( If the ball rolls without slipping, So, v 0 ω 0 h I cω 0 p I cω 0 M v 0 K r 5 km K CM K r 6 k correct K CM K r I CM ω and K CM M v CM Because there is no slipping, v CM ω, and K r I CM ω km ω K r K CM k km v CM kk CM KE atio 039 00 points A ball with I cm κm, mass M and radius rolls along a horizontal surface without slipping I c M 5 (083 cm 033 cm 5 KE Comparison 0 038 00 points A object of mass M and radius has a rotational inertia of I CM km, where k is a constant If the object rolls without slipping, what is the ratio of its rotational kinetic energy to its linear kinetic energy? 3 4 K r k K CM K r k + K CM k K r k K CM k + K r k + K CM What is the ratio K r K cm of rotational to center-of-mass kinetic energy? κ κ κ 3 4 κ 5 κ 6 κ correct 7 κ
Version 00 olling & Angular Momentum ramadoss (7 6 8 ; they have the same magnitude K cm M v cm, and for rolling without slipping v cm ω, so the rotational kinetic energy is K r I cmω ( κm ( v cm ( κ M v cm κk cm K r κ K cm Kinetic Energy of a olling Wheel 040 00 points If a steel, thin-shelled wheel of radius r and mass M is moving along the road at m/s, what is its total kinetic energy? 3M M 3 5M 4 M 5 4M correct The moment of inertia of a thin ring or hollow cylinder about its axis is I M r and the translational speed of the wheel is v ωr The wheel is rotating and translating, so its total kinetic energy is K K rot +K trans I ω + M v ( M r ( v + r M v M v 4M Maximize Torque 04 00 points Aforceistobeappliedtoawheel Thetorque can be maximized by: applying the force near the axle, radially outward from the axle applying the force at the rim, at 45 to the tangent 3 applying the force near the axle, parallel to the tangent to the wheel 4 applying the force at the rim, tangent to the rim correct 5 applying the force near the rim, radially outward from the axle Let slookathow torqueisdefined viacross product,wherewetake F tobethemagnitude of the force causing the torque, and we take r to be the vector pointing from the center of rotation of the object to the point at which F is applied: τ r F τ r F sinθ Where θ is the angle between the vectors r and F So, to maximize the force, we want F to be applied as far from the center of the wheel as possible, and we want it to be perpendicular to the wheel Thus, it should be near the rim, and tangent to the wheel Length atio 04 00 points A non-uniform horizontal beam of length l andmassm hasitscenter ofmass atdistance x from the left end, as shown It is supported bytworopes;theleft-endropemakesanangle of θ with the vertical and has tension T, and the right-hand rope makes an angle of φ with the vertical and has tension T T θ x cm l φ T
Version 00 olling & Angular Momentum ramadoss (7 7 If sinθ cosφ 3 5 and cosθ sinφ 4 5, andthe rodisstationary,what is theratio x l? 9 5 correct 6T +9T 5mg T 3 5 mg Taking torques about the left end of the rod, 5 8 3 3 8 4 5 5 6 5 6 4 7 8 5 9 3 7 0 3 5 Consider the free-body diagram T θ x cm m g l φ T τ mgx+t l cosφ 0 ( ( 3 3 x l T cosφ 5 mg 5 mg mg 9 5 Momentum of a Particle 043 00 points A particle whose mass is 3 kg moves in xyplane with a constant speed of m/s in the positive x-direction along y 4 m Find the magnitude of its angular momentum relative to the point (x 0,y 0, where x 0 m and y 0 m Correct answer: 4 kgm /s Apply L L r p m r v [ ] (y y 0 ĵ+(x x 0 î mvî mv(y y 0 ˆk F 0 for the rod to be in static equilibrium, so horizontally, and vertically, T sinθ +T sinφ 0 ( ( 3 4 T +T 0 5 5 T 4 3 T T cosθ +T cosφ mg 0 ( ( ( 4 4 3 3 T +T mg 5 5 Then the magnitude of the angular momentum is L mv(y y 0 (3 kg( m/s(4 m m 4 kgm /s Momentum of Airplane 044 00 points An airplane of mass 998 kg flies level to the groundataconstantspeedof75m/srelative to the Earth An observer on the ground along the path of the plane sees the plane a
Version 00 olling & Angular Momentum ramadoss (7 8 distance 98747 m away at an angle above the horizontal of 66453 What is the magnitude of the airplane s angular momentum relative to a ground observer directly below the airplane? Correct answer: 5785 0 9 kgm /s Basic Concepts: L r p L rmv sin(θ (98747 m(998 kg(75 m/ssin(66453 5785 0 9 kgm /s Momentum of Moon 045 00 points There is a moon orbiting an Earth-like planet The mass of the moon is 389 0 kg, the center-to-center separation of the planet and the moon is 46 0 5 km, the orbital period ofthemoonis9days,andtheradiusofthe moon is 530 km What is the angular momentum of the moon about the planet? Correct answer: 378 0 34 kgm /s The angular speed ω in radians per unit time, for a complete circle is ω π T Angular momentum L is L mvr mr ω πmr T π(389 0 kg [ (46 0 5 km 000 m (9 days km ( 4 hr day 378 0 34 kgm /s Projectile Hits a Disk ] 046 (part of 00 points A particle of mass m and speed v 0 collides with and sticks to the edge of a uniform disk of mass M m and radius That is, the distance between the trajectory of the particle and the center of the disk is If the disk is initially at rest and is pivoted about a frictionless axle through its center O perpendicular to the plane of the paper, the angular velocity of the disk plus particle system after the collision in terms of v 0 and? v 0 8 v 0 3 4 M v 0 correct v 0 4 5 3v 0 4 O By conservation of angular momentum L i L f The angular momentum before the collision is L i mv 0 The moment of inertia of the uniform disk is I M (m m The angular momentum after the collision is the sum of the angular momentum due to the mass and that due to the rotating disk: m
Version 00 olling & Angular Momentum ramadoss (7 9 L f mv+iω m ω +I ω m ω, where ω is the angular velocity of the system after the collision For the system of m plus disk, there is no external torque acting on the system, and angular momentum is constant before, during and after the collision, ie, L i L f, so ω v 0 047 (part of 00 points What is the loss of kinetic energy due to the collision process? where v ω, I m and ω v 0 The loss in kinetic energy is K Kf K i 4 mv 0 mv 0 4 mv 0 Projectile Sticks to a od 048 (part of 00 points A projectile of mass m 64 kg moves to the right with speed v 0 7 m/s The projectile strikes and sticks to the end of a stationary rod of mass M 30 kg and length d 85mthatispivotedaboutafrictionless axle through its center v 0 ω m 3 8 mv 0 O d O mv 0 3 4 mv 0 correct 4 3 6 mv 0 5 8 mv 0 The kinetic energy of the system before collision is and after the collision K i mv 0 K f Iω ( M +m Iω (m Iω m ω m ( v 0 4 mv 0, M Find the angular speed of the system right after the collision Correct answer: 557 rad/s The initial angular momentum of the projectileaboutthepivotoisl i d mv 0 With the projectile stuck to the end of the rod, the rotational inertia of the projectile and the rod combined about O is, ( d I m + M d [ m 4 + M ] d ( [ ] (64 kg (30 kg + (85 m 4 67 kgm Using conservation of angular momentum, we have L i I f ω ( r p d mv 0 (3
Version 00 olling & Angular Momentum ramadoss (7 0 Therefore, combining Eqs (, (, and (3, we have ω d mv 0 md + M d 4 6mv 0 (3m+M d 6(64 kg(7 m/s [3(64 kg+(30 kg] (85 m 557 rad/s 049 (part of 00 points Determine the ratio of the kinetic energy lost to the initial kinetic energy Correct answer: 037957 The initial energy of the system is M 3m+M (30 kg 3(64 kg+(30 kg 037957 Qualitative String on ope 050 00 points A particle, held by a string whose other end is attached to a fixed point C, moves counterclockwise in a circle on a horizontal frictionless surface If the string is cut, the angular momentum of the particle about the point C: does not change correct decreases 3 changes direction but not magnitude 4 increases E 0 mv 0 (64 kg(7 m/s 4538 J The final energy of the system is E Iω [( 3m+M [ mv 0 3m 3m+M d ]{ 6mv0 ] [3m+M] d (64 kg(7 m/s [ ] 3(64 kg 3(64 kg+(30 kg 655 J } Therefore, the fractional loss of the energy is f E 0 E E 0 mv 0 [ 3m mv 0 3m+M ] 5 none of these So long as no external work is done on the particle, the angular momentum of the particle must be constant One might wonder how the angular momentum could possibly remain constant when L r p, and the momentum is staying constant while r is changing But, remember that the cross product depends on the relative orientation of p and r Let s set up a coordinate system, so that the x and y directions are in the plane of the surface, and the z direction is perpendicular to the surface Let s also assume that the string is cut at a moment when the particle s instantaneous velocity is purely in the +x direction, and the +y direction is the direction from the particle to the center of the circle Then, we know that the particle s momentum forever has no y component and a constant x component Meanwhile, we have L r p L rpsinθ
Version 00 olling & Angular Momentum ramadoss (7 p(rsinθ pr y (The direction of the angular momentum vector is in the z direction Since all motion of the particle is only in the x direction, this means that the y position of the particle is constant, and therefore, we conclude that the angular momentum of the particle is constant as well ace Down a Plane 05 (part of 3 00 points A thin cylindrical shell and a solid cylinder have the same mass and radius The two are released side by side and roll down, without slipping, from the top of an inclined plane that is m above the ground Find the final linear velocity of the thin cylindrical shellthe acceleration of gravity is 98 m/s Correct answer: 45365 m/s 05 (part of 3 00 points Find the final linear velocity of the solid cylinder Correct answer: 5383 m/s Because there is no slipping, v ω and the rotational inertia of the solid cylinder is I m, so K rot +K trans H I ω + m (ω mgh ( m ω + mv mgh 4 v 3 gh 4 v 3 (98 m/s ( m 5383 m/s Let : H m, and g 98 m/s H h S Because there is no slipping, v ω and the rotational inertia of the cylindrical shell is I m Thus,fromconservationofenergy K rot +K trans H I ω + m(ω mgh ( v (m + mv mgh,so S v gh (98 m/s ( m 45365 m/s 053 (part 3 of 3 00 points When the first object reaches the bottom, what is the height above the ground of the other object? Correct answer: 055 m Because v > v, object reaches the bottom first Since acceleration is constant, then Since v i 0 Thus v f s t second object is v v f +v i v f and the final velocity of the v s t In that same time t the first object travels a distance s but its velocity is v < v (the speed at the buttom, so let
Version 00 olling & Angular Momentum ramadoss (7 v s t Dividing these two equations, one obtains s s v v 4 3 gh g(h h where h is the height of object when object reaches the bottom of the incline From similar triangle in the figure keywords: H H h s s 4 3 gh g(h h H H h 4H 3(H h H 4 H h 3 H H h 4 3 3H 4H 4h h 4 H ( m 4 055 m igid Body Torque 054 00 points A force F F 0 (î+ĵ+ˆk acts on a rigid body at a point r r 0 (î ĵ away from the axis of rotation What is the resulting torque on the body? r 0 F 0 (ˆk î ĵ r 0 F 0 (ˆk î ĵ 3 r 0 F 0 (ˆk î ĵ 4 r 0 F 0 (ˆk î ĵ 5 r 0 F 0 (ˆk î ĵ correct 6 r 0 F 0 (ĵ î ˆk 7 Zero 8 r 0 F 0 (î ˆk ĵ 9 r 0 F 0 (î+ĵ+ˆk τ r F F 0 r 0 (î ĵ+î ˆk ĵ î ĵ ˆk F 0 r 0 (ˆk ĵ+ˆk î F 0 r 0 (ˆk ĵ î ing and Disk ace 055 00 points A solid disk (thin cylindrical object with uniform density and a ring (all of its mass located at the radius of the ring roll down an incline without slipping Let m ring, m disk be the inertial masses and r ring and r disk the radii The ring is slower than the disk only if m ring > m disk m ring m disk and r ring r disk 3 r ring > r disk 4Theringisalwaysslowerregardlessofthe relative values of m and r correct Say the two objects start at the top of an incline of vertical height h Since v rω(no slipping conservation of energy gives mgh mv + Iω mv + (m+ I v r v Ir
Version 00 olling & Angular Momentum ramadoss (7 3 I ring mr and I disk mr so I Cmr and θ 40, and 009 m v (m+cm mgh mv (+C mgh, gh v +C Therefore, the speed is independent of both mass and radius and depends only on the shape of the object Since C ring > C disk and the constant C appears in the denominator, the ring is always slower keywords: olling Basketball 056 00 points A 80 g basketball has a 8 cm diameter and may be approximated as a thin spherical shell N mgsinθ mg The moment of inertia of the ball about the point of contact between the ball and the inclined plane is P I P I cm +md 3 m +m θ 5 3 m 608 m 80 g µ 059 40 Starting from rest, how long will it take a basketball to roll without slipping 608 m down an incline that makes an angle of 40 with the horizontal? The moment of inertia of a thin spherical shell of radius and mass m is I 3 m, the acceleration due to gravity is 98 m/s, and the coefficient of friction is 059 Correct answer: 79368 s Let : l 608 m, The net torque about the point of contact between the ball and the inclined plane is mg sinθ I P α 5 3 m α a mg sinθ 5 3 m 3 5 gsinθ Because the sphere starts from rest, its center of mass moves a distance l at l t a (608 m 377959 m/s 79368 s evolving Mass on a Spring 057 00 points A 5 kg mass is on a spring of length 9 cm
Version 00 olling & Angular Momentum ramadoss (7 4 and stiffness constant 33 N/cm It is spun at a speed of 08 /s in the absence of gravity How far will the spring stretch? Correct answer: 03797 cm The centripetal force is F c mv r m(rω r mrω Since the spring has an unstretched length L, and the rotation will cause it to stretch a distance x, the radius of rotationis L+x and F c m(l+xω mlω +mxω The unbalanced force on the mass is supplied by the spring, so F c F sp mlω +mxω kx mlω kx mxω x mlω k mω The terms in the denominator are not compatible: N ( cm kg kgm/s s cm kg s so a conversion is necessary: kgm/s cm 00cm m kg s kg s kg s System Angular Momentum 058 00 points Suppose a physics instructor seated himself at rest on a low-friction piano stool and heldaspinningbikewheelwithitsangularvelocity vector initially horizontal Suppose the instructor then turned the bike wheel so that its angular velocity vector pointed vertically downward When this turn was completed, the instructor would be spinning rapidly on the stool What would be the final, total angular momentum of the system of instructor plus bike wheel, projected along the vertical axis, the only axis along which the system can spin? Twice its initial value The same, nonzero value it had initially 3Equalbutoppositetowhatthewheelhad initially 4 Zero, as it was initially correct The system originally had zero angular momentum about a vertical axis, so no matter what happens within the system of person, bike wheel and stool, as long as no external torque is involved, the angular momentum along this axis, say z, is always zero When the bike wheel s angular momentum vector is twisted downward by an internal torque, the rest of the system develops an angular momentum vector of equal magnitude in the opposite direction to maintain a total angular momentum about the vertical axis of zero There is, of course, a third law for torques; when the man exerts a torque to twist the bike wheel, it exerts an equal and opposite torque on him and the stool Supernova Explosion 059 00 points A star of radius 55 0 5 km rotates about its axis with a period of 99 days The star undergoes a supernova explosion, whereby its core collapses into a neutron star of radius 67 km Estimate the period of the neutron star (assume the mass remains constant Correct answer: 0006933 s Given : T 99 days 85536 0 6 s, r i 55 0 5 km, and r f 67 km, We will assume that during the collapse of the star,
Version 00 olling & Angular Momentum ramadoss (7 5 ( no torque acts on it, ( it remains spherical, (3 its mass remains constant Since I is proportional to r, and ω π T, conservation of angular momentum gives T f T i ( rf r i (85536 0 6 s 0006933 s ( 67 km 55 0 5 km Yo Yo on table with friction 060 00 points Ayo-yo,arrangedasshown,restsonasurface the relative sizes of these two forces First, analyze the torques at the point of contact between the ground and the yo-yo, which will be instantaneously at rest if the yo-yo is not moving In this reference frame, we have (I + m α ( rf, where we used the parallel axis theorem to translate the center of rotation to the hub We can see from the direction of the torques that α must be counterclockwise Now, analyze the problem with the center of rotation being the center of the yo-yo We find that the frictional force must act to the left, because otherwise the angular acceleration would be in the wrong direction, as both torqueswouldbetotheleft, andwefoundthe angular acceleration already In this frame, When a force F is applied to the string as shown, the yo-yo: moves to the right and does not rotate moves to the left and rotates clockwise 3 moves to the left and rotates counterclockwise 4 moves to the right and rotates clockwise correct 5 moves to the right and rotates counterclockwise To solve this problem, the first thing we note is that there are two horizontal forces acting on the yo-yo: static friction and the applied force Let the radius of the yo-yo be, and the distance from the center of the yo-yo to the point of application of F to be r, and the moment of inertial of the yo-yo to be I In order to determine the net direction of motion, we need to determine Iα F s Fr I( rf I +m F s Fr ( I( r F s F I +m +r ( I Ir +Ir +m r F I +m ( I +m ( r F s F I +m So, we can see from this expression that, if r <, then the numerator is less than the denominator, and therefore F s < F, and the net force on the yo-yo is to the right Yo Yo N 07 06 (part of 4 00 points Assume: Drag (friction is negligible Given: g 98 m/s The density of this large Yo-Yo like solid is uniform throughout The Yo-Yo like solid has a mass of 9 kg
Version 00 olling & Angular Momentum ramadoss (7 6 And the moment of inertia are I M r r 4 r +r M, l l l Front View A cord is wrapped around the stem of the Yo-Yo like solid and attached to the ceiling The radius of the stem is 4 m and the radius of the disk is 9 m h 9 m 4 m 9 kg Cross sectional Side View Calculate the moment of inertia about the center of mass (axis of rotation Correct answer: 0904 kgm Let : g 98 m/s, m 9 kg, r 4 m, r 9 m, and h 3 m see, Part 4 The areal density of the Yo-Yo is σ A +A M π(r +r M Therefore the masses of the disks are M σa, so ω M r r M, and +r M σa, so I M r r 4 r M, and +r I I +I r 4 +r4 r M +r (4 m 4 +(9 m 4 (4 m +(9 m (9 kg 0904 kgm 06 (part of 4 00 points What is the vertical acceleration of the center of mass of the Yo-Yo? Correct answer: 30697 m/s Basic Concepts: General rotational kinematics, energy conservation In particular K rot Iω, and the moment of inertia for a uniform disk is I M Solution: On the center of mass the total force and torque is F : T M g M a, so T M (g a, ( τ : T r Iα r 4 +r4 r +r M a r, so M r r M +r T r 4 +r4 r M a ( (r +r
Version 00 olling & Angular Momentum ramadoss (7 7 Setting Eq equal to, we have M (g a g g r 4 +r4 r (r +r M a r (r +r +r4 +r4 r (r +r 3r 4 +r r +r4 r (r +r a r [r +r ] r 4 g (3 +[r +r ] (4 m [(4 m +(9 m ] (4 m 4 +[(4 m +(9 m ] (98 m/s 30697 m/s a a Find the velocity v of the center of mass of the disk at the height h 3 m Correct answer: 88 m/s By energy conservation we know that the initial energy equals the final energy In this case energy conservation implies M gh K linear +K rot where K linear M v We also know that K rot isequalto I ω wherei r 4 +r4 r M +r for our uniform disk, and we know ω v This implies 063 (part 3 of 4 00 points What is magnitude of the torque the cord exerts on the center of mass of the Yo-Yo? Correct answer: 7895 N m SubstitutingafromEq3intoEq,wethe tension is r 4 T +r4 r 4 M g (4 +[r +r ] (4 m 4 +(9 m 4 (4 m 4 +[(4 m +(9 m ] (9 kg(98 m/s 9548 N Therefore, the torque is τ r T r [r 4 +r4 ] r 4 M g (5 +[r +r ] (4 m[(4 m 4 +(9 m 4 ] (4 m 4 +[(4 m +(9 m ] (9 kg(98 m/s 7895 Nm Thus K rot Iω 4 r 4 +r4 r (r +r M v (6 M gh K linear +K rot M v + r 4 +r4 4 r [r +r ] M v 4 r 4 +r r +r4 +r4 r [r +r ] M v 3r4 +r r +r4 4r [r +r ] M v (7 Therefore, the velocity is 4r v [r +r ] 3r 4 gh (8 +r r +r4 4(4 m [(4 m +(9 m ] 3(4 m 4 +(4 m (9 m +(9 m 4 (98 m/s (3 m 88 m/s 064 (part 4 of 4 00 points The Yo-Yo is released from rest at height h Yo Yo MC 05 065 (part of 00 points
Version 00 olling & Angular Momentum ramadoss (7 8 You are playing with a Yo-Yo which you can describe as a uniform disk with mass m and radiusr The string oftheyo-yohas alength L The Yo-Yo rolls down vertically h Determine the tension T of the string at a height h below the original position h as indicated in the sketch T 3mg T r L mg 3 T L h r 4 T 3 5 T mg 6 T 4 mg 7 T mg 8 T 5 mg 9 T 7 4 mg 0 T 5 mg mg mg correct Apply F ma to the center of mass of the disk, m F net F : mg T ma ( Apply τ Iα to the Yo-Yo (where the torque is calculated about the center, T r Iα mr α r ω T ma (3 In the 3 rd step we used a αr Substituting Eq 3 into leads to, mg ma ma, so a g, and 3 T ma 3 mg 066 (part of 00 points After the Yo-Yo has been elastically reflected at length L, it moves upward Determine the speed of its center when it is at a height h below the original position h v 8 g h v 3g h 3 v g h gl 4 v g h 5 v g h 6 v g h 4 7 v g h correct 3 3 8 v g h 5 9 v 3 g h r 0 v L g h The application of work-energy equation gives mra, so ( mg h mv + Iω
Version 00 olling & Angular Momentum ramadoss (7 9 But So Iω ( mr ω 4 mv ( mg h + 4 v 4 3 g h mv Yo Yo MC 06 067 (part of 5 00 points A yo-yo consists of two identical, uniform disks whose total mass is M and radius connected between the disks by a shaft of radius r and negligible mass See the figure below r h The combined mass of the three disks (the yo-yo is M The yo-yo rolls down an inextensible, thin string of length h + l, where l Neglect the mass and moment of inertia of the shaft The string is initially wound around the shaft and unwinds as the yo-yo rolls down the string What is the moment of inertia, I, of the yo-yo about the axis through the centers of mass of the two disks? I 4 M (r+ I M r 3 I M ( +r 4 I M (r+ 5 I M ( +r 6 I 4 M ( r M ω 7 I M r 8 I M (r + 9 I M correct 0 I M Basic Concepts: F net m a τ net Iα No-slip condition (magnitudes: s rθ, v rω, a rα Solution: For a disk, the moment of inertia is I disk M Since there are two disks, each with mass M, the moment of inertia of the yo-yo is I ( M M emember: We are neglecting the mass and moment of inertia of the connecting spindle 068 (part of 5 00 points What is the ratio of the magnitude of the gravitational acceleration g to the magnitude of the acceleration a of the center of mass of the yo-yo? g a r g a r 3 g ( a r 4 g + ( correct a r 5 g a r 6 g + a r 7 g ( + a r
Version 00 olling & Angular Momentum ramadoss (7 30 8 g + a r 9 g + a r 0 g + a r We will take the upward direction as the positive y direction and clockwise as the positive sense of rotation y T θ M(a+g I a r g Ia M r a [ + I M r g + I a M r ] a, so The ratio of the moment of inertia, I, to M r is I M r M M r ( r W M g Looking at our free body diagram above, we see that the net force on the yo-yo is F net T W M a Similarly, the net torque on the yo-yo about the center of mass is τ net T rsin(90 T r Iα, where the sign of τ is given by the right-hand rule, in the sense that clockwise is positive here To relate the rotational and translational motion we have the fact that the string does not slip as it unrolls from the yo-yo This is expressed in the no-slip condition, a rα Substituting for α, we have T I a r (If we looked the yo-yo from the other side so the tension acted on the left side of the yo-yo, thesenseofrotationwouldbereversed; inthis case, τ and α would both have the opposite sign We can now substitute the tension T into our equation for the acceleration T M a+w M (a+g, so So, the ratio of accelerations is g + I a M r + ( r 069 (part 3 of 5 00 points Assume: The yo-yo started from rest What is the ratio of the final kinetic energy K f totherotationalkineticenergyk rot about the axis through the center of mass when the yo-yo has fallen the entire length of the string? K f r +( K rot K f r + K rot 3 4 5 6 7 8 K f K rot r K f K rot + r K f K rot ( r K f + r K rot K f r + K rot K f r +( correct K rot
Version 00 olling & Angular Momentum ramadoss (7 3 9 K f r K rot Here, we will use the principle of conservation of energy There are no dissipative forces This means that the total energy of the yo-yo plus string system is conserved: K f +U f K i +U i Sincetheyo-yo startsfrom rest, K i 0 Ifwe measure the potential energy from the starting height, U i 0 The height the yo-yo falls is equal to the length, l, of string unwound from the yo-yo Putting this together, we get K f mgl 0, so K f mgl But, this is the total kinetic energy of both translation and rotation! We need to relate the two type of motion to each other As before, this comes through a no-slip condition; this time it is v rω The ratio of translational kinetic energy to rotational kinetic energy is then K tr K rot M v I ω M r M ( r Krot K tr M v ( I v r This means that ( ] r K f K rot [+, so K f r +( K rot 070 (part 4 of 5 00 points If E is the total mechanical energy, p is the linear momentum of the center of mass and L is the angular momentum relative to the rotational axis through the center of mass, mark the correct statement Neither E, p, or L is conserved E is conserved but not p and L correct 3 E and p are conserved but not L 4 p and L are conserved, but not E 5 E, p, and L are all conserved 6 p is conserved, but not E and L 7 E and L are conserved, but not p 8 L is conserved, but not E and p p of the yo-yo is NOT conserved There is a net force acting on the yo-yo You can see this explicitly because the initial momentum is zero and the final momentum is not zero L of the yo-yo is NOT conserved There is a net torque acting on the yo-yo Again, you could also see this because the initial angular momentum is zero and the final angular momentum is nonzero E is conserved Although there is an external force acting on the string-yo-yo-system, it does not do any work to change the total energy of the system This force is exerted on thestringbythepersonholdingtheendofthe string fixed The displacement is zero, however, so no work is done If the person had instead pulled the end of the string up while the yo-yo fell, then the person would do work on the system and increase its energy The amountofenergyaddedwouldbeequaltothe amount of work done in lifting the string 07 (part 5 of 5 00 points What is the SI unit of torque τ? [τ] Nm/s [τ] kgm /s correct 3 [τ] N/m
Version 00 olling & Angular Momentum ramadoss (7 3 4 [τ] kgs /m 5 [τ] kgm/s 6 [τ] kgm/s 7 [τ] Nkg 8 [τ] kg s/m The SI unit of torque is Nm By writing Newtons in terms of fundamental quantities: [F] N kgm/s (compare, for example, F ma Therefore [τ] Nm kgm /s Yo Yo MC 03 07 (part of 00 points Given: M, g, and A string is wound around a uniform disk of radius and mass M The disk is released fromrest ataheight hwiththestringvertical and its tip end tied to a fixed support as in figure 6 T M g h 7 T M g+m g h 8 T M g 9 T M g h 0 T 3 M g Basic Concepts: General rotational kinematics, energy conservation In particular K rot Iω, and the moment of inertia for a uniform disk is I M Solution: On the center of mass the total force and torque is F : T M g M a, so T M (g a, ( τ : T Iα h ω M ( a, so T M a ( Derive an expression for the tension, T, on the string at height h + h, as the disk descends T M g ( h T M g 3 T 3 M g correct 4 T M g ( h h 5 T M g h M Adding Eq and, we have Therefore 3T M g T 3 M g (3 073 (part of 00 points What is the equation for the vertical accelerationofthecenterofmassintermsofthegiven parameters? a ( h a 4g g