Chapter 11 Rolling, Torque, Angular Momentum Copyright
11.2 Rolling as Translational and Rotation Combined Motion of Translation : i.e.motion along a straight line Motion of Rotation : rotation about a fixed axis Pure Translation Motion + Pure Rotation Motion = ROLLING MOTION Translation of COM Rotation around COM com Smooth Rolling of a Disk Although the center of the object moves in a straight line parallel to the surface, a point on the rim certainly does not.
11.2 Rolling as Translational and Rotation Combined We consider only objects that roll smoothly (no slip) The center of mass (com) of the object moves in a straight line parallel to the surface The object rotates around the com as it moves The rotational motion is defined by: Wheel moves through the arc length of s Wheel s COM moves through a translational motion PATHS ARE EQUAL s= R
11.2 Rolling as Translational and Rotation Combined The figure shows how the velocities of translation and rotation combine at different points on the wheel B D Every point on the wheel rotates about com with angular speed of Everypoint outmost part of the wheel has linear speed of ν; Where ν = R = ν com All points on the wheel move to the right with same linear velocity, ν com At the bottom of the wheel (point P), the portion of the wheel is stationary The portion at the top (point T) moving at a speed 2ν com At points B and D, the speeds are smaller than the point T.
11.3 The Kinetic Energy of Rolling Rolling can be considered as pure rotation around contact point P. (V p =0) com R (V p = 0) Rotational inertia about COM: I com Rotational inertia about point P: I P = I com + MR 2 M : mass of the wheel, I com : rotational inertia about an axis through its center of mass R : the wheel s radius, at a perpendicular distance h). I P = I com + MR 2 K = ½ I com 2 + ½ M R 2 2 inserting v com (v com = R), then; (Using the parallel-axis theorem ) K.E of ROLLING: Rotational kinetic energy + Translational kinetic energy
11.4 The Forces of Rolling: Friction and Rolling A wheel rolls horizontally without sliding while accelerating with linear acceleration a com. A static frictional force f s acts on the wheel at P, opposing its tendency to slide. The magnitudes of the linear acceleration a com, and the angular acceleration a can be related by; R :the radius of the wheel. α : angular acceleration Rolling is possible when there is friction between the surface and the rolling object. The frictional force provides the torque to rotate the object.
11.4 The Forces of Rolling: Friction and Rolling If the wheel slides when the net force acts on it, the frictional force that acts at P in Figure is a kinetic frictional force, f k. The motion then is not smooth rolling, and the above relation (a COM =ar) does not apply to the motion. Sliding of Wheel Motion is not Smooth Rolling f k
11.4 The Forces of Rolling: Rolling Down a Ramp A round uniform body of radius R rolls (smoothly f s ) down a ramp. We want to find an expression for the body s acceleration a com,x down the ramp. We do this by using Newton s second law in its linear version (F net =Ma) and its angular version (t net =Ia). For smooth rolling down a ramp: 1. F g : The gravitational force is vertically down 2. F N : The normal force is perpendicular to the ramp 3. f s :The force of friction points up the slope (opposing the sliding) 4. a com =ar where a com is ( x) and a is (+, ccw) (see 4)
11.4 The Forces of Rolling: Rolling Down a Ramp We can use this equation to find the acceleration of such a body Note that the frictional force produces the rotation Without friction, the object will simply slide K i + U i = K f + U f Mgh = ½ I COM ω 2 + ½ Mv COM 2 v cm = ωr
11.4 The Forces of Rolling: Rolling Down a Ramp Sample problem: Rolling Down a Ramp A uniform ball, of mass M=6.00 kg and radius R, rolls smoothly from rest down a ramp at angle =30.0. a) The ball descends a vertical height h=1.20 m to reach the bottom of the ramp. What is its speed at the bottom? Calculations: Where I com is the ball s rotational inertia about an axis through its center of mass, v com is the requested speed at the bottom, and is the angular speed there. Substituting v com /R for, and 2/5 MR 2 for I com, b) What are the magnitude and direction of the frictional force on the ball as it rolls down the ramp? Calculations: First we need to determine the ball s acceleration a com,x : We can now solve for the frictional force:
11.6 Torque Revisited Previously, torque was defined only for a rotating body and a fixed axis. Now we redefine it for an individual particle that moves along any path relative to a fixed point. The path need not be a circle; torque is now a vector. Direction determined with right-hand-rule. Figure (a) A force F, lying in an x-y plane, acts on a particle at point A. (b) This force produces a torque t = r x F on the particle with respect to the origin O. By the right-hand rule for vector (cross) products, the torque vector points in the positive direction of z. Its magnitude is given by in (b) and by in (c).
11.6 Torque Revisited The general equation for torque is: We can also write the magnitude as: Or, using the perpendicular component of force or the moment arm of F:
11.6 Torque Revisited Sample problem: Torque on a particle due to a force In Figure, three forces, each magnitude 2.0 N, act on a particle. The particle is in the xz plane at a point A given by position vector r, where r = 3.0 m and = 30 o. Force F 1 is parallel to the x axis, force F 2 is parallel to the z axis, and F 3 is parallel to the y axis. What is the torque, about the origion O, due to each force? Calculations: Because we want the torques with respect to the origin O, the vector required for each cross product is the given position vector r. To determine the angle q between the direction of r and the direction of each force, we shift the force vectors of Fig.a, each in turn, so that their tails are at the origin. Figures b, c, and d, which are direct views of the xz plane, show the shifted force vectors F1, F2, and F3. respectively. In Fig. d, the angle between the directions of r and t is 90. Now, we find the magnitudes of the torques to be
11.6 Torque Revisited
11.7 Angular Momentum Here we investigate the angular counterpart to linear momentum We write: Note that the particle need not rotate around O to have angular momentum around it. The unit of angular momentum is kg m 2 /s, or J s To find the direction of angular momentum, use the right-hand rule to relate r and v to the result To find the magnitude, use the equation for the magnitude of a cross product: Angular momentum has meaning only with respect to a specified origin Which can also be written as:
11.7 Angular Momentum Sample problem: Angular Momentum
11.8 Newton s 2 nd Law in Angular Form We rewrite Newton's second law as: Newton s 2 nd Law for translation PROOF: Newton s 2 nd Law for ANGULAR form The torque and the angular momentum must be defined with respect to the same point (usually the origin).
11.8 Newton s 2 nd Law in Angular Form Sample problem: Torque, Penguin Fall Calculations: The magnitude of l can be found by using The perpendicular distance between O and an extension of vector p is the given distance D. The speed of an object that has fallen from rest for a time t is v =gt. Therefore, To find the direction of we use the right-hand rule for the vector product, and find that the direction is into the plane of the figure. The vector changes with time in magnitude only; its direction remains unchanged. (b) About the origin O, what is the torque on the penguin due to the gravitational force? Calculations: Using the right-hand rule for the vector product we find that the direction of t is the negative direction of the z axis, the same as l.
11.9 The Angular Momentum of a System of Particles The total angular momentum L of the system is the (vector) sum of the angular momenta l of the individual particles (here with label i): With time, the angular momenta of individual particles may change because of interactions between the particles or with the outside. The rate of change of the net angular momentum is: Therefore, the net external torque acting on a system of particles is equal to the time rate of change of the system s total angular momentum L.
11.9 The Angular Momentum of a System of Particles Note that the torque and angular momentum must be measured relative to the same origin. If the center of mass is accelerating, then that origin must be the center of mass. a) A rigid body rotates about a z axis with angular speed. A mass element of mass Dm i within the body moves about the z axis in a circle with radius. The mass element has linear momentum p i and it is located relative to the origin O by position vector r i. Here the mass element is shown when is parallel to the x axis. b) The angular momentum l i, with respect to O, of the mass element in (a). The z component l iz is also shown.
11.9 The Angular Momentum of a System of Particles We can find the angular momentum of a rigid body through summation: The sum is the rotational inertia I of the body. Therefore this simplifies to:
11.11 Conservation of Angular Momentum Since we have a new version of Newton's second law, we also have a new conservation law: If the net external torque acting on a system is zero, the angular momentum L of the system remains constant, no matter what changes take place within the system. The law of conservation of angular momentum states that, for an isolated system, (net initial angular momentum) = (net final angular momentum) Depending on the torques acting on the system, the angular momentum of the system might be conserved in one or two directions but not in all directions. This means that if external torque along an axis is zero then L is constant.
11.11 Conservation of Angular Momentum If the component of the net external torque on a system along a certain axis is zero, then the component of the angular momentum of the system along that axis cannot change, no matter what changes take place within the system. (a) The student has a relatively large rotational inertia about the rotation axis and a relatively small angular speed. (b) By decreasing his rotational inertia, the student automatically increases his angular speed. The angular momentum of the rotating system remains unchanged.
11.11 Conservation of Angular Momentum Examples: Angular momentum conservation A student spinning on a stool: rotation speeds up when arms are brought in, slows down when arms are extended. A springboard diver: rotational speed is controlled by tucking her arms and legs in, which reduces rotational inertia and increases rotational speed. A long jumper: the angular momentum caused by the torque during the initial jump can be transferred to the rotation of the arms, by windmilling them, keeping the jumper upright.
11.11 Conservation of Angular Momentum Sample problem
11 Table 11.1
11 Table 10.3
Example:
11 Summary Rolling Bodies Eq. (11-2) Eq. (11-5) Eq. (11-6) Torque as a Vector Direction given by the righthand rule Eq. (11-14) Angular Momentum of a Particle Newton's Second Law in Angular Form Eq. (11-18) Eq. (11-23)
11 Summary Angular Momentum of a System of Particles Angular Momentum of a Rigid Body Eq. (11-31) Eq. (11-26) Eq. (11-29) Conservation of Angular Momentum Eq. (11-32) Precession of a Gyroscope Eq. (11-46) Eq. (11-33)
11 Solved Problems 1. In Figure, a solid cylinder of radius 10 cm and mass 12 kg starts from rest and rolls without slipping a distance L=6.0 m down a roof that is inclined at the angle ϴ=30 o. (a) What is the angular speed of the cylinder about its center as it leaves the roof? (b) The roof's edge is at height H=5.0 m. How far horizontally from the roof's edge does the cylinder hit the level ground?
11 Solved Problems
11 Solved Problems 2. At time t=0, a 3.0 kg particle with velocity v=(5.0 m/s)i (6.0 m/s)j is at x=3.0 m, y=8.0 m. It is pulled by a 7.0 N force in the negative x direction. About the origin, what are (a) the particle's angular momentum, (b) the torque acting on the particle, and (c) the rate at which the angular momentum is changing?
11 Solved Problems 3. In Figure, a small 50 g block slides down a frictionless surface through height h=20 cm and then sticks to a uniform rod of mass 100 g and length 40 cm. The rod pivots about point O through angle ϴ before momentarily stopping. Find ϴ.
11 Solved Problems
11 Solved Problems
11 Solved Problems 4. In Figure, a constant horizontal force F app of magnitude 12 N is applied to a uniform solid cylinder by fishing line wrapped around the cylinder. The mass of the cylinder is 10 kg, its radius is 0.10 m, and the cylinder rolls smoothly on the horizontal surface. (a) What is the magnitude of the acceleration of the center of mass of the cylinder? (b) What is the magnitude of the angular acceleration of the cylinder about the center of mass? (c) In unit vector notation, what is the frictional force acting on the cylinder?
11 Solved Problems (a) What is the magnitude of the acceleration of the center of mass of the cylinder? (b) What is the magnitude of the angular acceleration of the cylinder about the center of mass? (c) In unit vector notation, what is the frictional force acting on the cylinder?
11 Solved Problems 5. Figure shows a uniform disk, with mass M = 2.5 kg and radius R = 20 cm, mounted on a fixed horizontal axle. A block with mass m = 1.2 kg hangs from a massless cord that is wrapped around the rim of the disk. Find the acceleration (a) of the falling block, the angular acceleration (α) of the disk, and the tension (T ) in the cord. The cord does not slip, and there is no friction at the axle. (I = 1/2 MR 2 )
1-8 Scientific Notation Additional Materials
11.5 The Yo-Yo As a yo-yo moves down a string, it loses potential energy mgh but gains rotational and translational kinetic energy
11.12 Precession of a Gyroscope (a) A non-spinning gyroscope falls by rotating in an xz plane because of torque t. (b) A rapidly spinning gyroscope, with angular momentum, L, precesses around the z axis. Its precessional motion is in the xy plane. This rotation is called precession (c) The change in angular momentum, dl/dt, leads to a rotation of L about O.
11.12 Precession of a Gyroscope The angular momentum of a (rapidly spinning) gyroscope is: Eq. (11-43) The torque can only change the direction of L, not its magnitude, because of (11-43) The only way its direction can change along the direction of the torque without its magnitude changing is if it rotates around the central axis. Therefore it precesses instead of toppling over.
11.12 Precession of a Gyroscope The precession rate is given by: True for a sufficiently rapid spin rate. Independent of mass, (I is proportional to M) but does depend on g. Valid for a gyroscope at an angle to the horizontal as well (a top for instance).
11.11 Conservation of Angular Momentum Sample problem Its angular momentum is now - L wh.the inversion results in the student, the stool, and the wheel s center rotating together as a composite rigid body about the stool s rotation axis, with rotational inertia I b = 6.8 kg m 2. With what angular speed b and in what direction does the composite body rotate after the inversion of the wheel? Figure 11-20 a shows a student, sitting on a stool that can rotate freely about a vertical axis. The student, initially at rest, is holding a bicycle wheel whose rim is loaded with lead and whose rotational inertia I wh about its central axis is 1.2 kg m 2. (The rim contains lead in order to make the value of I wh substantial.) The wheel is rotating at an angular speed wh of 3.9 rev/s; as seen from overhead, the rotation is counterclockwise. The axis of the wheel is vertical, and the angular momentum L wh of the wheel points vertically upward. The student now inverts the wheel (Fig. 11-20b) so L wh that, as seen from overhead, it is rotating clockwise.
11-2 Rolling as Translation and Rotation Combined Learning Objectives 11.01 Identify that smooth rolling can be considered as a combination of pure translation and pure rotation. 11.02 Apply the relationship between the center-of-mass speed and the angular speed of a body in smooth rolling. Figure 11-2
11-3_4 Forces and Kinetic Energy of Rolling Learning Objectives 11.03 Calculate the kinetic energy of a body in smooth rolling as the sum of the translational kinetic energy of the center of mass and the rotational kinetic energy around the center of mass. 11.04 Apply the relationship between the work done on a smoothly rolling object and its kinetic energy change. 11.05 For smooth rolling (and thus no sliding), conserve mechanical energy to relate initial energy values to the values at a later point. 11.06 Draw a free-body diagram of an accelerating body that is smoothly rolling on a horizontal surface or up or down on a ramp.
11-3_4 Forces and Kinetic Energy of Rolling 11.07 Apply the relationship between the center-of-mass acceleration and the angular acceleration. 11.08 For smooth rolling up or down a ramp, apply the relationship between the object s acceleration, its rotational inertia, and the angle of the ramp.
11-5 The Yo-Yo Learning Objectives 11.09 Draw a free-body diagram of a yo-yo moving up or down its string. 11.10 Identify that a yo-yo is effectively an object that rolls smoothly up or down a ramp with an incline angle of 90. 11.11 For a yo-yo moving up or down its string, apply the relationship between the yoyo's acceleration and its rotational inertia. 11.12 Determine the tension in a yo-yo's string as the yo-yo moves up or down the string.
11-6 Torque Revisited Learning Objectives 11.13 Identify that torque is a vector quantity. 11.14 Identify that the point about which a torque is calculated must always be specified. 11.15 Calculate the torque due to a force on a particle by taking the cross product of the particle's position vector and the force vector, in either unit-vector notation or magnitude-angle notation. 11.16 Use the right-hand rule for cross products to find the direction of a torque vector.
11-7 Angular Momentum Learning Objectives 11.17 Identify that angular momentum is a vector quantity. 11.18 Identify that the fixed point about which an angular momentum is calculated must always be specified. 11.19 Calculate the angular momentum of a particle by taking the cross product of the particle's position vector and its momentum vector, in either unit-vector notation or magnitude-angle notation. 11.20 Use the right-hand rule for cross products to find the direction of an angular momentum vector.
11-8 Newton's Second Law in Angular Form Learning Objectives 11.21 Apply Newton's second law in angular form to relate the torque acting on a particle to the resulting rate of change of the particle's angular momentum, all relative to a specified point.
11-9 Angular Momentum of a Rigid Body Learning Objectives 11.22 For a system of particles, apply Newton's second law in angular form to relate the net torque acting on the system to the rate of the resulting change in the system's angular momentum. 11.23 Apply the relationship between the angular momentum of a rigid body rotating around a fixed axis and the body's rotational inertia and angular speed around that axis. 11.24 If two rigid bodies rotate about the same axis, calculate their total angular momentum.
11-11 Conservation of Angular Momentum Learning Objectives 11.25 When no external net torque acts on a system along a specified axis, apply the conservation of angular momentum to relate the initial angular momentum value along that axis to the value at a later instant.
11-9 Precession of a Gyroscope Learning Objectives 11.26 Identify that the gravitational force acting on a spinning gyroscope causes the spin angular momentum vector (and thus the gyroscope) to rotate about the vertical axis in a motion called precession. 11.27 Calculate the precession rate of a gyroscope. 11.28 Identify that a gyroscope's precession rate is independent of the gyroscope's mass.