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Chapter 11: Rotational Vectors and Angular Momentum Vector (cross) products Definition of vector product and its properties Axis of rotation unit vector normal to the plane defined by and

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Vector (cross) products (cont’d) Properties of vector product x y z : unit vector in x,y,z direction Consider three vectors: Then

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Vector (cross) products (cont’d) Properties of vector product (cont’d) x y z

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Torque Case for 1 point-like object of mass m with 1 force x y massless rigid rod r is constantonly tangential component of causes rotation 0 torque moment of inertia unit Nm Torque is a quantitative measure of the tendency of a force to cause or change the rotational motion.

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Torque Case for 1 point-like object of mass m with 1 force x y lever arm Define is aligned with the rotation axis

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Torque (cont’d) Case for 2 point-like objects with 2 forces increases decreases x y Example: A see-saw in balance Mm R r Mg mg

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Work & energy (I) A massive body on massless rigid rod x y Work done by the force: Also W= K

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Work & energy (I) (cont’d) Power in rotational motion Work done by the force: Power :

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Correspondence between linear & angular quantities linear angular displacement velocity acceleration mass force Newton’s law kinetic energy work

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Work & energy (II) A massive body in rotational & translational motion Kinetic energy: Now where is the velocity with respect to the center of mass (COM) and the velocity of COM. Then 0 com

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Work & energy (II) cont’d A massive body in rotational & translational motion (cont’d) kinetic energy due to translational motion kinetic energy due to rotational motion about a rotation axis through COM

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Work & energy (II) cont’d A massive body in rotational & translational motion (cont’d) Example: A rolling cylinder (without sliding due to friction) P P O O s s View point 1: + = OO O This view point was used in the last few slides rotation translation

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Work & energy (II) cont’d A massive body in rotational & translational motion (cont’d) Example: A rolling cylinder (without sliding) (cont’d) View point 2: Any point in the cylinder rotates around P rotation axis P From the parallel axis theorem rotational translational

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Angular momentum & torque x y lever arm Define angular momentum as: Since and 0 angular momentum net torque Angular momentum of a particle

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Angular momentum & torque As before let’s break down vectors into two components Then ang. mom. about COM + ang. mom. of COM 0 position vector for com velocity of com Angular momentum of a multi-particle system

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Conservation of angular momentum If the net torque is zero, the total angular momentum of the system is conserved. Conservation of angular momentum

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Angular momentum & torque x z plane A plane B plane A plane B In general, because of non-zero x/y component of angular momentumunless the object is symmetric about the axis of rotation. If the object is symmetric about the axis of rotation, then Angular momentum vector and axis of rotation

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Gyroscope Gyroscopic motion pivot support mg remove pivot precession

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Gyroscope Principle of gyroscopic pivot z y x

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Gyroscope x y Precession angular speed: Assumption: The angular momentum vector is associated only with the spin of the flywheel and is purely vertical. The procession is much slower than the rotation,

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Gyroscope

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Spinning Top

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Example A disk of mass M and radius R rotates around the axis with angular velocity i. A second identical disk, initially not rotating, is dropped on top of the first. There is friction between the disks, and eventually they rotate together with angular velocity f. 1) f = i 2) f = ½ i 3) f = ¼ i ii z ff z

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First realize that there are no external torques acting on the two-disk system. Angular momentum will be conserved! ii z 2 1 ff z

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Example You are sitting on a freely rotating bar-stool with your arms stretched out and a heavy glass mug in each hand. Your friend gives you a twist and you start rotating around a vertical axis though the center of the stool. You can assume that the bearing the stool turns on is frictionless, and that there is no net external torque present once you have started spinning. You now pull your arms and hands (and mugs) close to your body.

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What happens to the angular momentum as you pull in your arms? 1. it increases 2. it decreases 3. it stays the same L1L1 L2L2 CORRECT

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11 22 I2I2 I1I1 L L What happens to your angular velocity as you pull in your arms? 1. it increases 2. it decreases 3. it stays the same CORRECT

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What happens to your kinetic energy as you pull in your arms? 1. it increases 2. it decreases 3. it stays the same CORRECT 11 22 I2I2 I1I1 L L (using L = I )

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Example A student sits on a barstool holding a bike wheel. The wheel is initially spinning CCW in the horizontal plane (as viewed from above). She now turns the bike wheel over. What happens? 1. She starts to spin CCW. 2. She starts to spin CW. 3. Nothing CORRECT

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Since there is no net external torque acting on the student stool system, angular momentum is conserved. Remember, L has a direction as well as a magnitude! LL Initially: L INI = L W,I LLL Finally: L FIN = L W,F + L S L L W,F LLSLLS L L W,I LLL L W,I = L W,F + L S

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Example l A puck slides in a circular path on a horizontal frictionless table. It is held at a constant radius by a string threaded through a frictionless hole at the center of the table. If you pull on the string such that the radius decreases by a factor of 2, by what factor does the angular velocity of the puck increase? (a) 2 (b) 4 (c) 8

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l Since the string is pulled through a hole at the center of rotation, there is no torque: Angular momentum is conserved. L 1 = I 1 1 = mR 2 1 m R mR 2 1 = m R 2 2 1 = 2 2 = 4 1

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Example A uniform stick of mass M and length D is pivoted at the center. A bullet of mass m is shot through the stick at a point halfway between the pivot and the end. The initial speed of the bullet is v 1, and the final speed is v 2. What is the angular speed F of the stick after the collision? (Ignore gravity) v1v1 v2v2 M FF initialfinal m D D/4

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Set L i = L f using v1v1 v2v2 M FF initialfinal m D D/4

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Exercises Problem 1 Find the acceleration of an object of mass m. Solution m R M h x y T Mg n T mg For the object, Newton’s 2 nd law gives: For the cylinder, the total torque is: The tangential acceleration of the cylinder is equal to that of the object: (1) (2) (3) (2)+(3): (4) a y >0 in –y direction

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Exercises Problem 1 (cont’d) Solution (cont’d) m R M h x y T Mg n T mg (1)+(4): (5) (4)+(5): The final velocity of the object when it was at rest initially:

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Exercises Problem 2 Solution m R M h x y T Mg n T mg (a) The normal force on the cylinder is: (b) Compare n with (m+M)g: As the suspended mass is accelerating down the tension is less than mg. Therefore n is less than the total weight (m+M)g. (c) If the cylinder is initially rotating clockwise and so that the object gets initial velocity upward, what effect does this have on T and n? As long as the cable remains taut, T and n remain the same.

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Exercises Problem 3 Solution T1T1 T2T2 R M m1m1 m2m2 A glider of mass m 1 slides without friction on a horizontal air track. It is attached to an object of mass m 2 by a massless string. The pulley is a thin cylindrical shell with mass M and radius R, and the string turns the pulley without slipping or stretching. Find the acceleration of each body, the angular acceleration of the pulley, and the tension in each part of the string. m1m1 n1n1 T1T1 m1gm1g m2m2 m2gm2g T2T2 T1T1 T2T2 Mg n2n2 gliderhanging obj.pulley glider: object: pulley: no stretching and slipping: y x (1) (2) (3) (4) glider hanging object pulley + y x y x

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Exercises Solution cont’d Problem 3 (cont’d) (1)-(4):

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Exercises Solution Problem 4 A primitive yo-yo is made by wrapping a string several times around a solid cylinder with mass M and radius R. You hold the end of the string stationary while releasing the cylinder with no initial motion. The string unwinds but does not slip or stretch as the cylinder drops and rotates. Find the speed v cm of the center of mass of the solid cylinder after it has dropped a distance h. h 1 2 Energy conservation

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Exercises Solution Problem 5 B A C (Atwood’s machine) Find the linear accelerations of blocks A and B, the angular acceleration of the wheel C, and the tension in each side of the cord if there is no slipping between the cord and the surface of the wheel. I mAmA mBmB R The accelerations of blocks A and B will have the same magnitude. As the cord does not slip, the angular acceleration of the pulley will be. If we denote the tensions in cord as and, the equations of motion are:

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Exercises Solution Problem 6 h=50.0 m rough smooth A solid uniform spherical boulder starts from rest and rolls down a 50.0 m high hill. The top half of the hill is rough enough to cause the boulder to roll without slipping, but the lower half is covered with ice and there is no friction. What is the translational speed of the boulder when it reaches the bottom of the hill? 1st half (rough): 2nd half (smooth):

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Exercises Solution Problem 7 Occasionally, a rotating neutron star undergoes a sudden and unexpected speedup called a glitch. One explanation is that a glitch occurs when the crust of the neutron star settles slightly, decreasing the moment of inertia about the rotation axis. A neutron star with angular speed 0 =70.4 rad/s underwent such a glitch in October 1975 that increased its angular speed to , where =2.01x If the radius of the neutron star before the glitch was 11 km, by now how much did its radius decrease in the starquake? Assume that the neutron star is a uniform sphere. Conservation of angular momentum:

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Exercises Solution Problem 8 A small block with mass kg is attached to a string passing through a hole in a frictionless, horizontal surface. The block is originally revolving in a circle with a radius of m about the hole with a tangential speed of 4.00 m/s. The string is then pulled slowly from below, shortening the radius of the circle in which the block revolves. The breaking strength of the string is 30.0 N. What is the radius of the circle when the string breaks. The tension on the string: The radius at which the string breaks is obtained from:

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Exercises Problem 9 pivot catcher v r M,I A ball catcher whose mass is M and moment of inertia is I is hung by a frictionless pivot. A ball with a speed v and mass m is caught by the catcher. The distance between the pivot point and the ball is r which is much greater that the radius of the ball. (a)Find the angular speed of the catcher right after it catches the ball. (b) After the ball is caught, the catcher – ball system swings up as high as h. Find the angular speed at the maximum height h.

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Exercises Solution Problem 10 pivot catcher v r M,I (a)The initial angular momentum with respect to the pivot is: The final total moment of inertia is (b) The kinetic energy after the collision is: Since potential energy at height h h

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Exercises Problem 11 h= 58.0 m d= 42.0 cm g/cm A 42.0 cm diameter wheel, consisting of a rim and six spokes, is constructed from a thin rigid plastic material having a linear mass density of 25.0 g/cm. This wheel is released from rest at the top of a hill 58.0 m high. (a)How fast is it rolling when it reaches the bottom of the hill? (b) How would your answer change if the linear mass density and the diameter of the wheel were each doubled? No sliding

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Exercises Solution Problem 11 (cont’d) h= 58.0 m d= 42.0 cm g/cm (a) Conservation of energy: (b) Doubling the density would have no effect! As, doubling the diameter would reduce the angular velocity by half. But would be unchanged. No sliding

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