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Chapter 11: Rolling Motion, Torque and Angular Momentum • Rolling motion (axis of rotation is moving) • Kinetic Energy of rolling motion • Rolling motion on an incline • Torque • Angular momentum • Angular momentum is conserved 1/26 Rolling motion of a particle on a wheel (Superposition of rolling and linear motion) 2/26 11-2 Rolling motion Smooth rolling: There is no slipping Linear speed of center of mass: 5/12/2017 vCOM ds R d R dt dt Lecture notes by Dr. M. S. Kariapper KFUPM - PHYSICS 3/26 11-2 Rolling motion The angular velocity of any point on the wheel is the same. The linear speed of any point on the object changes as shown in the diagram!! For one instant (bottom), point P has no linear speed. 5/12/2017 Lecture notes by Dr. M. S. Kariapper KFUPM - PHYSICS For one instant (top), point P’ has a linear speed of 2·vCOM 4/26 11-3 Kinetic Energy of Rolling Superposition principle: Rolling motion = Pure translation + Pure rotation Kinetic energy 1 1 2 2 K Mv I CM 5/12/2017 Lecture notes by Dr. M. S. Kariapper KFUPM of rolling motion: 2 2 - PHYSICS 5/26 Sample Problem 11-1 Approximate each wheel on the car Thrust SSC as a disk of uniform thickness and mass M = 170 kg, and assume smooth rolling. When the car’s speed was 1233 km/h, what was the kinetic energy of each wheel? vcom 1233 km / h 342.5m / s vcom / R Icom 12 MR2 K 12 Icom2 12 M vcom2 ( 12 )( 12 MR2 )(vcom / R)2 2 1Mv com 2 43 .M vcom2 43 (170 kg )(342.5 m / s)2 1.50107 J 6/26 11-4 Forces of Rolling Friction and Rolling If the wheel rolls without sliding (smooth rolling) and is accelerating, then from vcom R , d vcom a Rd com dt dt acom R (smooth rolling) where acom is the linear acceleration of the center of mass and α is the angular acceleration. • • • • The force to provide for macom is the static frictional force (assuming the wheel rolls without sliding). Therefore, for a wheel to roll without sliding, the maximum static frictional force, s N between the wheel and the ground must be greater than macom. f s and acom point to the right if the wheel if the wheel rotates faster, for example, at the start of a bicycle race. Do not assume that f s is equal to the maximum value of s N 7/26 Rolling Down a Ramp Mg sin f s M acom F Ma (1) The positive direction here is chosen to be down the plane. y Do not assume that f s is at its maximum value of s N . The value of f s self-adjusts so the body rolls without sliding. x R f s Icom I (2) α is counterclockwise and positive. acom R where acom points down plane +ve Therefore from (2) and substituting this in (1) f s I com acom R2 acom g sin 1 Icom / M R2 Note that a positive acom points down plane. 8/26 Demo A ring and and disk of equal mass and diameter are rolling down a frictionless incline. Both start at the same position; which one will be faster at the end of the incline? 5/12/2017 9/26 Sample Problem 11-2 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? Wex K U Eth Icom (ball ) 25 MR2 vcom R 0 K U 0 0 12 Icom2 12 M vcom2 0 Mg h 2 2 2vcom v 11 2 2MR 2 2 com 1 M R 2 2 M vcom M g h 22 5 5 RR 2 M gh 7 M vcom 10 2 10 vcom (10 7 ) g h ( 7 )(9.8 m / s )(1.2 m) 4.1 m / s A positive vcom points down plane. 10/26 (b) What are the magnitude and direction of the friction force on the ball as it rolls down the ramp? acom g sin g sin 1 Icom / MR2 1 25 MR 2 / MR 2 2 (9.8 m / s ) 2sin30.0 1 5 3.50 m / s2 A positive acom points down plane. I f s Icom acom R2 25 MR2 acom R2 f s R I com acom R 25 Macom 25 (6.00 kg )(3.50 m / s2 ) 8.40 N A positive fs means that the direction we selected for fs (up) is correct! fsR is a clockwise torque (+ve) 11/26 11-5 Yo-Yo The yo-yo can be considered as a rolling down a ramp: • Instead of rolling down a ramp at angle with the horizontal, the yo-yo rolls down a string at angle = 90° with the horizontal. • Instead of rolling on its outer surface at radius R, the yo-yo rolls on an axle of radius Ro. • Instead of being slowed by frictional force fs, the yo-yo is slowed by the net force T on it from the string. So we would again get the same expression for the acceleration as for rolling with = 90°. acom g 1 Icom / M Ro2 12/26 11-6 Torque and the vector product r F magnitude : 5/12/2017 r F sin r F Fis the component of F perpendicular to r r F r is the moment arm of F (the perpendicular distance between O and the line of action of F ) Lecture notes by Dr. M. S. Kariapper KFUPM - PHYSICS 13/26 Sample Problem 11.3 In Fig. 11-11a, three forces, each of magnitude 2.0 N, act on a particle. The particle is in the xz plane at point A given by position vector r , where r = 3.0 m and = 30°. Force F1is parallel to the x axis, force F2 is parallel to the z axis, and force F3 is parallel to the y axis What is the torque, with respect to the origin O, due to each force? 1 r F1 sin 1 (3.0 m)(2.0 N )(sin150 ) 3.0 N m 2 r F2 sin 2 (3.0 m)(2.0 N )(sin120 ) 5.2 N m 3 r F3 sin3 (3.0 m)(2.0 N )(sin90 ) 6.0 N m To find the directions of the torques, we use the right hand rule and rotate r into F through the smaller of the two angles between their directions. 5/12/2017 Lecture notes by Dr. M. S. Kariapper KFUPM - PHYSICS 14/26 Sample Problem 11.3 5/12/2017 Lecture notes by Dr. M. S. Kariapper KFUPM - PHYSICS 15/26 11-7 Angular momentum of a particle l rp m( r v ) • The SI unit of angular momentum l is kg m2 / s J s . • Angular momentum is a “vector”, the direction is determined by the right hand rule. • The magnitude of angular momentum is l r m v sin • where φ is the angle between r and p when these two vectors are arranged tail to tail. l r p rmv magnitude : l r p rmv 5/12/2017 Lecture notes by Dr. M. S. Kariapper KFUPM - PHYSICS 16/26 Sample Problem 11-4 Figure 11-13 shows an overhead view of two particles moving at constant momentum along horizontal paths. Particle 1, with momentum magnitude p1 = 5.0 kg·m/s, has position vector r1 and will pass 2.0 m from point O. Particle 2, with momentum magnitude p2 = 2.0 kg·m/s, has position vector r2 and will pass 4.0 m from point O. What is the net angular momentum L about point O of the two-particle system? l1 r1 p1 (2.0 m)(5.0 kg m / s) 10 kg m2 / s The RHR indicates that l1 is positive. l1 10 kg m2 / s RHR = right hand rule l2 r2 p2 (4.0 m)(2.0 kg m / s) 8.0 kg m2 / s The RHR indicates that l2 is negative. l2 8.0 kg m2 / s L l1 l2 10 kg m2 / s ( 8.0 kg m2 / s) 2.0 kg m2 / s 17/26 11-8 Newton’s Second Law in Angular Form Fnet d p (single particle) dt net d l dt (single particle) • Note that the torque and angular momentum l must be defined with respect to the same origin. • Proof: l m (r x v ) d l m (r x d v d r x v ) m (r x a v x v ) dt dt dt Because v x v 0 , this leads to d l m ( r x a ) r x ma r x F (r x F ) net net dt Therefore, net d l dt 18/26 11-9 Angular momentum of a system of Particles n L l1 l 2 l 3 ... l n l i (L = total angular momentum) i 1 d L n d li dt i 1 d t • n net,i i 1 n net,i is the net torque on the particle. net,i is the sum of all the torque i 1 (internal and external) on the system. However the internal torques sums to zero. Let net represent the net external torque on the system. ith net d L dt ( system of particles ) • The net external torque net acting on a system of particles is equal to the time rate of change of the system's total angular momentum L . 19/26 11-10 Angular momentum of a rigid object rotating about a fixed axis We’ll consider an object that is rotating about the z-axis. The angular momentum of the object is given by: Lz I Note that in this case L and are along the z axis. Also note the analog formula for linear momentum p = m·v 20/26 11-11 Conservation of angular momentum The total angular momentum of a system is constant in both magnitude and direction if the resultant external torque acting on the system is zero. L constant If the system undergoes an internal “rearrangement”, then Li L f constant If the object is rotating about a fixed axis (say z-axis), then: 5/12/2017 I i i I f f constant Lecture notes by Dr. M. S. Kariapper KFUPM - PHYSICS 21/26 Demo A students stands still on a rotating platform and holds two texts on outstretched arms. He brings the arms closer. What happens? Discuss A students stands still on a rotatable platform and holds a spinning wheel. The bicycle wheel is spinning in the clockwise direction when viewed from above. He flips the wheel over. What happens? 22/26 TABLE 11-1 More Corresponding Variables and Relations for Translational and Rotational Motiona Translational Rotational Force F Torque Linear momentum p Angular momentum Linear momentumb Linear momentumb Newton's second lawb Conservation a b c d lawd P ( pi ) P M vcom dP Fnet dt P a constant ( r x F ) l ( r x p) Angular momentumb L ( li ) Angular momentumc L I Newton's second lawb Conservation lawd dL net dt L a constant See also Table 10-3. For systems of particles, including rigid bodies. For a rigid body about a fixed axis, with L being the component along that axis. For a closed, isolated system. 23/26 P22 Force F = 2i-3k acts on a particle with position vector r = 0.5j2.0k relative to the origin. In unit vector notation, what is the resulting torque on the pebble about (a) the origin and (b) the point (2.0, 0, -3.0)? P72 A uniform solid ball rolls smoothly along a floor and up a ramp inclined at 15.0°. It is momentarily stops when it has rolled 1.50 m along the ramp. What was its initial speed? 24/26 P85 In fig. 11.-62, a constant horizontal force Fapp of magnitude 12 N is applied to a uniform solid cylinder by a 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 com of the cylinder? (b) What is the magnitude of the angular acceleration of the cylinder about the com? (c) In unit vector notation, what is the frictional force acting on the cylinder? P90 A uniform rod rotates in a horizontal plane about a vertical axis through one end. The rod is 6.00 m long, weighs 10.0 N, and rotates at 240 rev/min. Calculate (a) its rotational inertia about the axis of rotation and (b) the magnitude of its angular momentum about the axis. 25/26 P48 A cockroach of mass 0.17 kg runs counterclockwise around the rim of a lazy Susan (a circular disk mounted on a veritcal axle) that has radius 15 cm, rotational inertia 5.0 x 10-3 kgm2, and frictionless bearings. The cockroach’s speed (relative to the ground) is 2.0 m/s, and the lazy Susan turns clockwise with angular velocity wo = 2.8 rad/s. The cockroach finds a bread crumb on the rim and, of course, stops. (a) What is the angular speed of the lazy Susan after the cockroach stops? 26/26