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Chapter 11 Rolling, Torque, and Angular Momentum In this chapter we will cover the following topics: -Rolling of circular objects and its relationship with friction -Redefinition of torque as a vector to describe rotational problems that are more complicated than the rotation of a rigid body about a fixed axis -Angular momentum of single particles and systems of particles -Newton’s second law for rotational motion -Conservation of angular momentum Applications of the conservation of angular momentum (11-1) t1 = 0 t2 = t Rolling as Translation and Rotation Combined Consider an object with circular cross section that rolls along a surface without slipping. This motion, though common, is complicated. We can simplify its study by treating it as a combination of translation of the center of mass and rotation of the object about the center of mass. Consider the two snapshots of a rolling bicycle wheel shown in the figure. An observer stationary with the ground will see the center of mass O of the wheel move forward with a speed vcom . The point P at which the wheel makes contact with the road also moves with the same speed. During the time interval t between ds the two snapshots both O and P cover a distance s, vcom (eq. 1). During t dt the bicycle rider sees the wheel rotate by an angle about O so that ds d s R R = (eq. 2). If we combine equation 1 with equation 2 dt dt we get the condition for rolling without slipping: vcom R (11-2) vcom R We have seen that rolling is a combination of purely translational motion with speed vcom and a purely rotational motion about the center of mass vcom with angular velocity . The velocity of each point is the vector sum R of the velocities of the two motions. For the translational motion the velocity vector is the same for every point (vcom ,see fig. b). The rotational velocity varies from point to point. Its magnitude is equal to r where r is the distance of the point from O. Its direction is tangent to the circular orbit (see fig. a). The net velocity is the vector sum of these two terms. For example, the velocity of point P is always zero. The velocity of the center of mass O is (11-3) vcom (r 0). Finally, the velocity of the top point T is equal to 2vcom . vT Rolling as Pure Rotation A vA vO B vB Another way of looking at rolling is shown in the figure. We consider rolling as a pure rotation about an axis of rotation that passes through the contact point P between the wheel and the road. The angular v velocity of the rotation is com . R In order to define the velocity vector for each point we must know its magnitude as well as its direction. The direction for each point on the wheel points along the tangent to its circular orbit. For example, at point A the velocity vector v A is perpendicular to the dotted line that connects point A with point B. The speed of each point is given by v r. Here r is the distance between a particular point and the contact point P. For example, at point T r 2 R. Thus vT 2 R 2vcom . For point O r R, thus vO R vcom . For point P, r 0 thus vP 0. (11-4) The Kinetic Energy of Rolling Consider the rolling object shown in the figure. It is easier to calculate the kinetic energy of the rolling body by considering the motion as pure rolling about the contact point P. The rolling object has mass M and radius R. 1 The kinetic energy K is then given by the equation K I P 2 . Here I P is the 2 rotational inertia of the rolling body about point P. We can determine I P using 1 I com MR 2 2 . 2 1 1 1 1 1 2 2 2 2 2 2 2 K I com Mvcom K I com MR I com MR 2 2 2 2 2 The expression for the kinetic energy consists of two terms. The first term corresponds to the rotation about the center of mass O with angular velocity . The second term is associated with the kinetic energy due to the translational motion of every point with speed vcom . . (11-5) the parallel axis theorem. I P I com MR 2 K Friction and Rolling When an object rolls with constant speed (see top figure) it has no tendency to slide at the contact point P and thus no frictional force acts there. If a net force acts on the acom 0 rolling body it results in a nonzero acceleration acom for the center of mass (see lower figure). If the rolling object accelerates to the right it has the tendency to slide at point P to the left. Thus a static frictional force f s opposes the tendency to slide. The motion is smooth rolling as long as f s f s ,max . The rolling condition results in a connection between the magnitude of the acceleration acom of the center of mass and its angular acceleration , vcom R. We take time derivatives of both sides acom acom R dvcom d R R . dt dt (11-6) Rolling Down a Ramp Consider a round uniform body of mass M and radius R rolling down an inclined plane of angle . We will calculate the acceleration acom of the center of mass acom along the x-axis using Newton's second law for the translational and rotational motion. Newton's second law for motion along the x-axis: f s Mg sin Macom (eq. 1) Newton's second law for rotation about the center of mass: Rf s I com acom R f s I com I com We substitute in the second equation and get Rf s I com acom R2 acom R (eq. 2). We substitute f s from equation 2 into equation 1 acom Mg sin Macom 2 R acom g sin I com 1 MR 2 (11-7) g sin | acom | I 1 com2 MR acom Cylinder Hoop MR 2 I1 2 g sin a1 1 I1 / MR 2 I 2 MR 2 g sin a1 1 MR 2 / 2MR 2 g sin a1 1 1/ 2 2 g sin a1 (0.67) g sin 3 g sin a2 1 MR 2 / MR 2 g sin a2 11 g sin a2 (0.5) g sin 2 a2 g sin 1 I 2 / MR 2 (11-8) Example: A uniform cylinder rolls down a ramp inclined at an angle of θ to the horizontal. What is the linear acceleration of the cylinder at the bottom of the ramp? Remember that: The friction force is used to rotate the object. Example: Consider a solid cylinder of radius R that rolls without slipping down an incline from some initial height h. The linear velocity of the cylinder at the bottom of the incline is vcm and the angular velocity is ω. We can also solve for angular velocity using the equation v cm R 4 gh 3R 2 Example: A bowling ball has a mass of 4.0 kg, a moment of inertia of 1.6×10**(−2) kg ·m2 and a radius of 0.10 m. If it rolls down the lane without slipping at a linear speed of 4.0 m/s, what is its total energy? Example: A bowling ball has a mass of 4.0 kg, a moment of inertia of 1.6×10**(−2) kg ·m2 and a radius of 0.10 m. If it rolls down the lane without slipping at a linear speed of 4.0 m/s, what is its total energy? The Yo - Yo Consider a yo-yo of mass M , radius R, and axle radius R0 rolling down a string. We will calculate the acceleration acom of the center of its mass along the y-axis using Newton's second law for the translational and rotational motion as we did in the previous problem. Newton's second law for motion along the y -axis: acom Mg T Macom (eq. 1). y Newton's second law for rotation about the center of mass: acom R0T I com . Angular acceleration . R0 We substitute in the second equation and get T I com acom R0 2 Mg I com (eq. 2). We substitute T from equation 2 into equation 1 acom Macom 2 R0 acom g . I 1 com2 MR0 (11-9) Torque Revisited In Chapter 10 we defined the torque of a rigid body rotating about a fixed axis with each particle in the body moving on a circular path. We now expand the definition of torque so that it can describe the motion of a particle that moves along any path relative to a fixed point. If r is the position vector of a particle on which a force F is acting, the torque is defined as r F . In the example shown in the figure both r and F lie in the xy -plane. Using the right-hand rule we can see that the direction of is along the z -axis. The magnitude of the torque vector rF sin , where is the angle between r and F . From triangle OAB we have r sin r r F , in agreement with the definition of Chapter 10. r F B (11-10) Torque Revisited ˆi ˆj r F (x x o ) ( y y o ) Fx Fy kˆ z z o Fz Example: (T72_Q17.) A force F 2.0 ˆi 3.0 ˆj N is applied to an object that is pivoted about a fixed axis aligned along the z-axis. If the force is applied at the point of coordinates (4.0, 5.0, 0.0) m, what is the applied torque (in N.m) about the z axis? ˆi r F (4 0) 2 2 kˆ N m ˆj (5 0) 3 kˆ 0 0 0 Example: (T71_Q19.). At an instant, a particle of mass 2.0 kg has a 2 position r 9.0 ˆi 15.0 ˆj m of and acceleration of a 3.0 ˆi m/s. What is the net torque on the particle at this instant about the point having the position vector:r 9.0 ˆi m ? o Example: (T71_Q19.). At an instant, a particle of mass 2.0 kg has a 2 position r 9.0 ˆi 15.0 ˆj m of and acceleration of a 3.0 ˆi m/s. What is the net torque on the particle at this instant about the point having the position vector:r 9.0 ˆi m ? o ˆi ˆj ˆi kˆ jˆ kˆ r F m r a m (x x o ) ( y y o ) z z o 2 (9 9) (15 0) 0 0 ax ay az 3 3 0 90 kˆ N m Angular Momentum The counterpart of linear momentum p mv in rotational motion is a new vector known as angular momentum. The new vector is defined as follows: r p. In the example shown in the figure both r and p lie in the xy -plane. Using the right-hand rule we can see that the direction of is along the z -axis. The magnitude of angular momentum B rmv sin , where is the angle between r and p. From triangle OAB we have: r sin r r mv. Note: Angular momentum depends on the choice of the origin O. If the origin is shifted, in general we get a different value of . SI unit for angular momentum: kg m2 / s. Sometimes the equivalent J s is used. r p mr v r mv (11-11) Example: (T052 Q#18) A stone attached to a string is whirled at 3.0 rev/s around a horizontal circle of radius 0.75 m. The mass of the stone is 0.15 kg. The magnitude of the angular momentum of the stone relative to the center of the circle is: L mvr mr 0.15 0.75 3 2 1.6 kg.m 2 / s 2 2 Newton's Second Law in Angular Form dp . Below we dt will derive the angular form of Newton's second law for a particle. Newton's second law for linear motion has the form: Fnet mr v net d m r a r ma r Fnet net dt d dp . Compare with: Fnet . dt dt v v 0 Thus: net d d dv dr m r v m r v m r a v v dt dt dt dt d dt (11-12) The Angular Momentum of a System of Particles z m1 ℓ1 m2 ℓ2 x We will now explore Newton's second law in angular form for a system of n particles that have ℓ3 m3 O angular momentum 1 , 2 , 3 ,..., n . y The angular momentum L of the system is L 1 2 dL The time derivative of the angular momentum is = dt 3 ... n n i. i 1 n d i . i 1 dt d i The time derivative for the angular momentum of the i-th particle net,i dt where net,i is the net torque on the particle. This torque has contributions from external as well as internal forces between the particles of the system. Thus n dL net,i net . Here net is the net torque due to all the external forces. dt i 1 By virtue of Newton's third law the vector sum of all internal torques is zero. dL net Thus Newton's second law for a system in angular form takes the form: dt (11-13) Angular Momentum of a Rigid Body Rotating About a Fixed Axis We take the z -axis to be the fixed rotation axis. We will determine the z -component of the net angular momentum. The body is divided into n elements of mass mi that have a position vector ri . The angular momentum Its magnitude is iz of i is iz i of the ith element is i ri pi . ri pi sin 90 = ri mi vi . The z -compoment i i sin ri sin mi vi ri mi vi . The z -component of the angular momentum Lz is the sum: n Lz i 1 n ri mi vi ri mi ri mi ri 2 i 1 i 1 i 1 n iz n n The sum 2 m r i i is the rotational inertia I of the rigid body. i 1 Thus: Lz I . Lz I (11-14) Conservation of Angular Momentum For any system of particles (including a rigid body) Newton's second law in angular form is dL net . dt dL If the net external torque net 0 then we have: 0 dt L a constant. This result is known as the law of the conservation of angular momentum. In words: Net angular momentum Net angular momentum at some later time t at some initial tim e t f i In equation form: Li L f Note: If the component of the external torque along a certain axis is equal to zero, then the component of the angular momentum of the system along this axis cannot change. (11-15) Example: The figure shows a student seated on a stool that can rotate freely about a vertical axis. The student, who has been set into rotation at an initial angular speed i , holds two dumbbells in his outstretched hands. His angular momentum vector L lies along the rotation axis, pointing upward. The student then pulls in his hands as shown in fig. b. This action reduces the rotational inertia from an initial value I i to a smaller final value I f . No net external torque acts on the student-stool system. Thus the angular momentum of the system remains unchanged. Angular momentum at ti : Li I ii . Li L f I ii I f f f Ii i . If Angular momentum at t f : L f I f f . Since I f I i The rotation rate of the student in fig. b is faster. Ii 1 f i . If (11-16) Sample Problem 11-7: I wh 1.2 kg m 2 wh 2 3.9 rad/s I b 6.8 rad/s b ? y-axis Li L f Lwh Lwh Lb Lb 2 Lwh I bb 2 I whwh b 2 I whwh 2 1.2 2 3.9 2 1.4 rad/s Ib 6.8 (11-17) Analogies Between Translational and Rotational Motion Translational Motion Rotational Motion x v a p mv 2 K 2 m F ma I 2 K 2 I I F P Fv P dp dt p mv net L I Fnet d dt (11-18)