P161Sp08Chapter_12

P161Sp08Chapter_12 - Chapter 12 Rotation of a Rigid Body...

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Unformatted text preview: Chapter 12 Rotation of a Rigid Body Rigid Object n A rigid object is one that is nondeformable n n The relative locations of all particles making up the object remain constant All real objects are deformable to some extent, but the rigid object model is very useful in many situations where the deformation is negligible Angular Position n n n Axis of rotation is the center of the disc Choose a fixed reference line Point P is at a fixed distance r from the origin Angular Position, 2 n n n n Point P will rotate about the origin in a circle of radius r Every particle on the disc undergoes circular motion about the origin, O Polar coordinates are convenient to use to represent the position of P (or any other point) P is located at (r, q) where r is the distance from the origin to P and q is the measured counterclockwise from the reference line Angular Position, 3 n n n As the particle moves, the only coordinate that changes is q As the particle moves through q, it moves though an arc length s. The arc length and r are related: n s=qr Radian n This can also be expressed as n n q is a pure number, but commonly is given the artificial unit, radian One radian is the angle subtended by an arc length equal to the radius of the arc Conversions n Comparing degrees and radians 1 rad = = 57.3 n Converting from degrees to radians q [rad] = [degrees] Angular Position, final n n We can associate the angle q with the entire rigid object as well as with an individual particle n Remember every particle on a rigid object rotates through the same angle The angular position of the rigid object is the angle q between the reference line on the object and the fixed reference line in space n The fixed reference line in space is often the xaxis Angular Displacement n The angular displacement is defined as the angle the object rotates through during some time interval n This is the angle that the reference line of length r sweeps out Average Angular Speed n The average angular speed, w, of a rotating rigid object is the ratio of the angular displacement to the time interval Instantaneous Angular Speed n The instantaneous angular speed is defined as the limit of the average speed as the time interval approaches zero Angular Speed n Units of angular speed are radians/sec n rad/s or s-1 since radians have no dimensions n n Angular speed will be positive if q is increasing (counterclockwise) Angular speed will be negative if q is decreasing (clockwise) Average Angular Acceleration n The average angular acceleration, a, of an object is defined as the ratio of the change in the angular speed to the time it takes for the object to undergo the change: Instantaneous Angular Acceleration n The instantaneous angular acceleration is defined as the limit of the average angular acceleration as the time goes to 0 Angular Acceleration, final n Units of angular acceleration are rad/s2 or s-2 since radians have no dimensions n Angular acceleration will be positive if an object rotating counterclockwise is speeding up n Angular acceleration will also be positive if an object rotating clockwise is slowing down Angular Velocity & Acceleration-signs Green arrows are angular velocity. The top arrow is the initial velocity and the bottom arrow is after the acceleration. ccw is +, cw is -. Angular Motion, General Notes n When a rigid object rotates about a fixed axis in a given time interval, every portion on the object rotates through the same angle in a given time interval and has the same angular speed and the same angular acceleration n So q, w, a all characterize the motion of the entire rigid object as well as the individual particles in the object Directions, details n n Strictly speaking, the speed and acceleration (w, a) are the magnitudes of the angular velocity and acceleration vectors The directions are actually given by the right-hand rule Hints for Problem-Solving n Similar to the techniques used in linear motion problems n With constant angular acceleration, the techniques are much like those with constant linear acceleration For rotational motion, define a rotational axis The object keeps returning to its original orientation, so you can find the number of revolutions made by the body n There are some differences to keep in mind n n Rotational Kinematics n Under constant angular acceleration, we can describe the motion of the rigid object using a set of kinematic equations n n These are similar to the kinematic equations for linear motion The rotational equations have the same mathematical form as the linear equations Rotational Kinematic Equations Comparison Between Rotational and Linear Equations Comparison Between Rotational and Linear Equations Clothes Washer Problem The tub of a washer goes into its spin cycle, starting from rest and gaining angular speed steadily for 8.00 s, at which time it is turning at 5.00 rev/s. At this point the person doing the laundry opens the lid, and a safety switch turns off the washer. The tub smoothly slows to rest in 12.0 s. Through how many revolutions does the tub turn while it is in motion? Note: 5.00 revs/s = 10.0p rad/s Clothes Washer Solution Speeding up: w f - w i 10.0p / s - 0 a= = Dt 8.00s Dq sp = w i Dt + 1 aDt 2 = 0 + 2 1 10p (8.00s) 2 = 40.0p 2 8.00 Slowing down: a = w - w = 0 -10.0p / s = f i Dt 12.00s 2 1 (-10ps -1 ) Dq sp = w i Dt + 1 aDt = 10.0ps (12.0s) + (12.0s) 2 = 60.0p 2 2 12.0s -1 Total revs: 40.0p + 60.0p =100p(1rev/2prad) = 50.0 rev Relationship Between Angular and Linear Quantities n Displacements n s=qr n Speeds vt=wr n n Accelerations Every point on the rotating object has the same angular motion Every point on the rotating object does not have the same linear motion Speed Comparison n n The linear velocity is always tangent to the circular path n called the tangential velocity The magnitude is defined by the tangential speed Acceleration Comparison n The tangential acceleration is the derivative of the tangential velocity Speed and Acceleration Note n n n All points on the rigid object will have the same angular speed, but not the same tangential speed All points on the rigid object will have the same angular acceleration, but not the same tangential acceleration The tangential quantities depend on r, and r is not the same for all points on the object Centripetal Acceleration n An object traveling in a circle, even though it moves with a constant speed, will have an acceleration n Therefore, each point on a rotating rigid object will experience a centripetal acceleration Resultant Acceleration n n n The tangential component of the acceleration is due to changing speed The centripetal component of the acceleration is due to changing direction Total acceleration can be found from these components Rotational Motion Example:CD Laser light is reflected off the cd and interpreted by computer. The reading is done at a constant rate http://electronics.howstuffworks.com/cd2.htm Rotational Motion Example CD n n For a compact disc player to read a CD, the tangential speed must be constant (vt = wr). The angular speed must vary to keep vt constant At the inner grooves the angular speed is increased compared to when reading the outer grooves http://electronics.howstuffworks.com/cd6.htm Start 4/18 Rotational Kinetic Energy n n A rigid object rotating about some axis with an angular speed w has rotational kinetic energy. Each particle of the rotating object has a kinetic energy of n n Since the tangential velocity depends on the distance, ri from the axis of rotation, we can substitute vi = w ri Ki =1/2mivi2 Rotational Kinetic Energy, cont n The total rotational kinetic energy of the rigid object is the sum of the energies of all its particles n Where I is called the moment of inertia Rotational Kinetic Energy, final n n n There is an analogy between the kinetic energies associated with linear motion (K = 1/2 mv 2) and the kinetic energy associated with rotational motion (KR= 1/2 Iw2) Rotational kinetic energy is not a new type of energy, the form is different because it is applied to a rotating object The units of rotational kinetic energy are Joules (J) Moment of Inertia n The definition of moment of inertia is The dimensions of moment of inertia are ML2 and its SI units are kg.m2 We can calculate the moment of inertia of an object more easily by assuming it is divided into many small volume elements, each of mass Dmi n n Moment of Inertia, cont n We can rewrite the expression for I in terms of Dm n n With the small volume segment assumption, dm = rdv If r is constant, the integral can be evaluated with known geometry, otherwise its variation with position must be known Notes on Various Densities n Volumetric Mass Density > mass per unit volume: r = M /V=dm/dV, then dm=rdV Face Mass Density > mass per unit area (A) of a sheet of uniform thickness: s = M/A=dm/dA, then dm = sdA=sdxdy Linear Mass Density > mass per unit length of a rod of uniform cross-sectional area: l = M /L=dm/dx, dm = l dx n n Moment of Inertia of a Uniform Rigid Rod About It's Center n The shaded area has a mass n dm = l dx n Then the moment of inertia is Moment of Inertia of a Rod About its End Rod rotates about a line along x = 0 I = xdm 0 L dm = ldx L M l= L L3 1 I = l x 2 dx = l = ML2 3 3 0 Moment of Inertia of a Slab Since the derivation of I for the rod didn't involve the width or thickness the same moment of inertia applies for a rod or slab of the same length and mass. Moment of Inertia of a Slab About its Center I = x + y dm +a /2 -a /2 +b /2 -b /2 ( 2 2 ) M s= ab dm = sdxdy I = dx dy s x + y 1 I = M (a 2 + b 2 ) 12 ( 2 2 ) Moment of Inertia of a Uniform Thin Hoop About the z Axis Through 0 n Since this is a thin hoop, all mass elements are to a good approximation the same distance from the center Moment of Inertia of a Uniform Solid Cylinder n n I = r 2 dm 0 R R Divide the cylinder into concentric shells with radius r, thickness dr and length L Then for I dm = rdV R dV = 2prLdr R4 I = r 2 r 2prLdr = 2pLr r 3 dr = 2pLr 4 0 0 I= 1 1 rpR 2 L R 2 = MR 2 2 2 ( ) where M=rpR2L Moments of Inertia of Cylindrical Rigid Objects Moment of Inertia of a Solid and Hollow Sphere Parallel-Axis Theorem n n n In the previous examples, the axis of rotation coincided with the axis of symmetry of the object For an arbitrary axis, the parallel-axis theorem often simplifies calculations The theorem states I = ICM + MD 2 n I is about any axis parallel to the axis through the center of mass of the object n ICM is about the axis through the center of mass n D is the distance from the center of mass axis to the arbitrary axis Parallel-Axis Theorem Example n n n The axis of rotation goes through O The axis through the center of mass is shown The moment of inertia about the axis through O would be IO = ICM + MD 2 Moment of Inertia for a Rod Rotating Around One End n The moment of inertia of the rod about its center is n n D is 1/2 L Therefore, Moment of Inertia for a Rod Rotating Around One End Note that this example is simple enough that we could have integrated the moment formula directly. I = x dm = x 2 ldx 0 0 L 2 L L3 M L3 I =l = 3 L 3 1 I = ML2 3 Most cases are not so simple. Pure Rolling Motion n n In pure rolling motion, an object rolls without slipping In such a case, there is a simple relationship between its rotational and translational motions Rolling Object, No Slipping n The velocity of the center of mass is n The acceleration of the center of mass is Rolling Motion Cont. n Rolling motion can be modeled as a combination of pure translational motion and pure rotational motion Rolling Object, Other Points n n A point on the rim rotates to various positions such as Q and P ' At any instant, the point on the rim located at point P is at rest relative to the surface since no slipping occurs Total Kinetic Energy of a Rolling Object n The total kinetic energy of a rolling object is the sum of the translational energy of its center of mass and the rotational kinetic energy about its center of mass n K = 1/2 ICMw2 + 1/2 MvCM2 K = 1/2 (ICM/R2)vCM2 + 1/2 MvCM2 K = 1/2 M(ICM/MR2+1)vCM2 Total Kinetic Energy, Example n Accelerated rolling motion is possible only if friction is present between the sphere and the incline n The friction produces the net torque required for rotation DEMO D2-01: RING, SPHERE AND DISC ON INCLINED PLANE Do they roll at the same speed? Analyze using the conservation of energy Total Kinetic Energy, Example cont n n n Despite the friction, no loss of mechanical energy occurs because the contact point is at rest relative to the surface at any instant, i.e. no sliding occurs Let U = 0 at the the ball is at the bottom of the plane Kf + U f = K i + Ui n n n n n vcm= [2gh/(1+(Icm/MR2))]1/2 Ki = 0 Kf = 1/2 (ICM/R 2) vCM2 + 1/2MvCM2 Ui = Mgh Uf = 0 Velocity at the bottom of the incline n vcm= [2gh/(1+(Icm/MR2))]1/2 2 2 n Solid Sphere: I = 2/5MR = 0.4MR n n Hollow Sphere: I = 2/3MR2 = 0.67MR2 n vcm = [2gh/(1+(2/5)]1/2 = 0.84[2gh]1/2 vcm = [2gh/(1+(2/3)]1/2 = 0.77[2gh]1/2 n Disk: I=1/2 MR2= 0.5MR2 n n Cylindrical Shell (Ring): I = MR2 n vcm = [2gh/(1+(1/2)]1/2 = 0.82[2gh]1/2 vcm = [2gh/(1+1]1/2 = 0.71[2gh]1/2 RING AND DISC ON INCLINED PLANE Analysis Work done by gravitational force, Wg = mgDy cos0 W = Kf - Ki = (1/2Iwf2 +1/2mv2cm,f) - (1/2Iwi2 +1/2mv2cm,i) If no slipping occurs w = vcm/R, also wi = 0. W = 1/2(I/R2+m) v2cm,f vcm,f = (2Wg/(I/R2+m)1/2 = (2gDy/(I/mR2+1)1/2 Disk: I/mR2 = 1/2 Ring: I/mR2 = 1 The bigger I the smaller vcm. Review Mass is the inertia associated with linear motion. Moment of inertia is the inertia associated with rotational motion. There is a difference in that mass is an inherent property of an object but moment of inertia depends on the choice of rotation axis. Vector Products c = axb The right-hand rule for vector products. (a) Swing vector a into vector b with the fingers of your right hand. Your thumb shows the direction of c. (b) Reversing the procedure shows that (b X a) = -(a X b). |c| = |axb| = absinf |c'| = |bxa| = absinf Torque n Torque, t, is the tendency of a force to rotate an object about some axis n n Torque is a vector (t = rxf) right hand rule t = r F sin f = Fd n n n F is the force f is the angle the force makes with the radial vector d is the moment arm (or lever arm) Torque, cont n The moment arm, d, is the perpendicular distance from the axis of rotation to a line drawn along the direction of the force n d = r sin f Torque, final n n The horizontal component of F (F cos f) has no tendency to produce a rotation Torque will have direction n n If the turning tendency of the force is counterclockwise, the torque will be positive right hand rule If the turning tendency is clockwise, the torque will be negative Net Torque n n n The force F1 will tend to cause a counterclockwise rotation about O The force F2 will tend to cause a clockwise rotation about O St = t1 + t2 = F1d1 F2d2 Torque vs. Force n Forces can cause a change in linear motion n Described by Newton's Second Law n Forces can cause a change in rotational motion n n The effectiveness of this change depends on the force and the moment arm The change in rotational motion depends on the torque Torque Units n The SI units of torque are N.m n n Although torque is a force multiplied by a distance, it is very different from work and energy The units for torque are reported in N.m and not changed to Joules Torque and Angular Acceleration n n Consider a particle of mass m rotating in a circle of radius r under the influence of tangential force Ft The tangential force provides a tangential acceleration: n Ft = mat Torque and Angular Acceleration, Particle cont. n n n The magnitude of the torque produced by Ft around the center of the circle is n t = Ft r = (mat) r The tangential acceleration is related to the angular acceleration 2 n t = (mat) r = (mra) r = (mr ) a Since mr 2 is the moment of inertia of the particle, n t = Ia n The torque is directly proportional to the angular acceleration and the constant of proportionality is the moment of inertia Torque and Angular Acceleration, Extended n n n Consider the object to consist of an infinite number of mass elements dm of infinitesimal size Each mass element rotates in a circle about the origin, O Assume each mass element has a tangential acceleration Torque and Angular Acceleration, Extended cont. n n n From Newton's Second Law n dFt = (dm) at The torque associated with the force and using the angular acceleration gives 2 dm n dt = r dFt = atr dm = ar Finding the net torque n n This becomes St = Ia Torque and Angular Acceleration, Extended final n n This is the same relationship that applied to a particle The result also applies when the forces have radial components n n The line of action of the radial component must pass through the axis of rotation These components will produce zero torque about the axis Demo: ROTATING ROD Why doesn't the ball move with the end of the rod? Falling Rod & Ball For the rod: I = 1/3 ML2 , t = -mg (L/2)cosq t = Ia -mg (L/2)cosq = 1/3 ML2 a a = -(3/2)(g/L)cosq at = La = -(3/2)gcosq ay =atcosq = -(3/2)gcos2q For the cup to fall faster than the ball ay -g cos2q 2/3 q 35.3 Demo: Falling Tower When a tall tower falls it often breaks somewhere along its length before it hits the ground. As it fall a = aR and the greater the distance from the axis of rotation the greater is a. At each pint along the tower there is a shear force from lower points that tries to make a = aR. At some position this shear force exceeds the breaking strength. Torque and Angular Acceleration, Wheel Example n n The wheel is rotating and so we apply St = Ia n The tension supplies the tangential force n Friction between the wheel and string causes the wheel to rotate The mass is moving in a straight line, so apply Newton's Second Law n SFy = may = -mg+T n What is a? Wheel Example Wheel: St = Ia +TRsin90 = Ia/R T = Ia/R2 Mass : T-mg = -ma Solve for a. a = -T/m +g a = -Ia/mR2+g a = g/(1+I/mR2) a < g for M >0 T = mg/(1+mR2/I) I = (1/2)MR2 T0 as M0, i.e. m is in free fall Torque and Angular Acceleration, Multi-body Ex., 1 n n n Both masses move in linear directions, so apply Newton's Second Law Both pulleys rotate, so apply the torque equation Identical pulleys Torque and Angular Acceleration, Multi-body Ex., 2 (1) -T1 + m1g = m1a (2) T3 - m2g = m2a Add (1) & (2) (3) T3 - T1 = (m1-m2)g -(m1+m2)a Torque and Angular Acceleration, Multi-body Ex., 2 (4) (T1 - T2)R = Ia Add (4) & (5) (5) (T2 -T3)R = Ia (6) (T1 - T3)=2Ia = 2Ia/R Torque and Angular Acceleration, Multi-body Ex., 3 (6) (T1 - T3)= 2Ia/R (3) (T3 - T1) = (m1-m2)g - (m1+m2)a Combine (6) & (3) to find a (m - m )g a= m + m + 2(I R ) 1 2 2 1 2 Work in Rotational Motion n Find the work done by F on the object as it rotates through an infinitesimal distance ds = r dq n dW = F . d s = (F sin f) r dq dW = t dq The radial component of F does no work because it is perpendicular to the displacement Power in Rotational Motion n The rate at which work is being done in a time interval dt is n This is analogous to P = Fv in a linear system Work-Kinetic Energy Theorem in Rotational Motion- Fixed Axis n The work-kinetic energy theorem for rotational motion states that the net work done by external forces in rotating a symmetrical rigid object about a fixed axis equals the change in the object's rotational kinetic energy dw W = tdq = Iadq = I wdt dt 1 2 1 2 W = Iwdw = Iw f - Iw i 2 2 w i wf Work-Kinetic Energy Theorem, General n The rotational form can be combined with the linear form which indicates the net work done by external forces on an object is the change in its total kinetic energy, which is the sum of the translational and rotational kinetic energies Energy Conservation n If no friction is involved total mechanical energy is conserved n Kf + Uf = Ki + Ui K includes both rotational and translational kinetic energies, if appropriate Rotating Rod What is the rod's angular speed when it reaches its lowest position? Two methods: 1) W = Kf - Ki MgL/2 = (1/2)Iwf2 & I = (1/3)ML2 -> wf = (3g/L)1/2 2) Since there is no friction mentioned assume energy is conserved. Kf + Uf = Ki + Ui (1/2)Iwf2 + mgL/2 = 0 + mgL -> wf = (3g/L)1/2 Energy in an Atwood Machine, Example n n n n Blocks released from rest: m1 drops, m2rises The blocks undergo changes in translational kinetic energy and gravitational potential energy The pulley undergoes a change in rotational kinetic energy What is v & w? Energy in an Atwood Machine, Example - 2 Assume Energy is conserved: DE = DK+DU=0 Ki = 0 DU=Uf-Ui = -m1gh +m2gh Kf = 1/2 (m1 + m2) v2 + 1/2 I w2 1/2 (m1 + m2) v2 + 1/2 I w2 + (m2-m1)gh = 0 v w= R v= 2(m2 - m1 )gh [m + m 1 2 + (I / R 2 ) = 2(m2 - m1 )gh [m1 + m2 + M ] If I = MR2 Summary of Useful Equations Ball in a Sphere A uniform solid sphere of radius r is placed on the inside surface of a hemispherical bowl with much larger radius R. The sphere is released from rest at an angle to the vertical and rolls without slipping. Determine the speed of the sphere when it reaches the bottom of the bowl. Ball in a Sphere Energy is conserved: Uf + Kf = Ui +Ki Ui = mg{R-(R-r)cosq} Uf = mgr Ki = 0 Kf = 1/2mv2 +1/2Iw2 = 1/2mv2 [1+I/mr2] Kf = 1/2mv2 [1+2/5] = 7/10mv2 mg{R-(R-r)cosq}- mgr = 7/10mv2 v= {(10/7)g(R-r)(1-cosq)}1/2 Wire Spool A spool of wire rests on a horizontal surface. As the wire is pulled, the spool does not slip at the contact point P. On separate trials, each one of the forces F1, F2, F3, and F4 is applied to the spool. For each one of these forces, determine the direction the spool will roll. Note that the line of action of F2 passes through P. Wire Spool Incline w/Massive Pulley Two blocks are connected by a string of negligible mass passing over a pulley of radius 0.250 m and moment of inertia I. The block on the frictionless incline is moving up with a constant acceleration of 2.00 m/s2. (a) Determine T1 and T2, the tensions in the two parts of the string. (b) Find the moment of inertia of the pulley. Incline w/ Pulley a) m1 : T1 - m1g sin 37.0 o = m1a T1 = 15.0kg(9.80 sin 37.0 o + 2.00)m / s 2 = 118N m2 : T2 - m2 g = -m2 a T2 = 20.0kg(9.80 - 2.00)m / s 2 = 156N b) T1 R -T2 R = Ia = -I T2 -T1 = I a 2 R a R a>0 a<0 a=-aR (T2 -T1 )R 2 (156N -118N)(0.250m)2 I= = = 1.19kgm 2 a 2.00m / s 2 Falling Disk A string is wound around a uniform disk of radius R and mass M. The disk is released from rest with the string vertical and its top end tied to a fixed bar. Show that (a) the tension in the string is one-third the weight of the disk, (b) the magnitude of the acceleration of the center of mass is 2g/3, and (c) the speed of the center of mass is (4gh/3)1/2 after the disk has descended through distance h. (d)Verify your answer to (c) using the energy approach. Falling Disk Fy = T - Mg = -Ma 1 2 a ^ t = -TR = -Ia = - MR ~ R 2 T = M (g - a) & T = Mg - 2T T= Mg 3 2T a= M Falling Disk Rolling Cylindrical Object n The red curve shows the path moved by a point on the rim of the object n This path is called a cycloid n The green line shows the path of the center of mass of the object ...
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This note was uploaded on 05/19/2008 for the course PHYS 161 taught by Professor Hammer during the Spring '07 term at Maryland.

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