* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Thursday, June 22, 2006
Inertial frame of reference wikipedia , lookup
Faster-than-light wikipedia , lookup
Center of mass wikipedia , lookup
Tensor operator wikipedia , lookup
Symmetry in quantum mechanics wikipedia , lookup
Classical mechanics wikipedia , lookup
Coriolis force wikipedia , lookup
Laplace–Runge–Lenz vector wikipedia , lookup
Modified Newtonian dynamics wikipedia , lookup
Photon polarization wikipedia , lookup
Relativistic mechanics wikipedia , lookup
Theoretical and experimental justification for the Schrödinger equation wikipedia , lookup
Mass versus weight wikipedia , lookup
Angular momentum operator wikipedia , lookup
Fictitious force wikipedia , lookup
Angular momentum wikipedia , lookup
Newton's theorem of revolving orbits wikipedia , lookup
Rotational spectroscopy wikipedia , lookup
Equations of motion wikipedia , lookup
Newton's laws of motion wikipedia , lookup
Jerk (physics) wikipedia , lookup
Moment of inertia wikipedia , lookup
Relativistic angular momentum wikipedia , lookup
Hunting oscillation wikipedia , lookup
Classical central-force problem wikipedia , lookup
PHYS 1443 – Section 001 Lecture #13 Thursday, June 22, 2006 Dr. Jaehoon Yu • • • • • • • • Thursday, June 22, 2006 CM and the Center of Gravity Fundamentals on Rotational Motion Rotational Kinematics Relationship between angular and linear quantities Rolling Motion of a Rigid Body Torque Torque and Vector Product Moment of Inertia PHYS 1443-001, Summer 2006 Dr. Jaehoon Yu 1 Announcements • Reading assignments – CH. 11.6, 11.8, 11.9 and 11.10 • Last quiz next Wednesday – Early in the class – Covers Ch. 10 – what we cover next Tuesday Thursday, June 22, 2006 PHYS 1443-001, Summer 2006 Dr. Jaehoon Yu 2 Center of Mass and Center of Gravity The center of mass of any symmetric object lies on an axis of symmetry and on any plane of symmetry, if object’s mass is evenly distributed throughout the body. How do you think you can determine the CM of objects that are not symmetric? Center of Gravity Dmi Axis of One can use gravity to locate CM. symmetry 1. Hang the object by one point and draw a vertical line following a plum-bob. 2. Hang the object by another point and do the same. 3. The point where the two lines meet is the CM. Since a rigid object can be considered as a collection of small masses, one can see the total gravitational force exerted on the object as Fg Fi Dmi g Mg i Dmig CM What does this equation tell you? Thursday, June 22, 2006 i The net effect of these small gravitational forces is equivalent to a single force acting on a point (Center of Gravity) with mass M. PHYS 1443-001, Summer 2006 3 The CoG is the point in an object asDr.ifJaehoon all the gravitational force is acting on! Yu Motion of a Group of Particles We’ve learned that the CM of a system can represent the motion of a system. Therefore, for an isolated system of many particles in which the total mass M is preserved, the velocity, total momentum, acceleration of the system are d 1 drCM dt M dt Velocity of the system vCM Total Momentum of the system pCM MvCM Acceleration of the system aCM External force exerting on the system If net external force is 0 Thursday, June 22, 2006 M 1 m r i i M i i i i 1 m v i i M Fext MaCM mi ai dptot dt dri mi vi dt M m v m v p p M d 1 dvCM dt M dt Fext 0 mi dptot dt ptot const PHYS 1443-001, Summer 2006 Dr. Jaehoon Yu tot mi dvi mi ai dt M What about the internal forces? System’s momentum is conserved. 4 Fundamentals on Rotation Linear motions can be described as the motion of the center of mass with all the mass of the object concentrated on it. Is this still true for rotational motions? No, because different parts of the object have different linear velocities and accelerations. Consider a motion of a rigid body – an object that does not change its shape – rotating about the axis protruding out of the slide. The arc length is l Rq l Therefore the angle, q, is q R . And the unit of the angle is in radian. It is dimensionless!! One radian is the angle swept by an arc length equal to the radius of the arc. Since the circumference of a circle is 2pr, 360 2pr / r 2p The relationship between radian and degrees is 1 rad 360 / 2p 180 / p Thursday, June 22, 2006 PHYS 1443-001, Summer 2006 Dr. Jaehoon Yu 180o 3.14 57.3o 5 Example 10 – 1 A particular bird’s eyes can barely distinguish objects that subtend an angle no smaller than about 3x10-4 rad. (a) How many degrees is this? (b) How small an object can the bird just distinguish when flying at a height of 100m? (a) One radian is 360o/2p. Thus 4 3 10 rad 3 10 rad 4 360 o 2p rad 0.017 (b) Since l=rq and for small angle arc length is approximately the same as the chord length. l rq 4 100m 3 10 rad 2 Thursday, June 22, 2006 3 10 m 3cm PHYS 1443-001, Summer 2006 Dr. Jaehoon Yu 6 o Rotational Kinematics The first type of motion we have learned in linear kinematics was under a constant acceleration. We will learn about the rotational motion under constant angular acceleration, because these are the simplest motions in both cases. Just like the case in linear motion, one can obtain Angular Speed under constant angular acceleration: Angular displacement under constant angular acceleration: One can also obtain Thursday, June 22, 2006 f i t 1 2 q f qi it t 2 2 q f qi 2 f PHYS 1443-001, Summer 2006 Dr. Jaehoon Yu 2 i 7 Angular Displacement, Velocity, and Acceleration Using what we have learned in the previous slide, how Dq q f q i would you define the angular displacement? q f qi How about the average angular speed? Unit? rad/s And the instantaneous angular speed? Unit? rad/s lim By the same token, the average angular acceleration Unit? rad/s2 And the instantaneous angular acceleration? Unit? rad/s2 t f ti Dt 0 t f ti Dt 0 Dq Dt Dq dq Dt dt f i lim qf qi D Dt D d dt Dt When rotating about a fixed axis, every particle on a rigid object rotates through the same angle and has the same angular speed and angular acceleration. Thursday, June 22, 2006 PHYS 1443-001, Summer 2006 Dr. Jaehoon Yu 8 Example for Rotational Kinematics A wheel rotates with a constant angular acceleration of 3.50 rad/s2. If the angular speed of the wheel is 2.00 rad/s at ti=0, a) through what angle does the wheel rotate in 2.00s? Using the angular displacement formula in the previous slide, one gets q f qi 1 2 t t 2 1 2 2.00 2.00 3.50 2.00 11.0rad 2 11.0 rev. 1.75rev. 2p Thursday, June 22, 2006 PHYS 1443-001, Summer 2006 Dr. Jaehoon Yu 9 Example for Rotational Kinematics cnt’d What is the angular speed at t=2.00s? Using the angular speed and acceleration relationship f i t 2.00 3.50 2.00 9.00rad / s Find the angle through which the wheel rotates between t=2.00 s and t=3.00 s. Using the angular kinematic formula qf 1 2 q i t t 11.0rad 2 1 At t=2.00s qt 2 2.00 2.00 2 3.50 2.00 1 2 q 2.00 3.00 3.50 3.00 21.8rad t 3 At t=3.00s 2 Angular displacement Thursday, June 22, 2006 Dq q q 2 10.8 rev. 1.72rev. 10.8rad 2p PHYS 1443-001, Summer 2006 Dr. Jaehoon Yu 10 Relationship Between Angular and Linear Quantities What do we know about a rigid object that rotates about a fixed axis of rotation? Every particle (or masslet) in the object moves in a circle centered at the axis of rotation. When a point rotates, it has both the linear and angular components in its motion. What is the linear component of the motion you see? Linear velocity along the tangential direction. The direction of follows a right-hand rule. How do we related this linear component of the motion with angular component? d dq dl The arc-length is l Rq So the tangential speed v is v rq r dt dt dt r What does this relationship tell you about Although every particle in the object has the same the tangential speed of the points in the angular speed, its tangential speed differs object and their angular speed?: proportional to its distance from the axis of rotation. Thursday, June 22, 2006 The farther away2006 the particle is from the center of PHYS 1443-001, Summer Dr. Jaehoon rotation, theYuhigher the tangential speed. 11 Is the lion faster than the horse? A rotating carousel has one child sitting on a horse near the outer edge and another child on a lion halfway out from the center. (a) Which child has the greater linear speed? (b) Which child has the greater angular speed? (a) Linear speed is the distance traveled divided by the time interval. So the child sitting at the outer edge travels more distance within the given time than the child sitting closer to the center. Thus, the horse is faster than the lion. (b) Angular speed is the angle traveled divided by the time interval. The angle both the children travel in the given time interval is the same. Thus, both the horse and the lion have the same angular speed. Thursday, June 22, 2006 PHYS 1443-001, Summer 2006 Dr. Jaehoon Yu 12 How about the acceleration? How many different linear accelerations do you see in a circular motion and what are they? Two Tangential, at, and the radial acceleration, ar. v r Since the tangential speed v is The magnitude of tangential a dv d r r d r t acceleration at is dt dt dt What does this relationship tell you? Although every particle in the object has the same angular acceleration, its tangential acceleration differs proportional to its distance from the axis of rotation. 2 v2 r The radial or centripetal acceleration ar is ar r r r 2 What does The father away the particle is from the rotation axis, the more radial this tell you? acceleration it receives. In other words, it receives more centripetal force. Total linear acceleration is Thursday, June 22, 2006 a at2 ar2 r 2 r PHYS 1443-001, Summer 2006 Dr. Jaehoon Yu 2 2 r 2 4 13 Example (a) What is the linear speed of a child seated 1.2m from the center of a steadily rotating merry-go-around that makes one complete revolution in 4.0s? (b) What is her total linear acceleration? First, figure out what the angular speed of the merry-go-around is. Using the formula for linear speed 1rev 2p 1.6rad / s 4.0s 4.0 s v r 1.2m 1.6rad / s 1.9m / s Since the angular speed is constant, there is no angular acceleration. Tangential acceleration is Radial acceleration is Thus the total acceleration is Thursday, June 22, 2006 at r 1.2m 0rad / s 2 0m / s 2 2 2 ar r 1.2m 1.6rad / s 3.1m / s 2 a a a 0 3.1 3.1m / s 2 2 t 2 r PHYS 1443-001, Summer 2006 Dr. Jaehoon Yu 2 14 Example for Rotational Motion Audio information on compact discs are transmitted digitally through the readout system consisting of laser and lenses. The digital information on the disc are stored by the pits and flat areas on the track. Since the speed of readout system is constant, it reads out the same number of pits and flats in the same time interval. In other words, the linear speed is the same no matter which track is played. a) Assuming the linear speed is 1.3 m/s, find the angular speed of the disc in revolutions per minute when the inner most (r=23mm) and outer most tracks (r=58mm) are read. Using the relationship between angular and tangential speed v r r 23mm v 1.3m / s 1.3 56.5rad / s r 23mm 23 10 3 9.00rev / s 5.4 10 2 rev / min r 58mm 1.3m / s 1.3 22.4rad / s 58mm 58 10 3 b) The maximum playing time of a standard music CD is 74 minutes and 33 seconds. How many revolutions does the disk make during that time? qf 540 210rev / min i f 2 2 Thursday, June 22, 2006 375rev / min 375 q i t 0 60 rev / s 4473s 2.8 104 rev l vt Dt c) What is the total length of the track past through the readout mechanism? d) What is the angular acceleration of the CD over the 4473s time interval, assuming constant ? 2.110 2 rev / min f i Dt PHYS 1443-001, Summer 2006 Dr. Jaehoon Yu 1.3m / s 4473s 5.8 103 m 22.4 56.5rad / s 4473s 7.6 10 3 rad / s 2 15 Rolling Motion of a Rigid Body What is a rolling motion? A more generalized case of a motion where the rotational axis moves together with the object A rotational motion about the moving axis To simplify the discussion, let’s make a few assumptions 1. 2. Limit our discussion on very symmetric objects, such as cylinders, spheres, etc The object rolls on a flat surface Let’s consider a cylinder rolling without slipping on a flat surface Under what condition does this “Pure Rolling” happen? The total linear distance the CM of the cylinder moved is s Rq R q s s=Rq Thursday, June 22, 2006 Thus the linear speed of the CM is vCM Condition PHYS 1443-001, Summer 2006 Dr. Jaehoon Yu ds dq R R dt dt for “Pure Rolling” 16 More Rolling Motion of a Rigid Body The magnitude of the linear acceleration of the CM is P’ CM aCM dvCM d R R dt dt As we learned in the rotational motion, all points in a rigid body 2vCM moves at the same angular speed but at a different linear speed. vCM CM is moving at the same speed at all times. At any given time, the point that comes to P has 0 linear speed while the point at P’ has twice the speed of CM P Why?? A rolling motion can be interpreted as the sum of Translation and Rotation P’ CM P vCM P’ vCM CM v=0 vCM Thursday, June 22, 2006 + v=R v=R 2vCM P’ = P PHYS 1443-001, Summer 2006 Dr. Jaehoon Yu CM vCM P 17 Torque Torque is the tendency of a force to rotate an object about an axis. Torque, t, is a vector quantity. F f r P Line of Action d2 d Moment arm F2 Consider an object pivoting about the point P by the force F being exerted at a distance r. The line that extends out of the tail of the force vector is called the line of action. The perpendicular distance from the pivoting point P to the line of action is called Moment arm. Magnitude of torque is defined as the product of the force exerted on the object to rotate it and the moment arm. When there are more than one force being exerted on certain points of the object, one can sum up the torque generated by each force vectorially. The convention for sign of the torque is positive if rotation is in counter-clockwise and negative if clockwise. Thursday, June 22, 2006 PHYS 1443-001, Summer 2006 Dr. Jaehoon Yu t rF sin f Fd t t 1 t 2 F1d1 F2d2 18 Example for Torque A one piece cylinder is shaped as in the figure with core section protruding from the larger drum. The cylinder is free to rotate around the central axis shown in the picture. A rope wrapped around the drum whose radius is R1 exerts force F1 to the right on the cylinder, and another force exerts F2 on the core whose radius is R2 downward on the cylinder. A) What is the net torque acting on the cylinder about the rotation axis? R1 F1 The torque due to F1 t1 R1F1 and due to F2 So the total torque acting on the system by the forces is R2 t t 1 t 2 R2 F2 t 2 R1F1 R2 F2 F2 Suppose F1=5.0 N, R1=1.0 m, F2= 15.0 N, and R2=0.50 m. What is the net torque about the rotation axis and which way does the cylinder rotate from the rest? Using the above result Thursday, June 22, 2006 t R1F1 R2 F2 5.0 1.0 15.0 0.50 2.5N m PHYS 1443-001, Summer 2006 Dr. Jaehoon Yu The cylinder rotates in counter-clockwise. 19 Torque and Vector Product z O r trxF Let’s consider a disk fixed onto the origin O and the force F exerts on the point p. What happens? p The disk will start rotating counter clockwise about the Z axis y The magnitude of torque given to the disk by the force F is q t Fr sin f F x But torque is a vector quantity, what is the direction? How is torque expressed mathematically? What is the direction? The direction of the torque follows the right-hand rule!! The above operation is called Vector product or Cross product What is the result of a vector product? Another vector Thursday, June 22, 2006 t rF C A B C A B A B sin q What is another vector operation we’ve learned? Scalar product PHYS 1443-001, Summer 2006 Dr. Jaehoon Yu C A B A B cosq Result? A scalar 20 Properties of Vector Product Vector Product is Non-commutative What does this mean? If the order of operation changes the result changes Following the right-hand rule, the direction changes Vector Product of two parallel vectors is 0. A B sin 0 0 C A B A B sin q A B B A A B B A Thus, A A 0 If two vectors are perpendicular to each other A B A B sin q A B sin 90 A B AB Vector product follows distribution law A B C A B A C The derivative of a Vector product with respect to a scalar variable is d A B dA dB B A dt dt dt Thursday, June 22, 2006 PHYS 1443-001, Summer 2006 Dr. Jaehoon Yu 21 More Properties of Vector Product The relationship between unit vectors, i, j and k i i j j k k 0 i j j i k j k k j i k i i k j Vector product of two vectors can be expressed in the following determinant form i A B Ax Bx j k Ay Az i By Bz Ay Az By Bz j Ax Az Bx Bz k Ax Ay Bx By Ay Bz Az B y i Ax Bz Az Bx j Ax B y Ay Bx k Thursday, June 22, 2006 PHYS 1443-001, Summer 2006 Dr. Jaehoon Yu 22 Moment of Inertia Measure of resistance of an object to changes in its rotational motion. Equivalent to mass in linear motion. Rotational Inertia: For a group of particles I mi ri 2 i What are the dimension and unit of Moment of Inertia? For a rigid body I r 2 dm ML 2 kg m 2 Determining Moment of Inertia is extremely important for computing equilibrium of a rigid body, such as a building. Thursday, June 22, 2006 PHYS 1443-001, Summer 2006 Dr. Jaehoon Yu 23 Example for Moment of Inertia In a system of four small spheres as shown in the figure, assuming the radii are negligible and the rods connecting the particles are massless, compute the moment of inertia and the rotational kinetic energy when the system rotates about the y-axis at angular speed . y m Since the rotation is about y axis, the moment of inertia about y axis, Iy, is b l M O l M x b m I mi ri2 Ml2 Ml 2 m 02 m 02 2Ml 2 i This is because the rotation is done about y axis, and the radii of the spheres are negligible. 1 2 1 K R I 2Ml 2 2 Ml 2 2 2 2 Why are some 0s? Thus, the rotational kinetic energy is Find the moment of inertia and rotational kinetic energy when the system rotates on the x-y plane about the z-axis that goes through the origin O. 2 2 2 I mi ri 2 Ml Ml 2 mb2 mb 2 2 Ml mb i Thursday, June 22, 2006 1 1 K R I 2 2Ml 2 2mb2 2 Ml 2 mb2 2 2 2 PHYS 1443-001, Summer 2006 Dr. Jaehoon Yu 24 Calculation of Moments of Inertia Moments of inertia for large objects can be computed, if we assume the object consists of small volume elements with mass, Dmi. 2 2 I lim r D m r i i The moment of inertia for the large rigid object is dm Dm 0 i It is sometimes easier to compute moments of inertia in terms of volume of the elements rather than their mass Using the volume density, r, replace dm in the above equation with dV. r i How can we do this? dm The moments of dm rdV inertia becomes dV I rr 2 dV Example: Find the moment of inertia of a uniform hoop of mass M and radius R about an axis perpendicular to the plane of the hoop and passing through its center. y O The moment of inertia is dm R x Thursday, June 22, 2006 What do you notice from this result? I r 2 dm R 2 dm MR 2 The moment of inertia for this object is the same as that of a point of mass M at the distance R. PHYS 1443-001, Summer 2006 Dr. Jaehoon Yu 25 Example for Rigid Body Moment of Inertia Calculate the moment of inertia of a uniform rigid rod of length L and mass M about an axis perpendicular to the rod and passing through its center of mass. M The line density of the rod is y L so the masslet is dm dx M dx L dx x x L The moment of inertia is M L L 3L 2 2 3 What is the moment of inertia when the rotational axis is at one end of the rod. Will this be the same as the above. Why or why not? Thursday, June 22, 2006 L/2 2 M 1 3 x M x I r dm dx L / 2 L 3 L / 2 L L/2 2 3 M L3 ML2 3 L 12 4 L M 1 I r 2 dm x M dx x3 0 L 3 0 L M M 3 ML2 3 L 0 L 3L 3L 3 2 L Since the moment of inertia is resistance to motion, it makes perfect sense for it to be harder to move when it is rotating about the axis at one end. PHYS 1443-001, Summer 2006 Dr. Jaehoon Yu 26 Torque & Angular Acceleration Ft r F r Let’s consider a point object with mass m rotating on a circle. What forces do you see in this motion? m The tangential force Ft and radial force Fr Ft mat mr 2 The torque due to tangential force Ft is t Ft r mat r mr I The tangential force Ft is What do you see from the above relationship? What does this mean? t I Torque acting on a particle is proportional to the angular acceleration. What law do you see from this relationship? Analogs to Newton’s 2nd law of motion in rotation. How about a rigid object? The external tangential force dFt is dFt dmat dmr dFt dt dFt r r 2 dm The torque due to tangential force Ft is dm The total torque is t r 2 dm I r Contribution from radial force is 0, because its What is the contribution due line of action passes through the pivoting O to radial force and why? point, making the moment arm 0. Thursday, June 22, 2006 PHYS 1443-001, Summer 2006 27 Dr. Jaehoon Yu Example for Torque and Angular Acceleration A uniform rod of length L and mass M is attached at one end to a frictionless pivot and is free to rotate about the pivot in the vertical plane. The rod is released from rest in the horizontal position. What are the initial angular acceleration of the rod and the initial linear acceleration of its right end? The only force generating torque is the gravitational force Mg L/2 t Fd F Mg L L Mg I 2 2 L Since the moment of inertia of the rod I r 2 dm 0 when it rotates about one end L We obtain MgL 2I MgL 3 g 2ML2 2 L 3 Thursday, June 22, 2006 L 0 3 M x ML2 2 x dx 3 L 3 0 Using the relationship between tangential and angular acceleration 3g at L 2 What does this mean? The tip of the rod falls faster than an object undergoing a free fall. PHYS 1443-001, Summer 2006 Dr. Jaehoon Yu 28 Rotational Kinetic Energy y vi What do you think the kinetic energy of a rigid object that is undergoing a circular motion is? 1 1 2 2 m v Kinetic energy of a masslet, mi, Ki m r i i i i 2 2 moving at a tangential speed, vi, is mi ri q O x Since a rigid body is a collection of masslets, the total kinetic energy of the rigid object is 1 1 2 2 K R Ki mi ri mi ri 2 i 2 i i Since moment of Inertia, I, is defined as I mi ri 2 i The above expression is simplified as Thursday, June 22, 2006 1 K R I 2 PHYS 1443-001, Summer 2006 Dr. Jaehoon Yu 29 Total Kinetic Energy of a Rolling Body What do you think the total kinetic energy of the rolling cylinder is? P’ CM Since it is a rotational motion about the point P, we can write the total kinetic energy 1 K I P 2 2 2vCM vCM Where, IP, is the moment of inertia about the point P. Using the parallel axis theorem, we can rewrite P 1 1 1 1 2 2 2 2 2 2 K I P I CM MR I CM MR 2 2 2 2 1 1 Since vCM=R, the above 2 2 K I CM MvCM relationship can be rewritten as 2 2 What does this equation mean? Rotational kinetic energy about the CM Translational Kinetic energy of the CM Total kinetic energy of a rolling motion is the sum of the rotational kinetic energy about the CM Thursday, June 22, 2006 And the translational PHYS 1443-001, Summer 2006 kinetic of the CM Dr. Jaehoon Yu 30 Kinetic Energy of a Rolling Sphere R h q vCM Since vCM=R What is the speed of the CM in terms of known quantities and how do you find this out? Let’s consider a sphere with radius R rolling down a hill without slipping. 1 1 2 2 2 K I CM MR 2 2 2 1 vCM 1 Mv 2 I CM CM 2 2 R 1 I CM 2 2 M vCM 2 R Since the kinetic energy at the bottom of the hill must be equal to the potential energy at the top of the hill 1 I CM 2 M 2 vCM Mgh K 2 R vCM Thursday, June 22, 2006 PHYS 1443-001, Summer 2006 Dr. Jaehoon Yu 2 gh 1 I CM / MR 2 31 Example for Rolling Kinetic Energy For solid sphere as shown in the figure, calculate the linear speed of the CM at the bottom of the hill and the magnitude of linear acceleration of the CM. Solve this problem using Newton’s second law, the dynamic method. What are the forces involved in this motion? Gravitational Force, Frictional Force, Normal Force Newton’s second law applied to the CM gives n f M h Mg F F q x Mg sin q f MaCM y n Mg cosq 0 Since the forces Mg and n go through the CM, their moment arm is 0 t and do not contribute to torque, while the static friction f causes torque CM We know that I CM 2 MR 2 5 aCM R We obtain 2 MR 2 I CM 2 f aCM MaCM 5 R R R 5 Substituting f in dynamic equations Thursday, June 22, 2006 fR I CM 7 Mg sin q MaCM 5 PHYS 1443-001, Summer 2006 Dr. Jaehoon Yu aCM 5 g sin q 7 32