Lecture 6 rotation sự quay

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Lecture ROTATION Lecturer Tran Thi Ngoc Dung dungttn@gmail.com Rotation in Reality • Rotational motion is all around us from molecules to galaxies - The earth rotates about its axis - Wheels, gears (sự truyền động bánh răng), propellers [( chân vịt (tàu), cánh quạt (máy bay) ], motors, a CD in its player, a pirouetting (múa xoay tròn) ice skater, all rotate Outline - Rotation about a fixed axis - Angular Velocity-Angular Acceleration - Torque - Equation of Rotation - Moment of Inertia - Kinertic Energy of Rotation Angular Velocity-Angular Acceleration Rigid object rotates about a fixed axis  The trajectory of each point (except those on the axis) is a circle a rigid body = system of particles, that the distance between any two particles that make up the object remains fixed During a time dt, all points move the same angle d Angular velocity  Angular Acceleration d M (t+dt) Mi(t) i O During a time dt, ith particle move a distance dsi dsi  ri d d M1(t) d  (rad / s ) dt d  (rad / s ) dt M1(t+dt) ri: distance from ith particle Mi to the rotation axis The speed of the ith particle: vi   dsi d  ri  ri dt dt The tangential acceleration of the ith particle : dvi d ai   ri  ri dt dt The centripetal acceleration of the ith particle : O d ri  ain  ai  vi v2 ain    ri ri The acceleration of the ith particle  a2i  ani  ri       vi    ri    ai    ri  2 ain   ri      O  ain     vi  ai O     ain vi  ai The object rotates counter-clockwise             ai  vi   ai  vi The rotation is accelerated The rotation is decelerated    vi    ri    ai    ri  2 ain   ri     ain O     vi         ai The object rotates clockwise  ain  a v i O    i       ai  vi   ai  vi The rotation is decelerated The rotation is accelerated Torque  Consider a force Fi exerting on the object, that can rotale about a fixed axis  F//  e O  ri  Fir  Fi  Fi  e : unit _ evector The torque of the Force F about the  axis going through the point O     F  F//  Fr  Ft Only the tangential force can rotate the object The torque of the Force F about a fixed point O:   F /O   F /   r F     ( F / O e ).e   F /O   F /O  F /        r  F  r  ( F // F r  Ft )        (r  F //)  (r  F r )  (r  Ft )            rF// e 0 rF e       ( M F / O e )e  rFt e  M F /  t Only the tangential force can rotate the object   F / F / t The torque of a parallel force F// is zero The torque of the force that has the line of action go through the axis is zero  0  0 F // /  F r / Dynamics of Rotation Let Force Fi exerts on the ith particle Force Fit causes a tangential acceleration ati of the ith particle: Fit  mi ait  mi ri (1) ati  ri Multiply the eq(1) by ri ri Fit  mi ri2 (2) Torque of Fi about the rotation axis  i  mi ri2 Summing over all the particles in the object (Angular acceleration is the same for all the particles of the object and can therefore be taken out of the sum )  : the net torque acting on the object: I: Moment of Inertia Equation of Rotation ( 2' )   ( m r  i  i i ) i i    i I   mi ri2 i   I i (3) Moment of Inertia I   mi ri2 (system of particles) ri is the distance of the ith particle from the rotation axis  (continuous object) r is the distance of the mass element dm from the rotation axis  i I   r dm   r1 m1 r1 r dm mN ri mi Moment of Inertia of Homogenous Rigid Objects about the axis through the Center of Mass  Solid cylinder/disk R I CM   Hoop or thin cylindrical shell R  mR 2 I CM  mR  Solid sphere R I CM  CM Hollow cylinder R2 I CM   Thin spherical shell I CM Thin rod I CM L R mR  R1  mR Sheet b a m( R12  R22 ) I CM   Through the end Thin rod  mL2 12 I   mL2 L m( a  b ) 12 The Parallel-Axis Theorem The moment of inertia about an axis is equal to the sum of moment of inertia about an axis through the center of mass Icm and Md2 I   ICM  md d : distance between through parallel axes  and  cm  CM I   I CM  md I CM  mR2 , d  R I   mR2  mR2  mR 2 Kinetic Energy of Rotation The kinetic energy of a rotating object is the sum of the kinetic energies of the individual particle s in t he object The kinetic energy of a mass element mi is K i  mi vi2 Summing over all the elements and using vi = riω gives K rot 1  2 2   K i   mi vi   mi ri     mi ri  2 i i i i  K rot  I KInectic Energy of Rotation Example Two blocks are connected by a string that passes over a pulley of radius R and moment of inertia I The block of mass m1 slides on a frictionless, horizontal surface; the block of mass m2 is suspended from the string Find the acceleration a of the blocks and the tensions T1 and T2 assuming that the string does not slip on the pulley Example A uniform thin stick of le ngth L and mass M is pivoted at one end It is held horizontal and released ( Figure 9-24) Assume the pivot is frictionless Find (a) the angular acceleration of the stick immediately after it is released, and (b) the force F0 exerted on the stick by the pivot at this time Example A 4-kg block resting on a frictionless horizontal ledge is attached to a string that passes over a pulley and is attached to a hanging 2-kg block (Figure 9-45 ) The pulley is a uniform disk of radius cm and mass 0.6 kg (a) Find the speed of the 2-kg block after it falls from rest a distance of 2.5 m (b) What is the angular velocity of the pulley at this time? Example1 Two blocks are connected by a string that passes over a pulley of radius R and moment of inertia I The block of mass m1 slides on a frictionless, horizontal surface; the block of mass m2 is suspended from the string Find the acceleration a of the blocks and the tensions T1 and T2 assuming that the string does not slip on the pulley 1 N T1 T’1 T’2 T2 P1 + m1 moves right + m2 moves down + Pulley rotates clockwise + The magnitude of acceleration :a1 = a2 = a + The mqgnitude of the tensions:T1 = T’1 , T2 = T’2 P2 N T1 T’1 T’2 T2 P1 P2 Equation of rotation of pulley Applying Newton’s 2nd law and projecting (1) and (2) on the direction of motion :     m1a1  P1  N  T1 (1)    m2 a2  P2  T2 (2) m1a1  T1 m2 a2  m2 g  T2 (4) (5) a  R (6) I (3) (4) I  T2' R  T1' R Retionship between a and  (3)+(4)+(7) (3) a R2  T2'  T1' (7) I    m1  m2  a  m2 g R   T1  m1a1  a m2 g m1  m2  I / R m1m2 g m1  m2  I / R T2  m2 g  m2 a  m1  I / R m1  m2  I / R m2 g Example A uniform thin stick of le ngth L and mass M is pivoted at one end It is held horizontal and released ( Figure 9-24) Assume the pivot is frictionless Find (a) the angular acceleration of the stick immediately after it is released, and (b) the force Fo exerted on the stick by the pivot at this time Fo  I   mL Applying the Newton’s 2nd law for the CM gives: macm  mg  Fo (1) CM P Eq of rotation: Remark Just after the stick is released, the pivot exerts an upward force equal to onefourth the weight of t he stick (1) L I   mg (2) 3g L   (rad / s ) mL   mg 2L L acm    g Fo  mg  macm  mg Ex A 4-kg block resting on a frictionless horizontal ledge is attached to a string that passes over a pulley and is attached to a hanging 2-kg block The pulley is a uniform disk of radius cm and mass 0.6 kg (a) Find the speed of the 2-kg block after it falls from rest a distance of 2.5 m (b) What is the angular velocity of the pulley at this time? N T1 From Ex 9-7 T’1 T’2 T2 P1 m1  m2  I / R The motion of m2 is linear with constant acceleration v  vo2  2as  0 P2 a m2 g The speed the angular velocity of the pulley v  2as   v/R .. .Rotation in Reality • Rotational motion is all around us from molecules to galaxies - The earth rotates about its axis - Wheels, gears (sự truyền động bánh răng), propellers... skater, all rotate Outline - Rotation about a fixed axis - Angular Velocity-Angular Acceleration - Torque - Equation of Rotation - Moment of Inertia - Kinertic Energy of Rotation Angular Velocity-Angular... object rotates counter-clockwise             ai  vi   ai  vi The rotation is accelerated The rotation is decelerated    vi    ri    ai    ri  2 ain   ri  