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Chapter 10 Dynamics of Rotational Motion PowerPoint® Lectures for University Physics, Thirteenth Edition – Hugh D. Young and Roger A. Freedman Copyright © 2012 Pearson Education Inc. Goals for Chapter 10 • To learn what is meant by torque • To see how torque affects rotational motion • To analyze the motion of a body that rotates as it moves through space • To use work and power to solve problems for rotating bodies • To study angular momentum and how it changes with time Copyright © 2012 Pearson Education Inc. Loosen a bolt • Which of the three equal-magnitude forces in the figure is most likely to loosen the bolt? • Fa? • F b? • Fc? Copyright © 2012 Pearson Education Inc. Loosen a bolt • Which of the three equal-magnitude forces in the figure is most likely to loosen the bolt? • F b! Copyright © 2012 Pearson Education Inc. Torque • Let O be the point around which a solid body will be rotated. O Copyright © 2012 Pearson Education Inc. Torque Line of action • Let O be the point around which a solid body will be rotated. • Apply an external force F on the body at some point. O • The line of action of a force is the line along which the force vector lies. Copyright © 2012 Pearson Education Inc. F applied Torque Line of action • Let O be the point around which a body will be rotated. • The lever arm (or moment arm) for a force L O is The perpendicular distance from O to the line of action of the force Copyright © 2012 Pearson Education Inc. F applied Torque Line of action • The torque of a force with respect to O is the product of force and its lever arm. t=FL L O • Greek Letter “Tau” = “torque” • Larger torque if • Larger Force applied • Greater Distance from O Copyright © 2012 Pearson Education Inc. F applied Torque t=FL Larger torque if • Larger Force applied • Greater Distance from O Copyright © 2012 Pearson Education Inc. Torque Line of action • The torque of a force with respect to O is the product of force and its lever arm. t=FL L O • Units: Newton-Meters • But wait, isn’t Nm = Joule?? F applied Copyright © 2012 Pearson Education Inc. Torque • Torque Units: Newton-Meters • Use N-m or m-N • OR...Ft-Lbs or lb - feet • Not… • • • • Lb/ft Ft/lb Newtons/meter Meters/Newton Copyright © 2012 Pearson Education Inc. Torque • Torque is a ROTATIONAL VECTOR! • Directions: • “Counterclockwise” (+) • “Clockwise” (-) • “z-axis” Copyright © 2012 Pearson Education Inc. Torque • Torque is a ROTATIONAL VECTOR! • Directions: • “Counterclockwise” • “Clockwise” Copyright © 2012 Pearson Education Inc. Torque • Torque is a ROTATIONAL VECTOR! • Directions: • “Counterclockwise” (+) • “Clockwise” (-) • “z-axis” Copyright © 2012 Pearson Education Inc. Torque • Torque is a ROTATIONAL VECTOR! • But if r = 0, no torque! Copyright © 2012 Pearson Education Inc. Torque as a vector • Torque can be expressed as a vector using the vector cross product: t= r x F • Right Hand Rule for direction of torque. Copyright © 2012 Pearson Education Inc. Torque as a vector • Torque can be expressed as a vector using the vector cross product: Line of action t= r x F • Where r = vector from axis of rotation to point of application of Force q = angle between r and F • Magnitude: t = r F sinq Copyright © 2012 Pearson Education Inc. Lever arm O r q F applied Torque as a vector Line of action • Torque can be evaluated two ways: Lever arm t= r x F t= r * (Fsin q) O r F applied remember sin q = sin (180-q) Copyright © 2012 Pearson Education Inc. q Fsinq Torque as a vector Line of action • Torque can be evaluated two ways: Lever arm r sin q t= r x F t= (rsin q) * F The lever arm L = r sin q Copyright © 2012 Pearson Education Inc. O r F applied q Torque as a vector Line of action • Torque is zero three ways: t= r x F Lever arm r sin q t= (rsin q) * F t= r * (Fsin q) If q is zero (F acts along r) If r is zero (f acts at axis) If net F is zero (other forces!) Copyright © 2012 Pearson Education Inc. O r F applied q Applying a torque • Find t if F = 900 N and q = 19 degrees Copyright © 2012 Pearson Education Inc. Applying a torque • Find t if F = 900 N and q = 19 degrees Copyright © 2012 Pearson Education Inc. Torque and angular acceleration for a rigid body • The rotational analog of Newton’s second law for a rigid body is tz =Iz. • What is for this example? Copyright © 2012 Pearson Education Inc. Torque and angular acceleration for a rigid body • The rotational analog of Newton’s second law for a rigid body is tz =Iz. t = I and t = Fr = Fr/I and a = r Copyright © 2012 Pearson Education Inc. so I = Fr Another unwinding cable – Example 10.3 • What is acceleration of block and tension in cable as it falls? Copyright © 2012 Pearson Education Inc. Rolling as Translation and Rotation Combined Identify that smooth rolling can be considered as a combination of pure translation and pure rotation. Copyright © 2012 Pearson Education Inc. Rolling as Translation and Rotation Combined Assume objects roll smoothly (no slipping!) The center of mass (com) of the object moves in a straight line parallel to the surface Object rotates around the com as it moves Rotational motion is defined by: Copyright © 2012 Pearson Education Inc. Rolling as Translation and Rotation Combined Copyright © 2012 Pearson Education Inc. Rolling without slipping • The condition for rolling without slipping is vcm = R. Copyright © 2012 Pearson Education Inc. 11-1 Rolling as Translation and Rotation Combined Copyright © 2012 Pearson Education Inc. Copyright © 2012 Pearson Education Inc. 11-1 Rolling as Translation and Rotation Combined Copyright © 2012 Pearson Education Inc. 11-1 Rolling as Translation and Rotation Combined Answer: (a) the same (b) less than Copyright © 2012 Pearson Education Inc. Rigid body rotation about a moving axis • Motion of rigid body is a combination of translational motion of center of mass and rotation about center of mass • Kinetic energy of a rotating & translating rigid body is KE = 1/2 Mvcm2 + 1/2 Icm2. Copyright © 2012 Pearson Education Inc. Forces and Kinetic Energy of Rolling Combine translational and rotational kinetic energy: Copyright © 2012 Pearson Education Inc. The race of the rolling bodies • Which one wins?? • WHY??? Copyright © 2012 Pearson Education Inc. A sphere smoothly rolls down a slope and accelerates… • What are acceleration & magnitude of friction on ball? • Use Newton’s second law for motion of center of mass and rotation about center of mass. Copyright © 2012 Pearson Education Inc. Acceleration of a rolling sphere For smooth rolling down a ramp: (If slip occurs, then the motion is not smooth rolling!) 1. Gravitational force is vertically down Copyright © 2012 Pearson Education Inc. Acceleration of a rolling sphere For smooth rolling down a ramp: 1. Gravitational force is vertically down 2. Normal force is perpendicular to ramp Copyright © 2012 Pearson Education Inc. Acceleration of a rolling sphere For smooth rolling down a ramp: 1. Gravitational force is vertically down 2. Normal force is perpendicular to ramp 3. Force of STATIC friction points up slope Copyright © 2012 Pearson Education Inc. Acceleration of a rolling sphere • Use Newton’s second law for the motion of the center of mass and the rotation about the center of mass. • What are acceleration & magnitude of friction on ball? Copyright © 2012 Pearson Education Inc. Acceleration of a rolling sphere • Use Newton’s second law for the motion of the center of mass and the rotation about the center of mass. • What are acceleration & magnitude of friction on ball? Note that static frictional force produces the rotation Without friction, the object will simply slide Copyright © 2012 Pearson Education Inc. Copyright © 2012 Pearson Education Inc. Answer: The maximum height reached by B is less than that reached by A. For A, all the kinetic energy becomes potential energy at h. Since the ramp is frictionless for B, all of the rotational K stays rotational, and only the translational kinetic energy becomes potential energy at its maximum height. Copyright © 2012 Pearson Education Inc. A yo-yo • What is the speed of the center of the yo-yo after falling a distance h? • Is this less or greater than an object of mass M that doesn’t rotate as it falls? Copyright © 2012 Pearson Education Inc. Acceleration of a yo-yo • We have translation and rotation, so we use Newton’s second law for the acceleration of the center of mass and the rotational analog of Newton’s second law for the angular acceleration about the center of mass. •What is a and T for the yo-yo? Copyright © 2012 Pearson Education Inc. As yo-yo moves down string, it loses potential energy mgh but gains rotational & translational kinetic energy 1. Rolls down a “ramp” of angle 90° 2. Rolls on an axle instead of its outer surface 3. Slowed by tension T rather than friction Copyright © 2012 Pearson Education Inc. Example Calculate the acceleration of the yo-yo o o M = 150 grams, R0 = 3 mm, Icom = Mr2/2 = 3E-5 kg m2 Therefore acom = -9.8 m/s2 / (1 + 3E-5 / (0.15 * 0.0032)) = - .4 m/s2 Copyright © 2012 Pearson Education Inc. Work and power in rotational motion • Tangential Force over angle does work • Total work done on a body by external torque is equal to the change in rotational kinetic energy of the body • Power due to a torque is P = tzz Copyright © 2012 Pearson Education Inc. Work and power in rotational motion • Electric Motor provide 10-Nm torque on grindstone, with I = 2.0 kg-m2 about its shaft. • Starting from rest, find work W down by motor in 8 seconds and KE at that time. • What is the average power? Copyright © 2012 Pearson Education Inc. Work and power in rotational motion • Analogy to 10.8: Calculating Power from Force • Electric Motor provide 10-N force on block, with m = 2.0 kg. • Starting from rest, find work W down by motor in 8 seconds and KE at that time. • What is the average power? • W = F x d = 10N x distance travelled • Distance = v0t + ½ at2 = ½ at2 • F = ma => a = F/m = 5 m/sec/sec => distance = 160 m • Work = 1600 J; Avg. Power = W/t = 200 W Copyright © 2012 Pearson Education Inc. Angular momentum • The angular momentum of a rigid body rotating about a symmetryaxis is parallel to the angular velocity and is given by L = I. • Units = kg m2/sec (“radians” are implied!) • Direction = along RHR vector of angular velocity. Copyright © 2012 Pearson Education Inc. Angular momentum • For any system of particles t = dL/dt. • For a rigid body rotating about the z-axis tz = Iz. Ex 10.9: Turbine with I = 2.5 kgm2; = (40 rad/s3)t2 What is L(t) & L @ t = 3.0 seconds; what is t(t)? Copyright © 2012 Pearson Education Inc. Angular Momentum Angular momentum has meaning only with respect to a specified origin and axis. L is always perpendicular to plane formed by position & linear momentum vectors Copyright © 2012 Pearson Education Inc. Angular Momentum Copyright © 2012 Pearson Education Inc. 11-5 Angular Momentum a) Rank particles according to magnitudes of Angular Momentum about O 1 & 3, 2 & 4, 5 b) Which have negative ang. Momentum about O? 2 and 3 (assuming counterclockwise is positive) Copyright © 2012 Pearson Education Inc. Conservation of angular momentum • When the net external torque acting on a system is zero, the total angular momentum of the system is constant (conserved). Copyright © 2012 Pearson Education Inc. A rotational “collision” Copyright © 2012 Pearson Education Inc. Angular momentum in a crime bust • A bullet (mass = 10g, @ 400 m/s) hits a door (1.00 m wide mass = 15 kg) causing it to swing. What is of the door? Copyright © 2012 Pearson Education Inc. Angular momentum in a crime bust • A bullet (mass = 10g, @ 400 m/s) hits a door (1.00 m wide mass = 15 kg) causing it to swing. What is of the door? • Lintial = mbulletvbulletlbullet • Lfinal = I = (I door + I bullet) • Idoor = Md2/3 • I bullet = ml2 • Linitial = Lfinal so mvl = I • Solve for and KE final Copyright © 2012 Pearson Education Inc. Gyroscopes and precession • For a gyroscope, the axis of rotation changes direction. The motion of this axis is called precession. Copyright © 2012 Pearson Education Inc. Gyroscopes and precession Copyright © 2012 Pearson Education Inc. Gyroscopes and precession • For a gyroscope, the axis of rotation changes direction. The motion of this axis is called precession. Copyright © 2012 Pearson Education Inc. A rotating flywheel • Magnitude of angular momentum constant, but its direction changes continuously. Copyright © 2012 Pearson Education Inc. A precessing gyroscopic • Example 10.13 Copyright © 2012 Pearson Education Inc.