Rolling, Torque, and Angular Momentum Chapter 11 Rolling, Torque, and Angular Momentum Copyright © 2014 John Wiley & Sons, Inc. All rights reserved. 1

© 2014 John Wiley & Sons, Inc. All rights reserved. 11-1 Rolling as Translation and Rotation Combined Learning Objectives 11.01 Identify that smooth rolling can be considered as a combination of pure translation and pure rotation. 11.02 Apply the relationship between the center-of-mass speed and the angular speed of a body in smooth rolling. Figure 11-2 © 2014 John Wiley & Sons, Inc. All rights reserved. 2

© 2014 John Wiley & Sons, Inc. All rights reserved. 11-1 Rolling as Translation and Rotation Combined We consider only objects that roll smoothly (no slip) The center of mass (com) of the object moves in a straight line parallel to the surface The object rotates around the com as it moves The rotational motion is defined by: Figure 11-3 © 2014 John Wiley & Sons, Inc. All rights reserved. 3

© 2014 John Wiley & Sons, Inc. All rights reserved. 11-1 Rolling as Translation and Rotation Combined The figure shows how the velocities of translation and rotation combine at different points on the wheel © 2014 John Wiley & Sons, Inc. All rights reserved. 4

© 2014 John Wiley & Sons, Inc. All rights reserved. 11-1 Rolling as Translation and Rotation Combined Figure 11-4 © 2014 John Wiley & Sons, Inc. All rights reserved. 5

© 2014 John Wiley & Sons, Inc. All rights reserved. 11-1 Rolling as Translation and Rotation Combined Figure 11-4 Answer: (a) the same (b) less than © 2014 John Wiley & Sons, Inc. All rights reserved. 6

© 2014 John Wiley & Sons, Inc. All rights reserved. 11-2 Forces and Kinetic Energy of Rolling Combine translational and rotational kinetic energy: © 2014 John Wiley & Sons, Inc. All rights reserved. 7

© 2014 John Wiley & Sons, Inc. All rights reserved. 11-2 Forces and Kinetic Energy of Rolling If a wheel accelerates, its angular speed changes A force must act to prevent slip © 2014 John Wiley & Sons, Inc. All rights reserved. 8

© 2014 John Wiley & Sons, Inc. All rights reserved. 11-2 Forces and Kinetic Energy of Rolling If slip occurs, then the motion is not smooth rolling! For smooth rolling down a ramp: The gravitational force is vertically down The normal force is perpendicular to the ramp The force of friction points up the slope © 2014 John Wiley & Sons, Inc. All rights reserved. 9

© 2014 John Wiley & Sons, Inc. All rights reserved. 11-2 Forces and Kinetic Energy of Rolling We can use this equation to find the acceleration of such a body Note that the frictional force produces the rotation Without friction, the object will simply slide © 2014 John Wiley & Sons, Inc. All rights reserved. 10

© 2014 John Wiley & Sons, Inc. All rights reserved. 11-2 Forces and Kinetic Energy of Rolling © 2014 John Wiley & Sons, Inc. All rights reserved. 11

© 2014 John Wiley & Sons, Inc. All rights reserved. 11-2 Forces and Kinetic Energy of Rolling 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. © 2014 John Wiley & Sons, Inc. All rights reserved. 12

11-3 The Yo-Yo As yo-yo moves down string, it loses potential energy mgh but gains rotational & translational kinetic energy Find linear acceleration of yo-yo accelerating down its string: Rolls down a “ramp” of angle 90° Rolls on an axle instead of its outer surface Slowed by tension T rather than friction 13

© 2014 John Wiley & Sons, Inc. All rights reserved. 11-3 The Yo-Yo Replacing the values leads us to: Example Calculate the acceleration of the yo-yo 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 © 2014 John Wiley & Sons, Inc. All rights reserved. 14

© 2014 John Wiley & Sons, Inc. All rights reserved. 11-4 Torque Revisited Previously, torque was defined only for a rotating body and a fixed axis Now we redefine it for an individual particle that moves along any path relative to a fixed point The path need not be a circle; torque is now a vector Direction determined with right-hand-rule Figure 11-10 © 2014 John Wiley & Sons, Inc. All rights reserved. 15

© 2014 John Wiley & Sons, Inc. All rights reserved. 11-4 Torque Revisited The general equation for torque is: We can also write the magnitude as: Or, using the perpendicular component of force or the moment arm of F: © 2014 John Wiley & Sons, Inc. All rights reserved. 16

© 2014 John Wiley & Sons, Inc. All rights reserved. 11-4 Torque Revisited © 2014 John Wiley & Sons, Inc. All rights reserved. 17

© 2014 John Wiley & Sons, Inc. All rights reserved. 11-4 Torque Revisited Answer: (a) along the z direction (b) along the +y direction (c) along the +x direction © 2014 John Wiley & Sons, Inc. All rights reserved. 18

Loosen a bolt Which of the three equal-magnitude forces in the figure is most likely to loosen the bolt? Fa? Fb? Fc?

Loosen a bolt Which of the three equal-magnitude forces in the figure is most likely to loosen the bolt? Fb!

Let O be the point around which a solid body will be rotated. Torque Let O be the point around which a solid body will be rotated. O

Let O be the point around which a solid body will be rotated. 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. The line of action of a force is the line along which the force vector lies. O F applied

Let O be the point around which a body will be rotated. 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 is The perpendicular distance from O to the line of action of the force L O 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 = F L Greek Letter “Tau” = “torque” Larger torque if Larger Force applied Greater Distance from O L O F applied

Torque t = F L Larger torque if Larger Force applied Greater Distance from O

Torque Line of action The torque of a force with respect to O is the product of force and its lever arm. t = F L Units: Newton-Meters But wait, isn’t Nm = Joule?? L O F applied

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

Torque Torque is a ROTATIONAL VECTOR! Directions: “Counterclockwise” (+) “Clockwise” (-) “z-axis”

Torque Torque is a ROTATIONAL VECTOR! Directions: “Counterclockwise” “Clockwise”

Torque Torque is a ROTATIONAL VECTOR! Directions: “Counterclockwise” (+) “Clockwise” (-) “z-axis”

Torque Torque is a ROTATIONAL VECTOR! But if r = 0, no torque!

Torque can be expressed as a vector using the vector cross product: 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.

Torque can be expressed as a vector using the vector cross product: Torque as a vector Torque can be expressed as a vector using the vector cross product: 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 Line of action Lever arm O r q F applied

Torque can be evaluated two ways: Torque as a vector Line of action Torque can be evaluated two ways: t= r x F t= r * (Fsin q) remember sin q = sin (180-q) Lever arm O r q F applied Fsinq

Torque can be evaluated two ways: Torque as a vector Line of action Torque can be evaluated two ways: t= r x F t= (rsin q) * F The lever arm L = r sin q Lever arm r sin q O q r F applied

Torque as a vector Torque is zero three ways: Line of action Torque is zero three ways: t= r x F 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!) Lever arm r sin q O q r F applied

Applying a torque Find t if F = 900 N and q = 19 degrees

Applying a torque Find t if F = 900 N and q = 19 degrees

© 2014 John Wiley & Sons, Inc. All rights reserved. 11-4 Torque Revisited Example Calculating net torque: Figure 11-11 © 2014 John Wiley & Sons, Inc. All rights reserved. 39

© 2014 John Wiley & Sons, Inc. All rights reserved. 11-5 Angular Momentum Angular counterpart to linear momentum Note that particle need not rotate around O to have angular momentum around it Unit of angular momentum is kg m2/s, or J s © 2014 John Wiley & Sons, Inc. All rights reserved. 40

© 2014 John Wiley & Sons, Inc. All rights reserved. 11-5 Angular Momentum To find direction of angular momentum, use right-hand rule to relate r & v to the result To find magnitude, use equation for magnitude of a cross product: This is equivalent to: © 2014 John Wiley & Sons, Inc. All rights reserved. 41

© 2014 John Wiley & Sons, Inc. All rights reserved. 11-5 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 © 2014 John Wiley & Sons, Inc. All rights reserved. 42

© 2014 John Wiley & Sons, Inc. All rights reserved. 11-5 Angular Momentum Angular momentum has meaning only with respect to a specified origin It is always perpendicular to the plane formed by the position and linear momentum vectors Answer: (a) 1 & 3, 2 & 4, 5 (b) 2 and 3 (assuming counterclockwise is positive) © 2014 John Wiley & Sons, Inc. All rights reserved. 43

© 2014 John Wiley & Sons, Inc. All rights reserved. 11-6 Newton's Second Law in Angular Form We rewrite Newton's second law as: Torque and angular momentum must be defined with respect to the same point (usually the origin) © 2014 John Wiley & Sons, Inc. All rights reserved. 44

© 2014 John Wiley & Sons, Inc. All rights reserved. 11-6 Newton's Second Law in Angular Form We rewrite Newton's second law as: Note the similarity to the linear form: © 2014 John Wiley & Sons, Inc. All rights reserved. 45

© 2014 John Wiley & Sons, Inc. All rights reserved. 11-6 Newton's Second Law in Angular Form © 2014 John Wiley & Sons, Inc. All rights reserved. 46

© 2014 John Wiley & Sons, Inc. All rights reserved. 11-6 Newton's Second Law in Angular Form Answer: (a) F3, F1, F2 & F4 (b) F3 (assuming counterclockwise is positive) © 2014 John Wiley & Sons, Inc. All rights reserved. 47

© 2014 John Wiley & Sons, Inc. All rights reserved. 11-7 Angular Momentum of a Rigid Body Sum to find angular momentum of a system of particles: The rate of change of the net angular momentum is: © 2014 John Wiley & Sons, Inc. All rights reserved. 48

© 2014 John Wiley & Sons, Inc. All rights reserved. 11-7 Angular Momentum of a Rigid Body The rate of change of the net angular momentum is: In other words, net torque is defined by this change: © 2014 John Wiley & Sons, Inc. All rights reserved. 49

© 2014 John Wiley & Sons, Inc. All rights reserved. 11-7 Angular Momentum of a Rigid Body We can find the angular momentum of a rigid body through summation: The sum is the rotational inertia I of the body © 2014 John Wiley & Sons, Inc. All rights reserved. 50

© 2014 John Wiley & Sons, Inc. All rights reserved. 11-7 Angular Momentum of a Rigid Body Therefore this simplifies to: © 2014 John Wiley & Sons, Inc. All rights reserved. 51

11-7 Angular Momentum of a Rigid Body Therefore this simplifies to: 52

© 2014 John Wiley & Sons, Inc. All rights reserved. 11-7 Angular Momentum of a Rigid Body © 2014 John Wiley & Sons, Inc. All rights reserved. 53

© 2014 John Wiley & Sons, Inc. All rights reserved. 11-7 Angular Momentum of a Rigid Body Answer: (a) All angular momenta will be the same, because the torque is the same in each case (b) sphere, disk, hoop © 2014 John Wiley & Sons, Inc. All rights reserved. 54

© 2014 John Wiley & Sons, Inc. All rights reserved. 11-8 Conservation of Angular Momentum Since we have a new version of Newton's second law, we also have a new conservation law: The law of conservation of angular momentum states that, for an isolated system, (net initial angular momentum) = (net final angular momentum) © 2014 John Wiley & Sons, Inc. All rights reserved. 55

11-8 Conservation of Angular Momentum Since these are vector equations, they are equivalent to the three corresponding scalar equations This means we can separate axes and write: 56

© 2014 John Wiley & Sons, Inc. All rights reserved. 11-8 Conservation of Angular Momentum If the distribution of mass changes with no external torque, we have: © 2014 John Wiley & Sons, Inc. All rights reserved. 57

© 2014 John Wiley & Sons, Inc. All rights reserved. 11-8 Conservation of Angular Momentum Examples A student spinning on a stool: rotation speeds up when arms are brought in, slows down when arms are extended A springboard diver: rotational speed is controlled by tucking her arms and legs in, which reduces rotational inertia and increases rotational speed A long jumper: the angular momentum caused by the torque during the initial jump can be transferred to the rotation of the arms, by windmilling them, keeping the jumper upright © 2014 John Wiley & Sons, Inc. All rights reserved. 58

© 2014 John Wiley & Sons, Inc. All rights reserved. 11-8 Conservation of Angular Momentum © 2014 John Wiley & Sons, Inc. All rights reserved. 59

© 2014 John Wiley & Sons, Inc. All rights reserved. 11-8 Conservation of Angular Momentum Answer: (a) decreases (b) remains the same (c) increases © 2014 John Wiley & Sons, Inc. All rights reserved. 60

The race of the rolling bodies Which one wins?? WHY???

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?

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?

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 = zz

Work and power in rotational motion Calculating Power from Torque 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?

Work and power in rotational motion 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

Angular momentum Units = kg m2/sec (“radians” are implied!) The angular momentum of a rigid body rotating about a symmetry axis 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.  

Angular momentum For any system of particles  = dL/dt. For a rigid body rotating about the z-axis z = Iz. Turbine with I = 2.5 kgm2; w = (40 rad/s3)t2 What is L(t) & L @ t = 3.0 seconds; what is t(t)?

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).

A rotational “collision”

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 w of the door?

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 w of the door? Lintial = mbulletvbulletlbullet Lfinal = Iw = (I door + I bullet) w Idoor = Md2/3 I bullet = ml2 Linitial = Lfinal so mvl = Iw Solve for w and KE final

Gyroscopes and precession For a gyroscope, the axis of rotation changes direction. The motion of this axis is called precession.

© 2014 John Wiley & Sons, Inc. All rights reserved. 11-9 Precession of a Gyroscope A non-spinning gyroscope falls © 2014 John Wiley & Sons, Inc. All rights reserved. 74

© 2014 John Wiley & Sons, Inc. All rights reserved. 11-9 Precession of a Gyroscope A spinning gyroscope instead rotates around a vertical axis This rotation is called precession © 2014 John Wiley & Sons, Inc. All rights reserved. 75

Gyroscopes and precession

Gyroscopes and precession For a gyroscope, the axis of rotation changes direction. The motion of this axis is called precession.

A rotating flywheel Magnitude of angular momentum constant, but its direction changes continuously.

© 2014 John Wiley & Sons, Inc. All rights reserved. 11-9 Precession of a Gyroscope Angular momentum of (rapidly spinning) gyroscope is: Torque can only change direction of L, not magnitude Only way direction can change along direction of torque without its magnitude changing is if it rotates around the central axis Therefore it precesses instead of toppling over © 2014 John Wiley & Sons, Inc. All rights reserved. 79

© 2014 John Wiley & Sons, Inc. All rights reserved. 11-9 Precession of a Gyroscope The precession rate is given by: True for a sufficiently rapid spin rate Independent of mass, (I is proportional to M) but does depend on g Valid for a gyroscope at an angle to the horizontal as well (a top for instance) © 2014 John Wiley & Sons, Inc. All rights reserved. 80