Chapter 11 Copyright 2014 John Wiley amp Sons Inc All rights reserved 111 Rolling as Translation and Rotation Combined 1101 Identify that smooth rolling can be considered as a combination of pure translation and pure rotation ID: 620162
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Slide1
Rolling, Torque, and Angular Momentum
Chapter 11
Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.Slide2
11-1
Rolling as Translation and Rotation Combined
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.
Learning Objectives
Figure 11-2
© 2014 John Wiley & Sons, Inc. All rights reserved.Slide3
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 surfaceThe object rotates around the com as it movesThe rotational motion is defined by:
Figure 11-3
© 2014 John Wiley & Sons, Inc. All rights reserved.Slide4
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.Slide5
11-1
Rolling as Translation and Rotation Combined
Figure 11-4
© 2014 John Wiley & Sons, Inc. All rights reserved.Slide6
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.Slide7
11-2
Forces and Kinetic Energy of Rolling
Combine translational and rotational kinetic energy:
© 2014 John Wiley & Sons, Inc. All rights reserved.Slide8
11-2
Forces and Kinetic Energy of Rolling
If a wheel accelerates, its angular speed changesA force must act to prevent slip
© 2014 John Wiley & Sons, Inc. All rights reserved.Slide9
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.Slide10
11-2
Forces and Kinetic Energy of Rolling
We can use this equation to find the acceleration of such a bodyNote that the frictional force produces the rotationWithout friction, the object will simply slide
© 2014 John Wiley & Sons, Inc. All rights reserved.Slide11
11-2
Forces and Kinetic Energy of Rolling
© 2014 John Wiley & Sons, Inc. All rights reserved.Slide12
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.Slide13
11-3
The Yo-Yo
As yo-yo moves down string, it loses potential energy mgh but gains rotational & translational kinetic energyFind linear acceleration of yo-yo accelerating down its string:Rolls down a “ramp” of angle 90°Rolls on an axle instead of its outer surfaceSlowed by tension
T rather than frictionSlide14
11-3
The Yo-Yo
Replacing the values leads us to:
Example
Calculate the acceleration of the yo-yo
M
= 150 grams,
R
0 = 3 mm, Icom = Mr2/2 = 3E-5 kg m2Therefore acom = -9.8 m/s2 / (1 + 3E-5 / (0.15 * 0.0032)) = - .4 m/s2© 2014 John Wiley & Sons, Inc. All rights reserved.Slide15
11-4
Torque Revisited
Previously, torque was defined only for a rotating body and a fixed axisNow we redefine it for an individual particle that moves along any path relative to a fixed pointThe path need not be a circle; torque is now a vectorDirection determined with right-hand-rule
Figure 11-10
© 2014 John Wiley & Sons, Inc. All rights reserved.Slide16
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.Slide17
11-4
Torque Revisited© 2014 John Wiley & Sons, Inc. All rights reserved.Slide18
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.Slide19
Loosen a boltWhich of the three equal-magnitude forces in the figure is most likely to loosen the bolt?Fa?Fb?
Fc?Slide20
Loosen a boltWhich of the three equal-magnitude forces in the figure is most likely to loosen the bolt?Fb!Slide21
TorqueLet O be the point around which a solid body will be rotated.
OSlide22
TorqueLet 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.F applied
Line of actionOSlide23
TorqueLet 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
Line of actionO
L
F
appliedSlide24
Torque
Line of action
OThe torque of a force with respect to O is the product of force and its lever arm. t = F LGreek Letter “Tau” = “torque”Larger torque if
Larger Force applied
Greater Distance from O
L
F
appliedSlide25
Torque t = F L
Larger torque ifLarger Force appliedGreater Distance from OSlide26
Torque
Line of action
OThe torque of a force with respect to O is the product of force and its lever arm. t = F LUnits: Newton-MetersBut wait, isn’t
Nm = Joule??
L
F
appliedSlide27
TorqueTorque Units: Newton-Meters
Use N-m or m-NOR...Ft-Lbs or lb - feet Not…Lb/ftFt/lbNewtons/meterMeters/NewtonSlide28
TorqueTorque is a ROTATIONAL VECTOR!Directions:“Counterclockwise” (+)“Clockwise” (-)“z-axis”Slide29
TorqueTorque is a ROTATIONAL VECTOR!Directions:“Counterclockwise”“Clockwise” Slide30
TorqueTorque is a ROTATIONAL VECTOR!Directions:“Counterclockwise” (+)“Clockwise” (-)“z-axis”Slide31
TorqueTorque is a ROTATIONAL VECTOR!But if r = 0, no torque!Slide32
Torque as a vectorTorque can be expressed as a vector using the vector cross product: t= r x FRight Hand Rule for direction of torque.Slide33
Torque as a vectorTorque can be expressed as a vector using the vector cross product: t= r x FWhere r = vector from axis of rotation to point of application of Force q
= angle between r and FMagnitude: t = r F sinq
F appliedLine of action
O
r
Lever
arm
qSlide34
Torque as a vectorTorque can be evaluated two ways: t= r x Ft= r * (
Fsin q) remember sin q = sin (180-q)
F applied
Line of action
O
r
Lever
arm
q
Fsin
qSlide35
Torque as a vectorTorque can be evaluated two ways: t= r x Ft= (rsin
q) * FThe lever arm L = r sin q
F applied
Line of action
O
r
Lever
arm
r sin
q
qSlide36
Torque as a vectorTorque is zero three ways: t= r x Ft= (
rsin q) * Ft= 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!)
F applied
Line of action
O
r
Lever
arm
r sin
q
qSlide37
Applying a torqueFind t if F = 900 N and q = 19 degreesSlide38
Applying a torqueFind t if F = 900 N and q = 19 degreesSlide39
11-4
Torque Revisited
Example Calculating net torque:
Figure 11-11
© 2014 John Wiley & Sons, Inc. All rights reserved.Slide40
11-5
Angular Momentum
Angular counterpart to linear momentumNote that particle need not rotate around O to have angular momentum around itUnit of angular momentum is kg m2/s, or J s
© 2014 John Wiley & Sons, Inc. All rights reserved.Slide41
11-5
Angular Momentum
To find direction of angular momentum, use right-hand rule to relate r & v to the resultTo find magnitude, use equation for magnitude of a cross product:This is equivalent to:
© 2014 John Wiley & Sons, Inc. All rights reserved.Slide42
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.Slide43
11-5
Angular Momentum
Angular momentum has meaning only with respect to a specified originIt 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.Slide44
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.Slide45
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.Slide46
11-6
Newton's Second Law in Angular Form© 2014 John Wiley & Sons, Inc. All rights reserved.Slide47
11-6
Newton's Second Law in Angular Form
Answer: (a) F3, F1, F2 & F
4 (b) F3 (assuming counterclockwise is positive)© 2014 John Wiley & Sons, Inc. All rights reserved.Slide48
Sum to find angular momentum of a system of particles:
The rate of change of the net angular momentum is:
11-7
Angular Momentum of a Rigid Body
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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.Slide50
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.Slide51
11-7
Angular Momentum of a Rigid Body
Therefore this simplifies to:
© 2014 John Wiley & Sons, Inc. All rights reserved.Slide52
11-7
Angular Momentum of a Rigid Body
Therefore this simplifies to:Slide53
11-7
Angular Momentum of a Rigid Body© 2014 John Wiley & Sons, Inc. All rights reserved.Slide54
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.Slide55
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.Slide56
11-8
Conservation of Angular Momentum
Since these are vector equations, they are equivalent to the three corresponding scalar equationsThis means we can separate axes and write:Slide57
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.Slide58
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 extendedA springboard diver: rotational speed is controlled by tucking her arms and legs in, which reduces rotational inertia and increases rotational speedA 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.Slide59
11-8
Conservation of Angular Momentum© 2014 John Wiley & Sons, Inc. All rights reserved.Slide60
11-8
Conservation of Angular Momentum
Answer: (a) decreases (b) remains the same (c) increases© 2014 John Wiley & Sons, Inc. All rights reserved.Slide61
The race of the rolling bodies Which one wins??WHY???Slide62
Acceleration of a yo-yoWe 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?Slide63
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?Slide64
Work and power in rotational motion Tangential Force over angle does workTotal 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 =
zzSlide65
Work and power in rotational motion Calculating Power from TorqueElectric 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?Slide66
Work and power in rotational motion Calculating Power from ForceElectric 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 travelledDistance = v0
t + ½ at2 = ½ at2F = ma => a = F/m = 5 m/sec/sec => distance = 160 mWork = 1600 J; Avg. Power = W/t = 200 WSlide67
Angular momentum 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.
Slide68
Angular momentum For any system of particles = dL/dt. For a rigid body rotating about the z-axis
z = Iz.Turbine with I = 2.5 kgm2; w = (40 rad/s3)t2What is L(t) & L @ t = 3.0 seconds; what is t(t)? Slide69
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).Slide70
A rotational “collision”Slide71
Angular momentum in a crime bustA 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?Slide72
Angular momentum in a crime bustA 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 finalSlide73
Gyroscopes and precession For a gyroscope, the axis of rotation changes direction. The motion of this axis is called precession.Slide74
11-9
Precession of a Gyroscope
A non-spinning gyroscope falls
© 2014 John Wiley & Sons, Inc. All rights reserved.Slide75
11-9
Precession of a Gyroscope
A spinning gyroscope instead rotates around a vertical axisThis rotation is called precession
© 2014 John Wiley & Sons, Inc. All rights reserved.Slide76
Gyroscopes and precession Slide77
Gyroscopes and precession For a gyroscope, the axis of rotation changes direction. The motion of this axis is called precession.Slide78
A rotating flywheel
Magnitude of angular momentum constant, but its direction changes continuously.Slide79
11-9
Precession of a Gyroscope
Angular momentum of (rapidly spinning) gyroscope is:Torque can only change direction of L, not magnitudeOnly way direction can change along direction of torque
without its magnitude changing is if it rotates around the central axisTherefore it precesses instead of toppling over
© 2014 John Wiley & Sons, Inc. All rights reserved.Slide80
11-9
Precession of a Gyroscope
The precession rate is given by:True for a sufficiently rapid spin rateIndependent of mass, (I is proportional to M) but does depend on
gValid for a gyroscope at an angle to the horizontal as well (a top for instance)
© 2014 John Wiley & Sons, Inc. All rights reserved.