Rotation Torque, Rolling, & Angular Momentum (Chapters 10 & 11, p.241-304)

RotationRotationTorque, Rolling, & Angular MomentumTorque, Rolling, & Angular Momentum

(Chapters 10 & 11, p.241-(Chapters 10 & 11, p.241-304)304)

Rotation: TerminologyRotation: Terminology Axis of rotation- the line/axis around which an object spinsAxis of rotation- the line/axis around which an object spins

r – the radius of a spinning thingr – the radius of a spinning thing

(theta) – a variable for the angle that something has rotated through(theta) – a variable for the angle that something has rotated through

s – a variable for the linear distance (arc length) that an object at radius “r” would spin s – a variable for the linear distance (arc length) that an object at radius “r” would spin throughthrough

- angular velocity (the angle-that-is-rotated-through per unit-of-time: radians/second)- angular velocity (the angle-that-is-rotated-through per unit-of-time: radians/second)

- angular acceleration (change in angular-velocity per unit-of-time: radians/second^2)- angular acceleration (change in angular-velocity per unit-of-time: radians/second^2)

v – “linear” (or “tangential”) velocityv – “linear” (or “tangential”) velocity

T - period (the time needed for object to rotate back to “start angle”)T - period (the time needed for object to rotate back to “start angle”)

K – as always, the “Kinetic energy”K – as always, the “Kinetic energy”

I – rotational inertia (I – rotational inertia (somewhatsomewhat analogous to typical, linear inertia) analogous to typical, linear inertia)

T T - torque (a sort of “rotational force”)- torque (a sort of “rotational force”)

W – workW – work

P – powerP – power

ll – angular momentum – angular momentum

The Basics of RotationThe Basics of Rotation 1 rev=3601 rev=360=2=2ππ radians radians Instantaneous angular velocity is derivative of Instantaneous angular velocity is derivative of

function for angular position: function for angular position: (t)(t) Angular acceleration is the derivative of velocity: Angular acceleration is the derivative of velocity:

(t)(t) The same acceleration-a/displacement-d/velocity-The same acceleration-a/displacement-d/velocity-

v equations (from kinematics) are used for v equations (from kinematics) are used for //// s (the arc length) = s (the arc length) = rr ……taking the derivative of above: v=taking the derivative of above: v=r, and a=r, and a=rr

Using centripetal equations (aUsing centripetal equations (acentripetal / radialcentripetal / radial=v=v22/r) /r) and and v=v=r, we get: ar, we get: aradialradial==22rr

James Rittner

angular velocity

Energy in RotationEnergy in Rotation Kinetic Energy= Kinetic Energy= ½I½I22

Every rotating particle has an “I” equal to its-mass Every rotating particle has an “I” equal to its-mass times its-radius-from-the-axistimes its-radius-from-the-axis

……thus a big object’s “I” is the integral of its thus a big object’s “I” is the integral of its particles: particles: I=I=∫∫rr22dmdm (all the infinitesimally (all the infinitesimally small masses, times each’s radius-from-axis)small masses, times each’s radius-from-axis)

If you know “I” for an object when it spins around a If you know “I” for an object when it spins around a certain axis that goes through the center-of-mass, certain axis that goes through the center-of-mass, BUT it is spinning around a new, different axis (that BUT it is spinning around a new, different axis (that happens to be parallel to the old one)… happens to be parallel to the old one)… I Inewnew=I=Iorigorig+(m+(mobjectobject))(distance-between-axes)(distance-between-axes)

d

Rotational Forces (Torque), Work, Rotational Forces (Torque), Work, and Powerand Power

Net Torque = INet Torque = I (sort of like (sort of like F=ma)F=ma)

Just as work=F*Just as work=F*ΔΔx=x=∫∫FdxFdx, , Work = Work = TTΔΔ = = ii∫∫ff TTdd = = ΔΔKK

Power=dW = Power=dW = TT dtdt

Rollin’, Rollin’, Rollin’!Rollin’, Rollin’, Rollin’! If a wheel spins through a certain angle, the distance it If a wheel spins through a certain angle, the distance it

moves on the ground equals the arc-length that moves on the ground equals the arc-length that corresponds to that angle… (if you get that good, if you corresponds to that angle… (if you get that good, if you don’t, forget it, it’s ok)don’t, forget it, it’s ok)

……thus, “distance-across-ground”=“arc-length”: thus, “distance-across-ground”=“arc-length”: “d”=“s”=“d”=“s”=rr

……taking the derivative of both sides, we get: taking the derivative of both sides, we get: v=v=rr (v=linear velocity along ground)(v=linear velocity along ground)

……deriving again: deriving again: a=a=rr

Note that d, v, and a refer to the motion of the rolling-object’s Note that d, v, and a refer to the motion of the rolling-object’s center-of-mass.center-of-mass.

Energy:Energy: Kinetic Energy=KKinetic Energy=Krotationalrotational+K+Klinearlinear= ½I= ½I22 + ½mv + ½mv22

Torque Related to Linear ForceTorque Related to Linear Force

The torque (or the-”rotational force”) The torque (or the-”rotational force”) being applied to a particle equals the cross being applied to a particle equals the cross product of that-particle’s-position-vector product of that-particle’s-position-vector (r) and the-force-on-the-particle (a vector (r) and the-force-on-the-particle (a vector called “F”): T=rcalled “F”): T=rxxF F

……by simply using the definition of a cross by simply using the definition of a cross product, we get: T=|r|*|F|*sinproduct, we get: T=|r|*|F|*sinΦΦ

……the basic point is that T equals the the basic point is that T equals the product of the PERPENDICULAR-product of the PERPENDICULAR-COMPONENTS-of-r-and-FCOMPONENTS-of-r-and-F

Angular MomentumAngular Momentum l l =“angular-momentum”= r=“angular-momentum”= rxxp = m(rp = m(rxxv) (…where “p” is v) (…where “p” is

linear-momentum)linear-momentum) ……for just for just ll’s magnitude: ’s magnitude: ll =|r|*m*|v|*sin=|r|*m*|v|*sinΦΦ, where , where ΦΦ is the is the

angle-between-vector-r-and-vector-pangle-between-vector-r-and-vector-p

For a system of particles, the system’s-total-angular-For a system of particles, the system’s-total-angular-momentum (called “L”) is just the sum of each particle’s-momentum (called “L”) is just the sum of each particle’s-angular-momentum: angular-momentum:

L= L= ll 11+ + ll 22+ + ll 33+…= +…= ΣΣll

For a rigid body, the component of its angular-momentum For a rigid body, the component of its angular-momentum that is parallel to its axis-of-rotation is found by: that is parallel to its axis-of-rotation is found by: LLparallelparallel=I=I

The total angular momentum is conserved; that is: L-initial=L-The total angular momentum is conserved; that is: L-initial=L-finalfinal

Question 1Question 1

What would be the axis of rotation What would be the axis of rotation for a car wheel?…for a spinning for a car wheel?…for a spinning quarter?quarter?

The axle; a line that is the diameter of The axle; a line that is the diameter of the quarter’s face and that cuts the quarter’s face and that cuts Washington’s face into two partsWashington’s face into two parts


While an object’s linear position is its While an object’s linear position is its x-coordinate (or coordinates), its x-coordinate (or coordinates), its angular position is __________.angular position is __________.

The angle that it is turned to.The angle that it is turned to.


Angular displacement is _____.Angular displacement is _____.

The angle through which the object The angle through which the object has been rotated…. Or “the final has been rotated…. Or “the final angular-position minus the initial angular-position minus the initial angular-position”angular-position”


(a) A wheel spins one full revolution (a) A wheel spins one full revolution every second. (b) That spinning causes every second. (b) That spinning causes that wheel’s center of mass (and the that wheel’s center of mass (and the car it is attached to) to travel 4 feet per car it is attached to) to travel 4 feet per second along the road. These are two second along the road. These are two types of velocities; what is each called?types of velocities; what is each called?

A>>angular velocity; B>>linear A>>angular velocity; B>>linear velocityvelocity

Question 5Question 5 You mark a point on a wheel’s edge. The You mark a point on a wheel’s edge. The

point is at an angle theta at time t, such point is at an angle theta at time t, such that that (t)=3t^3 + t^2 + t …which of (t)=3t^3 + t^2 + t …which of the following is a function describing the the following is a function describing the angular acceleration of the point? angular acceleration of the point? A) alpha=0 B) A) alpha=0 B) alpha=9t^2 + 2t C) alpha=9t^2 +2t +1 alpha=9t^2 + 2t C) alpha=9t^2 +2t +1 D) alpha=18t +2 D) alpha=18t +2

DD


If roll of toilet paper rolls along the If roll of toilet paper rolls along the ground and rotates an angle equal to ground and rotates an angle equal to “theta,” how much paper will roll off? “theta,” how much paper will roll off? (how far along the ground will it roll?)(how far along the ground will it roll?)

Length-of-paper-to-roll-off = arc-Length-of-paper-to-roll-off = arc-length-that-corresponds-to-theta-length-that-corresponds-to-theta-radians = radius-times-thetaradians = radius-times-theta


Linear velocity equals relates to Linear velocity equals relates to angular velocity in what way?angular velocity in what way?

Linear-velocity = angular-velocity * Linear-velocity = angular-velocity * radiusradius


How does Period relate to angular How does Period relate to angular velocity? And how does it relate to velocity? And how does it relate to linear velocity?linear velocity?

T = (2pi)/T = (2pi)/ T=(2pi*radius)/v T=(2pi*radius)/v


What is the rotational inertia of a What is the rotational inertia of a cylinder, rolling on it’s side?cylinder, rolling on it’s side?

I=.5Mr^2 where r is radius and M is I=.5Mr^2 where r is radius and M is massmass


What is the rotational inertia of a rod What is the rotational inertia of a rod spun like a baton?spun like a baton?

I=(1/12)*ML^2, where M is the mass I=(1/12)*ML^2, where M is the mass and L is the lengthand L is the length


What is the rotational inertia of a tire What is the rotational inertia of a tire shaped object, rotating as a tire does shaped object, rotating as a tire does on a car wheel?on a car wheel?

I=.5M*(a^2 + b^2), where a is the I=.5M*(a^2 + b^2), where a is the inner radius, and b is the outer radiusinner radius, and b is the outer radius


What is the main equation of What is the main equation of Newton’s Second Law as applied to Newton’s Second Law as applied to rotation?rotation?

Torque=I*alphaTorque=I*alpha


The force above is applied to a wheel’s The force above is applied to a wheel’s edge, but is not applied tangential to that edge, but is not applied tangential to that edge. How is the torque-applied calculated?edge. How is the torque-applied calculated?

Torque-applied = |r|*|F|*sinTorque-applied = |r|*|F|*sin (this is the same (this is the same as multiplying the magnitude-of-the-radius and the-as multiplying the magnitude-of-the-radius and the-magnitude-of-the-component-of-F-that-is-magnitude-of-the-component-of-F-that-is-perpendicular-to-the-radius)perpendicular-to-the-radius)

Force-Vector

Radius-VectorRadius-Vector

Question 14Question 14 Work is the integral of what function? (possible Work is the integral of what function? (possible

hint: when you integrate, you find the area of hint: when you integrate, you find the area of a function; in a way, you are sort of a function; in a way, you are sort of multiplying the x-values of the function by the multiplying the x-values of the function by the y-values… so whatever two things multiply to y-values… so whatever two things multiply to equal “Work,” those are the x and y of your equal “Work,” those are the x and y of your function)function)

YY XX

Work is the integral of Force as a function of Work is the integral of Force as a function of distance (or, “x”)distance (or, “x”)


Power is the integral of Power is the integral of what what functionfunction? What is the ? What is the independent independent variablevariable of this function? of this function?

WorkWork as a function of as a function of timetime


What is the the parallel axis theorem?What is the the parallel axis theorem?

If the rotational inertia is calculated for an If the rotational inertia is calculated for an object spinning around a certain axis that object spinning around a certain axis that goes through the center-of-mass, but then goes through the center-of-mass, but then a new, different axis-of-rotation is used a new, different axis-of-rotation is used (and the new axis is parallel to the old (and the new axis is parallel to the old axis, but does not go through the center of axis, but does not go through the center of mass), then: Imass), then: Inewnew= I= Icenter-of-center-of-

massmass+Mass*distance-between-axes+Mass*distance-between-axes


If a wheel rolls smoothly, the If a wheel rolls smoothly, the distance it will move (linearly, across distance it will move (linearly, across the ground) will equal the ___-length the ground) will equal the ___-length that corresponds to the angle the that corresponds to the angle the wheel rotated through.wheel rotated through.

The The arcarc-length-length


A rolling wheel experiences which A rolling wheel experiences which kind of friction? A)static or kind of friction? A)static or B)kineticB)kinetic

A) A) StaticStatic (because the wheel never (because the wheel never actually moves, or “skids” along the actually moves, or “skids” along the surface… if it were perfect it would surface… if it were perfect it would simply touch down on the ground, simply touch down on the ground, then lift up as it rolled)then lift up as it rolled)


What is one way to calculate angular-What is one way to calculate angular-momentum (momentum (l l )? (give an equation))? (give an equation)

It equals “the cross-product of a It equals “the cross-product of a radius-vector and a linear-radius-vector and a linear-momentum-vector”. It also equals momentum-vector”. It also equals “mass multiplied by the cross-“mass multiplied by the cross-product of radius and linear-product of radius and linear-velocity”.velocity”.


If you take the derivative of angular-If you take the derivative of angular-momentum (with respect to time), momentum (with respect to time), what do you get? (possible hint: what do you get? (possible hint: what do you get when you take the what do you get when you take the derivative of linear-momentum as a derivative of linear-momentum as a function of time?)function of time?)

You get the net-torqueYou get the net-torque

Problem 1Problem 1

A wheel (of radius=25cm) makes 2.5 A wheel (of radius=25cm) makes 2.5 revolutions every second as it rolls revolutions every second as it rolls smoothly. What is it’s linear velocity? How smoothly. What is it’s linear velocity? How long will it take to roll a linear distance of long will it take to roll a linear distance of “3.25*pi” meters? And what is it’s “3.25*pi” meters? And what is it’s rotational period?rotational period?

1.25*pi (meters per second)1.25*pi (meters per second)

2.5 seconds2.5 seconds

.4 seconds.4 seconds

Problem 2Problem 2

At any time “t” (in seconds) a rotating At any time “t” (in seconds) a rotating object is at an angle “theta” (in object is at an angle “theta” (in radians) and Theta= t^4 + 4t^3 radians) and Theta= t^4 + 4t^3 +2t^2 +5. What is the instantaneous +2t^2 +5. What is the instantaneous angular acceleration at second-2? angular acceleration at second-2? What is the average angular What is the average angular acceleration between seconds 1 and 3?acceleration between seconds 1 and 3?

100 radians per second-squared100 radians per second-squared187 radians per second-squared187 radians per second-squared

Problem 3Problem 3

Through what angle will an object Through what angle will an object rotate given that: it starts at an rotate given that: it starts at an angular-velocity of pi radians-per-angular-velocity of pi radians-per-second, it has a constant angular second, it has a constant angular acceleration of 3*pi radians-per-second-acceleration of 3*pi radians-per-second-squared, and it goes for 2 seconds ? Be squared, and it goes for 2 seconds ? Be sure not to over-think this one.sure not to over-think this one.

8*pi radians8*pi radians

Problem 4Problem 4

At the bottom of a ramp, a smoothly At the bottom of a ramp, a smoothly rolling tire is traveling at an angular rolling tire is traveling at an angular speed of 3 rad-per-sec. The inner speed of 3 rad-per-sec. The inner radius of the tire is 50cm; the outer radius of the tire is 50cm; the outer is 1m. It’s mass is 12kg. How high is 1m. It’s mass is 12kg. How high will the rolling tire go up the ramp?will the rolling tire go up the ramp?

0.066 meters0.066 meters

h?

Problem 5Problem 5 Gary is on a miniature, “spinable” merry-go-round Gary is on a miniature, “spinable” merry-go-round

on the play ground, standing .5 meters from the on the play ground, standing .5 meters from the center. Larry then applies a force of 11 Newtons center. Larry then applies a force of 11 Newtons to the edge of the merry-go-round to spin Gary. to the edge of the merry-go-round to spin Gary. The edge is 1.5 meters in radius from the center. The edge is 1.5 meters in radius from the center. BUT, the force is applied not tangentially (at an BUT, the force is applied not tangentially (at an angle of 90-degrees from the radius-vector), but angle of 90-degrees from the radius-vector), but at an angle of 30-degrees to the radius-vector. If at an angle of 30-degrees to the radius-vector. If Gary and the merry-go-round altogether masses Gary and the merry-go-round altogether masses at 300kg, then how fast, linearly, will Gary at 300kg, then how fast, linearly, will Gary accelerate?accelerate?

.014 m/s^2.014 m/s^2

Force applied

Gary 30°

Radius-vector

Problem 6Problem 6 Force= 4Force= 4^2 (where ^2 (where =0 radians just as Larry =0 radians just as Larry

starts)starts) What will be the overall change in energy What will be the overall change in energy of the merry-go-round-AND-Gary system from of the merry-go-round-AND-Gary system from problem #5, if the above equation gives the Force problem #5, if the above equation gives the Force applied by Larry (still applied at 30-degrees) as a applied by Larry (still applied at 30-degrees) as a function of the merry-go-round’s angular position function of the merry-go-round’s angular position “theta” AND IF Larry spins it two full revolutions?“theta” AND IF Larry spins it two full revolutions?

+64*pi^3 Joules+64*pi^3 Joules

Problem 7Problem 7 A boy in a marching band is smoothly spinning a .75-meter-A boy in a marching band is smoothly spinning a .75-meter-

long baton of uniformly distributed mass about an axis that long baton of uniformly distributed mass about an axis that is perpendicular to the baton and goes through it’s center. is perpendicular to the baton and goes through it’s center. It spins at 3*pi rad-per-sec. He lets go of it quickly, without It spins at 3*pi rad-per-sec. He lets go of it quickly, without getting in its way or altering its rotation. He then quickly getting in its way or altering its rotation. He then quickly grabs the end of it and spins the baton about a new axis grabs the end of it and spins the baton about a new axis that is parallel to the old one, but that goes through the that is parallel to the old one, but that goes through the rod’s end. The baton quickly ends up having a new angular rod’s end. The baton quickly ends up having a new angular speed (with respect to the new axis) of 1*pi rad-per-sec. speed (with respect to the new axis) of 1*pi rad-per-sec. The baton has a mass of 0.3kg. What is the minimal amount The baton has a mass of 0.3kg. What is the minimal amount of work the rod did upon the band-member’s hand? of work the rod did upon the band-member’s hand? (ignoring gravity and air resistance…)(ignoring gravity and air resistance…)

0.35 Joules0.35 Joules

Problem 8Problem 8

At one instant, force-vector “F=4.0At one instant, force-vector “F=4.0ĵ N” ĵ N” acts on a 0.25 kg object that has position-acts on a 0.25 kg object that has position-vector “r=(2.0î-2.0k) meters” and velocity-vector “r=(2.0î-2.0k) meters” and velocity-vector “v=(-5.0î+5.0k) m/s”. About the vector “v=(-5.0î+5.0k) m/s”. About the origin and in unit-vector notation, what are origin and in unit-vector notation, what are (a) the object’s angular momentum and (a) the object’s angular momentum and (b) the torque acting on the object?(b) the torque acting on the object?

00(8.0 Newton-meters)î + (8.0 Newton-(8.0 Newton-meters)î + (8.0 Newton-

meters)kmeters)k

Problem 9Problem 9 A wheel is rotating freely at an angular A wheel is rotating freely at an angular

speed of 800 rev/min on a shaft whose speed of 800 rev/min on a shaft whose rotational inertia is negligible. A second rotational inertia is negligible. A second wheel, initially at rest and with twice the wheel, initially at rest and with twice the rotational inertia of the first, is suddenly rotational inertia of the first, is suddenly coupled to the same shaft. (a) What is the coupled to the same shaft. (a) What is the angular speed of the resultant combination angular speed of the resultant combination of the shaft and two wheels? (b) What of the shaft and two wheels? (b) What fraction of the original rotational kinetic fraction of the original rotational kinetic energy is lost?energy is lost?

a) 267 rev/mina) 267 rev/minb) 0.0667b) 0.0667

Problem 10Problem 10 Two 2.00 kg balls are attached to the ends of a thin rod of Two 2.00 kg balls are attached to the ends of a thin rod of

length 50.0 cm and negligible mass. The rod is free to length 50.0 cm and negligible mass. The rod is free to rotate in a vertical plane without friction about a horizontal rotate in a vertical plane without friction about a horizontal axis through its center. With the rod initially horizontal (see axis through its center. With the rod initially horizontal (see figure), a 50.0 g wad of wet putty drops onto one of the figure), a 50.0 g wad of wet putty drops onto one of the balls, hitting it with a speed of 3.00 m/s and then sticking to balls, hitting it with a speed of 3.00 m/s and then sticking to it. (a) What is the angular speed of the system just after it. (a) What is the angular speed of the system just after the putty wad hits? (b) What is the ratio of the kinetic the putty wad hits? (b) What is the ratio of the kinetic energy of the system after the collision to that of the putty energy of the system after the collision to that of the putty wad just before? (c) Through what angle will the system wad just before? (c) Through what angle will the system rotate before it momentarily stops?rotate before it momentarily stops?

a) 0.148 rad/sa) 0.148 rad/sb) 0.0123b) 0.0123c) 181 degreesc) 181 degrees

Axis of rotation

putty

Closing NoteClosing Note

Always remember the Always remember the difference between difference between a positive and a positive and negative torque. It negative torque. It is not that one is is not that one is “up” and one’s “up” and one’s “down.” Rather, “down.” Rather, remember that they remember that they are “clockwise” and are “clockwise” and “counterclockwise.”“counterclockwise.”

Positive torque (that happens to point down)

Negative torque (that happens to point down)

Documents

Rotation Torque, Rolling, & Angular Momentum (Chapters 10 & 11, p.241-304)