1. Chapter 8 Rotational Motion

2. Units of Chapter 8

Angular Quantities

Constant Angular Acceleration

Rolling Motion (Without Slipping)

Torque

Rotational Dynamics; Torque and Rotational Inertia

Solving Problems in Rotational Dynamics

Rotational Kinetic Energy

Angular Momentum and Its Conservation

3. 8-1 Angular Quantities In purelyrotationalmotion, all points on the object move incirclesaround theaxisof rotation ( O ). Theradiusof the circle isr . All points on a straight line drawn through the axis move through the sameanglein the sametime .When P (at radius r) travels an arc length , OP sweeps out an angle . Angular Displacement of the body 4. 8-1 Angular Quantities

Angular Displacement

Commonly, measurein degrees.

Mathof rotation: Easier ifis measured inRadians

1 Radian Angle swept out

when thearc length = radius

When r, 1 Radian

in Radiansisdefinedas:

• = ratio of 2 lengths ( dimensionless )

• MUSTbe in radians for this to be valid!

In doing problems in this chapter,need calculators inRADIAN MODE!!!! 5.

in Radiansfor a circle of radiusr , arc lengthis defined as: ( /r)

Conversionbetween radians & degrees:

for afull circle =360=( /r)radians

Arc length for afull circle =2 r

for afull circle =360=2 radians

Or1 radian (rad) =(360/2 )57.3

Or1 = (2 /360) rad0.017 rad

• In doing problems in this chapter, put yourcalculators inRADIAN MODE!!!!

8-1 Angular Quantities 6. 3 10 -4rad = ? r = 100 m,= ? a)= (3 10 -4rad) [(360/2 )/rad] = 0.017 For small angles:The chordarc length b)= r = (100)(3 10 -4 ) = 0.03 m = 3 cm MUSTbe in radians in partb Example 8-2 : 7. 8-1 Angular Quantities Angulardisplacement: The average angularvelocityis defined as the total angulardisplacementdivided bytime : Theinstantaneousangular velocity: (8-2a) (8-2b) (Units =rad/s ), ValidONLYif is inrad ! 8. 8-1 Angular Quantities The angularaccelerationis the rate at which the angular velocity changes with time: Theinstantaneousacceleration: (8-3a) (8-3b) (Units =rad/s 2 ) ValidONLY if is inrad & is inrad/s ! 9.

From Ch. 5 (circular motion): A mass moving in a circle has a linear velocityv& a linear accelerationa .

Weve just seen that it also has an angular velocity and an angular acceleration.

Every point on a rotating

body has anangular

velocity and alinear

velocity v .

ThereMUSTbe relations

between the linear & the

angular quantities!

10. Connection Between Angular & Linear Quantities v = ( / t), = r v = r( / t) = r Radians! v = r Depends on r ( is the same for all points!) v A= r A A v B= r B B v B> v Asince r B> r A Therefore, objectsfartherfrom the axis of rotation will movefaster . 11. 8-1 Angular Quantities If the angular velocity of a rotating objectchanges , it has atangentialacceleration: a tan= ( v/ t), v =r = r ( / t) a tan= r Even if the angular velocity is constant, each point on the object has acentripetalacceleration: 12. 8-1 Angular Quantities Here is thecorrespondencebetweenlinearandrotationalquantities: 13. 8-1 Angular Quantities Thefrequencyis the number of completerevolutionsper second: or= 2 f (angular frequency) Frequencies are measured inhertz .Theperiodis the time one revolution takes: 14. 8-2 Constant Angular Acceleration The equations of motion forconstantangular acceleration are the same as those forlinearmotion, with the substitution of theangularquantities for thelinearones. NOTE: These areONLY VALIDif all angular quantities are in radian units!! 15. 8-3 Rolling Motion (Without Slipping) In (a), a wheel isrollingwithout slipping. The point P, touching the ground, is instantaneously atrest , and the center moves with velocity v. In (b) the same wheel is seen from areference framewhere C is at rest. Now point P is moving with velocity v. Thelinear speedof the wheel is related to itsangular speed : 16. Example 8-7

Bicycle:v 0= 8.4 m/s . Comes to rest after115 m . Diameter of wheel = 0.68 m ( r = 0.34m )

a) 0=( v 0 /r)= 24.7 rad/s

b)total=( /r)

= (115m)/(0.34m)

= 338.2 rad = 53.8 rev

c)= ( 2 0 2)/(2 )Stopped = 0 = -0.902 rad/s 2 d)t = ( 0 )/ Stopped = 0 t = 27.4 s 17. 8-4 Torque To make an objectstartrotating, aforceis needed; thepositionanddirectionof the force matter as well. Theperpendiculardistance from the axis of rotation to the line along which the force acts is called thelever arm . 18. 8-4 Torque Alongerlever arm is very helpful in rotating objects. 19.

Lever Arm r = distance of the axis of rotation from the line of action of forceF

r = Distance which is to both the axis of rotation and to an imaginary line drawn along the direction of the force (line of action).

Find: Angular acceleration

(force) (lever arm) = Fr

Define: TORQUE

Fr causes

(Just as in the linear motion case, F causesa )

Lower caseGreek tau 8-4 Torque 20. 8-4 Torque Here, the lever arm for F Ais the distance from theknobto thehinge ; the lever arm for F Diszero ; and the lever arm for F Cis as shown. 21. 8-4 Torque Thetorqueis defined as: F =F sin F = F cos = rF sin Units of torque: Newton-meters (N m) 22. Example 8-9 r =r 2 sin60 2 = -r 2 F 2 sin60 1 = r 1 F 1 = 1 + 2= -6.7 m N Always use the following sign convention ! Counterclockwise rotation+ torque Clockwise rotation- torque 23. Section 8-5: Rotational Dynamics

Goal:Newtons 2 ndLaw for rotational motion.

Weve just seen:

The angular acceleration isthe net torque or

net= =sum of torques

• Analogous to Newtons 2 ndLaw :

• aF net= F=sum of forces

• Recall, Newtons 2 ndLaw: F = ma

• a , F

Question:What plays the role of massmfor rotational problems?

24. Simplest Possible Case A massmmoving in aCircle of radiusr , one forceF TANGENTIAL tothe circle = rF Newtons 2 ndLaw + relation (a = r ) between tangential & angular accelerations F = ma = mr So = mr 2 Newtons 2 ndLaw for Rotations Proportionality constant between&ismr 2 (point mass only!) 25.

Newtons 2 ndLaw, rotational motionis:

=I

The net torque = (moment of inertia)(angular acceleration)

Moment of Inertiaof the body:

I = (mr 2 )

Moment of Inertia I = A measure of the rotational inertia of the body. Analogous to the massm= A measure of translational inertia of the body.

26.

For one particle, moving in a circle,Newtons 2 ndLaw for rotational motion is:

= mr 2

Extrapolate to a rigid body, made of many particles. Newtons 2 ndLaw, rotational motionbecomes

= (mr 2 )

• Since is the same for all points in the body, it comes out of the sum.

Write =I ,where the proportionality constantI = (mr 2 ) = m 1 r 1 2+ m 2 r 2 2+ m 3 r 3 2+

Moment of Inertia of the body

27. 8-5 Rotational Dynamics; Torque and Rotational Inertia The quantityis called therotational inertiaof an object. Thedistributionof mass matters here these two objects have the same mass, but the one on the left has a greaterrotational inertia , as so much of its mass is far from the axis of rotation. 28. 8-5 Rotational Dynamics; Torque and Rotational Inertia Therotational inertiaof an object depends not only on itsmass distributionbut also the location of theaxisof rotation compare (f) and (g), for example. 29. Example 8-10

I = (mr 2 )

a) I = (5)(2) 2+(7)(2) 2

= 48 kg m 2

b) I = (5)(0.5) 2

+(7)(4.5) 2

= 143 kg m 2

30. 8-6 Solving Problems in Rotational Dynamics

Drawa diagram.

Decide what thesystemcomprises.

Draw afree-body diagramfor each object under consideration, including all theforcesacting on it and where they act.

Find theaxis of rotation ; calculate thetorquesaround it.

31. 5. Apply Newtons second law forrotation . If therotational inertiais not provided, you need to find itbeforeproceeding with this step. 6. Apply Newtons second law fortranslationand other laws and principles as needed. 7.Solve . 8.Checkyour answer for units and correct order of magnitude. 8-6 Solving Problems in Rotational Dynamics 32. Example 8-11

M = 4 kg, F T= 15 N

Frictional torque:

fr= 1.1 m N, t = 0, 0 =0

t=3 s,=30 rad/s, I = ?

Compute angular accel. :

= 0 + t,

= ( - 0 )/t

= 10 rad/s 2

=I

F T R - fr= I

I = [(15)(0.33) -1.1]/10

I = 0.385 kg m 2

33. 8-7 Rotational Kinetic Energy Thekinetic energyof a rotating object is given byBy substituting the rotational quantities, we find that the rotational kinetic energy can be written: An object that has bothtranslationalandrotationalmotion also has both translational and rotationalkinetic energy : (8-15) (8-16) 34. Example 8-13:

Sphere rolls down incline

(no slipping or sliding).

KE+PEconservation :

()Mv 2+ ()I 2+MgH

=constant

or

(KE) 1+(PE) 1= (KE) 2+ (PE) 2

whereKE has 2 parts :

(KE) trans= ()Mv 2

(KE) rot=()I 2

35. 8-7 Rotational Kinetic Energy When usingconservationof energy, bothrotationalandtranslationalkinetic energy must be taken into account. All these objects have the samepotentialenergy at the top, but the time it takes them to get down the incline depends on how muchrotationalinertiathey have. 36. 8-7 Rotational Kinetic Energy The torque doesworkas it moves the wheel through an angle : (8-17) Torque:= Fr Work:W = F = Fr = 37. 8-8 Angular Momentum and Its Conservation In analogy with linear momentum, we can defineangular momentum L :

Similar to momentum case, can show, rotational version of Newtons 2 ndLaw:

= ( L/ t)(=I )

if= 0, ( L/ t) = 0&L =constant

Angular Momentum is conserved if total torque = 0

Example:t = 0 , angular velocity 0 timet , angular velocity

L = L 0 orI = I 0 0

38. 8-8 Angular Momentum and Its Conservation Therefore, systems that canchangetheir rotational inertia through internal forces will also change theirrateof rotation: 39. Example 8-15

v 1= 2.4 m/s

r 1= 0.8m

r 2= 0.48 m

v 2= ?

I 1 1= I 2 2

m(r 1 ) 2 1= m(r 2 ) 2 2

2= 1(r 1 ) 2 /(r 2 ) 2

1= (v 1 /r 1 )

v 2= r 2 2= r 2 (v 1 /r 1 )[(r 2 ) 2 /(r 1 ) 2 ] = v 1(r 1 /r 2 )

v 2= 4.0m/s

40. Translation-Rotation

TranslationRotation

Displacement x

Velocity v

Acceleration a

Force (Torque) F

Mass (moment of inertia) m I

Newtons 2 ndLaw F =ma =I

Kinetic Energy (KE) ()mv 2 ()I 2

Work(constant F, ) Fd

Momentum mv I

CONNECTIONS :v = r a tan = r

a R= (v 2 /r) = 2 r=r F

I = (mr 2 )

41. Summary of Chapter 8

Angles are measured in radians; a whole circle is 2 radians.

Angular velocity is the rate of change of angular position.

Angular acceleration is the rate of change of angular velocity.

The angular velocity and acceleration can be related to the linear velocity and acceleration.

The frequency is the number of full revolutions per second; the period is the inverse of the frequency.

42. Summary of Chapter 8, cont.

The equations for rotational motion with constant angular acceleration have the same form as those for linear motion with constant acceleration.

Torque is the product of force and lever arm.

The rotational inertia depends not only on the mass of an object but also on the way its mass is distributed around the axis of rotation.

The angular acceleration is proportional to the torque and inversely proportional to the rotational inertia.

43. Summary of Chapter 8, cont.

An object that is rotating has rotational kinetic energy. If it is translating as well, the translational kinetic energy must be added to the rotational to find the total kinetic energy.

Angular momentum is

If the net torque on an object is zero, its angular momentum does not change.