Relating Translational and Rotational Variables

Rotational position and distance moved

€

s = θ r (only radian units)

Rotational and translational speed

€

v =d s→

dt=

dsdt

=d θdt

r

v = ω r

Relating period and rotational speed

€

T =2πrv

=2πω

units of time (s) €

distance = rate× time[ ]

ωT = 2π

Relating Translational and Rotational Variables

Rotational and translational acceleration

“acceleration” is a little tricky

from change in angular speed

tangential acceleration

€

at =dωdt

r = α r

b) from before we know there’s also a “radial” component

€

ar =v2

r= ω

2r radial acceleration

€

from v = ω r

dvdt

=dωdt

r

a)

€

a tot

2= a r

2+ a t

2

= ω 2r2

+ αr 2

c) must combine two distinct rotational accelerations

Relating Translational and Rotational Variables Rotational position and distance moved

€

s = θ r (only radian units)

Rotational and translational speed

€

v =d r dt

=dsdt

=d θdt

r

v = ω r

Rotational and translational acceleration

tangential acceleration

€

at =dωdt

r = α r

€

ar =v2

r= ω

2r radial acceleration

€

a tot

2= a r

2+ a t

2

= ω 2r2

+ αr 2

€

v = ω × r

€

a t = α × r

€

a r = ω ×

ω × r ( )

€

a tot = a t + a r

€

θ (t) Δ

θ = θ 2 − θ 1( )

€

ω (t) =

d θ (t)dt

€

α (t) =

d ω (t)dt

=d 2 θ (t)dt 2

€

v =dθdt

r = ω r€

s = θ r

€

a tot

2= a r

2+ a t

2

= ω 2r2

+ αr 2

€

v = ω × r

€

a t = α × r

€

a r = ω ×

ω × r ( )

€

a tot = a t + a r

Rotational Kinematics: ( ONLY IF α = constant)

€

ω =ω 0 +αtθ = θ +ω 0t + 1

2αt

Example #1

A beetle rides the rim of a rotating merry-go-round.

If the angular speed of the system is constant, does the beetle have a) radial acceleration and b) tangential acceleration?

If the angular speed is decreasing at a constant rate, does the beetle have a) radial acceleration and b) tangential acceleration?

€

If ω = constant, α = 0

€

If α = neg constant, ω =ω 0 +αt

€

v

€

a t

€

a r

€

a tot

€

ˆ r

€

a tot

2= a r

2+ a t

2

= ω 2r2

+ 0 2 ⇒ a tot =ω 2r(−ˆ r )only radial

€

a tot

2= a r

2+ a t

2

= ω 2r2

+ αr 2

atot = (ω 0 +αt)2r( )2 + αr( )2 both radial and tangential

€

a tot = a t + a r

Checkpoint #1 A ladybug sits at the outer edge of a merry-go- round, and a gentleman bug sits halfway between her and the axis of rotation. The merry-go-round makes a complete revolution once each second. The gentleman bug’s angular speed is

1. half the ladybug’s. 2. the same as the ladybug’s. 3. twice the ladybug’s. 4. impossible to determine

Checkpoint #2

A ladybug sits at the outer edge of a merry-go-round, that is turning and speeding up. At the instant shown in the figure, the tangential component of the ladybug’s (Cartesian) acceleration is:

1.  In the +x direction 2.  In the - x direction 3.  In the +y direction 4.  In the - y direction 5.  In the +z direction 6.  In the - z direction 7.  zero

Checkpoint#3 A ladybug sits at the outer edge of a merry-go- round that is turning and is slowing down. The vector expressing her angular velocity is

1.  In the +x direction 2.  In the - x direction 3.  In the +y direction 4.  In the - y direction 5.  In the +z direction 6.  In the - z direction 7.  zero

Problem10‐7:Thewheelinthepicturehasaradiusof30cmandisrota6ngat2.5rev/sec.Iwanttoshoota20cmlongarrowparalleltotheaxlewithouthi?nganspokes.(a)Whatistheminimumspeed?(b)DoesitmaGerwherebetweentheaxleandrimofthewheelyouaim?Ifsowhatisthebestposi6on.

The arrow must pass through the wheel in less time than it takes for the next spoke to rotate

Δt=

18rev

2.5rev / s= 0.05s

(a) The minimum speed is

vmn= 20cm0.05s

= 400cm / s = 4m / s

(b) No there is no dependence upon the radial position.

Problem10‐30:WheelAofradiusrA=10cmiscoupledbybeltBtowheelCofradiusrC=25cm.TheangularspeedofwheelAisincreasedfromrestataconstantrateof1.6rad/s2.Findthe6meneededforwheelCtoreachanangularspeedof100rev/min.

If the belt does not slip the tangential acceleration of each wheel is the same.rAαA = rCαC

αC =rArCαA = 0.64rad / s2

The angular velocity of C is.

ωC = αCt so t=ωC

αC

=ωCrCrAαA

with ω= 100rev/sec t=16s c

Kinetic energy of Rotation TRANSLATION Review:

Newton’s 2nd Law ⇒

€

F net ∝

a proportionality is inertia, m ⇒ constant of an object

€

KEtrans =12mv 2

energy associated with state of translational motion

“motion” of particles with same v

→ mass is translational inertia ←

What about rotation? What is energy associated with state of rotational motion?

€

= 12 mi∑ ωri( )2

= 12 miri

2( )∑[ ]ω 2 €

KEsystem = 12 mi∑ vi

2 where vi =ωritrans→ rot

All particles have same ω

€

≡ I rotational inertia (moment of inertia) about some axis of rotation

€

KErot = 12 Iω

2

€

KEtrans = 12 mv

2[ ] Energy of rotational motion

Moment of Inertia I

- Rotational inertia (moment of inertia) only valid about some axis of rotation.

€

I ≡ miri2

all mass∑For a discrete number

of particles distributed about an axis of rotation

- For arbitrary shape, each different axis has a different moment of inertia.

- I relates how the mass of a rotating body is distributed about a given axis.

units of kg·m2

€

I = miri2

i=1

4

∑ = m1r12 + m2r2

2 + m2r22 + m1r1

2

= 2m1r12 + 2m2r2

2

what about other axis?

- r is perpendicular distance from mass to axis of rotation

Simple example:

Moment of inertia: comparison

€

I2 = (5 ⋅ kg)(0.5 ⋅m)2 + (7 ⋅ kg)(4.5 ⋅m)2

= 1.3+142( ) ⋅ kgm 2 =144 ⋅ kgm 2

Note: 5 kg mass contributes <1% of total

- mass close to axis of rotation contributes little to total moment of inertia.

If rigid body = few particles

€

I = miri2∑

If rigid body = too-many-to-count particles Sum→ Integral

€

I1 = miri2∑

= (5 ⋅ kg)(2 ⋅m)2 + (7 ⋅ kg)(2 ⋅m)2

= 52 ⋅ kgm 2

Moment of inertia: continuum mass

€

I = miri2∑ ⇒ r 2∫ dm

€

with ρ =mV

⇒ I = ρ r 2∫ dV

I = ρA x2

-L2

+L2

∫ dx

I = r2∫ dm = ρ r2dV∫ where dV = area ⋅dx = A ⋅dx

I = ρA r2dx ∫ r2 = x2 in this case

− L 2 → x→ + L 2

= ρA 13 x

3( )-L

2

+L2 = ρA 1

3 (18 L

3) − 13 (−

18 L

3)( )

I =mV

A

L3

12

=112

m ⋅ AA ⋅ L

L3( ) = 1

12 mL2

INDEPENDENT OF AREA

Example: moment of inertia of thin rod with perpendicular rotation through center

Some Rotational Inertias

EACH OF THESE Rotational Inertias GO THROUGH THE Center of Mass !

10-7: Parallel-Axis Theorem

Proof:

I= r2 dm = { x − a( )∫∫2+ y − b( )2}dm

I = x2 + y2( )dm − 2a xdm∫∫ − 2b ydm + a2 + b2( )∫∫ dm

I = ICOM + 0 + 0 + h2M

I = ICOM + Mh2

Ifhisaperpendiculardistancebetweenagivenaxisandtheaxisthroughthecenterofmass(thesetwoaxesbeingparallel).Thentherota6onaliner6aIaboutthegivenaxisis

COM

Example #1

MomentofIner4a:LookatthedrawingofthesimplerodoflengthL.(a)WhatisIthroughanaxisparalleltotherod?(b)WhatisIthroughanaxisperpendiculartotherodthroughtheCM?(c)WhatisIforanaxisattheendbutperpendiculartotherod?

(a) I through the axis parallel to the rod.

I=mR2

2 for a solid cylinder

R → 0, I→ 0

(b) I through the axis perpendicular to the rod.

I=mL2

12 from table

(C) I through the axis perpendicular to the rod at the end.

I=mL2

12+ mh2 =

4mL2

3

Example #2: Moment of inertia of a Pencil

It depends on where the rotation axis is considered…

I = 0.3×10−6 kg ⋅m2

I = 12 MR

2

I = 112 ML

2I = 1

3ML2

Consider a 20g pencil 15cm long and 1cm wide …

I = 38 ×10−6 kg ⋅m2I = 150 ×10−6 kg ⋅m2

… somewhat like MASS, you can feel the difference in the rotational inertia

Example #3

What is moment of inertia about COM?

→ What is Ispoke (parallel-axis)?

€

Ispoke = Irod .com + Mh 2 = 112 ML

2 + M 12 L( )

2= 1

3 ML2

= 13 0.01kg( ) 0.33m( )2 ≅ 3.6 ×10−4 kg ⋅m 2

A bicycle wheel has a radius of 0.33m and a rim of mass 1.2 kg. The wheel has 50 spokes, each with a mass 10g. What is the moment of inertial about axis of rotation?

€

Itot ,com = Irim ,center + 50Ispoke

→ Putting together

€

Itot ,com = Irim ,center + 50Ispoke= Mwheel R

2 + 50Ispoke= 1.2kg( ) 0.33m( )2 + 50(3.6 ×10−4 kg ⋅m 2 ) = 0.149kg ⋅m 2

Example #4

What is the kinetic energy of the earth’s rotation about its axis?

€

KErot = 12 Iω

2 Energy of rotational motion is found from:

What is earth’s moment of inertia, I?

€

Iearth = Isphere = 25 Mr

2

= 25 6 ×1024 kg( ) 6.4 ×106m( )2 ≅1×1038 kg ⋅m 2

What is earth’s angular velocity, ω?

€

ω = 2π radiansday = 2π

3600 ×24( ) = 7.3×10−5rad /s

Now plug-’n-chug:

€

KErot = 12 1×1038 kg ⋅m 2

7.3×10−5rad /s( )2

= 2.6 ×1029 J €

From T = 2πω ⇒