Rolling •  translational+rotational

•  smooth rolling acom =αR & vcom =ωR

•  Equations of motion from: - Force/torque -> a and α - Energy -> v and ω

€

L sinθ = hh

θ

€

vCOM =2gL sinθ

1+I

mr 2

€

KEtot = KEtrans + KErot = KEtrans 1+Imr 2

slowest

fastest €

Δtbottom =2L

gsinθ1+

Imr 2

Hoop Disk Sphere Block

€

12 Icomω

2

€

+ 12 Mvcom

2

€

= KEtot

acom

RollingDownaRamp:Accelera2onbyfs

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ˆ x : fs − Fg sinθ = −macom

ˆ y : N − Fg cosθ = 0

ˆ z : Icomα = τ = Rfs

€

→ acom =αR

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α =acom

R=RfsIcom€

acom =R2 mgsinθ −macom( )

Icom

€

acom =gsinθ

1+ Icom mR2

Newton’s 2nd Law Linear version

Angular version

- Acceleration downwards

- Friction provides torque

- Static friction points upwards

Question:

1. How much work is done by fs?

A: 0

2.  Is mechanical energy conserved?

A: YES

Problem 11-HW7: A constant horizontal force is applied to a uniform solid cylinder by fishing line wrapped around the cylinder. The mass of the cylinder is M, its radius R and it rolls on a smooth surface. (a) What is the magnitude of α?(b) What is the acceleration of the COM? (c)Whatisthefric2onalforce‐vectornota2on.

Find Torque about point Pτ=Iατ = 2RFapp

I =MR2

2(com) + MR2 =

3MR2

2

(a) find a.

2RFapp =3MR2

2

α

a =4Fapp3M

(b) find α. α= aR

α =4Fapp3MR

(c) Find

€

F→

f

€

F∑ = Fapp→

+Ff→

= m a→

€

Ff→

= (M4Fapp3M

−Fapp ) i→

=Fapp3

i→

P

PhysicsgoneWrong??

€

r

Angular momentum with respect to point O the angular momentum is defined as:

where is the position vector of the particle with respect to O.

€

= r × p = m r × v ( )

Linear momentum when particle passes through point A it has linear momentum:

with components in x-y plane

€

p = m v

• Units kg⋅m2/s • components in z-direction only !

(⊥ to x-y plane)

Just as in τ

, has meaning only with respect to a given axis

ccw – “+” angular momentum cw – “-” angular momentum

Newton’s2ndLawinAngularForm(singlepar>cle)

Relationship between torque and angular momentum

Relationship between force and linear momentum

€

F net =

F i∑ = m a = d p

dtNewton’s 2nd Law in linear form

€

τ net =

d

dtThe vector sum of all the torques acting on a particle is equal to the time rate change of the angular momentum of that particle.

Then

d

dt= m

dr

dt x v + r x dv

dt

d

dt= m v

x v + r x a ( )

But

v x v

( ) = 0

d

dt= m r

x a

( ) = r x ma

d

dt== r x ma= r x F= τ

Proof:

AngularMomentumofaRigidBodyRota>ngaboutaFixedaxis

€

τ net = I α

Example #3 (Problem 11-39) Three particles are fastened to each other with massless string and rotate around point O. What is the total angular momentum about point O?

€

Itot = m d( )2 + m 2d( )2 + m 3d( )2

=14md 2

€

L tot = Iω = 14mωd 2 ⋅ ˆ z

[kg⋅m2/s]

€

L = I ω

€

d L

dt= = I d ω

dt

Conserva>onofAngularMomentum

If the net external torque acting on a system is zero ( ), the angular momentum of the system remains constant, no matter what happens within the system

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Δ L = 0

Conservation Law of Linear Momentum (closed, isolated system = no net external forces)

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Δ P = 0

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L i =

L f

-  Total angular momentum of system at all times is equal

- Vector {conserved in all three directions, x-y-z}

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Iiω i = I fω f

- Initially rigid body redistributes mass relative to rotational axis

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τ net =

d

dt

Demo:Conserva>onofAngularMomentum

No external torques act

Li = Lf

Iiω i = I fω f

I person + 2MR2( )iω i = I person + 2Mr

2( ) f ω f

€

if Δr ↓ then ω f >ω i

if Δr ↑ then ω f <ω i

Demo:Conserva>onofAngularMomentum

Linitial= Lfinal= + =

No external torques act

Newton’s2ndLawinAngularForm‐‐Systemofpar-cles

Relationship between torque and angular momentum for system of particles

€

τ net =

d L

dt

€

τ net represents the net external torque.

Total angular momentum

€

L = l 1 + l 2 + l 3 + ....+

l n =

l i

i=1

n

∑

Linearandangularrela>ons

Force

Linear momentum (one)

Linear momentum (system)

Linear momentum (system)

Newton’s second law (system)

Conservation Law (closed,isolated)

€

F

m v = p p ∑ = P

M v com = P

d P

dt= F net

Macom = F net

0 = Δ P

τ =r ×F

l = r × pL =

l∑

L = I

ω

τ net =

dLdt

τ net = I

α

ΔL = 0

has no meaning unless the net torque , and the total rotational momentum , are defined with respect to the same origin

€

L

€

τ nethas no meaning unless the net torque , and the total rotational

momentum , are defined with respect to the same origin

€

L

€

τ net

€

τ net = d

L dt

Torque

Angular momentum (one)

Angular momentum (system)

Angular momentum (system, fixed axis)

Newton’s second law (system)

Conservation Law (closed,isolated)

L= Iω

τ=dL

dt= I

dω

dtv = rω

I(disk) = MR2

2

I(Sphere) = 2MR2

5I(Hoop) = MR2

Example:Problem11‐27

Two particles are moving as shown. a)  What is their total moment of inertia about point O? b)  What is their total angular momentum about point O? c)  What is their net torque about point O?

a) The total moment of inertia about point O is found from:

b) The total angular momentum about point 0 is:

c) The net torque about point 0 is:

€

τ net =

d L tot

dt

€

Itot = Ii∑ = miri2∑ = (6.5 ⋅ kg)(1.5 ⋅m)2 + (3.1⋅ kg)(2.8 ⋅m)2 = 38.9 ⋅ kg ⋅m 2

#1

#2

Ltot =

i∑ = mi

ri ×vi( )∑ = m1 r1v1 sinθ1( )(− z) + m2 r2v2 sinθ2( )(+ z)

€

L tot = −(6.5 ⋅ kg)(1.5 ⋅m)(2.2 ⋅m /s) + (3.1⋅ kg)(2.8 ⋅m)(3.6 ⋅m /s)[ ] ˆ z

= 9.8 ⋅ kg ⋅m 2 /s ⋅ ˆ z

closed, isolated system

€

= 0

A wheel is rotating freely with angular speed ωi on a shaft whose rotational inertia is negligible. A second wheel, initially at rest and with twice the rotational inertia of the first, is suddenly coupled to the same shaft.

What is the angular speed of the the resultant combination of the shaft and two wheels What fraction of the original rotational kinetic energy is lost?

€

Li = L f

Iω i( ) smallwheel

+ 2Ii (0)( ) largewheel

= Iω f( ) smallwheel

+ 2Iω f( ) largewheel

€

Fraction of KE lost :

KEi − KE f

KEi

%

Example#3:Clutch

€

Iω i = 3Iω f

ω f = 13ω i

€

=12 Iinitialω i

2 − 12 I finalω f

2

12 Iinitialω i

2

% =

12 Iω i

2 − 12 3I( ) 1

3ω i( )2

12 Iω i

2

%

=1− 3( ) 1

3( )2

1

% = 2

3( )% = 66.7%

Before After

Problem11‐66 A particle of mass m slides down the frictionless surface through height h and collides with the uniform vertical rod (of mass M and length d), sticking to it. The rod pivots about point O through the angle θ before momentarily stopping. Find θ.

What do we know? Where are we trying to get to?

There’s a collision! What kind? What do we know about this collision? What is constant?

What do we know about the ramp? What does this give us?

Now what? After the collision, what holds?

Only Conservation of Angular Momentum holds! (choose a point wisely)

Conservation of Mechanical Energy going down ramp!

€

L initial =

L final

€

( r × p m ,0 + 0) = ( r × p m, f + Irodω)

d(mvm,0 )( ) = md 2 ωbottom( ) + 13Md

2( )ωbottom( )

This gives us the speed of the mass right before hitting the rod: €

KEinit +Uinit( ) = KE final +U final( )

€

0 + mgh( ) = 12mvfinal

2 + 0( )

€

vm ,bottom = 2gh

This gives the angular speed of the rod/mass just after hitting:

€

ω bottom =mvm,0

dm + 13 Md

=m 2ghdm + 1

3 Md

Now Conservation of Mechanical energy holds again!

€

KEinit +Uinit( ) = KE final +U final( )

€

12 Itotω bottom

2 + 0( ) = 0 + mgd(1− cosθ final ) + Mg(12 d)(1− cosθ final )( )

Solve for θ:

€

θ = cos−1 1− m 2hd m + 1

2 M( ) m + 13 M( )

Problem 11-40: The figure plots the torque τthatactsonaini2allysta2onarydiskthatcanrotateaboutitscenter.Τs=4.0N.m.Whatistheangularmomentumofthediskat(a)t=7.0s,(b)t=20s?

τ =dLdt

L(t) = τ dt0

t

∫

At 7.0 Sec

L(7) = 2 4( ) + 2 3( ) + 12

2 + 3 3( )

L(7) = 24kgim2

s

At 20.0 Sec

L(20) = 24 +3 2( )

2−

3 3( )2

− 3 6( ) − 32

2

= 7.5