lecturenotes-March3

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Unformatted text preview: 10.5‐‐Rela*ng
Transla*onal
and
Rota*onal
Variables
 Rota*onal
posi*on
and
distance
moved
 s = θ r 


(only
radian
units)
 Rota*onal
and
transla*onal
speed
 dr ds d θ v= = = r dt dt dt v= ω r Rota*onal
and
transla*onal
accelera*on
 dv d ω | at |= = r dr dt at = α r Rela*ng
period
and
rota*onal
speed
[distance = rate × time] T= 2 πr 2 π = v ω units
of
*me
(s)
 € ω T = 2π € 10.7‐‐Moment
of
iner*a
 For
a
discrete
number

 of
par*cles
distributed

 about
an
axis
of
rota*on
 I≡ ∑m r all mass 2 ii units
of
kg∙m2
 Simple
example:
 € 4 I = ∑ m i ri 2 = m1r12 + m 2 r22 + m 2 r22 + m1r12 i= 1 what
about

 other
axis?
 = 2 m1r12 + 2 m 2 r22 ‐
Rota*onal
iner*a
(moment
of
iner*a)
only
valid
about
some
axis
of
rota*on.
 ‐
For
arbitrary
shape,
each
different
axis
has
a
different
moment
of
iner*a.
 ‐
I
relates
how
the
mass
of
a
rota*ng
body
is
distributed
about
a
given
axis.
 ‐
r
is
perpendicular
distance
from
mass
to
axis
of
rota*on
 € 10.7‐‐Some
Rota*onal
Iner*as
 rdv dr EACH
OF
THESE
Rota*onal
Iner*as
GO
THROUGH
THE
Center
of
Mass
!

 10‐9:
Newton’s
2nd
law
for
rota*on
 τ net = ∑ τ i = Iα {Fnet = ∑ Fi = ma} € 10‐10:
Work
and
Rota*onal
Kine*c
Energy
 € W = ΔKE rot = 1 I (ω 2 − ω i2 ) f 2 θ2 NET
Work
done

 

ON
system
 W = ΔKE trans 2 2 1 = 2 m (v f − vi ) Rota*onal
work,
 fixed
axis
rota*on
 € 


(if
torque
is
const)
 W = ∫ τ net dθ θ1 € = τ (θ f − θ i ) dW d P= = (τθ ) = τω dt dt W = x2 x1 ∫ Fdx 1 − D motion Power,
 € fixed
axis
rota*on
 € dW d = ( F • x ) = F • v P = dt dt € € Problem
10‐101
 A
thin
uniform
rod
(mass
3.0
kg,
length
4.0
m)
rotates
freely
 about
a
horizontal
axis
A
that
is
perpendicular
to
the
rod
 and
passes
through
a
point
at
a
distance
d=1.0
m
from
the
 end
of
the
rod.

The
Kine*c
energy
of
the
rod
as
it
passes
 through
the
ver*cal
posi*on
is
20J.
(a)
What
is
the
 rota*onal
iner*a
of
the
rod
about
axis
A?
(b)
What
is
the
 liner
speed
of
the
end
B
of
the
rod
as
the
rod
passes
 through
the
ver*cal
posi*on?
(c)
At
what
angle
θ
will
the
 rod
momentarily
stop?
 Lets define the zero of potential when the rod is vertical E mech = Kω + mgh Emech (θ = 0 ) = 20 J o ML2 L I= + M − d 2 12 2 (b ) find v of end at θ =0 2(60 J ) I v = ωr = r (c) find θ where K ω = 0 mg(l / 2 − d ) sin θ = 60 J 60 sin θ = mg(l / 2 − d ) Iω 2 E mech = + mg(l / 2 − d ) sin θ 2 Emech (θ = 0 ) = 60 J θ = 0 o at the bottom Problem
10‐101
 A
thin
uniform
rod
(mass
3.0
kg,
length
4.0
m)
rotates
freely
 about
a
horizontal
axis
A
that
is
perpendicular
to
the
rod
 and
passes
through
a
point
at
a
distance
d=1.0
m
from
the
 end
of
the
rod.

The
Kine*c
energy
of
the
rod
as
it
passes
 through
the
ver*cal
posi*on
is
60J.
(a)
What
is
the
 rota*onal
iner*a
of
the
rod
about
axis
A?
(b)
What
is
the
 liner
speed
of
the
end
B
of
the
rod
as
the
rod
passes
 through
the
ver*cal
posi*on?
(c)
At
what
angle
θ
will
the
 rod
momentarily
stop?
 Lets define the zero of potential when the rod is vertical E mech = Kω + mgh Emech (θ = 0 ) = 60 J o ML2 L I= + M − d 2 12 2 (b ) find v of end at θ =0 2(60 J ) I v = ωr = r (c) find θ where K ω = 0 mg(l / 2 − d ) sin θ = 60 J 60 sin θ = mg(l / 2 − d ) Iω 2 E mech = + mg(l / 2 − d ) sin θ 2 Emech (θ = 0 ) = 60 J θ = 0 o at the bottom Problem:
Newton’s
second
Law:
10‐9
 Ch.
10
#
55:


Block
1
has
mass
m1=460
g,
block
2
has
mass
m2=500
g,
and
 the
pulley,
has
a
radius
R‐5.00
cm.

When
released
from
rest,
block
2
fall
 75.0
cm
in
5.00
s
without
the
cord
slipping
on
the
pulley.


 (a)  What
is
the
magnitude
of
the
accelera*on
of
the
blocks?
 (b)  What
are
the
tensions
T2
and
T1?
 (c)  What
is
the
magnitude
of
the
angular
accelera*on?
 (d)  What
is
the
rota*onal
iner*a?

 Problem:
Newton’s
second
Law:
10‐9
 Ch.
10
#
55:


Block
1
has
mass
m1=460
g,
block
2
has
mass
m2=500
g,
and
 the
pulley,
has
a
radius
R‐5.00
cm.

When
released
from
rest,
block
2
fall
 75.0
cm
in
5.00
s
without
the
cord
slipping
on
the
pulley.


 (a)  What
is
the
magnitude
of
the
accelera*on
of
the
blocks?
 (b)  What
are
the
tensions
T2
and
T1?
 (c)  What
is
the
magnitude
of
the
angular
accelera*on?
 (d)  What
is
the
rota*onal
iner*a?

 If m 2 > m1 T1 − m1g = m1a m2 g − T2 = m2 a Torque--Rotation τ 2 − τ 1 = Iα a (T2 − T1 ) R = I R Subsitute in Tensions a m2 ( g − a ) − m1 ( g + a ) = I 2 R (c) Angular Accelearation I a(− m2 + m1 − 2 ) = − g( m2 + m1 ) R a g( m2 + m1 ) α= = (d ) What is I g( m2 + m1 ) I R a= R m2 − m1 + 2 2 I R τ 2 − τ 1 R (T2 − T1 ) m2 − m1 + 2 I= = R α a Problem:
Newton’s
second
Law:
10‐9
 Ch.
10
#69:

Two
6.2
kg
blocks
are
connected
by
a
massless
 string
over
a
pulley
of
radius
2.40
cm
and
I=7.40x10‐4
kg.m.


 It
is
not
known
whether
there
is
fric*on
between
the
table
 and
the
sliding
block;
the
pulley’s
axis
is
fric*onless.

 When
it
is
released
the
pulley
turns
through
1.30
rad
in
 91.0
ms
and
the
accelera*on
of
the
blocks
is
constant.

 What
is
the
magnitude
of
the
pulley’s
angular
accelera*on
 (a),
the
magnitude
of
the
block
accelera*on,
the
tensions
 T1
and
T2?


 Problem:
Newton’s
second
Law:
10‐9
 Ch.
10
#69:

Two
6.2
kg
blocks
are
connected
by
a
massless
 string
over
a
pulley
of
radius
2.40
cm
and
I=7.40x10‐4
kg.m.


 It
is
not
known
whether
there
is
fric*on
between
the
table
 and
the
sliding
block;
the
pulley’s
axis
is
fric*onless.

 When
it
is
released
the
pulley
turns
through
1.30
rad
in
 91.0
ms
and
the
accelera*on
of
the
blocks
is
constant.

 What
is
the
magnitude
of
the
pulley’s
angular
accelera*on
 (a),
the
magnitude
of
the
block
accelera*on
(b),
the
 tensions
T1
and
T2?


 Lets solve this assuming no friction T2 = ma, mg-T1 = ma (c) Tensions Ia T1 = aI+T2 =a ( I+m ) (T1 − T2 ) R = Iα = R (T1 − T2 ) R I α (T − T2 ) (b ) a= = 1 R I (a ) α = We
can
compare
linear
variables
with
rota*onal
variables
 x v a Δt F m θ ω α Δt τ I s = rθ v T = rω aT = rα The
same
can
be
done
for
work
and
energy:
 € € € For
transla*onal
systems
 For
rota*onal
systems
 W =F⋅x 12 KE = mv 2 W = τ ⋅θ 12 KE = Iω 2 € € Problems:

Kine*c
Energy
10‐10
 Ch
10,
P.
90:
A
rigid
body
is
made
of
three
iden*cal
 thin
rods,
each
with
a
length
L=0.600
m,
fastened
 together
in
the
form
of
a
leher
H.
The
body
is
free
to
 rotate
about
the
axis
shown
through
on
of
the
rods.
 The
body
is
allowed
to
fall
rota*ng
around
the
axis.

 What
is
the
angular
velocity
when
it
is
ver*cal?

 Problems:

Kine*c
Energy
10‐10
 Ch
10,
P.
90:
A
rigid
body
is
made
of
three
iden*cal
 thin
rods,
each
with
a
length
L=0.600
m,
fastened
 together
in
the
form
of
a
leher
H.
The
body
is
free
to
 rotate
about
the
axis
shown
through
on
of
the
rods.
 The
body
is
allowed
to
fall
rota*ng
around
the
axis.

 What
is
the
angular
velocity
when
it
is
ver*cal?

 1
 2
 3
 How do we solve this problem? --Find Moment of Inertia. --Then what? Torque, Kinetic Energy? Can't do Torque--use conservation of energy! ΔKω = ΔU g = Iω L = mg + mgL 2 2 2 Moment of Inertia? I = I1 + I 2 + I 3 I1 = 0 why? mL2 mL2 L I2 = +M = 2 12 3 I 3 = mL2 4 mL2 I= 3 2 Iω 2 4 mL2 2 3L = ω = mg 2 3 2 9L ω= g 4L Chapter 11 Rolling, Torque, and Angular Momentum In this chapter we will cover the following topics: -Rolling of circular objects and its relationship with friction -Redefinition of torque as a vector to describe rotational problems that are more complicated than the rotation of a rigid body about a fixed axis -Angular momentum of single particles and systems of particles -Newton’s second law for rotational motion - 
11‐2:


What
Is
Rolling?
 Understanding
rolling
with
wheels
 Wheel
moving
forward
with
constant
speed
vcom
 s = θR vcom = a com = displacement:


transla*on→
rota*on
 € ds d (θR) = = ωR dt dt These
rela*onships
 € define
“smooth
rolling
mo*on”
 € dvcom d (ωR) = = αR dt dt Only
if
 


NO
SLIDING
 [smooth
rolling]
 The
top
of
the
wheel
has
a
 velocity
twice
vCOM
 At
point
P
(point
of
contact),
 wheel
does
not
move
 Understanding
rolling
with
wheels
II
 
All
points
on
wheel
move
 with
same
ω.

All
points
on
outer
 rim
move
with
same
linear
speed
 v
=
vcom.

 
All
points
on
wheel
move
 
Combina*on
of
“pure
 to
the
right
with
same
linear
 rota*on”
and
“pure
transla*on”
 velocity
vcom
as
center
of
wheel
 v =ω ×r Note
at
point
P:
 








at
point
T:
 € 
vector
sum
of
velocity
=
0 
 
(point
of
sta*onary
contact)
 
vector
sum
of
velocity
=
2vcom
(top
moves
twice
as
fast
as
com)
 11‐3:
Kine*c
Energy
of
Rolling
 1 2 I comω + 2 1 2 Mv 2 com = KE tot Note:
rota*on
about
COM
and
transla*on
of
COM
combine
for
total
KE
 € Remember:


vcom=ωr

 € acom Rolling Down a Ramp Consider a round uniform body of mass M and radius R rolling down an inclined plane of angle θ . We will calculate the acceleration acom of the center of mass along the x-axis using Newton's second law for the translational and rotational motion. Newton's second law for motion along the x-axis: f s − Mg sin θ = Macom (eq. 1) Newton's second law for rotation about the center of mass: τ = Rf s = I comα (eq. 2) acom R acom R α=− We substitute α in the second equation and get Rf s = − I com acom 2 → f s = − I com − I com acom R2 (eq. 3). We substitute f s from equation 3 into equation 1 → R vf v 2 = 2 ah t= g si n θ f a − Mg sin θ = Macom acom = − I 2 Lg sin θ I 1 + com2 v f = 2 L 1 + com2 MR I MR 1 + com2 v = − MR f g sin θ acom Rolling Down a Ramp Consider a round uniform body of mass M and radius R rolling down an inclined plane of angle θ . We will calculate the acceleration acom of the center of mass along the x-axis using Newton's second law for the translational and rotational motion. Newton's second law for motion along the x-axis: f s − Mg sin θ = Macom (eq. 1) Newton's second law for rotation about the center of mass: τ = Rf s = I comα (eq. 2) acom R acom R α=− We substitute α in the second equation and get Rf s = − I com acom 2 → f s = − I com − I com acom R2 (eq. 3). We substitute f s from equation 3 into equation 1 → R vf v 2 = 2 ah t= g si n θ f a − Mg sin θ = Macom acom = − I 2 Lg sin θ I 1 + com2 v f = 2 L 1 + com2 MR I MR 1 + com2 v = − MR f g sin θ 11.4:
Example:
Rolling
down
a
ramp
with
fric*on
 Object,
with
mass
m
and
radius
r,

roles
 from
top
of
incline
plane
to
bohom.

What
 is
v,
a,
and
Δt
at
bohom
 ΔE mech = 0 ΔKE tot = −ΔU h
 L sin θ = h θ
 (KE rot + trans,COM 1 2 ) final 2 mvCOM € − (0) init = − (0) final − ( mgh ) init € + 1 I comω 2 = mgL sin θ 2 2 [ Using
1‐D
kinema*cs
 1 2 mv 2 COM v 1 + 2 I COM = mgL sin θ r I COM m + 2 = 2 mgL sin θ r vCOM = 2 gL sin θ I 1 + mr 2 v 2 = v02 + 2 aL v2 g sin θ a= = I 2L 1 + 2 mr AND 2L I t= 1 + 2 g sin θ mr v 2 COM AT BOTTOM € acom g si n θ | acom |= I 1 + com2 MR The
bigger
I,

 The
smaller
a,
 The
smaller
vf,
 The
longer
t
 Cylinder MR 2 I1 = 2 g sin θ a1 = 1 + I1 / MR 2 g sin θ a1 = 1 + MR 2 / 2 MR 2 g sin θ a1 = 1+ 1/ 2 2 g sin θ a1 = = (0.67 ) g sin θ 3 Hoop I 2 = MR 2 a2 = g sin θ 1 + I 2 / MR 2 g sin θ a2 = 1 + MR 2 / MR 2 g sin θ a2 = 1+ 1 g sin θ a2 = = (0.5) g sin θ 2 Compare
Different
Objects
 Assuming
same
work
done
(same
change
in
U),
 objects
with
larger
rota*onal
iner*al
have
larger
KErot
 and
during
rolling,
their
KEtrans
is
smaller.
 vCOM = KE tot = KE trans + KE rot I com = KE trans 1 + mr 2 € 2 gL sin θ I com 1 + mr 2 Roll
a
hoop,
disk,
and
solid
sphere
down
a
ramp
‐
what
wins?
 € Moment
of
iner*a
 


large
→
small
 


 

Rota*onal













Frac*on
of
Energy
in
 Object 



Iner*a,
Icom












Transla*on 


Rota*on
 Hoop
 Disk
 Sphere
 1mr 2 1 2 0.5 





 




0.5

 0.67
















0.33
 0.71
















0.29
 slowest
 mr 2 mr 2 Δt bottom = 2L I 1 + g sin θ mr 2 2 5 sliding
block

 (no
fric*on)
 € 0 1 
















0
 fastest
 € Question A ring and a solid disc, both with radius r and mass m, are released from rest at the top of a ramp. Which one gets to the bottom first? 1.  

Solid
disc
 2.  
Ring
(hoop)
 3.  
both
reach
bo9om
at
 same
;me
 Checkpoint:

Disks
A
and
B
are
iden*cal
and
roll
across
a
floor
 with
equal
speeds.

Then
disk
A
rolls
up
an
incline,
reaching
a
 maximum
height
h,
and
disk
B
moves
up
an
incline
that
is
 iden*cal
except
that
it
is
fric*onless.

Is
the
maximum
height
 reached
by
disk
B
greater
than,
less
than,
or
equal
to
h?
 Question #2 Two solid disks of equal mass, but different radii, are released from rest at the top of a ramp. Which one arrives at the bottom first? 1.  

The
smaller
radius
disk.
 2.  
The
larger
radius
disk.
 3.  
Both
arrive
at
the
same
*me.
 Using
1‐D
kinema*cs
 v 2 = v02 + 2 aL v2 g sin θ a= = I 2L 1 + 2 mr AND 2L I t= 1 + 2 g sin θ mr The
equa*on
for
the
speed
of
the
a
disk
at
the
bohom
of
the
 4 ramp
is






 gl sin θ No*ce,
it
does
not
depend
on
the
radius
or
the
mass
of
the
 disk!!

 3 € Ch.
8
Problem
 A
small
block
of
mass
m
can
slide
along
the
fric*onless
loop‐ the‐loop.

The
block
is
released
from
rest
at
point
P,
at
a
 height
h
=
5R
above
the
bohom
of
the
loop.


 How
much
work
does
the
gravita*onal
force
do
on
the
block
as
 the
block
travels
from
point
P
to
point
Q?
 j j j j j Wg = Fg • d = mg(− ˆ ) • ( Rˆ − 5 Rˆ) = mg(− ˆ ) • (−4 Rˆ ) = 4 mgR ( )( )( )( ) If
U
=
0
at
the
bohom,
what
is
the
poten*al
energy
when
the
block
is
at
the
top
of
the
loop?

 =
0
 € Wg = Fg €d = −ΔU • ⇒ U top − U bottom = −Wg = − mg(− ˆ ) • (2 Rˆ − 0 Rˆ ) = 2 mgR j j j [( )( )] € If,
instead
of
being
released,
the
block
is
given
some
ini*al
speed
downward
along
the
track,
 how
do
the
above
answers
change?
 Do
the
answers
depend
on
velocity?

NO

‐>
Answers
don’t
change
 € What
is
the
minimum
height
h
so
that
the
block

 ΔU = − ΔKE 
makes
it
around
the
loop?
 From
forces
(see
prior
lecture)
 v2 ˆ y : −Fg = −( mg) = m− r or vmin = gr ( mghmin ) − ( mg(2 R)) = 1 m ( vt2op − vi2nitial ) 2 hmin = hmin 12 vtop − 0 + 2 R 2g 1 5 = ( gR ) + 2 R = R 2g 2 ( ) € Inclined
plane
again
 A
solid
cylinder
starts
from
rest
at
the
upper
end
of
the
track
as
 shown.

What
is
the
angular
speed
of
the
cylinder
about
its
 center
when
it
is
at
the
top
of
the
loop?
 Inclined
plane
again
 A
solid
cylinder
starts
from
rest
at
the
upper
end
of
the
track
as
 shown.

What
is
the
angular
speed
of
the
cylinder
about
its
 center
when
it
is
at
the
top
of
the
loop?
 Using conservation of mechanical energy:
 0 = ΔE mech = ΔKE tot + ΔU grav 0= [( 1 2 2 mvcom + 1 I com ω 2 ) final − (0) init + ( mg(2 R) final − ( mg(h )) init 2 [ 2 If the disc does not fall off v2 m > mg R 2 g(h − 2 R ) m > mg R h > 3R mg(h − 2 R) = 1 m (ωr ) + 1 ( 1 mr 2 )ω 2 2 22 Rearranging yields 2 mg(h − 2 R) = ω 2 1 + 1 2 2 mr ω= 4 g(h − 2 R ) 3r 2 NOTE:

Compare
with
before
 Sliding ⇒ ⇐ Rolling v = ω r = 2 g(h − 2 R) = 2 gd y v = ωr = 4 3 g (h − 2 R ) = 4 3 gd y A
solid
cylinder
of
radius
10
cm
and
mass
12
kg
starts
from
rest
 and
rolls
without
slipping
a
distance
of
6
m
down
a
house
roof
 that
is
inclined
at
30º.
 Where
does
it
hit?
 Using conservation of mechanical energy:
 11‐4:
Forces
on
Rolling
 0 = ΔEmech = ΔKEtot + ΔU grav 0= ( 1 2 2 mvcom + 1 I comω 2 2 ) 1 2 final − ( 0 )init + ( 0 ) final − ( mgh )init 2 1 2 mgL sin θ = m ( v ) + ( 1 2 v mr r 2 2 ) Rearranging yields 2 mgL sin θ = v 2 1 + 1 2 m v f , COM = 4 3 gl sin θ Then
just
use
kinema*cs
(vox,
voy…)
 11.5:
Yo‐Yo
 A
yo‐yo
has
a
rota*onal
iner*a
of
Icom
and
mass
of
m.

Its
axle
radius
is
R0
 and
string’s
length
is
h.

The
yo‐yo
is
thrown
so
that
its
ini*al
speed
 down
the
string
is
v0.
 a)  How
long
does
it
take
to
reach
the
end
of
the
string?
 1‐D
kinema*cs
given
acom
 −h = Δy = −v0 t − 1 a com t 2 ⇒ solve for t (quadradic equation) 2 b)  As
it
reaches
the
end
of
the
string,
what
is
its
total
KE?
 Conserva*on
of
mechanical
energy
 2 vcom ,0 2 1 1 KE f = KE i + U = 2 mvcom ,0 + 2 I com + mgh € R0 c)  As
it
reaches
the
end
of
the
string,
what
is
its
linear
speed?
 1‐D
kinema*cs
given
acom
 € − vcom = −v0 − a com t ⇒ solve for vcom d)  As
it
reaches
the
end
of
the
string,
what
is
its
transla*onal
KE?
 Knowing
|vcom|
 € a com = g e)  1 + I com mR02 downwards As
it
reaches
the
end
of
the
string,
what
is
its
angular
speed?
 Knowing
|vcom|
 v 12 KE trans = mvcom 2 € f)  Two
ways:
 ω= com R As
it
reaches
the
end
of
the
string,
what
is
its
rota*onal
Kω?
 €Kω = 1 I comω 2 or Kω = K Ef , tot − KEtrans 2 11‐5:
Which
way
will
it
roll??


 1
 1
 3
 Problem.
 11‐13


 NON-smooth rolling motion A bowler throws a bowling ball of radius R along a lane. The ball slides on the lane, with initial speed vcom,0 and initial angular speed ω0 = 0. The coefficient of kinetic friction between the ball and the lane is µk. The kinetic frictional force fk acting on the ball while producing a torque that causes an angular acceleration of the ball. When the speed vcom has decreased enough and the angular speed ω has increased enough, the ball stops sliding and then rolls smoothly. a)  During the sliding, what is the ball’s angular acceleration? d) What is the speed of the ball when smooth rolling begins? e) How long does the ball slide? Problem.
 11‐13


 NON-smooth rolling motion A bowler throws a bowling ball of radius R along a lane. The ball slides on the lane, with initial speed vcom,0 and initial angular speed ω0 = 0. The coefficient of kinetic friction between the ball and the lane is µk. The kinetic frictional force fk acting on the ball while producing a torque that causes an angular acceleration of the ball. When the speed vcom has decreased enough and the angular speed ω has increased enough, the ball stops sliding and then rolls smoothly. a)  [After it stops sliding] What is the vcom in terms of ω ? Smooth rolling means vcom =Rω b)  During the sliding, what is the ball’s linear acceleration? acom = − fk m ˆ fk = µ k N From 2nd law: x : − f k = ma com But So ˆ : N − mg = 0 = − µk g = µ k mg (linear) y c)  During the sliding, what is the ball’s angular acceleration? ˆ fk = µ k N From 2nd law: τ = Rf k (−z ) But So Iα = Rfk = R ( µ k mg ) € € Rµ k mg ˆ ˆ Iα (− z ) = τ = Rf k (−z ) = µ k mg (angular) α= I What is the speed of the ball when smooth rolling begins? vcom I vcom€= v0 + a com t vcom = v0 − µ k gt R When does vcom =Rω ? € vcom = v0 − µ k g ω = ω 0 + αt ω = Iω Rµ k mg From kinematics: t= α Rµ k mg v0 vcom = How long does the ball slide? € (1 + I mR 2 ) d) e) t= v0 − vcom µkg € € Conservation of: •  •  •  •  •  Linear momentum (closed, isolated system) KE (elastic collisions) Mechanical energy (only conservative forces) Total energy Angular momentum (closed, isolated system) Physics
gone
Wrong??
 Example
of:
 
1)
Conserva*on
of
angular
momentum
 
 
2)
Impulse/Collision
 
 
3)
Stress
and
Strain 

 Linear momentum when particle passes through point A it has linear momentum: Angular momentum with respect to point O the angular momentum is defined as: where r is the position vector of the particle with respect to O. p = mv = r × p = m (r × v ) with components in x-y plane € € € • 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 11‐8:
Newton’s
2nd
Law
in
Angular
Form
 (single
par;cle)
 Relationship between force and linear momentum dp Fnet = ∑ Fi = ma = Newton’s 2nd Law in linear form dt Relationship between torque and angular momentum € d τ net = ∑ τ i = dt The vector sum of all the torques acting on a particle is equal to the time rate change of the angular momentum of that particle. € € τ net = d dt has no meaning unless the net torque τ net , and the rotational momentum , are defined with respect to the same origin € € 11‐8:
Newton’s
2nd
Law
in
Angular
Form
 τ net d = ∑τ i = dt Proof: Start with definition of angular momentum =m r x v ( ) Then dv dr d = m xv+rx dt dt dt d =m vxv+rxa dt ( ( ) But vxv =0 d = m r x a = r x ma dt ) ( ) d == r x m a = r x F = τ dt 11‐10:
Angular
Momentum
of
a
Rigid
Body

 Rota*ng
about
a
Fixed
axis
 dL dω = τ net = Iα = I dt dt € € L = Iω 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? € I tot = m ( d ) + m (2 d ) + m ( 3d ) = 14 md 2 2 2 2 € ˆ L tot = Iω = 14 mωd 2 ⋅ z [kg⋅m2/s] € 11‐11:
Conserva*on
of
Angular
Momentum
 Conservation Law of Linear Momentum (closed,isolated system = no net external forces) ΔP = 0 ΔL = 0 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 -  Total angular momentum of system at all times is equal - Vector {conserved in all three directions, x-y-z} Li = L f € I iω i = I f ω f € - Initially rigid body redistributes mass relative to rotational axis Demo:
Conserva*on
of
Angular
Momentum
 No external torques act Li = L f I iω i = I f ω f (I person + 2 MR 2 i ω i = I person + 2 Mr 2 f ω f ) ( ) if Δr ↓ then ω f > ω i if Δr ↑ then ω f < ω i € Demo:
Conserva*on
of
Angular
Momentum
 No external torques act Linitial= Lfinal= + = Newton’s
2nd
Law
in
Angular
Form
 System
of
par;cles
 Total angular momentum n L = l1 + l 2 + l 3 + .... + l n = ∑ l i i= 1 Relationship between torque and angular momentum for system of particles € dL τ net = dt τ net represents the net external torque. € € Example
#2
 (Problem
11‐27)
 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? #1 #2 a) The total moment of inertia about point O is found from: I tot = ∑ I i = ∑ m i ri 2 = (6.5 ⋅ kg)(1.5 ⋅ m ) 2 + ( 3.1 ⋅ kg)(2.8 ⋅ m ) 2 = 38.9 ⋅ kg ⋅ m 2 b) The total angular momentum about point 0 is: ˆ ˆ Ltot = ∑ i = ∑ mi ( ri × vi ) = m1 ( r1v1 sin θ1 ) (− z ) + m2 ( r2 v2 sin θ 2 ) (+ z ) € ˆ L tot = [−(6.5 ⋅ kg)(1.5 ⋅ m)(2.2 ⋅ m / s ) + ( 3.1 ⋅ kg)(2.8 ⋅ m )( 3.6 ⋅ m / s )] z ˆ = 9.8 ⋅ kg ⋅ m 2 / s ⋅ z c) The net torque about point 0 is: € dL tot =0 τ net = dt closed, isolated system € € Linear
and
angular
rela*ons
 Force Linear momentum (one) Linear momentum (system) Linear momentum (system) Newton’s second law (system) Conservation Law (closed,isolated) F mv = p ∑p=P Mvcom = P dP = Fnet dt Ma com = Fnet 0 = ΔP τ =r×F l =r×p L = ∑l L = Iω dL τ net = dt τ net = Iα ΔL = 0 Torque Angular momentum (one) Angular momentum (system) Angular momentum (system, fixed axis) Newton’s second law (system) Conservation Law (closed,isolated) τ net τ has € meaning unless the net torque n , and the total rotational no has no meaning unless the net torque τ netet ,and the total rotational = dL dt momentum , are defined with respect to the same origin L momentum L , are defined with respect to the same origin € € € € L = Iω dL dω τ= =I dt dt v = rω MR 2 I (disk ) = 2 2 MR 2 I (Sphere) = 5 I ( Hoop ) = MR 2 Example
#3
:
Clutch
 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. Before After What is the angular speed of the the resultant combination of the shaft and two wheels? Li = L f ( Iω i ) small + (2 I i (0)) large = ( Iω f ) small + (2 Iω f ) large wheel wheel wheel wheel Iω i = 3 Iω f € What fraction of the original rotational kinetic energy is lost? Fraction of KE lost : 1 Iω 2 − 1 ( 3 I )( 1 ω ) 2 1 I initial ω i2 − 1 I final ω 2 KE i − KE f i i f 2 2 3 2€ 2 % % = % = 2 2 1 1 I ωi Iω i 2 initial 2 KE i 1 − ( 3)( 1 ) 2 3 % = ( 2 )% = 66.7% = 3 1 ω f = 1 ωi 3 € Problem
11‐66
 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? What do we know about the ramp? What does this give us? Conservation of Mechanical Energy going down ramp! (KE init + U init ) = (KE final + U final ) (0 + mgh ) = ( 12 mv 2final + 0) This gives us the speed of the mass right before hitting the rod: There’s a collision! What kind? What do we know about this collision? What is constant? Only Conservation of Angular Momentum holds! (choose a point wisely) This gives the angular speed of the rod/ mass just after hitting: € € vm ,bottom = 2 gh L initial = L final (r × pm ,0 + 0) = (r × pm , f + I rod ω ) € € 2 m, 0 bottom € ( d (mv )) = ( md (ω ω bottom = ) + ( 13 Md 2 )ω bottom ) mvm ,0 m 2 gh = dm + 1 Md dm + 1 Md 3 3 Now what? After the collision, what holds? (KE init + U init ) = (KE final + U final ) Now Conservation of Mechanical energy holds again! Solve for θ: 2 I tot ω bottom + 0) = (0 + mgd (1 − cos θ final ) + Mg( 1 d )(1 − cos θ final )) 2 € m 2h −1 θ = cos 1 − € d ( m + 1 M )( m + 1 M ) 2 3 ( 1 2 € Summary:
Forces
of
Rolling
 1)

If
object
is
rolling
with
acom=0
(i.e.
no
net
forces),
then
vcom=ωR
=
constant
(smooth
roll)
 …if
constant
speed,
it
has
no
tendency
to
slide
at
point
of
contact
‐
no
fric*onal
forces
 acom
=
0


















α
=
0





















τ

=
0






















f
=
0
 acom
=
αR
 2)  Icomα
=
τnet
 τnet
=
|R||f|sinθ
 If
object
is
rolling
with
acom≠
0
(i.e.
there
are
net
forces)
and
no
slipping
occurs,
 





then
α
≠
0


⇒

τ
≠
0

 …
sta*c
fric*on
needed
to
supply
torque
!
 kinetic friction - during sliding (acom
≠
αR) static friction smooth rolling (acom
=
αR)
 Summary:
Effects
of
rota*on
 From
before
 done
with
both
 With
Rota*on
 m
=
M
 a= 2g 3 2 g( m 2 − m1 sin θ ) a= M w + 2( m1 + m 2 ) done
via
force/torque
 v= 4 gd ( M − m) M W + 2( m + M ) v= 4 3 gd done
via
energy
 € € remember
this
for
later
 € In
each
of
these
cases:
“transla*on”
was
separate
from
“rota*on”
 
 












Pure
transla*on
+
Pure
rota*on


 Quick
Review:
Rolling
=
Rota*on
+
Transla*on
 Wheel
moving
forward
with
constant
speed
vcom
 displacement:


transla*on→
rota*on
 s = θR ds d (θR) vcom = = = ωR dt dt Only
if
 a com € dvcom d (ωR) = = = αR dt dt € 


NO
SLIDING
 [smooth
rolling]
 € 1 2 I comω + 2 1 2 Mv 2 com = KE tot Note:
rota*on
about
COM
and
transla*on
of
COM
combine
for
total
KE
 Rolling
down
a
ramp:
accelera*on
 ‐
Accelera*on
downwards
 ‐
Fric*on
provides
torque
 ‐
Sta*c
fric*on
points
 upwards
 Yo‐Yo
 Newton’s
2nd
Law
 


Linear
version
 ˆ x : f s − Fg sin θ = − ma com ˆ y : N − Fg cos θ = 0 


Angular
version
 → a com = αR € ˆ z : I com α = τ = Rf s a com R 2 ( mg sin θ − ma com ) = I com ‐
Tension
provides
 torque
 ‐
Here
θ
=
90°
 ‐
Axle:
R
⇒
R0
 α= a com Rf s = R I com € a com = € g sin θ 1 + I com mR 2 € a com = g 1 + I com mR02 € Quick
Review:
Angular
momentum
 l =r ×p L = Iω dL τ net = dt τ net = Iα ΔL = 0 Angular
momentum
(one)
 Angular
momentum
(system,
fixed
axis)
 
Newton’s
second
law
(system) 
 Conserva*on
Law
(closed,isolated)
 Quick
Review:
Linear
and
angular
rela*ons
 Force 
 Linear
momentum
(one) 
 Linear
momentum
(system) 
 Linear
momentum
(system) 
 Newton’s
second
law
(system) 
 Conserva*on
Law
(closed,isolated) 
 F mv = p ∑p=P Mvcom = P dP = Fnet 
 
 dt 
 Ma com = Fnet 0 = ΔP τ = r × F Torque
 l = r × p Angular
momentum
(one)
 L = ∑l Angular
momentum
(system)
 L = Iω Angular
momentum
(system,
fixed
axis)
 dL τ net = 
 dt 
 
Newton’s
second
law
(system) τ net = Iα 
 ΔL = 0 Conserva*on
Law
(closed,isolated)
 
 τ net n τ has
no

meaning
unless
the
net
torque






,
and
the
total
rota*onal
 has
no

meaning
unless
the
net
torque






,
and
the
total
rota*onal
 € τ netet = dL dt momentum



,
are
defined
with
respect
to
the
same
origin
 L € L momentum



,
are
defined
with
respect
to
the
same
origin
 € € € € 11‐10:
Angular
Momentum

 P#11.40: The figure plots the torque τ
that
acts
on
an
ini*ally
sta*onary
disk
 that
can
rotate
about
its
center.
Τs=4.0
N.m.

What
is
the
angular
momentum
of
 the
dist
at
(a)
t=7.0
s,
(b)
t=20s? dL τ= dt t L (t ) = ∫ τ dt 0 At 7.0 Sec L ( 7 ) = 2 ( 4 ) + 2 ( 3) + m2 L ( 7 ) = 24 kgi s 1 2 + 3( 3) 2 At 20.0 Sec L (20 ) = 24 + = 7.5 3( 2 ) 3( 3) 3 − − 3( 6 ) − 2 2 2 2 11‐10:
Angular
Momentum

 Problem #55: A uniform thin rod of length L and mass M can rotate in horizontal plane about a vertical axis through the CM. The rod is at rest when a bullet of mass m is fired at the rod. The velocity of the bullet makes and angle θ with
respect
to
the
rod.

It
strikes
and
s*cks
at
the
 end
of
the
rod.
Find
the
ini*al
velocity
of
the
bullet: Conservation of angular momentum L i = m(r x v)i L Li = m v sin θ 2 Conservation of angular momentum L f = Li ML mLv mL sin θ = + ω 2 4 12 2 2 Conservation of angular momentum L2 L f = Iω + m ω 4 ML2 mL2 Lf = + ω 4 12 1 Lω M v= + 6 m 2 sin θ 11‐10:
Angular
Momentum

 Problem#11-60: The uniform rod of length L and mass M rotates about an axis at the end. As it rotates through its lowest position, it collides with a putty wad of mass m that sticks to the end. If the angular velocity right before the collision was ω0, what
is
the
angular
velocity
right
aAer
the
collision? Conservation of Angular Momentum ML2 ML2 ML2 L i = Iω = + ω = 3 ω 4 12 Conservation of Angular Momentum L f = L f (rod ) + L f ( putty) ML2 Lf = ω f + mL2ω f 3 Li = L f ML2 ML2 ω= ω f + mL2ω f 3 3 M ωf = ω M + 3m 11‐10:
Angular
Momentum

 Problem #66: a small block of mass m slides down frictionless surface through a height h and then sticks to a uniform rod of mass M and length L. The rod pivots through as axis at the end. Find the angle θ
 where
the
two
stop
momentarily. First consider the block-conservation of E mech Emech = mgh = vb = 2 gh mv 2 2 b Li = L f ML2 mL 2 gh = ω + mL2ω 3 3m 2 gh ω= ML + 3mL Conservation of energy Iω 2 L = mgh1 + Mgh2 = mgL + Mg sin θ 2 2 gh ( M + 3m ) 3m L = mgL + Mg sin θ M + 3m 3 2 2 Conservation of angular momentum Li = mLvb = mL 2 gh L f = L f (rod ) + L f (block ) ML2 Lf = ω + mL2ω 3 6m2h = sin θ L ( M + 3m ) ( 2 m + M ) 11‐10:
Angular
Momentum

 Problem # 97: A particle of mass M is dropped from a point that is at a height h, and a distance s from the point O. What is the magnitude of the angular momentum of the particle with respect to point O when the particle has fallen h/2? First lets find the velocity of the object. h v2f = 2 g = gh 2 v f = gh r θ θ L= rmvr L = rmv − k h s + 2 2 2 (−k ) Lets find the angle between r and v L = m r x v = rv sin θ − k ( ) () sin θ = 2 s h s + 2 2 () Ch.
11:
Torque,
Accelera*on
 Problem 11-HW7: A constant horizontal force is applied to a uniform solid cylinder by a 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 acceleration of the COM? (b) What is the magnitude of α?
(C)
 What
is
the
fric*onal
force‐vector
nota*on. Find Torque about point P τ =Iα τ = 2 RFapp MR 2 3MR 2 2 I= (com ) + MR = 2 2 (c) find F f . Watch out the force (a) find a. 2 RFapp acom = 3MR 2 = α 2 4 Fapp 3M a R of friction keeps the wheel from slipping F app + F f = M a com Fapp Ff = + i 3 (b) find α . α = α= 4 Fapp 3MR ...
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