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Unformatted text preview: Chapter
10:
Useful
rela1onships
 Translational Motion Rotational Motion x↔θ v ↔ω a v = v0 + at at 2 x = xo + v0t + 2 2 v 2 − v0 = 2 a x − xo ↔α ↔ ω = ω0 + αt (eq. 1) ( ) αt 2 ↔ θ = θ + ω 0t + (eq. 2) 2 2 ↔ ω 2 − ω 0 = 2α θ − θ 0 (eq. 3) ( ) 10.5‐‐Rela1ng
Transla1onal
and
Rota1onal
Variables
 Rota1onal
posi1on
and
distance
moved
 s = θ r 


(only
radian
units)
 Rota1onal
and
transla1onal
speed
 dr ds d θ v= = = r dt dt dt v= ω r Rota1onal
and
transla1onal
accelera1on
 dv d ω | at |= = r dr dt at = α r Rela1ng
period
and
rota1onal
speed
[distance = rate × time] T= 2 πr 2 π = v ω units
of
1me
(s)
 € ω T = 2π € 10.5‐‐Rela1ng
Transla1onal
and
Rota1onal
Variables
 “accelera1on”
is
a
liFle
tricky
 Rota1onal
and
transla1onal
accelera1on
 a)
 from v= ω r dv d ω = r dt dt at = dω r= α r dt from
change
in
angular
speed
 tangen&al
accelera&on

 b)

from
before
we
know
there’s
also
a
“radial”
component
 v2 2 ar = = ω r r 2 2 a tot = a r + a t 2 2 radial
accelera&on
 c)

must
combine
two
dis1nct
rota1onal
accelera1ons
 € = ω 2 r + αr 2 Rela1ng
Transla1onal
and
Rota1onal
Vectors!
 Rota1onal
posi1on
and
distance
moved
 s = θ r 


(only
radian
units)
 Rota1onal
and
transla1onal
speed
 € dr ds d θ v= = = r dt dt dt v= ωr v =ω ×r Rota1onal
and
transla1onal
accelera1on
 €at = € € dω r = α r tangen&al
accelera&on

 at = α × r dt v2 2 ar = = ω r radial
accelera&on
 a r = ω × (ω × r ) r 2 2 2 a tot = a r + a t € 2 2 a tot = a t + a r = ω 2 r + αr € € 10.7‐‐Moment
of
iner1a
 For
a
discrete
number

 of
par1cles
distributed

 about
an
axis
of
rota1on
 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 ‐
Rota1onal
iner1a
(moment
of
iner1a)
only
valid
about
some
axis
of
rota1on.
 ‐
For
arbitrary
shape,
each
different
axis
has
a
different
moment
of
iner1a.
 ‐
I
relates
how
the
mass
of
a
rota1ng
body
is
distributed
about
a
given
axis.
 ‐
r
is
perpendicular
distance
from
mass
to
axis
of
rota1on
 € 10.7‐‐Moment
of
iner1a:
con$nuum
mass
 m 2 2 with ρ = ⇒ I = ρ ∫ r dV I = ∑ m i ri ⇒ ∫ r dm 2 V € € Example:
moment
of
iner&a
of
thin
rod
with
perpendicular
 
rota&on
through
center

 I = ∫ r 2 dm = ρ ∫ r 2 dV where dV = area ⋅ dx = A ⋅ dx I = ρ A ∫ r 2 dx +L 2 r 2 = x 2 in this case −L2 → x→ + L2 I = ρA -L 2 ∫ x dx 2 = ρA ( x) 1 3 3 +L 2 -L 2 = ρA ( 1 3 ( 1 L3 ) − 1 (− 1 L3 ) 8 3 8 ) m L3 1 m ⋅ A 3 I= = A L V 12 12 A ⋅ L () 1 = 12 mL2 INDEPENDENT

 OF
AREA
 10.7‐‐Moment
of
iner1a:
con$nuum
mass
 Sample
Problem
10‐7:

Calculate
I
for
the
thin
 uniform
rod
of
mass
m
and
length
L.

 10.7‐‐Moment
of
iner1a:
con$nuum
mass
 Sample
Problem
10‐7:

Calculate
I
for
the
thin
 uniform
rod
of
mass
m
and
length
L.

 I = ∫ r dm 2 M dm = dx L dm = ρdx + L /2 I = ∫ r 2 dm = − L /2 ∫ x 2 ρdx (b) Through the end ρ 3 ML2 I = ∫ x 2 ρ dx = L = 3 3 0 ML2 ML2 ML2 I= + = 12 4 3 L 3 3 ρ 3 L /2 ρ L − L I = x − L / 2 = − 2 2 3 3 (b) Through the end use Parrell-Axis 2ρ L 2M I= = 3 2 3L 3 L3 ML2 8 = 12 10.7‐‐Some
Rota1onal
Iner1as
 rdv dr EACH
OF
THESE
Rota1onal
Iner1as
GO
THROUGH
THE
Center
of
Mass
!

 10‐7:
Parallel‐Axis
Theorem
 If
h
is
a
perpendicular
distance
between
a
given
 axis
and
the
axis
through
the
center
of
mass
(these
 two
axes
being
parallel).
Then
the
rota1onal
iner1a
 I
about
the
given
axis
is
 I = I COM + Mh 2 Proof: I= ∫ r dm = ∫ {( x − a ) + ( y − b ) }dm 2 2 2 I= ∫( x 2 + y 2 dm − 2 a ∫ x dm − 2b ∫ y dm + ∫ a 2 + b 2 dm ) ( ) I = I COM + 0 + 0 + h 2 M 10‐7:
Parallel‐Axis
Theorem
 Moment
of
Iner&a:

Look
at
the
drawing
of
the
 simple
rod
of
length
L.

(a)
What
is
I
through
an
axis
 parallel
to
the
rod?

(b)
What
is
I
through
an
axis
 perpendicular
to
the
rod
through
the
CM?

(c)
 What
is
I
for
an
axis
at
the
end
but
perpendicular
to
 the
rod?

 10‐7:
Parallel‐Axis
Theorem
 Moment
of
Iner&a:

Look
at
the
drawing
of
the
 simple
rod
of
length
L.

(a)
What
is
I
through
an
axis
 parallel
to
the
rod?

(b)
What
is
I
through
an
axis
 perpendicular
to
the
rod
through
the
CM?

(c)
 What
is
I
for
an
axis
at
the
end
but
perpendicular
to
 the
rod?

 (a) I through the axis parallel to the rod. mR 2 I= for a solid cylinder 2 R → 0, I → 0 (b) I through the axis perpendicular to the rod. mL2 I= from table 12 (C) I through the axis perpendicular to the rod at the end. mL2 4 mL2 2 I= + mh = 12 3 Example:
Moment
of
iner1a
of
a
Pencil
 It
depends
on
where
the
rota1on
axis
is
considered…
 1 3 2 I = ML I= 1 12 ML2 I = 1 MR 2 2 I = 0.3 × 10 −6 kg ⋅ m 2 I = 150 × 10 kg ⋅ m −6 2 I = 38 × 10 −6 kg ⋅ m 2 Consider
a
20g
pencil
15cm
long
and
1cm
wide
…
 …

somewhat
like
MASS,
you
can
feel
the

 
difference
in
the
rota1onal
iner1a
 10‐7:
Moment
of
Iner1a
 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
iner1al
about
axis
of
rota1on?
 What
is
moment
of
iner1a
about
COM?

 I tot ,com = I rim ,center + 50 I spoke 





→
What
is

Ispoke
(parallel‐axis)?

 2 € 1 Ispoke = Irod .com + Mh 2 = 12 ML2 + M ( 1 L ) = 1 ML2 2 3 = 1 (0.01kg)(0.33m) ≅ 3.6 × 10 −4 kg ⋅ m 2 3 2 





→
Pubng
together

 € I tot ,com = I rim ,center + 50 I spoke = M wheel R 2 + 50 I spoke = (1.2 kg)(0.33m) + 50( 3.6 × 10 −4 kg ⋅ m 2 ) = 0.149 kg ⋅ m 2 2 € Example:
Moment
of
iner1a
of
a
Pencil
 Problem
10‐36:

Find
the
moment
 of
iner1a
for
the
configura1on
 shown
in
the
picture.
 I = I1 + I 2 + I 3 I = md + m ( 2 d ) + m ( 3d ) 2 2 2 I = 14 md 2 Example:
Moment
of
iner1a
of
a
Pencil
 Problem
10‐36:

Find
the
moment
 of
iner1a
for
the
configura1on
 shown
in
the
picture.
 I = I1 + I 2 + I 3 I = md + m ( 2 d ) + m ( 3d ) 2 2 2 I = 14 md 2 10‐7:
Rota1onal
Iner1a
 Problem
10‐39:
In
the
figure
two
par1cles,
 each
with
mass
m,
are
fastened
to
each
 other,
and
to
a
rota1on
axis
at
0,
by
two
thin
 rods,
each
with
length
d
and
mass
M.

The
 combina1on
rotates
around
the
rota1on
axis
 with
angular
speed
ω.

About
O
what
is
the
 rota1onal
iner1a
and
Kine1c
energy?
 First let us find I: masses treated as points: Look at table for Rods 2 I=I1 + I 2 + I 3 + I 4 Md 2 d I1 = + M I 2 = md 2 Iω 2 2 12 K= 2 2 2 Md 3d I3 = + M I 4 = 4 md 2 2 12 8 Md 2 I= + 5 md 2 3 10‐8:

Rota1onal
dynamics
‐
Torque
 τ =r ×F vector

















=















magnitude












+











direc&on
 τ = r F sin ϑ vector τ is always perpendicular to both vectors r and F € € If
Ft

is
tangen1al
component
of
force
 If
r⊥
is
moment
arm

 τ = r F sin ϑ = rFt = ( r sin ϑ ) F = r⊥ F ( ) Units
of
torque:
N∙m

(not
Joule)
 € 10‐8:

Torque
 The
figure
shows
an
overhead
view
of
a
meter
s1ck
that
 can
pivot
about
the
dot
at
the
posi1on
marked
20
(for
 20
cm).

All
five
horizontal
forces
on
the
s1ck
have
the
 same
magnitude.

Rank
those
forces
according
to
the
 magnitude
of
the
torque
that
they
produce,
greatest
 first.
 

Checkpoint

 |
τ
|
=
r
F
sinθ
 |τ1|
=

|τ3|

>

|τ4|

>
|τ2|

=
|τ5|
=
0
 You
are
using
a
wrench
and
trying
to
loosen
a
rusty
nut.
Which
of
the
arrangements
 shown
is
most
effec1ve
in
loosening
the
nut?
 List
in
order
of
descending
efficiency
the
following
arrangements:
 1. 

A
>
B
>
C
>
D
 2. 
B
>
A
=
C
>
D
 3. 
B
=
D
>
A
>
C
 4. 
B
>
A
=
D
>
C
 A





























B

















C

























D


 Problem
10‐47:
Torque
 A
body
is
pivoted
at
O
and
two
forces
act
on
it
as
shown.
 Find
an
expression
for
the
net
torque
on
the
body
about
the
pivot
 What
is
vector
equa1on
for
net
torque?

 τ net = ∑ τ i =∑ ri × Fi = r1 × F1 + r2 × F2 





→
What
is
magnitude
of
torque?

 Thus
expression
is:


 τ = τ = rF sin θ € ˆ τ net = [r1 F1 sin θ1 − r2 F2 sin θ 2 ] k (out − of − page) € € 















Use
angle
between
F
and
r
which
is
less
than
180°
 Note:

When
using
the
right‐hand
rule
…
 θ2 θ1 F € τ = rF sin θ Which
angle?


 The
one
less
than
180°

→

θ1
 r € 10‐9:
Rota1onal
dynamics
‐
Newton’s
2nd
Law
 Force
about
a
point

 causes
torque
 Torque
causes
rota1on
 τ =r ×F {Fnet = ∑ Fi = ma} τ net = ∑ τ i = Iα € • torque
is
posi1ve
when
the
force
tends
to
produce
a
counterclockwise
rota1on
about
an
axis
 












is
nega1ve
when
the
force
tends
to
produce
a
clockwise
rota1on
about
an
axis
 
 
(→
VECTOR!

posi1ve
if
poin1ng
in
+z,
nega1ve
if
poin1ng
in
‐z
)
 € € • net
torque
and
angular
accelera1on
are
parallel
 Sample
Problem
11‐7
 A
uniform
disk
of
mass
M
and
radius
R
is

 mounted
on
a
fixed
horizontal
axle.

A
block
of
mass
m

 hangs
from
the
rim
of
the
disk.


 Find
the
accelera1on
of
block
m
(assume
no
fric1on
on

 







axle
and
cord
does
not
slip).
 













 



 Example
#1
A
uniform
disk,
with
mass
M
and
radius
R,
is
mounted
on
a
 fixed
horizontal
axle.

A
block
of
mass
m
hangs
from
a
massless
cord
that
is
 wrapped
around
the
rim
of
the
disk.

a)
Find
the
accelera1on
of
the
disk,
and
 the
tension
in
the
cord.
 b)
Assume
now
that
m
=
M.

Ini1ally
at
rest,
what
is
the
speed
of
the
 block
aoer
dropping
a
distance
d?
 Equal
 masses
 M a
 M d
 A
wheel
(mass
M,

radius
 R,
and
I
=1/2MR2)
is
 Using



force














+












torque
 aFached
to
block
with
 ˆ y : T − Mg = − Ma ˆ z : τ net = r × F equal
mass
M
(i.e.
 Mdisk=Mblock).

Ini1ally
at
 ⇒ T = M (g − a ) ˆ = ( RT )(−z ) rest,
what
is
the
speed
of
 the
block
aoer
dropping
 τ = Iα a
distance
d?
 € ˆ (− ˆ ( RT )€ z ) = ( Iα )(−z ) Example
#1
 Using

conserva1on
of
Energy
 ΔKE tot = − ΔU a t = αr & ( € 1 2 Mv 2 + 1 Iω 2 2 ) final − (0 + 0) init = − (0) final − (Mgd ) init [ € I = 1 MR 2 ⇒ 2 € a R[ M ( g − a )] = ( 1 MR 2 ) 2 R a ( 1 M + M ) = Mg 2 a= Mg 2 =g 3 M3 2 2 1 Subs1tu1ng
 v = ωr & I = 2 MR 1 2 Mv + 2 1 2 ( 1 2 v MR ) = Mgd R 2 2 Now
use
1‐D
kinema1cs
 € 2 v 2 + ( 1 )(v) = 2 gd 2 2 € v 2 = v02 + 2 ad v 2 = 0 + 2 d ( 2 g) 3 v= 4 3 v (1 + € 1 2 )= 3 2 v = 2 gd 4 3 2 v= gd gd € 10.10—Work
and
Rota1onal
Kine1c
Energy
 where
τ
is
the
torque
doing
the
work
W,
and
θi
and
θf
are
the
body’s
angular
 posi1ons
before
and
aoer
the
work
is
done,
respec1vely.
When
τ
is
constant,
 The
rate
at
which
the
work
is
done
is
the
power
 10‐9:
Newton’s
2nd
law
for
rota1on
 τ net = ∑ τ i = Iα {Fnet = ∑ Fi = ma} € 10‐10:
Work
and
Rota1onal
Kine1c
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 ) Rota1onal
work,
 fixed
axis
rota1on
 € 


(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
rota1on
 € dW d = ( F • x ) = F • v P = dt dt € € Chapter
10:
Useful
rela1onships
 Block
and
Pulley
Accelera1ng
 m,
r
 T
 Probelme
10‐54:
A
wheel
of
radius
r
m
is
mounted
on
 a
fric1onless
horizontal
axis.

The
rota1onal
iner1a
of
 the
wheel
is
I
kg/m2.

A
massless
cord
wrapped
around
 the
wheel
is
aFached
to
a
M
kg
block
that
slides
on
a
 horizontal
fric1onless
surface.

If
a
horizontal
force
of
 magnitude
P
is
applied,
what
is
the
magnitude
of
the
 angular
accelera1on
of
the
wheel?

Assume
that
the
 string
does
not
slip
on
the
wheel.
 M
 Force
about
a
point

 causes
torque
 τ = F P − T = maT Tr = τ = Iα = IaT r Torque
causes
rota1on
 τ net = ∑ τ i = Iα Trans
‐
rota1onal
 




rela1onship
 € aT = rα IaT P − 2 = MaT r P aT = I M+ 2 r mr 2 For a disc I= 2 P aT = m M+ 2 Check
M=0,
m=0
 Example:
Three
masses
M,
2M
and
3M
are
shown
in
 T 
 1 the
figure
with
two
pulley
of
moment
of
iner1al
I
 and
radius
r.
 (a)
What
is
the
linear
Accelera1on?
 T2
 T3
 T4
 T1 − Mg = Ma Ia (T2 − T1 ) r = Iα = r T3 − T2 = 2 Ma Ia (T4 − T3 ) r = r 3Mg − T4 = 3Ma Solve for T4 , then T3 , -----If I did it right 4 Mg a= 2I + 6M r2 Check
I=0
 F = 3Mg − Mg = 6 Ma 2g a= 3 Example
x?
 A
50kgwoman
is
in
a
rotor
with
a
2m
radius,
laying
against
the
inner
surface.
 The
fric1on
between
the
woman's
sweater
and
the
surface
is
μs
=
0.5.
(a)
The
 rotor
begins
spinning
with
a
constant
angular
accelera1on,
and
when
the
 rotor
reaches
enough
angular
speed,
the
floor
drops
but
the
woman
stays
 s1cking
to
the
wall.
(a)
What
is
the
minimum
angular
velocity
that
will
allow
 the
woman
not
to
fall
down
when
the
floor
drops?
(b)
What's
the
maximum
 angular
accelera1on
that
will
not
make
the
woman
slide
along
the
wall
when
 the
rotor
is
speeding
up
to
the
angular
velocity
calculated
in
(a)?
 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
Kine1c
energy
of
the
rod
as
it
passes
 through
the
ver1cal
posi1on
is
20J.
(a)
What
is
the
 rota1onal
iner1a
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
ver1cal
posi1on?
(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 (b ) find v of end at θ =0 v = ωr = r 2(60 J ) I ML2 L I= + M − d 2 12 2 Iω 2 E mech = + mg(l / 2 − d ) sin θ 2 θ = −90 o at the bottom (c) find θ where K ω = 0 mg(l / 2 − d ) sin θ = 60 J Block
and
Pulley
Accelera1ng:
10‐9
 In the figure one block has a mass M, the other has mass m , and the pulley (I), which is mounted in horizontal frictionless bearings, has a radius r. When released from rest, the heavier block falls a distance d in time t. (a) What is the magnitude of the block's acceleration? (b) What is the tension in the part of the cord that supports the heavier block? (c) What is the tension in the part of the cord that supports the lighter block? (d) What is the magnitude of the pulley's angular acceleration? 
 Force
about
a
point

 causes
torque
 T1
 T2
 τ = F Basic Equations Mg-T2 = Ma T1 − mg = ma Ia (T2 − T1 ) r = Iα = r Ia + ( M + m)a = ( M − m) g 2 r ( M − m) g a= I + ( M − m) +
 2 r Torque
causes
rota1on
 τ net = ∑ τ i = Iα Trans
‐
rota1onal
 




rela1onship
 aT = rα Check:
I=0
 Check:
M=m
 Check:
m=0
 Block
and
Pulley
Accelera1ng:
10‐9
 In the figure one block has a mass M, the other has mass m , and the pulley (I), which is mounted in horizontal frictionless bearings, has a radius r. When released from rest, the heavier block falls a distance d in time t. (a) What is the magnitude of the block's acceleration? (b) What is the tension in the part of the cord that supports the heavier block? (c) What is the tension in the part of the cord that supports the lighter block? (d) What is the magnitude of the pulley's angular acceleration? 
 Force
about
a
point

 causes
torque
 T1
 T2
 τ = F Basic Equations Mg-T2 = Ma T1 − mg = ma Ia (T2 − T1 ) r = Iα = r Ia + ( M + m)a = ( M − m) g 2 r ( M − m) g a= I + ( M + m) 2 r Torque
causes
rota1on
 τ net = ∑ τ i = Iα Trans
‐
rota1onal
 




rela1onship
 aT = rα Check:
I=0
 Check:
M=m
 Check:
m=0
 Two
mass,
m
and
M
(M
>
m),
are
aFached
around
a
pulley
 with
radius
R
and
mass
MW
(IW=1/2MWR2).

Originally
 at
rest,
what
is
the
velocity
of
the
blocks
aoer
M
has
 fallen
a
distance
d?
 Use
conserva1on
of
mechanical
energy:

 ΔE mech = 0 ⇒ ΔKE tot = −ΔU v2 = € ( ΔKE + ΔKE + ΔKE ) = − ΔU ( mv + Mv + Iω ) = − ( m − M ) gd m M Mw grav , m & M 1 2 2 1 2 2 1 2 2 4 gd ( M − m ) MW + 2 (m + M ) v mv 2 + Mv 2 + 1 M W R 2 = −2 ( m − M ) gd 2 R 2 ( ) How
do
we
get
back
to
 the
equa1on
on
the
 previous
page???
 2 ( m + M ) v 2 + ( M W ) v 2 = 4 ( M − m ) gd d = θr Example:
Torque
or
Kine1c
Energy
 Massless
cord
wrapped
around
a
pulley
of
radius
R
and
 mass
MW
(fric1onless
surface/bearings)
and
IW=1/2MWR2.

 What
is
angular
accelera1on,
α,
of
pulley
(disc)?
 1)
What
are
forces
on
m1?

 ˆ x : T1 − m1 g sin θ = m1 a ˆ y : N − m1 g cos θ = 0 ⇒ T1 = m1 (a + g sin θ ) α
‐
into
board
 a
 2)
What
are
forces
on
m2?

 ˆ y : T2 − m 2 g = − m 2 a ⇒ T2 = m 2 (g − a ) NOTE:
T1
&
T2
are
NOT
equal
 3)
What
are
torques
about
wheel?


 € τ net = τ1 + τ 2 € τ net = RT1 sin (90 ) + RT2 sin (−90 ) ˆ = R(T1 − T2 )(+ z ) NOTE:
angular
accelera1on
vector
is
in
nega1ve‐z
direc1on

 € 4)
Solve
for
α
?


 € here a t = αR ⇒ IW α = R[(T2 ) − (T1 )] IW α = ( 1 M W R 2 )α = R[( m 2 ( g − a )) − ( m1 ( a + g sin θ ))] 2 ( 1 2 M W R 2 α = Rg ( m2 − m1 sin θ ) − R (α R ) ( m2 + m1 ) ) € € α= 2 g m2 − m1 sin θ R M w + 2 ( m1 + m2 ) a)  If
m1
=
0,
same
as

 














earlier
problem
 b)  If
θ
=90º,
same
as

 














table
problem
 We
can
compare
linear
variables
with
rota1onal
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 € € Summary:
Effects
of
rota1on
 From
before
 done
with
both
 With
Rota1on
 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:
“transla1on”
was
separate
from
“rota1on”
 
 












Pure
transla1on
+
Pure
rota1on


 Example
#2

 What
is
the
kine1c
energy
of
the
earth’s
rota1on
 about
its
axis?
 Energy
of
rota1onal
mo1on
is
found
from:

 KE rot = 1 Iω 2 2 What
is
earth’s
moment
of
iner1a,
I?

 € Iearth = Isphere = 2 Mr 2 5 = 2 5 (6 × 10 24 kg)(6.4 × 10 6 m ) ≅ 1 × 10 38 kg ⋅ m 2 2 What
is
earth’s
angular
velocity,
ω?

 € From T = 2 π ω ⇒ ω = 2 π radians day = 2 π (3600 × 24 ) = 7.3 × 10 −5 rad / s Now
plug‐’n‐chug:

 € KE rot = 1 1 × 10 38 kg ⋅ m 2 ( 7.3 × 10 −5 rad / s ) = 2.6 × 10 29 J € 2 2 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
accelera1on
of
the
blocks?
 (b)  What
are
the
tensions
T2
and
T1?
 (c)  What
is
the
magnitude
of
the
angular
accelera1on?
 (d)  What
is
the
rota1onal
iner1a?

 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
accelera1on
of
the
blocks?
 (b)  What
are
the
tensions
T2
and
T1?
 (c)  What
is
the
magnitude
of
the
angular
accelera1on?
 (d)  What
is
the
rota1onal
iner1a?

 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
fric1on
between
the
table
 and
the
sliding
block;
the
pulley’s
axis
is
fric1onless.

 When
it
is
released
the
pulley
turns
through
1.30
rad
in
 91.0
ms
and
the
accelera1on
of
the
blocks
is
constant.

 What
is
the
magnitude
of
the
pulley’s
angular
accelera1on
 (a),
the
magnitude
of
the
block
accelera1on,
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
fric1on
between
the
table
 and
the
sliding
block;
the
pulley’s
axis
is
fric1onless.

 When
it
is
released
the
pulley
turns
through
1.30
rad
in
 91.0
ms
and
the
accelera1on
of
the
blocks
is
constant.

 What
is
the
magnitude
of
the
pulley’s
angular
accelera1on
 (a),
the
magnitude
of
the
block
accelera1on
(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 = ...
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