Download Lecture 24 - LSU Physics

Physics 2101
Section
S cti n 3
March 24th
Announcements:
• Review today
• Exam tonight at 6 PM,
LockettLockett
-6
Class Website:
http://www.phys.lsu.edu/classes/spring2010/phys2101--3/
http://www.phys.lsu.edu/classes/spring2010/phys2101
http://www.phys.lsu.edu/~jzhang/teaching.html
Chap. 10: Rotation
θ (t ) ( Δθ = θ 2 − θ1 )
dθ (t)
ω (t) =
dt
s=θ r
v =
dω (t ) d 2θ (t )
=
α (t ) =
dt
dt 2
dθ
r=ωr
dt
atot = ar + at
2
2
Rotational Kinetic energy:
Moment of inertia:
Torgue:
2
I≡
2
ar = ω × (ω × r )
at = α × r
ω = ω0 + αt
θ = θ + ω0t + 12 αt
ω2 = ω02 + 2αθ
KE rot = 12 Iω 2
∑
all mass
τ =r ×F
v =ω×r
a tot = a t + a r
2
= ω r + αr
2
Rotational Kinematics: Rotational
Kinematics:
( ONLY IF α = constant)
mi ri 2 = ∫ r 2 dm
I = I COM + Mh 2
τ = r F sinϑ
θ2
Work done by a torgue: W = ∫ τ net dθ = τ (θ f − θi ) = ΔKErot = 12 I (ω 2f − ωi2 )
θ1
Chap. 11: Rotational dynamics
2nd – Law for rotation: Law for rotation:
τ nett = ∑ τ i = Iα
{F
net
= ∑ Fi = ma
ma }
1
1
2
2
I
ω
+
Mv
Rolling bodies:
com
com
2
2
The friction is always against the tendency of sliding
vcom = ω R acom = α R
Angular momentum:
= r × p = m ( r × v ) τ net = ∑ τ i =
dL
dω
= τ net = Iα = I
dt
dt
L = Iω
Angular momentum Angular
momentum
conservation:
KEtot =
Li = L f
if
τ net = 0;
d
dt
(a particle)
(a rigid body)
I iωi = I f ω f
Summary: Effects of rotation
From before
With Rotation
m = M
a = 23 g
2g(m 2 − m1 sin θ )
a=
M w + 2(m1 + m 2 )
v=
4gd (M − m )
M W + 2(m + M )
v=
4
3
gd
d
Done with both
Done via force/torque
Done via energy
remember this for later
IIn each of these cases: “translation” was separate from “rotation”
h f th
“t
l ti ”
t f
“ t ti ”
Pure translation + Pure rotation SHW#6: Three masses M, 2M and 3M are shown in the figure with two pulley of moment of inertial I and radius r. The coefficient of friction is μk and the table is L m long.
T2
T1
T1 − Mg = Ma
⎞
⎛ I
T2 = gM (3 − 2 μ k ) − a ⎜ 2 + 5M ⎟
⎠
⎝r
Ia
a
(T2 − T1 )r = Iα =
r
T3 − T2 − 2 μ k Mg = 2Ma
⎡I
⎤
gM ⎢ 2 (2 − μ k ) + M (4 − μ k )⎥
⎣r
⎦
T2 =
⎡I
⎤
+
3M
⎢⎣ r 2
⎥⎦
⎞
⎛ I
T3 = −a ⎜ 2 + 3M ⎟ + 3Mg
⎠
⎝r
Ia
(T4 − T3 )r =
r
3M − T4 = 3Ma
3Mg
3M
Need v = rω & a = rα
T3 = −Mg
M (1 − μ k ) + 3Mg
3M = Mg
M (2 + μ k )
Mg (1 − μ k )
a=
⎡I
⎤
⎢⎣ r 2 + 3M ⎥⎦
T3
T4
SHW#6: Three masses M, 2M and 3M are shown in the figure with two pulley of moment of inertial I and radius r. The coefficient of friction is μk and the table is L m long.
Let us use Energy.
T2
T3
T4
T1
We can define zero so E mech ( start ) = 0
After block 3 has moved a distance d we have
6 Mv 2 2 I ω 2
+
− (3M − M ) gd
E mech = 0 − 2 μk Mgd =
2
2
3Mv 2 Iv 2
− μk Mgdd =
+ 2 − Mgdd (use
(
v = rω )
2
2r
2 Mgd (1 − μk )
2
:
v =
Mgg (1 − μ k )
2
I
v
=
2ad
:
a
=
3M + 2
I
r
3M + 2
r
v f + vi
2
t=d: t=
I⎞
⎛
2d ⎜ 3M + 2 ⎟
⎝
r ⎠
Mgd (1 − μ k )
Same as before
Mg (1 − μ k )
a=
⎡I
⎤
+
3M
⎢⎣ r 2
⎥⎦
Rolling down a ramp : Energy considerations
An object at rest, with mass m
An
object at rest with mass m and radius r, and radius r
roles from top of incline plane to bottom. What is v at bottom and a throughout.
ΔE mech = 0
ΔKE tot = −ΔU
(KE
rot +trans,COM
1
2
)
final
[
h
L sin θ = h
θ
− (0)init = − (0) final − (mgh )init
]
2
mvCOM
+ 12 ICOM ω 2 = mgL sin θ
Using v 2 = v02 + 2aL
Using vcom = ωr
vCOM
2gL sin θ
=
⎛
I com ⎞
⎜1 +
⎟
2
⎝ mR ⎠
Speed
at Bottom
Using 1‐D
a com
kinematics
(const accel)
v2
g sin θ
=
=
I com ⎞
2L ⎛
1
+
⎜
⎟
⎝ mR 2 ⎠
Constant
Acceleration
Friction and Rolling
The static friction is always opposing the tendency to slide
1) If acom = 0, it has no tendency to slide at point of contact ‐ no frictional force
2)
If aacom > 0 (i.e. there are net forces) and
If
> 0 (i e there are net forces) and no slipping
no slipping occurs, τ
occurs τ ≠ 0 ≠0
provided by static friction force
fs
3) If a
3)
If acom > 0, and
> 0 and no slipping
no slipping occurs, τ
occurs τ ≠ 0 provided by weight and ≠ 0 provided by weight and
static friction force
θ
Rolling down a ramp: acceleration
‐ Acceleration downwards
‐ Friction provides torque
Yo‐Yo
‐ Static friction points upwards
Newton’s 2nd Law
Linear version
xˆ : f s − Fg sin θ = −ma com
yˆ : N − Fg cos θ = 0
Angular version
zˆ : I com α = τ = Rf s
a com
→ a com = αR
α=
R 2 (mg sin θ − ma com )
=
I com
a com Rf s
=
R
I com
a com =
g sin θ
1 + I com mR 2
‐ Tension provides torque
‐ Here θ = 90°
‐ Axle: R ⇒
A l R R0
a com =
g
1 + I com mR
R02
g sin θ
| acom |=
I
1 + com2
MR
acom
Cylinder
Hoop
MR 2
I1 =
2
g sin θ
a1 =
1 + I1 / MR 2
I 2 = MR 2
g sin θ
a1 =
1 + MR 2 / 2MR 2
g sin θ
a1 =
1 + 1/ 2
2 g sin θ
= (0.67) g sin θ
a1 =
3
g sin θ
a2 =
1 + MR 2 / MR 2
g sin θ
a2 =
1+1
g sin θ
= (0.5) g sin θ
a2 =
2
a2 =
g sin θ
1 + I 2 / MR 2
Prob. 11‐64 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
g
g,
acceleration?
fk = μ k N
From 2nd law: xˆ : − f k = ma com
But So a com = − f k m
(linear) yˆ : N − mg = 0
= μ k mg
= −μ k g
c))
During the sliding, what is the ball’s angular
g
g,
g
acceleration?
Iα = Rf k = R(μ k mg)
fk = μ k N
From 2nd law: τ = Rf k (−zˆ)
But So (angular) Iα (−zˆ ) = τ = Rf k (−zˆ)
= μ k mg
α = Rμ k mg
I
d))
What is the speed of the ball when smooth rolling begins?
p
g g
vcom = v0 + a com t
When does vcom =Rω ?
ω = ω 0 + αt
From kinematics: e)
How long does the ball slide? t=
v0 − vcom
μkg
vcom = v0 − μ k gt
t = ω α = Iω Rμ mg
k
vcom
vcom
⎛ ⎛vcom ⎞ ⎞
⎟
⎜ I⎜⎝
R⎠ ⎟
= v0 − μ k g⎜
⎟
R
μ
mg
k
⎟
⎜
⎝
⎠
v0
=
(1 + I mR 2 )
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 (
g
)
g
p
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)
vm,bottom = 2gh
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)
L initial = L final
(r × pm,0 + 0) = (r × pm, f + I rod ω )
(d(mv ))= (md (ω
2
m,0
This gives the angular speed of the rod/mass just after hitting:
rod/mass just after hitting: ω bottom =
bottom
)+ (13 Md 2 )ω bottom )
mvm,0
m 2gh
=
dm + 13 Md dm + 13 Md
Now what? After the collision, what holds? (KE init + U init ) = (KE final + U final )
Now Conservation of Mechanical energy holds again! (
1
2
2
I tot ω bottom
+ 0) = (0 + mgd(1
g ( − cos θ final ) + Mg(
g( 12 d)(1
)( − cos θ final ))
⎡
⎤
m 2h
θ = cos ⎢1 −
⎥
1
1
M
m
+
M
d
m
+
(
)
(
)
⎣
⎦
2
3
−1
Solve for θ:
Sample Problem* 12‐9
L = r × mv
( angular version)
v
Four thin, uniform rods, each with mass M and length d, are attached rigidly
to a vertical axle to form a turnstile (ts). The turnstile rotates ccw with an initial angular velocity ωi. A mud ball (mb) of mass m = 1/3 M and initial speed vi is thrown along the path shown and sticks to one end of one rod. What is the final angular velocity of the ball‐turnstile system?
1)
2)
3)
4)
To find ωf of the combined system, what is conserved?
Using Conservation of angular momentum about axis:
Using Conservation of angular momentum about axis:
⎡⎣ Li = L f ⎤⎦
about axle
Lts ,i + Lmb ,i = Lts , f + Lmb , f
[ Itsωi ]ts ,i + Lmb,i = ⎡⎣ I tsω f ⎤⎦ ts , f
Is KEtot conserved.
Is Emech conserved. P
Is conserved. Is L conserved
Is conserved.
Lts ,i = I tsωi
I ts = 4
+ Lmb , f
(
1
12
NO!
NO!
NO! YES!
Lts , f = I tsω f
&
Md 2 + M
r
( d) ) =
1
2
2
4
3
Md 2
Lmbb, f = I mbbω f = ( md 2 ) ω f = 13 Md 2ω f
⎡⎣ 43 Md 2ωi ⎦⎤ + Lmb ,i = ⎣⎡ 34 Md 2ω f ⎦⎤ + Lmb , f
ts ,i
ts , f
Lmb ,i = rmvi sin ϑ = d
⎡⎣ 34 Md 2ωi ⎤⎦ + Lmb ,i = ⎡⎣ 43 Md 2ω f ⎤⎦ + ⎡⎣ 13 Md 2ω f ⎤⎦
ts ,i
ts , f
mb , f
(
1
3
)
M vi sin ( 90 + 60 ) = 16 Mdvi
⎡⎣ 43 Md 2ωi ⎤⎦ + ⎡⎣ 16 Mdv
Md i ⎤⎦ = ⎡⎣ 43 Md 2ω f ⎤⎦ + ⎡⎣ 13 Md 2ω f ⎤⎦
= 53 Md 2ω f
ts ,i
mb ,i
ts , f
mb , f
ωf
(
=
4
3
Md 2ωi + 16 Mdvi
)
5
3
Md
2
= 54 ωi +
vi
10d
= ωf
Chap. 1212-13: Equilibrium, Gravitation
Static equilibrium: Gravitation law:
Fnet = 0 (balance of forces)
τ net = 0 (balance of torgue) Virtual rotation for torgues
m1 m 2
F12 = G 2 rˆ12
r12
F = ma g ;
Gravitational potential energy:
n
F1,net = F12 + F13 + F14 + ... + F1n = ∑ F1i
i=1
ag = G
U (r ) = −G
M
r2
m1m2
; ΔU fi = −W = − ∫ Fg ⋅ dr
r12
U total = −G ∑
i< j
Kepler’s Laws:
Conservative force
i f
mi m j
rij
dA L
=
= constant (2nd -Law)
dt 2m
(many objects)
⎛ 4π 2
T =⎜
⎝ GM
2
⎞ 3
⎟r
⎠
(3rd - Law)
Problem 37
37::
A uniform plank, with a length L of 6.10 m and a weight of mg = 445 N, rests on the ground
and against frictionless roller at the top of a wall of height h = 3.05 m. The plank remains in equilibrium for
any value of θ ≥ 70° but slips if θ < 70°. Find the coefficient of static friction between the plank and the
g
ground.
fs = μs N 2
What do we know: θ = 70° is critical angle:
∑F =0
i
∑F
∑F
ix
= 0 : f s − N1 sin θ = 0
(1)
iy
= 0 : N2 +N1 cosθ − mg = 0
(2)
N1
N2
y
x
Choose a convenient VIRTURAL rotation axis ∑ τ i = 0 N1 ⋅
h
L
− mg ⋅ cos θ = 0
sin θ
2
(3)
f s mg
Ch. 13 # 11: Two spheres of mass m and a third of mass M form an equilateral triangle, and a forth of mass m4 is at the center of the triangle. If the net force on the center is zero what is M? If m4 is doubled what is the answer?
The angle at each corner is 60 0 , so the distance to the center mass is
l
l
r= cos 30 0 =
2
3
All the x forces cancel, the downward force on m 4 is
⎛ Gm4 m ⎞
⎛ Gm4 m ⎞
0
sin
30
Fy (net) = 2 ⎜
=
3
⎜⎝ 2 ⎟⎠
⎝ r 2 ⎟⎠
l
The downward force must equal the upward force by M,
M so
⎛
⎞
⎛ Gm4 m ⎞ ⎜⎜ Gm4 M ⎟⎟
3⎜ 2 ⎟ =
⎝ l ⎠ ⎜ ⎛ l ⎞2 ⎟
⎝⎜ ⎜⎝ 3 ⎟⎠ ⎟⎠
m=M
Ch. 13 #46: Four masses M are located on each corner
of a square with side d. If d is reduced by ½ what is the
change in the potential energy?
U total = −G ∑
i< j
= −G (
mi m j
rij
1
2
4
3
(many objects)
m1m2 m1m3 m1m4 m2 m3 m2 m4 m3m4
+
+
+
+
+
)
r12
r13
r14
r23
r24
r34
m2 m2 m2 m2 m2 m2
= −G (
+
+
+
+
+
)
d
d
2d d
2d d
4m 2
2m 2
m2
) = −((4 + 2))
= −G (
+
d
d
d
1
If d → d ;
2
U total
m2
= −2(4 + 2)
d
Double in magnitude!
Ch. 13 #64: Two satellites of the same mass are orbiting
the earth at a radius r.r But they are going in opposite
directions and collide. (a) Find the total mechanical
energy for A and B and the earth before the collision. (b)
If the collision is completely inelastic (mass =2m) what is
the mechanical energy right after the collision? (c) Just
after the collision is the wreckage falling directly toward
Earth’s center or orbiting around the Earth?
(a) From equation 13-40 we see that the energy of each satellite is
GM E m
GM E m
Total energy: E A + EB = 2r
r
(b) The speed of the wreckage is zero
zero--right
right.
GM E ( 2m )
E=−
2r
(c) It will fall straight down!!!!
Ch. 13 # 93: A triple‐star system is shown in the
picture. The two orbiting stars are always at opposite
ends of the diameter of the orbit. Find the period?
The magnitude of the net gravitational force on one of the smaller stars (mass m) is
GMm G
GM
Gmm G
Gm ⎡
m⎤
+
=
+
M
2
4 ⎥⎦
r2
( 2r ) r 2 ⎢⎣
This supplies the centripetal force needed for the motion of the star.
Gm ⎡
m⎤
v2
M + ⎥=m
2 ⎢
4⎦
r ⎣
r
2π r
v=
T
T=
2π r 3/2
G [M + m / 4 ]