Physics 2101
Section
S cti n 3
Apr 12th
Announcements:
• Finish Ch. 15 today
• Start Ch. 16 today too
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
Torsion Pendulum
Spring
ma = −kx
k
a=− x
m
k
ω2 =
m
What is period of torsion pendulum? Torque
τ = Iα = −κθ
“kappa” is torsion spring constant
κ
α =− θ
I
ω =
2
κ
I
⇒ω =
κ
I
⇒ T = 2π
I
κ
Simple Pendulum
2) Simple Pendulum : gravity is “restoring force” τ = Iα = −L (mg sin θ ) ⇒ α ≅ −
SHM : θ ((t)) = θ max cos(ωt + ϕ ) ⇒ ω =
mgL
θ
I
mgL
=
I
mgL
=
2
(mLL )
• Independent
n p n n of
f amplitude
mp u
and
n m
mass ((in
n small
m
angle
ng approximation)
pp
m
n) !
• Dependent only on L and g
g
L
Clicker Question
Question 10‐4
A bowling ball and a ping‐pong ball are each tied to a string and hung from the ceiling. The distance from the ceiling to the CM of each object is the same Which object would have a longer period of
object is the same. Which object would have a longer period of motion if they were set swinging? Neglect air resistance and frictional effects.
1. Bowling ball
2 Ping‐pong ball
2.
Ping‐pong ball
3. Both have the same period ω = √(g/L)
Physical Pendulum
S i i
Swinging mass under gravity
d
it
“spring” restoring force is gravity τ pivot pt. = rF⊥ = −h (mg sin θ )
⇒ τ = I rot α
⇒α ≅−
mghrot−com
θ
I rot
SHM : θ (t)
( ) = θ max cos(ωt + ϕ )
⇒ω =
mgh rot−com
I rot
⇒ T = 2π
I rot
mghrot−com
15‐49: A stick of length L oscillates as a physical pendulum. (a) What value of distance x between the stick’s COM ant is pivot point O gives p
p
g
the least period? (b) What is the least period?
General Solution of 15‐49
ω=
mghh
I
ω=
gx
I = I COM
L2
+ x2
T = 2π 12
gx
L2
+ x2
12
12gT
g 2
Find the minimal T Take derivative
L
x=
12
2
mL
L
+ mh 2 =
+ mx 2
12
(2π )
2
L2
=
+ 12x
12
x
⎞
d ⎛ 12gT
12 T 2 ⎞ d ⎛ L2
L2
+ 12x ⎟ = − 2 + 12 = 0
⎜
2 ⎟ =
⎜
dx ⎝ (2π ) ⎠ dx ⎝ x
x
⎠
Problem 16‐53: In an overhead view, a long uniform rod of length L and mass m is free to rotate in a horizontal axis through its center. A spring with force constant k is connected g
p g
to the rod and the fixed wall. When the rod is in equilibrium, it is parallel to the wall. What is the period of small oscillations that results when the rod is rotated slightly and released? TEST QUESTION LAST YEAR
τ pivot pt.
Problem
xˆ
⎛L⎞
⎛ L⎞
= rF⊥ = −⎜ ⎟(Fspring ) = −⎜ ⎟(kx)
⎝2⎠
⎝2⎠
[( ) ]
Lk
Lk L
L2 k
=−
x=−
θ =−
θ
2
2
2
4
2
⎛1
⎞
⎛
⎞
1
L
k
2
2
τ = I rot α = ⎜ mL ⎟α ⇒ ⎜ mL ⎟α = −
θ
⎝12
⎠
⎝12
⎠
4
SHM : θ (t) = θ max cos(ωt + ϕ )
⎛ 3k ⎞
L2 k ⎛ 12 ⎞
⇒α =−
⎜ 2 ⎟θ = −⎜ ⎟θ
⎝m⎠
4 ⎝ mL ⎠
⇒ ω =
3k
m
Problem: The pendulum consists of a uniform disk with radius R and mass M attached to a uniform rod of length L and mass m.
(a) what is the rotational inertia of the pendulum?
(b) What is the distance between the pivot and the COM?
(c) What is the period?
(a) The disc has an I=MR2/2 about the COM so.
MR 2
MR 2
2
2
I(Di ) =
I(Disc)
+ Mh =
+ M (r + L )
2
2
I (Rod) =
(b) COM
2
mL
nL
mL
⎛ L⎞
+ mh 2 =
+ m⎜ ⎟ =
⎝ 2⎠
12
12
3
2
dCOM
(c) The period 2
Mld + mlr
=
M +m
With ld and lr being the the COM of the disc and rod. g
I
T = 2π
(M + m)gd
15-106: A solid Cylinder attached to a horizontal spring (k=3.0 N/m) rollos
without slipping along a horizontal surface. If the system is released from rest
when
h the
h spring
i iis stretched
h db
by d=0.25
d 0 2 m, find
fi d
(a) The translational kinetic energy
(b) The rotational kinetic energy of the cylinder as it passes through x=0.
(c) Show that under these conditions the COM is SHM with
3M
T = 2π
2k
Solution:
(a)
Emech
2
Mv 2 I CM
ω2
= Us +
+
2
2
2
2
2
I CM
ω 2 Mvcm
kd 2 Mvcm
1 ⎛ MR 2 ⎞ ⎛ vcm ⎞
=
+
=
+ ⎜
⎟⎜
⎟
2
2
2
2
2 ⎝ 2 ⎠⎝ R ⎠
2
kd 2 3Mvcm
1
1
2
=
; so that KEtran = Mvcm
= kd 2 = 66.25
25 ×10−2 J
2
4
2
3
2
(b)
( )
(c) KErot
1 2 1
1 2
2
= I ω = Mvvcom = kxm = 3.
3.13
3 × 10
0−2 J
2
4
6
2
dEmechh d ⎛ 3Mvcm
kx 2 ⎞ 3Mvcm acm
= ⎜
+
+ kxvcm = 0
⎟=
dt
dt ⎝ 4
2 ⎠
2
acm
⎛ 2k ⎞
= −⎜
x
⎟
⎝ 3M ⎠
ω=
2k
3M
Chap. 15: Oscillations
ω = 2 πf → T =
2π
→ ωT = 2 π
ω
x (t ) = x max cos (ω t + ϕ )
Position
dx
= − x max ω sin (ω t + ϕ )
dt
dv
a (t ) =
= − x max ω 2 cos (ω t + ϕ )
dt
v(t ) =
Velocity
Acceleration
a ( t ) = −ω 2 x ( t )
In SHM, the acceleration is proportional to the displacement but opposite sign. The proportionality is the square of the angular frequency.
ω spring =
k
m
ω simple pendulum =
ω torsion pendulum =
g
L
ω physical pendulum
κ
I rot
mgh rot−com
=
I rot
“Phase Diagram” of SHO
Knowing:
U (t) = kx = kx
2
1
2
1
2
2
max
x2
⇒ 2 = cos 2 (ωt + ϕ )
x max
cos (ωt + ϕ )
2
2
KE(t) = 12 mv 2 = 12 mω 2 x max
sin 2 (ωt + ϕ )
v2
⇒ 2 2 = sin
i 2 (ωt + ϕ )
ω x max
E mech = U (t) + KE(t)
= kx
1
2
2
max
sin 2 (ωt + ϕ ) + cos 2 (ωt + ϕ ) = 1
v2
x2
+
=1
⎛2E mech ⎞ ⎛2E mech ⎞
⎜
⎟ ⎜
⎟
k⎠ ⎝
m⎠
⎝
= mv
1
2
2
max
x2
x2
=
2
2E mech
x max
k
2
v
v2
=
2
2ω 2 E mech
ω 2 x max
=
k2
v2
2E mech
v
m
x
v(t) = vmax
x2
1− 2
x max