1 7/26/04 midterm 2 – next friday (7/30/04) material from chapters 7-12 i will post a practice...
TRANSCRIPT
17/26/04
Midterm 2 – Next Friday (7/30/04)Material from Chapters 7-12
I will post a practice exam on Monday
Announcements
27/26/04
Last Time…
Rotation
Position x
Velocity = d/dtv = dx/dt
Acceleration = d/dta = dv/dt
Translation
Kinetic Energy K = ½I2K=½mv2
Mass I = miri2m
F = ma = INewton’s 2nd Law
37/26/04
Example: (Problem 11.49)
During a launch from the board, a diver’s angular speed about her center of mass changes from zero to 6.20 rad/s in 220 ms. Her rotational inertia about her center of mass is 12.0 kg m2. During the launch, what are the magnitudes of
a) her average angular acceleration and b) the average external torque on her from the board?
47/26/04
Example:
20 /2.28220.0
/0/20.6srad
s
sradsrad
tf
a) Average angular acceleration
b) Average external torque
NmsradmkgI 338)/2.28)(0.12( 22
57/26/04
Example: (Problem 11.53)
The world’s heaviest hinged door has a mass of 44,000 kg, a rotational inertia about a vertical axis through its hinges of 8.7x104 kg m2, and a width of 2.4 m. Neglecting friction, what steady force applied to the door’s outer edge and perpendicular to the plane of the door can move it from rest through an angle of 90° in 30 s?
67/26/04
Solution:
00
221
0 tt
2
2
t
radf 290
Initial and final positions will be:
Using a standard kinematics equation,
Finally,
IFr 90sin
77/26/04
Solution:
IFr
2
2
rt
I
r
IF
2
24
)30)(4.2(
)107.8)(2/(2
sm
mkgradF
NF 130
Plugging in our value for :
7/26/04 8
Chapter 12
Rolling, Torque, and Angular Momentum
97/26/04
Rolling Without Slipping
How do we describe an object that is rolling?
Consider the case of pure translation:
vcm
vcm
vcm
2
2
1cmMvK
107/26/04
Rolling Without Slipping
Consider the case of pure rotation:
vcm
vcm
2
2
1 cmIK
117/26/04
Rolling Without Slipping
Combining the two:
vcm
vcm
vcm
22
2
1
2
1 cmcm IMvK
vcm
vcm
+ =vcm
2 vcm
v=0
127/26/04
Rolling: Another View
Rωdt
dθR
dt
dsvcm
Rα acm
At any instantthe wheel rotates aboutthe point of contact
vc
vc
s = R
P
137/26/04
2
2
1ωIK P
2
22
2
1
2
1
R
vMRIK cm
cm
R
vω cm
I about point of contact
Parallel Axis Theorem
K = ½IP2
IP = Icm + MR2
P
22
2
1
2
1cmcm MvIK
Rolling: Another View
222
2
1ωMRωIK cm
147/26/04
Rolling Down a Hill
h
vc
Conservation of Energy:
MghvMR
Icm
cm
2
22
1
Ui = Kf2
21
cmMvMgh
For a point mass:
ghvcm 2
For an extended object:
21
2
/MRI
ghv
cmcm
157/26/04
Rolling Down a Hill
2
2
1MRIcm
2MRIcm
gh/MRMR
ghvcm 3
4
1
222
21
gh/MRMR
ghvcm
221
2
For a disc:
For a ring:
vc
h
Notice there is no mass or radial dependence!
167/26/04
How to Solve Torque Problems
1) Draw a picture2) Pick an origin (axis of rotation usually a good
choice)3) Sum torques about the origin
net= I4) Sum the forces in every direction (if
necessary)
Fnet = ma5) Solve for unknowns
177/26/04
The Forces of Rolling
Constraint of Rolling without Slipping:
α Racm
Mg
Mg sin
Mg cos
FN
FFBefore when we drew a free body diagram, we drew all the forces from the center of mass. Cannot do that with torques!
Note: Friction causes rolling (otherwise it would just slide)
187/26/04
The Forces of Rolling
Pick the center of mass asthe origin. Sum the forces and torques.
cmfrxnet MaFθMgF sin,αIRF cmfrnet
Mg
Mg sin
Mg cos
FN
FFFind the linear acceleration of the center of mass.
cmfr MaθMgF sin
cmcm IRMaθMg )sin(
197/26/04
The Forces of Rolling
Mg
Mg sin
Mg cos
FN
FF
cmcm IMRaθMRg sin
α Racm Recall:
cm
cmcm I
aMRθgMRa
22 sin
θgMRMRIa cmcm sin)( 22
2
2 sin
MRI
θgMRa
cmcm
2/1
sin
MRI
θga
cmcm
Linear acceleration doesn’t depend on mass or radius!!
207/26/04
Another Way to Solve the Problem
What if we pick a different origin? Pick point of contact.
No torque from normal forceor friction!
21
sin
/MRI
θga
cmcm
Same result! (Don’t need to sum forces)
Mg
Mg sin
Mg cos
FN
FF
Pnet IMRg sin2MRII cmP
)(sin 2MRIMRg cm
2
sin
MRI
MRg
cm
217/26/04
Angular Momentum
Recall: linear motion
Using the correspondence with linear motion, angular momentum should have the form:
If no torque, then L is a constant.
We have conservation of angular momentum!
vmp
amdt
IL dt
LdItot
227/26/04
Conservation of Angular Momentum
0 fi LL
237/26/04
Conservation of Angular Momentum
IL
If I changes, then must change to keep L constant (no external torques)
Example: Ice skaters pulling arms in
Li=Iii Lf=Iff
247/26/04
Angular Momentum of a Particle
Can also define angular momentum for a particle with a linear velocity v
r is vector from origin to particle
rv r v L is big
r || v L is 0
(Example: something circling the origin)
)( vrmprL
257/26/04
Angular Momentum of a Particle:
dt
pdrp
dt
rdpr
dt
dL
dt
d
)(
Fr
netLdt
d
Is this new definition of L consistent with the old one?
Same as before!
dt
pdrvmv
267/26/04
Example:Constant velocity particle: Is L really constant?
Direction is constant as well!
)( vrmprL
r
v
θd
const)cos()90sin( mvdrmvmrvL
277/26/04
Torque and Angular Momentum
A) No torque: L is constant
L = Iωif you change I, ω changes to keep L constant
This allows skaters and divers to spin reallyreally fast (they studied their physics!)
L=Iω =dL/dt
B) If I is constant and ω changes, there must have been a torque
287/26/04
Angular Momentum of a Group of Particles:
For a group of particles:
...321 LLLLtot
The net torque on the system will be:
dt
Ld totnet
No need to worry about internal torques
297/26/04
A merry-go-round with radius 3 m and Imgr=90 kg m2 is spinning at 0=5 rad/s. A 100 kg pig is then dropped on it at a radius of 2 meters. His landing takes t=1s.
3 m
2 m
Example: Merry Go Round
307/26/04
Example: Merry Go Round
What is the final angular velocity of the system?
f
iif I
I 2
22
490
)2)(100()90(
mkg
mkgmkg
III pigmgrf
sradmkg
sradmkg
I
I
f
iif /918.0
490
)/5)(90(2
2
ffii
fi
II
LL
317/26/04
What is the average torque on the merry-go-round during the landing?
Example: Merry Go Round
t
Lmgravg
t
I fmgravg
)( 0
Nms
sradsradmkgavg 367
1
)/5/918.0)(90( 2
Note: This is an internal torque
327/26/04
What is the average linear tangential acceleration of the rim of the merry-go-round?
Example: Merry Go Round
2
0
/08.4
1
/5/918.0
srad
s
sradsradt
f
22 /2.12)/08.4)(3( smsradmRat
337/26/04
Example:
FOR SALE
m = 5 gM = 2.2 kgv = 300 m/sℓ = 0.2 m
vℓ/2
ℓθ
A bullet hits a sign: how high does it go?
347/26/04
Example:
m = 5 gM = 2.2 kgv = 300 m/sℓ = 0.2 m
Before bullet hits:
After bullet hits:
vℓ/2
ℓ
θsmkgmvlLi /15.0)( 221
ILL if
sradmkg
smkg
I
L f /10.5029.0
/15.02
2
222
029.023
mkgl
mMl
I
357/26/04
Example:
Ki+Ui = Kf+Uf
ℓ/2
ℓ θ
h
221 IK i
ghmMU i )(
ghmMI )(221
)005.2.2)(/8.9(2
)/10.5)(029.0(
)(2 2
222
kgkgsm
sradmkg
mMg
Ih
cmmh 75.10175.0
367/26/04
Precession
FrLdt
d
Right hand rule dL/dt out of the page!
mg
hL
Consider a spinning gyroscope:
377/26/04
Precession
Iω
mgh
L
mgh
Δt
Δφωp
L=I
= mgh
ΔL = Δt
ΔL = mgh Δt
L
tmgh
L
ΔLΔφ
Δ
Lf
LiΔL
mg
hL
387/26/04
The Bicycle
Why is it hard to fall over?
L gets larger as the bike goes faster
A torque is needed to change L
If we lean to one side (i.e. fall over)
= dmg
x
yz
mg
d
in the x direction
Turns the bike!
d