Transcript p250c09

Chapter 9: Rigid Bodies and Rotational Motion
Angular velocity
an object which rotates about a fixed axis has an average angular velocity wav :
   
wav  2 1 
usually rad/s but sometime rpm, rps
t 2  t1
t
instantaneous angular velocity is given by:
 d 
w  lim

t  0 t
dt
since s  r
ds
d
r
dt
dt
Phys 250 Ch9 p1
or v  rw
r

s
Angular Acceleration: the rate of change of angular speed
ω
ω dω d 2θ
αav 
;
α  lim

 2
t 0 t
t
dt
dt
related to linear accelerati on in circular motion :
atan 
dv
dω
r
 at  rα
dt
dt
v2
ac 
 w 2r
r
ac=w2r
aT=ar
total linear accelerati on
a  ac  aT
2
2
Example: In a hammer throw, a 7.25 kg shot is swung in a circle 5 times and then released.
The shot moves with an average radius of 2.1 m and an average angular acceleration of 2.3
rad/s2. What is the average tangential force and what is the maximum centripetal force on the
hammer?
Phys 250 Ch9 p2
Rotation with constant angular acceleration (just like linear 1-d)
Angular
Linear
1 2
  0  w0t  at
2
 w0  w 
  0  
t
 2 
1 2
x  x0  v0t  at
2
 v0  v 
x  x0  
t
 2 
w  w0  at
v  v0  at
w2  w0  2a(  0 )
v 2  v0  2a ( x  x0 )
2
watch units consistency!!!
Phys 250 Ch9 p3
2
Example: The wheel on a moving car slows uniformly from 70m rad/s to 42 rad/s in 4.2 s.
What is the angular acceleration of the wheel? What angle does the wheel rotate in those 42 s?
How far does the car go if the radius of the wheel is 0.32 m.
Phys 250 Ch9 p4
Torque: the rotational analogue of
force
Torque = force x moment arm
t = FL=F r sin 
moment arm = perpendicular
distance through which the
force acts
L
L
F
F
L
F
Example: The bolts holding a head gasket are to be “torqued down” to 90 N-m. If a 45 cm
wrench is used, what force should be applied perpendicular to the wrench handle?
Phys 250 Ch9 p5
Example: The crank arm of a bicycle pedal is 16.5 cm long. If a 52.0 kg woman puts all her
weicht on one pedal, how much torque is developed when the crank is horizontal? How much
torque is developed when the pedal is 15º from the top?
Phys 250 Ch9 p6
Equilibrium: stability, steadiness, balance etc.
Mechanical Equilibrium: absence of change in motion

=> Net Force = 0 !
F  0
(usually, no motion)
sum of x force components =
sum of y force components =
F
F
x
y
0
0
With Rotational Equilibrium
Rotational Equilibrium: absence of change in rotation (usually: no rotation)
=> net torque is zero
Sti = 0
about any axis!
t
i
0
for all torques lying in the same plane
Watch signs for torque
F
L
Positive torque for counterclockwise
rotation: t = F L
Phys 250 Ch9 p7
L
F
Negative torque for clockwise
rotation: t =  F L
Center of Gravity (CG) aka Center of Mass:
the point of an object from which it could be suspended without tending to rotate.
The point where all the mass of an object can be considered to be located.
CG does not need to be located within the physical object!
Horseshoe, for example
usually easily identified from symmetry.
Example: A 5 kg mass hangs from the 5 cm mark on a 1 meter long rod. An unknown mass hangs from the
85 cm mark. The rod has a mass of 2.0 kg and is balanced at the 35 cm mark. What is the unknown mass?
Phys 250 Ch9 p8
Example: A sign weighing 400 N is suspended from the end of a 350 N horizontal
uniform beam. What is the tension in the cable?
35o
w
Phys 250 Ch9 p9
Elasticity
“stretchiness/springiness”
-how materials respond to stress
compression
tension
shear
“Stretch-ability” = amount of stress (applied force) produces a strain
(elongation/compression/shear)
Hooke’s Law: the amount of stretching is proportional to the applied force.
F=kx
The details of such springiness depends
upon the size and shape of the material as
well as how the forces are applied
1
Ton
x
x
2
Tons
Phys 250 Ch9 p10
Elastic Limit: the maximum stress (force) which can be applied
to an object without resulting in permanent deformation.
Plastic Deformation: the permanent deformation which results
when a materials elastic limit has been exceeded.
Ultimate strength: greatest tension (or compression or shear) the
material can withstand. *snap*
A malleable or ductile material has a large range of plastic
deformation.
Fatigue: small defects reduce materials strength well below
original strength.
Phys 250 Ch9 p11
Young’s Modulus: how things stretch (elastically)
stress: force per area = F/A
A
L0
L0
compression
L
A
tension
A
L
strain: fractional change in length
= change in length per original length = L/Lo
Elastic modulus = stress/strain
Young’s modulus (for stretching in one direction)
Y
Phys 250 Ch9 p12
F A
L L0
A
Example: A steel elevator cable supports a load of 900 kg. The cable has a diameter of
2.0 cm and an initial length of 24 m. Find the stress and the strain on the cable and the
amount that it stretches under this load.
Phys 250 Ch9 p13
Torque and Moment of Inertia
For a single mass:
FT= maT
FTr =maTr
t = mar r = mr2 a
moment of inertia I = mr2
t= I a looks like F = ma
for a system of objects ( a rigid object)
I = Smiri2
Phys 250 Ch9 p14
ac=w2r
L
R2
L
R2
R
1
I  ML2
3
Thin Rod (axis at end)
1
ML2
12
Thin Rod
I
1
MR 2
2
Solid Disk
I
a
1
2
2
M ( R1  R2 )
2
Hollow Cylinder
I
a
R
b
b
1
I  M (a 2  b 2 )
3
Thin Rectangula r Plate (about edge)
1
I  M (a 2  b 2 )
12
Rectangula r Plate (through center)
R
2
MR 2
5
Solid Sphere
I
Phys 250 Ch9 p15
I  MR 2
Thin Walle d Hollow Cylinder
R
2
MR 2
3
Thin Walle d Hollow Sphere
I
Example: A cylindrical winch of radius R and moment of inertia I is free to rotate
without friction. A cord of negligible mass is wrapped about the shaft and attached to a
bucket of mass m. What is the acceleration of the bucket when it is released?
Phys 250 Ch9 p16
Angular Momentum
L=Iw
(like p = m v )
+ Angular momentum is conserved in the absence of external torques
Lstart = Lend
for a point mass moving in a circle L = mvr = mr2w
conservation of angular momentum implies Kepler’s 3rd law!
Example: Ann ice skater starts spinning at a rate of 1.5 rev/s with arms extended. He
then pulls his arms close to his body, decreasing his moment of inertia to ¾ of its initial
value. What is the skater’s final angular velocity?
Phys 250 Ch9 p17
Rotational Kinetic Energy
for a single point particle
KE 
1 2 1 2 2
mv  mr w
2
2
for a solid rotating object
1
1
2
2
m1v1  m2 v2  
2
2
1
1
2
2
 m1r1 w 2  m2 r2 w 2  
2
2
1
2
2
 (m1r1  m2 r2  )w 2
2
1
1
  mr 2 w 2  Iw 2
2
2
KE 

Phys 250 Ch9 p18

I
 mr 
KE 
2
1 2
Iw
2
Combined Translation and Rotation
KE = KEtranslation + KErotation
KE 
1 2 1 2
mv  Iw
2
2
when rolling without slipping
s= r
v=wr
a : angular acceleration
The Great Race
lost PE = gained KE
same radius, object with
the smallest I has most v
=> wins race
a = a r,
mgh  KE 
1 2 1 2
mv  Iw
2
2
2
1
1 v
1
1 I 2
 mv 2  I    mv 2 
v
2
2
2 r
2
2r
1
I 
  m  2 v 2
2
r 
Phys 250 Ch9 p19