Download lecture ch7-8-Circles

11/14/2011
Quantities and Units
• Angular Position: θ (radians or degrees)
∆θ = θ f − θ i
∆x = x f − xi
Rotational Motion
• Angular Velocity: ω (radians/sec)
v=
Phy 121
Eyres
∆x
∆t
ω=
∆θ
∆t
• Angular Acceleration: α (rad/sec^2)
a=
Angular Position
∆v
∆t
α=
∆ω
∆t
Circles
C = 2πr
v=
2πr
T
T is time to go
around once
Problem 7.8
• A tire placed on a
balancing machine
in a service station
starts from rest and
turns through 4.7
revolutions in 1.2 s
before reaching its
final angular speed.
Calculate its angular
acceleration.
t=
θ=
ω=
Problem 7.8
α=
t=
θ=
ω=
ω=
∆θ
∆t
α=
∆ω
∆t
∆θ = ωi t + 12 αt 2
• A tire placed on a
balancing machine
in a service station
starts from rest and
turns through 4.7
revolutions in 1.2 s
before reaching its
final angular speed.
Calculate its angular
acceleration.
t =1.2 s
θ =4.7 revs
ω=
α=
t =0
θ =0
ω =0
ω=
∆θ
∆t
α=
∆ω
∆t
∆θ = ωi t + 12 αt 2
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Problem 7.8
 2π rads 
4.7 revs
 = 29.53 rads
 1 rev 
29.53 rads
ω=
1.2 sec
ω = 24.6 rads
sec
t =1.2 s
θ =4.7 revs
ω=
Circular or Angular?
Velocity
α=
v=
∆ω
∆t
49.2 rad
sec
α=
= 41.0 rad
s2
1 .2 s
α=
α=
ω=
∆θ
∆t
α=
∆ω
∆t
• Acceleration:
t =0
θ =0
ω =0
ω f = 49.2 rads
sec
2πr
∆t
a=
∆ω
∆t
v2
r
∆θ = ωi t + 12 αt 2
Bug
Torque
• (a)What is the tangential
acceleration of a bug on the
rim of a 10-in.-diameter disk
if the disk moves from rest
to an angular speed of 78
rev/min in 3.0 s? (b) When
the disk is at its final speed,
what is the tangential
velocity of the bug? (c) One
second after the bug starts
from rest, what are its
tangential acceleration,
centripetal acceleration,
and total acceleration?
Interpreting Torque
τ = F⊥ × r
τ = F × r⊥
• A steel band exerts a
horizontal force of 80.0
N on a tooth at point B.
What is the torque on
the root of the tooth
about point A?
A Second Interpretation of Torque
Torque is due to the component of the force
perpendicular to the radial line.
τ = rF⊥ = rF sin φ
τ = r⊥ F = rF sin φ
Slide 7-16
Slide 7-17
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11/14/2011
Torque
Example
Which force would be most effective in opening the door?
Revolutionaries attempt to pull down a statue of the Great
Leader by pulling on a rope tied to the top of his head. The
statue is 17 m tall, and they pull with a force of 4200 N at an
angle of 65°to the horizontal. What is the torque t hey exert on
the statue? If they are standing to the right of the statue, is the
torque positive or negative?
Slide 7-15
Lifting Problem
Lifting
• A person bending forward to lift a load “with his back” rather
than “with his knees” can be injured by large forces exerted on
the muscles and vertebrae. The spine pivots mainly at the fifth
lumbar vertebra, with the principal supporting force provided by
the erector spinalis muscle in the back. Consider the model of a
person bending forward to lift a 200-N object. The spine and
upper body are represented as a uniform horizontal rod of
weight 350 N, pivoted at the base of the spine. The erector
spinalis muscle, attached at a point two-thirds of the way up the
spine, maintains the position of the back. The angle between the
spine and this muscle is 12.0°. Find the tension in the back
muscle and the compressional force in the spine.
τ = F⊥ × r
Στ = 0
Slide 7-18
ΣF = 0
τ = F⊥ × r
• Consider the model of a
person bending forward
to lift a 200-N object.
The spine and upper
body are represented as
a uniform horizontal rod
of weight 350 N, pivoted
at the base of the spine.
The angle between the
spine and this muscle is
12.0°. Find all Forces
Στ = 0 ΣF =0
Answers
• Lifting T=2710 N,
Rx=2650 N
• Bug:
a) tan accel = 0.35 m/s2
b) 1.0 m/s
c) 0.35 m/s2, 0.95 m/s2,
1.0 m/s2 at 20° forward
from radial axis
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