Transcript Circular Motion PowerPoint

Circular Motion
1
the motion or spin
on an internal axis
the motion or spin
on an external axis
• Number of rotations per unit of time
Rpm or Rps
• All objects that rotated
on same axis have the
same rotational speed.
• Also called Frequency
(cycles/s or Hertz)
• Period (seconds) is the inverse of Frequency
Gymnast on a High Bar
A gymnast on a high bar
swings through two
rotations or cycles in a
time of 1.90s. Find the
average rotational speed
(in rps) or frequency (in
Hz) of the gymnast.
7
Given: t = 1.90 s & 2 rotation (cycle)
Find the average rotational speed (in rps)
Rps = rotations
second
= 2 rotation
1.90 seconds
= 1.05 rps = 1.05 cycles/second
A Helicopter Blade
Find the rotational
speed or frequency at
#1 if it takes 0.154 s
for one rotation
(cycle)?
9
Given: t = 0.154 s & 1 rotation (cycle)
Find the average rotational speed (in rps)
Rps = rotations
second
= 1 rotation
0.154 seconds
= 6.49 rps = 6.49 cycles/second
A Helicopter Blade
Find the rotational
speed or frequency at
#2 if takes 0.154 s for
one rotation (cycle)?
11
Given: t = 0.154 s & 1 rotation (cycle)
Find the average rotational speed (in rps)
Rps = rotations
second
= 1 rotation
0.154 seconds
= 6.49 rps = 6.49 cycles/second
Do Frequency/
Rotational Speed
Problems
• The speed in m/s
of something
moving along a
circular path.
• It always tangent
to the circle.
Linear Speed
• The distance moved
per unit of time.
• Linear speed is
greater on the outer
edge of a rotating
object than it is
closer to the axis.
Distance traveled in one
period is the
circumference 2πr
Time for one “cycle” is
the “period” (T)
Tangential Speed = Circumference / Period
Tangential Speed = 2πr
T
But remember that
period is the inverse of
frequency
So instead of dividing by
period you multiply by
frequency
Tangential Speed = Circumference x Frequency
Tangential Speed = 2πr x cycle
cycle
s
A Helicopter Blade
A helicopter blade
has an angular speed
of 6.50 rps. For
points 1 on the blade,
find the tangential
speed
18
Given: r = 3.00 m &
Angular speed = 6.50 rps
Tangential Speed = 2πr x cycle
cycle
s
= 2π 3.00m x 6.50 cycle
cycle
s
= 122 m/s
A Helicopter Blade
A helicopter blade
has an angular speed
of 6.50 rps. For
points 2 on the blade,
find the tangential
speed
20
Given: r = 6.70 m &
Angular speed = 6.50 rps
Tangential Speed = 2π 6.70m x 6.50 cycle
cycle
s
= 273 m/s
Notice that the tangential speed at 3 meter
is 122 m/s while at 6.70 meters is 273 m/s
Do Tangential Speed
Linear Velocity
Problems
Centripetal Acceleration
2
T
v
ac 
r
(centripetal
acceleration)
23
A Helicopter Blade
A helicopter blade
has an angular speed
of 6.50 rps. For
points 1 on the blade,
find the tangential
acceleration
24
Given: r = 3.00 m &
Tangential Speed = 122 m/s
2
T
v
ac 
r
Tangential
Acceleration = (122 m/s)2 / 3.00m
= 4,960 m/s2
= 4.96 x 103 m/s2
A Helicopter Blade
A helicopter blade
has an angular speed
of 6.50 rps. For
points 2 on the blade,
find the tangential
acceleration
26
Given: r = 6.70 m &
Tangential Speed = 273 m/s
2
T
v
ac 
r
Tangential
Acceleration = (273 m/s)2 / 6.70m
= 11,200 m/s2
= 1.12 x 104 m/s2
Do Centripetal
Acceleration
Problems
Centripetal Force
Fc = mac
Fc = mvT2
r
Centrifugal force: Center-fleeing, away form
center
Vertical drum rotates, you’re pressed
against wall
Friction force against wall matches gravity
Seem to stick to wall, feel very heavy
The forces real and
perceived
Real Forces:
Friction; up
Centripetal; inwards
Gravity (weight); down
Perceived Forces:
Centrifugal; outwards
Gravity (weight); down
Perceived weight; down and out
Weight
the force due to gravity on an object
Weight = Mass  Acceleration of Gravity
W=mg
Weightlessness - a conditions wherein
gravitational pull appears to be lacking
Examples:
Astronauts
Falling in an Elevator
Skydiving
Underwater
• Just like spinning
drum in amusement
park, create gravity
in space via rotation
• Where is the “floor”?
• Where would you
still feel weightless?
• Note the windows
on the face of the
wheel
From 2001: A Space Odyssey
rotates like bicycle tire
Do Centripetal
Force Problems
What makes something rotate?
TORQUE
How do I apply a force to make the rod rotate
about the axle? Not just anywhere!
AXLE
Torque = force times lever arm
Torque = F  L
Torque example
F
L
What is the torque on a bolt
applied with a wrench that
has a lever arm of 30 cm
with a force of 30 N?
Torque = F x L
= 30 N x 0.30 m
=9Nm
For the same force, you get more torque
with a bigger wrench  the job is easier!
Net Force = 0 , Net Torque ≠ 0
10 N
10 N
> The net force = 0, since the forces are applied in
opposite directions so it will not accelerate.
.
Net Force = 0 , Net Torque ≠ 0
10 N
10 N
> However, together these forces will make the rod
rotate in the clockwise direction.
Net torque = 0, net force ≠ 0
The rod will accelerate upward under these
two forces, but will not rotate.
Balancing torques
20 N
10 N
1m
0.5 m
Left torque = 10 N x 1 m = 10 n m
Right torque = 20 N x 0.5 m = 10 N m
Balancing torques
20 N
10 N
1m
0.5 m
Left torque = 10 N x 1 m = 10 n m
Right torque = 20 N x 0.5 m = 10 N m
How much force is exerted up by the Fulcrum?
Torque = force times lever arm
Torque = F  L
Equilibrium
• To ensure that an object
does not accelerate or
rotate two conditions must
be met:
 net force = 0
 net torque = 0
Example 1
Given M = 120 kg.
Neglect the mass of the beam.
Find the Torque
exerted by the mass
Torque = F L
= 120 kg (9.8 m/s2) (7 m)
= 8232 N m
Example
Tleft
A
Tright
Wbeam
B
Given:
8m
Wbox=300 N
Find: FTR = FCC = ? N
TorqueC = TorqueCC
FCL = FCCL
300 N (6 m) = FCC (8 m)
225 N = FCC
D
C
2m
Wbox
A
C
D
Example
Tleft
A
Tright
Wbeam
B
Given:
8m
Wbox=300 N
Find: FTL = FCC = ? N
TorqueC = TorqueCC
FCL = FCCL
FC (8 m) = 300N (2 m)
FC = 75 N
D
C
2m
Wbox
A
C
D
Example
Tleft
Tright
Wbeam
B
A
Given:
Wbox=300 N
D
C
8m
2m
Does this make sense?
W
FTL = 75 N
A
FTR = 225 N
C
Does the FUP = FDOWN ?
D
FUP = 75 N + 225 N = 300 N =FDOWN(Box)
box
Another Example
Given: W=50 N,
L=0.35 m,
x=0.03 m
Find the tension in the muscle
TorqueC = TorqueCC
FCL = FCCL
50N (0.350 m) = FCC (0.030m)
50N (0.350 m) / (0.030m) = FCC
583 N = FCC
W
x
L
Stability
CM &Torque
Condition for stability
CG
If the CG is above
the edge, the object
will not fall
when does it fall over?
STABLE
NOT STABLE
If the vertical line
extending down from
the CG is inside the
edge the object will
return to its upright
position  the torque
due to gravity brings
it back.
Stable and Unstable
stable
torque due to gravity
pulls object back
unstable
torque due to gravity
pulls object down
Stable structures
Structures are
wider at their
base to lower their
center of gravity
Playing with your blocks
CG
If the center of gravity
is supported, the
blocks do not fall over
Rotational Inertia
The rotational “laziness” of an object
Recall : inertia
A measure of the “laziness” of an
object because of
Fnet
a
m
Quantified by the mass (kg) of object

Rotational Inertia (I)
A measure of an object’s “laziness” to
changes in rotational motion
Depends on mass
AND
distance of mass from axis of rotation
Balancing Pole increases Rotational Inertia
QuickTime™ and a
TIFF (Uncompressed) decompressor
are needed to see this picture.
Angular Momentum
Angular Momentum
Momentum resulting
from an object moving in
linear motion is called
linear momentum.
Momentum resulting
from the rotation (or
spin) of an object is
called angular
momentum.
Conservation of Angular
Momentum
Angular momentum is
important because it
obeys a conservation
law, as does linear
momentum.
The total angular
momentum of a closed
system stays the same.
Calculating angular momentum
Angular momentum is calculated in a similar way to
linear momentum, except the mass and velocity are
replaced by the moment of inertia and angular velocity.
Angular
momentum
(kg m/sec2)
L=Iw
Moment of
inertia
(kg m2)
Angular
velocity
(rad/sec)
Gyroscopes Angular Momentum
A gyroscope is a device that contains a spinning object
with a lot of angular momentum.
Gyroscopes can do amazing tricks because they
conserve angular momentum.
For example, a spinning gyroscope can easily balance
on a pencil point.
Gyroscopes Angular Momentum
A gyroscope on the space shuttle is mounted at the
center of mass, allowing a computer to measure
rotation of the spacecraft in three dimensions.
An on-board computer is able to accurately measure
the rotation of the shuttle and maintain its
orientation in space.
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