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Transcript
Uniform circular motion is the motion of
an object in a circle with a constant speed.
What force or forces provides the circular
motion in the following pictures?
What direction are these forces?
Can it be more than one force?
Could it be components of a force?
The force or forces needed to
bend the normally straight path
of a particle into a circular or
curved path is called the
CENTRIPETAL FORCE.
It is a pull on the body and is
directed toward the center of
the circle.
Force
on car
Force on
passenger
Tendency for
passenger to
go straight
Without a
centripetal force, an
object in motion
continues along a
straight-line path.
Remember
Newton’s 1st Law?
With a centripetal
force, an object in
motion will be
accelerated and
change its direction.
What is the
centripetal force?
-not a real force
-feeling due to
inertia
Circular Motion Vocabulary
•
•
•
•
•
•
•
•
•
•
•
r = radius
m= mass
v = velocity
Fc = centripetal force
FT = tension force
(sometimes written as T, not to be confused with
the T for period)
Ff = friction force
t = time
T = period = “sec / rev”
linear (tangential) velocity = 2πr/T “m / s”
rotational (angular) velocity (ώ) = “rev / sec” = 1 / T
ac = centripetal acceleration
Why does the ball stay in a circular path?
Top view
Centripetal: (“c”) “center seeking” variables
that exist when curved or circular path occurs.
Linear (tangential) velocity is the
velocity tangent to the curve at a point in
time. (Newton’s 1st law)
V
┴
What would happen to the
ball if I cut the string?
Fnet-c , ac
An object is moving in a clockwise direction around a circle at
constant speed.
Which vector below represents the direction of the velocity vector
when the object is located at point A, B, and C on the circle?
the acceleration vector?
the force vector?
e.
Rotation → axis inside object
(Earth rotates around its axis)
Revolution → axis outside object
(Earth revolves around the Sun)
Rotational speed → rot. speed = # of rev / time (rev/s)
Period
→
Linear velocity →
T
=
time / # of rev (s/rev)
v = 2πr / T
(m/s)
What affects the speed of a ball tied to a
string moving in a circular path?
Tension, mass, radius
Linear = Polar = Angular = Rotational
2πr
360º
2π rad
1 Rev
Convert rotational speed to linear velocity:
Example: The spinning ball (in the above picture) has a
string radius of 0.5 m with the tube. There are 5
revolutions in 2.5 seconds (2 rev/s).
Convert to linear velocity …
Example: The spinning ball (in the
picture at left) has a string radius of
0.5 m with the tube. The ball
completes 5 revolutions in 2.5 seconds.
(Rotational velocity)
r = 0.5 m
t = 2.5 seconds for 5 revolutions = T (period)
T = time/ # of rev = 2.5 s/ 5 rev = 0.5 s/rev
Convert rotational speed of 5 rev / 2.5 s to linear
speed in m/s. Remember T = 1 / rot. speed
Linear velocity: v = 2 π r / T
v = (2 π 0.5 / 0.5) = 6.28 m/s
Purpose: Find the relationship between the
_________variable and the velocity of the mass.
Manipulate either Tension, m, r
Control the other two variables
Plot Tension, m, r vs. velocity
Procedure: (data table, plot and calculation on separate paper
• Find the time for 5 revolutions for each data point.
• Do 3 trials for each data point.
• Do 5 data points (15 total time trials)
• Convert rotational speed to linear velocity before plotting.
•Determine the relationship of the graph – explain!
Groups 3 masses, 3 tensions,
2 radius
Data table
FT,r,m
units
Time for 5 revolutions
Trial 1
Trial 2
Trial 3
Period
velocity
s/1 rev
m/s
Graph
Control Hints
r = .5 m
0
0
0
0
0
0
FT vs v
m = .05 kg
r vs v
T=2N
m vs v
T=2N
m = .05 kg
r = .5 m
Do not start with a (0,0) data point
1.) Find the Period:
T (s/1rev) = time / # of rev.
2.) Find Linear Velocity:
v (m/s) = 2πr / T
Remember: Plot velocity on the x-axis
Tension
FT
FT
v2
v
Radius
FT α v2
r
Fc = mv2/r
r
r α v2
v2
v
Mass
m
m
v
m
1/v
For all
Centripetal
Forces:
m α 1/v2
1/v2
3 Important Equations
v = 2pr/T
Period (T) = time for 1 rev
•Centripetal: (“c”) “center seeking”
•Fc and ac always act toward center
•Velocity is tangent to the curve.
Magnitude is constant, direction is not.
•V
┴
Fnet-c, ac
•Radius is the ┴ distance from the axis of rotation to the
object
•Never put Fc on the F.B.D.. Ie. Fc = Ff , FT , FN
•Projectile Motion is not true Circular motion, Why?
In circular motion the force is always ┴ to the velocity; the
magnitude of velocity is not effected by the force
1. What does centripetal mean?
A ball is attached to a string and whirled around at a constant
speed in a circle with a radius 3 meters. (Questions 2-6)
2. What is the direction of the net force on the ball?
3. What is the direction of the accel. of the ball? Is it constant?
4. What is the direction of the velocity of the ball? Is it constant?
5. What are the 3 important equations? Does rotational speed
depend upon the radius? Does linear speed depend upon the radius?
6. If it takes 50 seconds for the ball to make 20 revolutions
around a 3 m radius, calculate the rotational speed and linear
velocity of the ball.
7. What were the relationships we found in the lab?
4. Tied to a post and moving in a circle at
constant speed on a frictionless horizontal
surface. Coming straight out of the paper.
FN
FT
ac
Fg
ΣFx = mac
-FT = m(-ac)
FT = mac
ΣFy = ma = 0
FN - Fg = 0
FN = Fg
5. Tied to point A by a string. Moving in a horizontal
circle at constant speed. Not resting on a solid
surface. No Friction. Coming straight out of paper.
A
FT
ac
Fg
ΣFx = mac
-(FTcosθ) = m(-ac)
FTcosθ = mac
ΣFy = ma = 0
FTsinθ - Fg = 0
FTsinθ = Fg
10. Riding on a horizontal disk that is rotating at
constant speed about its vertical axis. Friction
prevents rock from sliding. Rock is moving straight
out of the paper.
FN
Ff
ac
Fg
ΣFx = mac
-Ff = m(-ac)
Ff = mac
ΣFy = ma = 0
FN - Fg = 0
FN = Fg
11. Resting against a wall with friction a cone is
rotating about its vertical axis at a constant speed.
Not accelerating vertically. Moving straight
out of the paper.
FN
ac
ө
ө
Ff
Fg
ΣFx = mac
ΣFy = ma = 0
-(FNsinθ) – (Ffcosө) = m(-ac) FNcosθ - Ffsinө - Fg = 0
FNcosθ - Ffsinө - Fg = 0
FNsinθ + Ffcosө = mac
12. Stuck by friction against the inside wall of a
drum rotating about its vertical axis at constant
speed. Rock is moving straight out of the paper.
Ff
FN
ac
ΣFx = mac
-FN = m(-ac)
FN = mac
Fg
ΣFy = ma = 0
Ff - Fg = 0
Ff = Fg
A ball held by a string is coasting around in a
large horizontal circle. The string is then pulled in
so the ball coasts in a smaller circle. When it is
coasting in the smaller circle its speed is …
(Assume tension and mass stay constant)
a) greater
b) less
c) Unchanged
Explain.
Problem #1
If the radius of a circle is 1.5 m and it
takes 1.3 seconds for a mass to swing
around it (1 rev).
a) What is the speed of the mass?
b) Find the tension if the mass is 2 kg.
s = 7.25 m/s
FT = 70.1 N
Problem #2
A 1200 kg car traveling at 8 m/s is
turning a corner with a 9 m radius.
a) How large a force is needed to keep the
car on the road?
b) b) Find the coefficient of friction.
Ff = 8533.3 N
μ = .726
Problem #3
A car travels around a circular flat
track with a speed of 20 m/s. The
coefficient of friction between the tires
and the road is 0.25. Calculate the
minimum radius needed to keep the car
on the track.
r = 163.27 m
What speed must a 1.5 kg pendulum bob swing in
the circular path of the accompanying figure if
the supporting cord is 1.2 m long and  is 30?
Also find the tension in the cord.
ө
Answer:
v = 1.84 m/sec;
T = 16.97 N
The Gunslinger is modeled after the famed Flying
Dutchmen rides of carnival midways. Guests ride in
individual chairs suspended by tempered steel chains.
The arms tilt to a 25 angle. As a safety engineer for
the Six Flags of America Corporation, you are asked
to determine the maximum allowable rotation rate
for the Gunslinger if the breaking strength of the
steel chains are 1000 N.
Other data:
length of chain and swing: 4.5 m
distance from center of rotation to chain attachment: 6.7 m
1. Draw a FBD of a rider and the swing 2nd Law Equations.
2. What is the source of the centripetal force acting on a
rider and the swing?
3. Which will ride higher: an empty swing or one with someone
in it? Explain.
4. Determine the maximum allowable rotation rate.
Banked Road problem
FN
ө
ө
Fg
ac
Ff
Prove that the scale reading is greater at
the poles than at the equator.
201N
200N
Bill the Cat, tied to a rope, is twirled around in a
vertical circle. Draw the free-body diagram for Bill in
the positions shown. Then sum the X and Y forces.
ΣFy = mac
FT + mg = mv²/r
FT = mv²/r - mg
ac
FT = m ((v²/r) - g)
ΣFy = mac
ac
mg
FT
FT
FT - mg = mv²/r
FT = mv²/r + mg
FT = m ((v²/r) + g)
mg
Minimum velocity needed for an object to
continue moving in a vertical circle. Any less
velocity and the object will fall.
At this point, FT = 0, so…
ΣFy = mac
FT + mg = mv²/r
0 + mg = mv²/r
g = v2/r
rg = v2 or, vc = rg
Suppose a car moves at a constant speed
along a mountain road. At what places
does it exert the greatest and least
forces on the road?
a) the top of the hill
b) at the dip between two hills
c) on a level stretch near the bottom of the hill
Explain each case with a free body diagram and
sum the forces.
1) Draw a FBD of the Greezed Lightning at the top of
the loop.
2) Write the 2nd Law Equation for the FBD.
3) List the assumptions made and calculate the
minimum velocity of the Greezed Lightning given the
following picture.
4) What if you had designed
a Greezed Lightning type
rollercoaster and the coaster
was moving too slow, how
could you change your design
to correct the problem?
7m
49 m
y
60
x
An automatic tumble dryer has a 0.65 m diameter
basket that rotates about a horizontal axis (x). As
the basket turns, the clothes fall away from the
basket’s edge and tumble over. If the clothes fall
away from the basket at a point 60 from the
vertical (y), what is the rate of rotation in
revolutions per minute?
Quiz
1. A race car travels around a flat, circular track with a radius of 180 m.
….The coefficient of friction between the tires and the pavement is 1.5.
a. Draw all forces on the car, including direction of acceleration.
b. Write the summation and net force equations (x & y).
c. Calculate the maximum velocity the car can go and stay on the track.
2. A ball is swung in a horizontal circle. The 1.5 m
20°
long string makes a 20° to the horizontal.
a. Draw all forces on the picture, including direction of acceleration.
b. Write the summation and net force equations (x & y)
c. Determine the velocity of the ball?
Quiz #2
1.
A loop-the-loop rollercoaster has a radius of 20 m.
Draw a FBD (at the top of the loop) showing all forces
and calculate the minimum velocity the roller coaster
must have in order to stay on the track.
v = __________
2.
A satellite of mass 1500 kg is in low orbit around a
planet of mass = 5.7 x 1022 kg and radius 8.5 x 105 m.
The altitude of the satellite is 4.7x 104 m. What is the
gravitational attraction between the two? (G = 6.672 x
10-11 N-m2/kg
Fg = __________