Transcript 6.1 Circular motion - Uplift North Hills

Data booklet reference:
• 𝑣 = 𝜔𝑟
•𝑎 =
•𝐹 =
𝑣2
𝑟
=
𝑚𝑣 2
𝑟
4  2𝑟
𝑇2
= 𝑚2𝑟
Goal is to obtain an expression for the magnitude and
direction of centripetal acceleration.
An object is moving in circular path at constant speed.
∆v = v2 – v1
P
v2 = v1 + ∆v
v1
v1
v2
r1
∆v
Velocities at two different
positions P and Q have the
equal magnitude (v1 = v2 = v),
but different directions.
ac =
v1
∆v
ac
lim  Δv 


Δt  0  Δt 
a𝑠 ∆𝑡 → 0, 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 ∆𝑣
𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ𝑒𝑠 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛
𝑡𝑜𝑤𝑎𝑟𝑑 𝑡ℎ𝑒 𝑐𝑒𝑛𝑡𝑟𝑒.
𝑎𝑠 𝑎𝑐 𝑖𝑠 𝑖𝑛 𝑡ℎ𝑒 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 ∆𝑣,
𝑤𝑒 𝑐𝑜𝑛𝑐𝑙𝑢𝑑𝑒 𝑎𝑐 𝑝𝑜𝑖𝑛𝑡𝑠 𝑡𝑜𝑤𝑎𝑟𝑑
𝑡ℎ𝑒 𝑐𝑒𝑛𝑡𝑟𝑒 𝑜𝑓 𝑐𝑖𝑟𝑐𝑢𝑙𝑎𝑟 𝑚𝑜𝑡𝑖𝑜𝑛
centripetal or radial acceleration
P
v1
Velocities at two different
positions P and Q have the
equal magnitude (v1 = v2 = v),
but different directions.
v1
r1
∆v
v2
∆v = v2 – v1
ac =
lim  Δv 


Δt  0  Δt 
∆ r , r , Δr  ~ ∆  v , v , Δv 
1
Δv
v
1
2
Δr
=
r
 Δv = v
magnitude of ac
v2
ac =
r
2
Δr
r
lim  v Δr 
ac =
Δt  0  r Δt 


 ac 
m/s 
v=
-1 2
=
m
= m/s2
lim
Δr
Δt  0 Δt
A “slightly” misleading sketch
∆𝑣 should be added to 𝑣1 to change it to 𝑣2 but if we
add it at point P 𝑎𝑐 would not point toward center !!!!!.
v2 = v1 + ∆v
v1
P
Q
∆r
r1
∆v
r2
∆θ
O
v2
v2
∆θ
_v
1
you must be able to sketch this.
v2
ac =
r
Centripetal force and acceleration
A particle is said to be in uniform circular motion if it
travels in a circle (or arc) with constant speed v.
 velocity – tangent to the circle
 centripetal acceleration – points toward the center

velocity and centripetal acceleration vectors are
always perpendicular to each other
If centripetal acceleration were suddenly to disappear, there would
be nothing to change the direction of the velocity. The object would
continue to move in a straight line with constant speed. In that case
that object moves “inertially” (because of inertia of the body).
mv 2
Fc = mac =
r
Object undergoing uniform circular motion
is accelerating with centripetal acceleration
ac , so it has a force acting upon it:
it is not separate force – it is simply one of our familiar
forces acting in the role of causing circular motion
Many forces can force an object to move in circular path,
therefore becoming centripetal force:
Moon around the Earth ………
gravity
object sitting on a rotating table
(strawberries sitting on a seat of a turning car)
(a car moving in a circular path) …………
a ball whirling in a circle
at the end of a string
………
friction
tension in the string
a person pressed against the inner wall of a rapidly rotating circular
room in an amusement park …………. normal force from the wall
Ferris–wheel rider passes through
the lowest point of the ride …….
normal force from the seat
and the force of gravity
useful relations: - speed(linear) and angular speed
 period T:
time required for one complete revolution (s)
speed v = distance/time
2πR
v=
T
constant speed
angular speed ω = angle swept/time
ω=
2π
T
Variable
distance
tangential
velocity
tangential
acceleration
Tran.
Variable
angle
swept (rad)
𝑠 angular
𝑣=
𝑡 velocity (rad/s)
angular
∆𝑣
acceleration (rad/s2)
a=
𝑡
 = 2 / T = 2f
v=s/t=(r)/t
=r
s
Rot.
Relationship
s= r

𝜔=
𝛼=
Speed depends on position (r)
but angular speed does not
𝜃
𝑡
∆𝜔
𝑡
v=rω
a=r
A CAR ON A LEVEL (UNBANKED) SURFACE
Think of what would happen if there were no friction. Have you ever found
yourself driving on ice? If you turn steering wheel nothing happens because
there is little or no friction between the ice-covered road and the car's tires. You
need frictional force to turn.
So if you turn the steering wheel you are pushing the ground
in the black direction and the ground is pushing back on
you in the orange direction.
That is the friction force that the road exerts on the
car's tires pointing toward the center of the turn,
responsible for turning the car.
A 1200 kg car rounds the curve on a flat road of radius 45 m. If the
coefficient of static friction between tires and the road is s = 0.82, what
is the greatest speed the car can have in the curve without skidding?
Fn = mg
Ffr = s Fn = s mg
v2
Ffr = ma = mac = m
r
v2
μsg =
r
v = 19 m/s
A CAR ON A BANKED SURFACE…
can a car negotiate a curve without friction (on ice)?
YES, but the curve has to be banked!
If a roadway is banked at the proper angle, a
car can round the corner without any
assistance from friction between the tires and
the road. The horizontal component of the
normal (reaction) force is the net force. This
force is centripetal force to turn the car.
Find the appropriate banking angle for a 900 kg car traveling at 20.5 m/s in a
turn of radius 85.0 m so that NO friction is required.
𝐹𝑛 cos𝜃 = m𝑔
⇒ 𝐹𝑛 =
𝑣2
𝐹𝑛 sin 𝜃 = m
𝑟
𝑚𝑔
cos 𝜃
𝑣2
⇒ tan 𝜃 =
𝑟𝑔
for given speed v and radius r
of the turn we get angle of the bank
independent of mass m
for given θ and r:
v2
θ = arctan
= 26.70
rg
if v < rg tan θ the car would slide down a frictionless banked curve
if v >
rg tan θ the car would slide off the top
HORIZONTAL CIRCULAR MOTION………
centripetal force = tension
cord cannot be horizontal, only for mg << T, ∑ F ≈ T
T will act nearly horizontally and provide the force necessary to give the
ball its centripetal acceleration
v2
T=m
r
 Tmax
2
mvmax
=
r

vmax =
Tmaxr
m
CONICAL PENDULUM…
L
r
T cos θ = mg
2
v
Tsinθ = m
r
v2
tan  
rg
𝑣=
𝑟 = 𝐿 sin 𝜃
𝑔𝑟 tan 𝜃 =
period TP =
𝑔𝐿 sin 𝜃 tan 𝜃
2πr
L cosθ
= 2π
v
g
independent of m
ROLLER COASTER
Top of a Hill
mv2
mg - Fn =
R
mv2
 Fn = mg R
minimum speed required to stay alive is
mathematically given with Fn = 0 (the rider
just barely looses contact with the seat
v T = Rg
𝑚𝑣 2
𝑚𝑣 2
𝐹𝑛 + 𝑚𝑔 =
 𝐹𝑛 =
− 𝑚𝑔
𝑅
𝑅
Fn = 0 danger when feeling weightless
min speed required to stay alive
and not to fall down:
Bottom of a Valley
mv2
Fn - mg =
R
Upside-down at the Top of a loop
v T = Rg
mv2
 Fn = mg +
R
The rider's hands and arms are hard to move. The rider's blood is even hard to
move. Airplane pilots are in this situation as they pull out of a dive. Apparent
weight of 6-7 times one's real weight -- can mean that not enough blood will be
circulated to the brain and a pilot -- or other passenger -- may pass out.
AIRPLANE MAKING A TURN - BANKING
As the plane banks (rolls), the lift vector
begins to have a horizontal component.
Lift force (L) generated by the airplane wings is not
equal to mg anymore, but greater.
𝐿 cos 𝜃 = 𝑚𝑔
𝑣2
𝐿 sin 𝜃 = 𝑚
𝑅
𝑣2
𝑅=
𝑔 tan 𝜃
ARTIFICIAL SATELLITE
G
mmE
r2
mv 2
=
r

v= G
mE
r
given r (we want satellite there)
v precise and independent of m
- we launch satellites above the atmosphere with rockets, then tilt them over and give them
enough horizontal speed that it can orbit the Earth.
- for slightly higher speed but the same distance the orbit will be elliptical
- if the speed is too high, the spacecraft will not be confined by the Earth’s gravity, and will
escape never to return (v ≈ 11.1 km/s ) - if the speed is too low, it will fall back to the Earth
A geostationary satellite (communications satellite)
is placed at an altitude of approximately 35,800
kilometers (22,300 miles) directly over the equator,
that revolves in the same direction the earth
rotates (west to east)
m
v= G E
r
2πr
v=
T

3
r =
GmET 2
4π 2
r = 4.23x107 m = 42,3000 km
rE = 6,380 km
from the Earth center
h = 36,000 km
EXAMPLE: Explain how an object can remain in orbit yet always
be falling.
SOLUTION:
Throw the ball at progressively larger speeds.
In all instances the force of gravity will draw the ball toward the
center of the earth.
When the ball is finally thrown at a great enough speed, the
curvature of the ball’s path will match the curvature
of the earth’s surface.
The ball is effectively falling around
the earth!
PRACTICE: Find the angular speed of the minute hand of a clock, and the
rotation of the earth in one day.
SOLUTION:
The minute hand takes 1 hour to go around one time. Thus
 = 2 / T = 2 / 3600 s = 0.00175 rad s-1.
The earth takes 24 h for each revolution so that
 = 2 / T
= ( 2 / 24 h )( 1 h / 3600 s )
= 0.0000727 rad s-1.
This small angular speed is why we can’t really feel the earth as it spins.
EXAMPLE: The Foucault pendulum is
a heavy pendulum on a very long cable that is set
in oscillation over a round reference table.
Explain how it can be used to tell time.
SOLUTION:
The blue arcs represent the motion
of the pendulum bob relative to the
universe at large.
The the green lines represent the
plane of motion of the pendulum
relative to the building.
 Since the building is rotating with the earth at
 = 0.0000727 rad s-1, each hour the green line
rotates by
 = t = 0.0000727(3600)
= 0.262 rad (360/ 2 rad) = 15.0.
FYI This solution only works when the pendulum is at one of the poles. See the
Wiki for a general solution.
EXAMPLE: Find the apparent weight of
someone standing on an equatorial scale if his
weight is 882 N at the north pole.
SOLUTION: Recall that  = 0.0000727 rad s-1
anywhere on the earth.
The blue arcs represent the lines of latitude.
The white line R represents the earth’s radius.
The yellow line r represents the radius of the
circle a point at a latitude of  follows.
Note that r = R cos , and that at the equator, 
0˚
= 0˚ and at the pole,  = 90˚.
Thus, at the equator, r = R, and at the pole, r = 0.
Furthermore, R = 6400000 m.
Then, at the equator,
ac = r 2 = 6400000 0.00007272 = 0.0338 ms-2.
Then, at the pole,
ac = r 2 = 0 0.00007272 = 0.000 ms-2.
90˚ 
r
R
Make a free-body diagram at the equator…
Scales read the normal force R:
F = ma
R – W = - mac
R = W – mac
Then, R = 882 – ( 882 / 9.8 )  0.0338 = 879 N.
The man has apparently “lost” about 3 N!
W R
ac
 F = kx (k = CONST).
 kx = FC = mv 2/ r implies that as v increases, so does the centripetal
force FC needed to move it in a circle.
 Thus, x increases.
kx = F  k = F / x = 18 / 0.010 = 1800 Nm-1.
FC = kx = 1800( 0.265 – 0.250 ) = 27 N.
FC = v 2/ r  v 2 = r FC = 0.265(27) = 7.155
v = 2.7 ms-1.
Use v = r  ( = CONST).
Use a = r  2 ( = CONST).
At P
r=R
v = R
a = R 2
At Q
r = 2R
v = 2R = 2v
a = 2R 2 = 2a
Objects moving in uniform circular motion feel a centripetal (centerseeking) force.