Dynamics of Uniform Circular Motion
• Uniform Circular Motion is the motion of an
object traveling at a constant (uniform)
speed on a circular path.
The period (T) is the time it takes an object to make
complete circle or revolution.
Because velocity = distance / time
the velocity of an object around a circle is
v = 2πr (the distance around a circle)
T (time it takes for one revolution)
This fan has a radius of .7m and is
rotating at 90 rev/min. How fast is
the end tip of the fan blade going?
This fan has a radius of .7m and is rotating at 90 rev/min.
How fast is the end tip of the fan blade going?
Convert 90 rev / min to the time it takes for
one revolution.
V = 2πr = 2(3.14)(.7m) = m/s
T
1.67 sec
r = .7m
Centripetal Acceleration
• Uniform circular motion emphasizes that velocity vector
is constant – but the direction of this constant velocity is
not constant.
• By definition, acceleration is the change in velocity over
time. The “change in velocity” for an object in UCM is
the change in the velocity vector.
• The acceleration that occurs is toward
the center of the circle. It is the force
that is pulling the object “in”
Centripetal Acceleration
• The centripetal acceleration (the acceleration toward the
center of a circle) of an object depends on the velocity of
the object and the radius of the circle and is expressed
as:
ac = v2
r
How fast would the outer most part
of the space station have to spin in
order to have a centripetal
acceleration of 9.8 m/s2
d = 100m
How fast would the outer most part of the space station
have to spin in order to have a centripetal acceleration of
9.8 m/s2 ?
a=
v2
/r
d = 100m
a = 9.8 m/s
r = 100m/2 = 50m
Solving for v:
v = \/~(a r) = \/~(9.8m/s2)(50m)
= 22.1 m/s
A bobsled goes down a track, moving at a
constant velocity throughout all three turns.
At which turn will there be the greatest
centripetal acceleration?
If the bobsled moves at 35 m/s throughout the course
and turn B has a radius of 20m and turn C is 50m how
many G’s of acceleration is the sledder feeling through
turns B & C ?
A
B
D – All the same
C
Centripetal Force
• The net force required to keep an object of
mass m, moving at a constant velocity v,
on a circular path.
D
F = mv2
r
Note that v2/r = acceleration
A
C
At what point will a person feel the most centripetal
force?
A
B
C
D
E= all the same
A 50 kg person and a 100 kg person ride
the “Enterprise” at Valleyfair. The ride has a
radius of 7 meters and takes 4 seconds per
revolution. How much centripetal force
does each person feel at point A & C?
B
“Gondolas start out hanging from the ride, then
go to a horizontal position by centrifugal force
from the spinning ride, which then lifts to a
vertical position turning riders upside down.”
Centripetal Force Lab Review
A person on a centrifuge with a radius of 6m
spins around once in 2 seconds. The
person is in contact with a scale that is
reading 1,100 lbs. What is the mass of the
person being spun?
How many G’s is this
person experiencing?
1N = .22 lb 2.2lbs = 1kg
A person on a centrifuge with a radius of 6m spins around once in 2
seconds. The person is in contact with a scale that is reading 1,100
lbs.
What is the mass of the person being spun?
How many G’s is this person experiencing?
Start by finding the velocity of the centrifuge: V = 2πr / T
T=2sec
r= 6m
v= 2(3.14)(6) / 2 = 18.8m/s
Then, using Fc = mv2/r, rearrange the equation solving for m
m = r Fc / v2
convert 1,100 lbs to Newtons
1,100 lbs x 1 N = 5000N
1
.22 lbs
m = 6 m(5000 N) / 18,82 = 84.9 kg = mass
ac = v2 / r = 18.82 / 6 = 58.9 m/s2
58.9m/s2 / 9.8m/s2 = 6.0 G’s of acceleration
Is it harder for the man to hold his partner when the partner
is hanging straight down and is stationary or when the
partner is swinging through the straight down position?
It is harder for the man to hold
his partner when the partner
is swinging through.
When stationary, the man only
needs to hold the force of the
partner against gravity .
Ex. 50kg x 9.8m/s2 = 490 N
When the partner swing, the
man needs to hold with the
force from gravity, PLUS the
centripetal force of the
partner because they are
accelerating (changing v)
When a car turns a corner, there is a centripetal force toward
the center of the circle. The force that is being applied to the
center of the circle is coming from static friction between the
car and the ground. If the rubber on the car cannot handle the
centripetal force, the car will slide.
A soapbox racer is making a turn on
asphalt at 20 m/s. The coefficient of
frictionstatic is .9. The radius of the
turn is 40 m. Will the soapbox crash
into the hay? Why or why not?
Static Friction is keeping the car from sliding
Fc = μFN = mv2/r
μ mg = mv2/r
V2 = μmgr / m = μgr
V = sqrt .9(9.8)(40) = 18.8 m/s max speed around the turn
Banked Turns
When a car turns a flat corner, static friction provides the
centripetal force. The need of static friction can be
eliminated by making a bank in the turn.
The centripetal force becomes a vector component of the
normal force. In the case above, it is = FN(cos Ɵ).
If there is a set amount of bank, the can be a perfect speed
to take the turn at without needing static friction. But, if
you go too slow, you may slip, too fast and you may skid
Banked Turns
*this equation indicates that for a
given speed, the centripetal force
needed for a turn of radius r can
be obtained by banking the turn at
an angle, independent of the mass
of the vehicle.
Tan Ɵ = v 2
rg
A racetrack is designed with a 40o bank angle
and a radius of 240 m at the ends. At what
speed is the track designed for cars to make an
ideal turn?
V = sqrt (tan Ɵ)rg = sqrt (tan 40)(240m)(9.8m/s2)
= 44.4 m/s
A racecar enters a banked turn that has a 40o angle. The car is going 34 m/s
into the turn. At what distance should the driver locate his car if he wishes
to take the turn without depending on friction?
140.5m
mg
d = 140.5
sin 50
d = 183.4 m
50o
40o
Using the equation tan
Ɵ = v2 / rg
we want to find the radius of the turn.
Once we have the radius, we’ll us trig
to find d
r = v2 / tan Ɵ(g) = 342 / tan 40 (9.8) = 140.5 m
A racecar enters a banked turn that has a 40o angle. The car is going 34 m/s
into the turn. At what distance should the driver locate his car if he wishes
to take the turn without depending on friction?
FN sin 50 – mg = 0 (no vertical acceleration)
50o
FN
50o
2
F140.5m
c =FN(Cos Ɵ) = mv / r
d = 140.5
sin 50 =
d
mg
183.4 m
40o
Fc = FN cos Ɵ = mv2 / r
Solving for r, r =
r=
mv2
mgcosƟ
(sin50)
if we can find the radius of the turn, we can find d using trig.
mv2
FN cos Ɵ
=
since FN = mg/sin50, we’ll substitute this in
v2 (sin50) =
g cosƟ
342 (sin50) = 885 = 140.5m
9.8(cos 500)
6.3
Satellites in Orbit
There is only one speed
that a satellite can
have if the satellite is
to remain in an orbit
with a fixed radius.
The centripetal force acting on satellites is gravity:
Fc = G m1m2 = mv2
r2
r
v = GME
r
T = 2πr3/2
GME
A rocket is launched to a height of
50 miles above the earth’s
surface. How fast does the
rocket need to be going in
order to stay in orbit?
(1 mile = 1.6 km)
(Earth’s radius = 6,380,000 m)
(Earth’s mass = 5.98 x 1024)
V = GME
r
How many miles per hour is the
earth orbiting the sun?
(93,000,000 miles radius)
(mass of sun = 2 x 1030)
V = GME
r
Synchronous
satellites are
satellites that
orbit the earth
with no
movement
relative to the
earth’s surface.
Artificial Gravity
Artificial gravity can be created by
centripetal force.
If this spaceship has a radius of
3000 m, how fast does the
outside of the spaceship have
to spin in order to simulate a
gravitational acceleration of
9.8m/s2?
A = v2 / r
Vertical Circular Motion
Vertical circular motion occurs when an object moves in a
vertical path. The speed of the object often changes
from moment to moment as a result of gravity.
If an object is in freefall, the
centripetal force is
gravitational force (mg).
If an airplane does a parabolic downward
turn to simulate artificial weightlessness,
what would the radius of the turn have to
be if the airplane flew the turn with a
speed of 100 m/s (220 mi/hr)?
mg
mg