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Transcript
Uniform Circular
Motion
SPH4U – Grade 12 Physics
Unit 3
Circular Motion & Roller coasters

Roller coasters are exciting not just because
they go fast, but because of the accelerations
that are involved in the ride.

Roller coasters are able to produce feelings of
weightlessness and weightiness which can be
fun. (Or terrifying, depending on who you are…)
Circular Motion & Roller coasters

They are also able to accelerate us in one
direction one moment, and then quickly change
course to accelerate us in another direction the
next.

Best of all, roller coasters seem dangerous,
(even when they’re not), which is a thrilling
experience for most people.
Circular Motion & Roller coasters

There are a lot of importance concepts related to
circular motion employed in roller coasters.
Roller coasters employ the physics of circular
motion in their loops, banked turns, and in the
small dips and hills found along otherwise
straight sections of track.

Understanding circular motion is imperative if we
are going to understand roller coasters.
Circular Motion & Roller coasters

The curvy sections of a rollercoaster
can actually be thought of as parts of
circles.
Uniform Circular Motion

Uniform motion is constant speed in a straight line.

Uniform circular motion is motion at a constant speed
in a circular path.

Since the direction (velocity vector) if constantly
changing during uniform circular motion, this type of
motion is accelerated motion (even though the speed
doesn’t change).
Centripetal Acceleration

Centripetal Acceleration
is the instantaneous
acceleration of an object
directed toward the center
of a circular path. All
objects moving in a circle
will experience centripetal
(center-seeking)
acceleration.
Centripetal Acceleration

Recall: Acceleration is defined as the change

in velocity over the change in time.

v
aav 
t

For an object travelling in a circle with uniform
circular motion, the tangential speed might not
be changing, but the direction of the velocity is
changing continuously with time.
Centripetal Acceleration
As you can see from the above diagrams, as v1 and v2 get closer
together, Δv is directed more and more toward the exact center of the
circular path. Since when we have the instantaneous acceleration, v1
and v2 should be about a fraction of a second apart, at that moment
the direction is in fact directed directly toward the center of the circle.
Centripetal Acceleration

The direction of
centripetal acceleration
is always toward the
center of the circular
path.
Centripetal Acceleration

There are three equations that determine
centripetal acceleration:
2
v
ac 
r
4 r
ac  2
T
2
a c  4 rf
2
ac = centripetal acceleration
f = frequency
r = radius of circular path
T = period
v = tangential velocity
2
Centripetal Force

Since the acceleration is directed toward the
center, we know that the net force on the object
must be directed toward the center as well.
Centripetal Force

The net force that causes uniform circular
motion is called the centripetal force.

The centripetal force might actually be made up
of other forces, such as the force of gravity,
tension, the normal force, or some other force.
Centripetal Force

From Newton’s 2nd law, we know that the
following must be true:
F  ma



c
If the centripetal force is the only force acting on
the object, then by Newton’s second law:
Fc = mac
It can be shown that Fc is also equal to the mass
of the object times it’s velocity squared over r:
mv 2
Fc 
r
Horizontal Circular Motion
- a car going around a curve
Horizontal Circular Motion
- a car going around a curve
Horizontal Circular Motion
- a car going around a curve
Horizontal Circular Motion
- a car going around a curve

Race car
drivers will
hug the
inside of
curves in
order to
increase
the radius
of
curvature
which in
turn
reduces
the
centripetal
force they
need from
the friction
of their
tires.

For the car that produced
these skid marks, the
frictional force did not
provide enough centripetal
force to keep the car on the
road. The car was going to
fast for this curve and
therefore required more
centripetal force for the
circular path than friction
could provide.
Practice Problem 1
Vertical Circular Motion
At the top of the loop
FN
Fc = Fnet = FN + Fg
Where FN = normal force
Fg = force of gravity
At the bottom of the loop
Fc = Fnet = FN - Fg
FN
Where FN = normal force
Fg = force of gravity
Minimum Speed of a Loop
What is the minimum speed required
by the object in order to complete the
loop?
F  mac
v2
FN  mg  m
r
At the minimum speed, the Normal
force will drop to zero. So, this
formula will become….
Minimum Speed of a Loop
F  mac
2
v
0  mg  m
r
2
v
g
r
v  gr
Thus the minimum speed required to complete
the loop can be calculated with this equation.
Implications for Roller coasters

Your roller coasters will have curved horizontal
turns in them. You need to consider what speed
your marble is going when designing these
turns. If your marble is going too fast it will not
be able to make it around the curve.

Similarly, if your marble is going over top of a
circular hill, if it has too much energy and speed
it will fly off the track.
Implications for Roller coasters

How would you redesign your coaster to solve
these problems?
Implications for Roller coasters

How would you redesign your coaster to solve
these problems?
make the radius of curvature greater / add
friction to slow the marble down
Implications for Roller coasters

Your roller coasters will have vertical loops in
them.

We know that the loops ought to have an
elliptical shape, rather than a circular one.
Implications for Roller coasters

You need to make sure the marble has enough
kinetic energy to have the minimum speed
required to make the loop.

(Granted.. Our equation is for a circular loop..
but for now you can use it as an estimation. In
fact, if you make your loops like clothoid loops
then you will need less speed).
Homework



Read Sections 3.2-3.3.
Add in extra information from your textbook to
supplement your notes
Answer the following questions:
 Pg. 119 # 1, 6, 7, 9
 Pg. 124 # 1, 2, 3, 4