Transcript Uniform Circular Motion

Uniform Circular
Motion
Physics 12
Uniform Circular Motion
object is moving at a constant speed but
changing directions
 acceleration occurs due to direction
changes

Circular Motion is Periodic

The motion occurs at a regular time
interval called the period (T). The period
is the time for one complete cycle.

(sinusoidal curve)
Calculating Speed
Constant speed: v = d/t
 In a circle, d = circumference and t is the
period

2r
v
T
Example

A mass on a string is swung in a circle of
radius 1.0m at a speed of 5.0 m/s. What
is the period of the revolution?

1.3 s
Centripetal Acceleration
In order for an object to follow a circular
path, a force needs to be applied in order
to accelerate the object
 Although the magnitude of the velocity
may remain constant, the direction of the
velocity will be constantly changing
 As a result, this force will provide a
centripetal acceleration towards the centre
of the circular path

Centripetal Acceleration

caused by a centre-seeking (centripetal)
force that acts toward the centre of the
circular path
Diagrams…


r1 and r2 represent POSITION at two different time
v1 and v2 represent VELOCITY at two difference
positions, each tangent to the path.
v  v2  v1
v  v2  (v1 )


Now use vector addition**
Note that the resultant vector is direction TOWARDS the
CENTRE of the circle.
How can we calculate centripetal
acceleration (Derivation)? r v

Acceleration on a v-t graph = slope
r
v
d  vt
r  vt
vt v

r
v
2
v
v

r
t
v2
ac 
r
Centripetal Acceleration (derivation
on page 552)
2
v
ac 
r
Where v is the magnitude of the velocity
(or speed) in m/s
 r is the radius of the circle (in m)
 and ac is the centripetal acceleration
(m/s2)

Example

A mass on a string is swung in a circle of
radius 0.75m at 7.0 m/s. What is its rate
of acceleration?

65 m/s2 (towards centre of circle)
Centripetal Force
Circular motion occurs due to the
presence of a centripetal (centre-seeking)
force that pulls toward the centre of the
path.
 This force is UNBALANCED, therefore
causing acceleration (Newton’s 2nd Law)
directed toward the centre of the circle.

Centripetal Force
The centripetal force works to oppose the
object’s inertia.
 NOTE: At one time inertia was mistakenly
referred to as centrifugal force. There is
no such thing as centrifugal force – it is an
illusion caused by inertia.
 www.physicsclassroom.com/mmedia/circm
ot/cf.cfm

Notes: Centripetal Force is not
really a specific force
Any force that causes circular motion is a
centripetal force – it isn’t one type of force.
 Examples: Mass on a string is really
Tension.
 Planetary Orbit is really Fg.
 Car rounding a bend is really Ff.

Centripetal Force


Like the centripetal
acceleration, the
centripetal force is always
directed towards the
centre of the circle
The centripetal force can
be calculated using
Newton’s Second Law of
Motion


F  ma
2
v
ac 
r
2
mv
Fc 
r


F  ma
v2
ac 
r
mv 2
Fc 
r
 2r 
m

 T 
r
 4 2 r 2 

m
2
 T 
r
m 4 2 r
T2
4 2 rm
T2
2
Fc 
Fc 
Fc 
Fc 

NOTE: T is the inverse of f (remember
from waves) so you can also manipulate
the equation that way as well!

T = 1/f
Problem – horizontal circle

A student attempts to spin a rubber
stopper (m=.050kg) in a horizontal circle
with a radius of .75m. If the stopper
completes 2.5 revolutions every second,
determine the following:
 The
centripetal acceleration
 The centripetal force
Remember

You only need to look at the forces that are
affecting the acceleration.

Gravity does not affect a horizontal circular
motion as it acts 90’ to the plane of motion.




The stopper will cover
a distance that is 2.5
times the
circumference of the
circle every second
Determine the
circumference
Multiply by 2.5
Use the distance and
time (one second) to
calculate the speed of
the stopper
C  2r
C  4.7 m
d  2.5(4.7 m)
d  12m
d
v
t
12m
v
1. 0 s
v  12m / s


Use the speed and
radius to determine
the centripetal
acceleration
Then use the
centripetal
acceleration and
mass to determine
the centripetal force
v  12m / s
v2
ac 
r
(12m / s ) 2
ac 
.75m
2
2
ac  1.9 x10 m / s
Fc  mac
Fc  (0.050kg)(1.9 x10 2 m / s 2 )
Fc  9.3 N
Try This
A certain string can withstand a maximum
tension of 55N before breaking. What is
the maximum speed at which we can
safely swing a 1.0kg mass on a piece of
this string 0.50 m long in a horizontal
circle?
 5.2m/s

Road Design

You are responsible to determine the
speed limit for a turn on the highway. The
radius of the turn is 55m and the
coefficient of static friction between the
tires and the road is 0.90.
 Find
the maximum speed at which a vehicle
can safely navigate the turn
 If the road is wet and the coefficient drops to
0.50, how does this change the maximum
speed
Diagrams
The maximum speed at which a
vehicle can safely navigate the turn
2
mv
Fc 
r
2
mv
mg 
r
v2
g 
r
v  gr
v  22m / s
Coefficient drops to 0.50, how does
this change the maximum speed
2
mv
Fc 
r
2
mv
mg 
r
v2
g 
r
v  gr
v  16m / s
Try This
A car of mass 1250kg goes around a
corner of radius 15.0m at 12.0 m/s. Find
the minimum coefficient of friction between
the tires and the road required to safely
navigate the turn at that speed.
 u = 0.979

Page 559

15, 16 ,18
Problem – vertical circle
Vertical Circle
While travelling in a vertical circle, gravity
must be considered in the solution
 While at the top of the circle, gravity acts
towards the centre of the circle and
provides some of the centripetal force
 While at the bottom of the circle, gravity
acts away from the centre of the circle and
the force applied to the object must
overcome both gravity and provide the
centripetal force

Vertical Circle

To determine the minimum velocity required,





Look at the force at the TOP of the circle
use the centripetal force equal to the gravitational force (as any
slower than this and the student would fall to the ground).
This means consider the centripetal force to ONLY be Fg (no
Ftens)
Fc = Fg
To determine the maximum force the student
experiences,


Consider the bottom of the ride when gravity must be overcome.
Fnet = Fg + Fc
Problem – vertical circle

A student is on a carnival ride that spins in
a vertical circle.
 Determine
the minimum speed that the ride
must travel in order to keep the student safe if
the radius of the ride is 3.5m.
 Determine the maximum force the student
experiences during the ride (in terms of
number of times the gravitational force)
Fg  Fc


At the top of the
circle, set the
gravitational force
(weight) equal to the
centripetal force
Solve for velocity
mv
mg 
r
2
v
g
r
2
2
v
9.81m / s 
3.5m
2
2
2
v  34m / s
v  5. 9 m / s
2


At the bottom of the
circle, the net force is
equal to the sum of
the gravitational force
and the centripetal
force
Solve for number of
times the acceleration
due to gravity
Fnet  Fc  Fg
Fnet
Fnet
Fnet
Fnet
mv 2

 mg
r

v2 
 m g  
r 

2


(
5
.
9
m
/
s
)
2

 m 9.81m / s 
3.5m 

 m 9.81m / s 2  9.81m / s 2

Fnet  2mg
Fnet  2 Fg

Practice Problems

Page 559
 15-19