Transcript Circular Motion

SPH4U
Centripetal Acceleration and
Circular Motion
Uniform Circular Motion
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What does it mean?
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How do we describe it?
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What can we learn about it?
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Circular Motion Question
B
A
C
v
Answer: B
A ball is going around in a circle attached to a string. If the string
breaks at the instant shown, which path will the ball follow?
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What is Uniform Circular Motion?
Motion in a circle with:
Constant Radius R
Constant Speed v = |v|
y
v
(x,y)
R
x
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How can we describe UCM?
In general, one coordinate system is as good as any other:
Cartesian:
» (x,y) [position]
y
» (vx ,vy) [velocity]
v
(x,y)
Polar:
R
» (R,) [position]

x
» (vR ,) [velocity]
Polar coordinates are a natural way to describe UCM! We will
be working in the cartesian coordinate system.
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Acceleration in Uniform
Circular Motion
v
v2
R
R
v2
a
R
v1
v2
v1
Centripetal acceleration
Centripetal force: Fc = mv2/R
aave= v / t
Acceleration inward
Acceleration is due to change in direction, not speed. Since
turns “toward” center, acceleration is toward the center.
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Definitions
Uniform Circular Motion: occurs when an object has
constant speed and constant radius
Centripetal Acceleration (or radial acceleration, ac): the
instantaneous acceleration towards the centre of the circle
Centrifugal Force: fictitious force that pushes away from
the centre of a circle in a rotating frame of reference
(which is noninertial)
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Equations to know
2
v
ac 
R
2
4 R
ac 
2
T
2
2
ac  4 Rf
1
f 
T
These are the
equations for
centripetal
acceleration (which we
will derive this class)
T – period (not to be
confused with tension)
f - frequency
R – radius
v – speed
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Dynamics of Uniform Circular Motion
Consider the centripetal acceleration aR
of a rotating mass:
The magnitude is constant.
The direction is perpendicular to the
velocity and inward.
The direction is continually changing.
Since aR is nonzero, according to
Newton’s 2nd Law, there must be a force
involved.
2
v
FR  maR  m
R
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Consider a ball on a string:
There must be a net force
force in the radial direction
for it to move in a circle.
Other wise it would just fly
out along a straight line, with
unchanged velocity as stated
by Newton’s 1st Law
Don’t confuse the outward
force on your hand
(exerted by the ball via
the string) with the inward
force on the ball (exerted
by your hand via the
string).
That confusion leads to
the mis-statement that
there is a “centrifugal” (or
center-fleeing) force on
the ball. That’s not the
case at all!
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Deriving centripetal
acceleration equations
A particle moves from
position r1 to r2 in time Δt
Because v is always
perpendicular to r, the
angle between v1 and v2 is
also θ.
Start with equation for
magnitude of instantaneous
acceleration
a  lim
t 0
v2
v1
θ
v2
r2
θ
r1
v1
v
t
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Deriving centripetal
acceleration equations
For Δr, find r2 - r1  For Δv, find v2 - v1
Notice:
r1  r2  R
Δr
-r1
θ
r2
θ
r1
v1  v2  v
-v1
v2
θ θ v1
Δv
Similar triangles!
The ratios of sides are the same for both triangles!
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Deriving centripetal
acceleration equations
v
v

 v 
r
Ratios of similar
Triangles are the same
R
v  r
R
 Sub this into our original
acceleration equation.
a  lim
t 0
v
t
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Deriving centripetal
acceleration equations
a  lim
t 0
v
t
 v  r
 a  lim 
t 0 
R

 1
  
  t 

 r 
v
 

 a  lim    
t 0 R
   t 


 r 
v

 a    lim 
 R  t 0  t 


Okay, everything is
straightforward now
except this thing.
But hey! That’s just the
magnitude of the
instantaneous velocity!
(also called speed,
which is constant for
uniform circular motion)
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Still deriving centripetal
acceleration
v
 a  v
R
2
v
 ac 
R
Finally…
a much nicer equation.
But what if we don’t
know speed v?
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Almost done now
The period (T) is the time it take to make a full rotation
dist. 2 R
v

time
T
v2
ac 
R
 2 R   1 
 ac  
  
 T  R
2
4 2 R
 ac 
T2
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And here’s the last equation
Frequency is the number of rotations in a given time. It is
often measured in hertz (Hz)
If the particle has a frequency of 100Hz, then it makes 100
rotations every second
1
1
T
or f 
f
T
 ac  4 rf
2
2
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Consider the following situation: You are driving a car with constant speed
around a horizontal circular track. On a piece of paper, draw a Free Body
Diagram (FBD) for the car. How many forces are acting on the car?
A) 1
B) 2
C) 3
D) 4
E) 5
FN
correct
f
R
W
“Fn = Normal Force, W = Weight, the force
of gravity, f = centripetal force.”
F = ma = mv2/R
“Gravity, Normal, Friction”
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Consider the following situation: You are driving a car with constant speed
around a horizontal circular track. On a piece of paper, draw a Free Body
Diagram (FBD) for the car. The net force on the car is
FN
f
A. Zero
B. Pointing radially inward
C. Pointing radially outward
R
W
correct
F = ma = mv2/R
If there was no inward force then the car would
continue in a straight line.
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Suppose you are driving through a valley whose bottom has a
circular shape. If your mass is m, what is the magnitude of the
normal force FN exerted on you by the car seat as you drive
past the bottom of the hill
A. FN < mg
B. FN = mg
C. FN > mg
a=v2/R
R
correct
FN
v
F = ma
FN - mg = mv2/R
mg
FN = mg + mv2/R
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Roller Coaster Example
What is the minimum speed you must have at the
top of a 20 meter diameter roller coaster loop,
to keep the wheels on the track.
Y Direction: F = ma
-N – mg = -m a
N
2
mg
-N – mg = -m v /R
Let N = 0, just touching
-mg = -m v2/R
g = v2 / R
v = (gR)
v = (9.8)(10) = 9.9 m/s
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Merry-Go-Round
Bonnie sits on the outer rim of a merry-go-round with
radius 3 meters, and Klyde sits midway between the
center and the rim. The merry-go-round makes one
complete revolution every two seconds.
Klyde
Bonnie
Klyde’s speed is:
(a) the same as Bonnie’s
(b) twice Bonnie’s
(c) half Bonnie’s
Bonnie travels 2  R in 2 seconds
1
VKlyde  VBonnie
2
vB = 2  R / 2 = 9.42 m/s
Klyde travels 2  (R/2) in 2 seconds vK = 2  (R/2) / 2 = 4.71 m/s
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Merry-Go-Round ACT II
Bonnie sits on the outer rim of a merry-go-round, and Klyde
sits midway between the center and the rim. The merry-goround makes one complete revolution every two seconds.
Klyde’s angular velocity is:
Klyde
Bonnie
(a) the same as Bonnie’s
(b) twice Bonnie’s
(c) half Bonnie’s
The angular velocity  of any point on a solid object
rotating about a fixed axis is the same.
Both Bonnie & Klyde go around once (2 radians) every
two seconds.
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Problem: Motion in a Circle
A boy ties a rock of mass m to the end of a string and
twirls it in the vertical plane. The distance from his hand
to the rock is R. The speed of the rock at the top of its
trajectory is v.
What is the tension T in the string at the top of the
rock’s trajectory?
v
T
R
Motion in a Circle...
Draw a Free Body Diagram (pick y-direction to be
down):
y
We will use FNET = ma (surprise)
First find FNET in y direction:
mg
FNET = mg +T
T
Motion in a Circle...
FNET = mg +T
v
Acceleration in y direction:
ma = mv2 / R
F = ma
mg
y
T
mg + T = mv2 / R
R
T = mv2 / R - mg
Motion in a Circle...
What is the minimum speed of the mass at the top of the
trajectory such that the string does not go limp?
i.e. find v such that T = 0.
v
mv2 / R = mg + T
v2 / R = g
mg
T= 0
v  Rg
Notice that this does
not depend on m.
R
Lecture 6, Act 3
Motion in a Circle
A skier of mass m goes over a mogul having a radius of
curvature R. How fast can she go without leaving the
ground?
v
mg
N
R
Rg
(a) v = mRg (b) v =
m
(c) v = Rg
Lecture 6, Act 3
Solution
mv2 / R = mg - N
For N = 0:
v  Rg
v
mg
N
R
Example: Force on a Revolving Ball
As shown in the figure, a ball of mass 0.150 kg
fixed to a string is rotating with a period of
T=0.500s and at a radius of 0.600 m.
What is the force the person holding the ball
must exert on the string?
+x
As usual we start with the
free-body diagram.
Note there are two forces
gravity or the weight, mg
tensional force exerted
by the string, FT
We’ll make the
approximation that the
ball’s mass is small enough
that the rotation remains
horizontal, f=0. (This is that
judgment aspect that’s
often required in physics.)
Looking at just the x
component then we have
a pretty simple result:
FX  ma X
v2
FX  m
r
(2 r / T ) 2
FX  m
r
4 2 mr
F
T2
4 2 (0.15kg )(0.60m)

(0.50 s) 2
 14 N
Example : A Vertically Revolving Ball
Now lets switch the
orientation of the ball to
the vertical and lengthen
the string to 1.10 m.
For circular motion
(constant speed and
radius), what’s the speed
of the ball at the top?
What’s the tension at
the bottom if the ball is
moving twice that speed?
+x
So to the free-body diagram, at the top, at point
A, there are two forces:
tensional force exerted by the string, FTA
gravity or the weight, mg
2
v
In the x direction:
A
FR  maR  m
r
Let’s talk about the
FR  FTA  mg
dependencies of this
equation.
vA2
m
 FTA  mg
Since mg is constant,
r
the tension will be larger
should vA increase.
This seems intuitive.
vA2
m
 0  mg
Now the ball will fall if
r
the tension vanishes
vA  gr
or if FTA is zero
 9.80m / s 2 1.10m
 3.28m / s
At point B there are also two forces
but both acting in opposite directions.
Using the same coordinate system.
vB 2
FR  maR  m
r
FR  FTB  mg
vB 2
m
 FTB  mg
r
vB 2
FTB  m(
 g)
r
Now since we were given vB  6.56m / s,
(6.56m / s) 2
FTB  0.150kg (
 9.80m / s 2 )
1.10m
FTB  7.34 N
Note that the tension still provides the
radial acceleration but now must also
be larger than maR to compensate for
gravity.
+x
Forces on a Swinging Weight
Part 1
A mass is hanging off of two ropes, one vertical and one at an
angle θ of 30°. The mass is 20 kg. What is the tension in the
angled rope?
30
Gravity is the force pulling down (vertical).
Therefore the matching force pulling up is the
tension in the vertical rope. The angled rope
will have zero tension (it plays no role in
holding up the mass).
Forces on a Swinging Weight
Part 2
A mass is hanging off of two ropes, one vertical and one at an
angle θ of 30°. The mass is 20 kg. What is the tension in the
angled rope the instant the vertical rope is cut?
Gravity is the force
pulling down (vertical).
Therefore the
matching force pulling
up is the tension in the
angled rope.
FT  FG cos  30 
 mg cos  30 

N2 
  20kg   9.8
 cos  30 
kg 

 169.7 N
30
30
FG
FG cos  30
Designing Your Highways!
Turns out this stuff is actually
useful for
civil engineering such as road
design
A NASCAR track
Let’s consider a car taking a
curve, by now it’s pretty clear
there must be a centripetal
forces present to keep the car
on the curve or, more
precisely, in uniform circular
motion.
This force actually comes from
the friction between the
wheels of the car and the
road.
Don’t be misled by the outward
force against the door you feel
as a passenger, that’s the door
pushing you inward to keep
YOU on track!
Example: Analysis of a Skid
The setup: a 1000kg
car negotiates a
curve of radius 50m
at 14 m/s.
The problem:
If the pavement is
dry and ms=0.60, will
the car make the
turn?
How about, if the
pavement is icy and
ms=0.25?
First off, in order to maintain
uniform circular motion the
centripetal force must be:
+x
+y
Looking at the car head-on
the free-body diagram shows
three forces, gravity, the
normal force, and friction.
We see only one force offers
the inward acceleration
needed to maintain circular
motion - friction.
v2
FR  maR  m
r
(14m / s ) 2
 1000kg 
50m
 3900 N
To find the frictional force
we start with the normal
force, from Newton’s second
law:
Fy  0  FN  mg
FN  mg  1000kg  9.8m / s 2
 9800 N
Back to the analysis
of a skid.
Since v=0 at contact,
if a car is holding the
road, we can use the
static coefficient of
friction.
If it’s sliding, we use
the kinetic
coefficient of
friction.
Remember, we need
3900N to stay in
uniform circular
motion.
Static friction force
first:
Ffr (max)  m s FN
 0.60  9800 N  5900 N
Holds the road!
Now kinetic,
Ffr  m K FN
 0.25  9800 N  2500 N
Off it goes!
The Theory of Banked Curves
The Indy picture shows that
the race cars (and street
cars for that matter) require
some help negotiating curves.
By banking a curve, the car’s
own weight, through a
component of the normal
force, can be used to provide
the centripetal force needed
to stay on the road.
In fact for a given angle
there is a maximum speed for
which no friction is required
at all.
From the figure this is given
by
2
v
FN sin   m
r
Example: Banking Angle
Problem: For a car traveling at speed
v around a curve of radius r, what is
the banking angle  for which no
friction is required? What is the
angle for a 50km/hr (14m/s) off ramp
with radius 50m?
To the free-body diagram! Note that
we’ve picked an unusual coordinate
system. Not down the inclined plane,
but aligned with the radial direction.
That’s because we want to determine
the component of any force or forces
that may act as a centripetal force.
We are ignoring friction so the only
two forces to consider are the weight
mg and the normal force FN . As can
be seen only the normal force has an
inward component.
As we discussed earlier in the
horizontal or + x direction,
Newton’s 2nd law leads to:
FN sin   m
2
v
r
In the vertical direction we
have:
Fy  FN cos  mg
FN cos  mg  0
Since the acceleration in this
direction is zero, solving for FN
mg
FN 
cos 
Note that the normal force is
greater than the weight.
This last result can be
substituted into the first:
mg
v2
sin   m
cos 
r
v2
mg tan   m
r
v2
g tan  
r
v2
tan  
gr
For v=14m/s and r= 50m
v2
(14m / s ) 2
tan  

 0.40
2
gr 9.8m / s  50m
  22o
Summary of Concepts
Uniform Circular Motion
Speed is constant
Direction is changing
Acceleration toward center a = v2 / r
Newton’s Second Law F = ma
Uniform Circular Acceleration Kinematics
Similar to linear!
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