Transcript Chapter 5

Chapter 5: Uniform Circular Motion
Chapter Goal: To learn
how to solve problems about
uniform circular motion.
Uniform circular motion:
the motion of an object
traveling at a constant speed
on a circular path.
The speed may be constant,
but the velocity is not.
Uniform Circular Motion
•The velocity vector shows the
direction of motion at any point on the
circle.
•For uniform circular motion the
velocity vectors at each point are
tangent to the circle and are the same
length
•Period (T) – the time it takes to make
one revolution around the circle.
•T is in seconds, v is in m/s.
Example Problem
Your roommate is working on his bicycle. The bicycle
wheel has a radius of 0.3 meters. As he spins the
bicycle wheel, you notice that a stone is stuck in the
tread. It goes by 3 times every second.
a. What is the period of the stone?
b. What is the speed of the stone?
Hint: the stone makes 3 revolutions per second (rps).
The period is how many seconds it takes to make
one revolution.
Example Problem - Answer
Your roommate is working on his bicycle. The bicycle
wheel has a radius of 0.3 meters. As he spins the
bicycle wheel, you notice that a stone is stuck in the
tread. It goes by 3 times every second.
a. What is the period of the stone? T = 1/3 rps = .333s.
Note that period, T is the reciprocal of rps.
b. What is the speed of the stone?
v = 2πr/T = 2π (.3m)/(.333s) = 5.7m/s
Determining the direction of acceleration for
uniform circular motion using vector subtraction
Draw a velocity vector on
top of each even dot,
tangent to the circle.
Starting with 0, take
each even number and
do a graphical vector
subtraction to find the
direction of ∆v and

therefore a . Locate the
∆v vector at the odd
number midway
between the 2 velocity
vectors.
Acceleration of Uniform Circular motion
• If some net force results in
uniform circular motion, the
resulting acceleration always
points towards the center of the
circle.
• This is called centripetal (“centerseeking”) acceleration.
• Is centripetal acceleration equal to
∆v/ ∆t? Yes, but if the magnitude
of v remains constant, how do we
evaluate ∆v ?
• Using trig, it can be shown that
ac = ∆v/ ∆t = v2/r.
Rank in order, from largest to smallest,
the centripetal accelerations (ac)a to (ac)e
of particles a to e.
Determining the direction of acceleration for
uniform circular motion using Newton’s 2nd Law
ΣF = ma
• tells us the net force vector and the acceleration vector
are in the same direction (since mass is a scalar).
• Take a ball with a string attached and roll it around on
the table in a perfect (!) circle at constant speed. How
do we draw a free body diagram of something that is
constantly changing direction?
• The radius becomes the x axis (actually the radial
axis). This means “toward the center of the circle” is
always considered positive.
• The y axis is perpendicular to the plane of motion.
FBD of a ball on a string in uniform circular
motion
• Draw a freebody diagram of the ball on a string in
uniform circular motion.
– In which direction is Fnet?
– In which direction is a?
– How does this change as the ball goes around
the circle?
FBD of a ball on a string in uniform circular
motion
ΣFy = may = 0
y
n
T
x
Fg
a
(ball is not hopping
up or crashing
down)
ΣFx = max = T
Since m is a scalar, T
and a must point in
the same direction,
toward the center
of the circle.
FBD of a ball on a track in uniform circular
motion
• Draw a freebody diagram of the ball on a track in
uniform circular motion.
– In which direction is Fnet?
– In which direction is a?
– How does this change as the ball goes around
the circle?
FBD of a ball on a track in uniform circular
motion
ΣFy = may = 0
y
n1
n2
Fg
a
x
(ball is not hopping
up or crashing
down)
ΣFx = max = n2
Since m is a scalar,
n2 and a must
point in the same
direction, toward
the center of the
circle.
Car turning a “circular” corner at constant speed
Is there a net force on the car
as it negotiates the turn?
Where did it come from?
Car turning a “circular” corner at constant speed
When the driver turns the wheel
the tires turn. To continue
along a straight line, the car
must overcome static friction
and slide. If the static friction
force is less than the
maximum, the tire cannot
slide and so has no choice but
to roll in the direction of the
turn.
fs-max = μs |n|
skid….
blue arrow
represents
direction car
would slide in the
absence of friction
Car turning a “circular” corner at constant speed –
example problem
What is the maximum speed at which a 1500 kg car can make a
turn around a curve of radius 50 m on a level road without
sliding out of the turn (skidding)?
Recall toward the center of the circle is positive, even if it is to
the left in this example
Car turning a “circular” corner at constant speed –
example problem
What is the maximum speed at which a 1500 kg car can make a
turn around a curve of radius 50 m on a level road without
sliding out of the turn (skidding)?
1. draw a free body diagram with appropriate axes and lists
knowns and “finds”
Car turning a “circular” corner at constant speed –
example problem
2. Newton’s 2nd Law in
component form. Start
with y (z) axis since there
is no acceleration:
ΣFy = may = 0
n – FG = 0, therefore n = mg
Car turning a “circular” corner at constant speed –
example problem
2. Newton’s 2nd Law in component form.
Now do forces in the x (r) direction:
3. ΣFx = mac = mv2/r. The only force in the
x direction is the frictional force, therefore:
fs = mv2/r
vmax occurs at fs-max. Therefore:
μs |n| = mv2/r or
μs mg = mv2/r
Car turning a “circular” corner at constant speed –
example problem
Note the mass drops out, as all terms
in the equation include mass.
Solve for vmax :
v = (us rg)1/2 = 22 m/s, about 45
mph
Slow down!
Example – comparing speeds
Car A uses tires for which the coefficient of static
friction is 1.1 on a particular unbanked curve. The
maximum speed at which the car can negotiate this
curve is 25 m/s. Car B uses tires for which the
coefficient of static friction is 0.95 on the same curve.
What is the maximum speed at which car B can
negotiate the curve?
2
v
ac 
r
v  ac r
Use Newton’s Law
to find a in terms of
µs
Example – comparing speeds
Car A uses tires for which the coefficient of static
friction is 1.1 on a particular unbanked curve. The
maximum speed at which the car can negotiate this
curve is 25 m/s. Car B uses tires for which the
coefficient of static friction is 0.95 on the same curve.
What is the maximum speed at which car B can
negotiate the curve?

 Fx  mac
f s max  mac
mg  mac
ac  g
Known
vA = 25m/s
µA = 1.1
µB = 0.95
Find
vB
Example – comparing speeds
Car A uses tires for which the coefficient of static
friction is 1.1 on a particular unbanked curve. The
maximum speed at which the car can negotiate this
curve is 25 m/s. Car B uses tires for which the
coefficient of static friction is 0.95 on the same curve.
What is the maximum speed at which car B can
negotiate the curve?
v  ac r
v  gr
Since the radius is the
same in both cases, and
g is a constant, use ratio
reasoning to calculate
vB
Example – comparing speeds
Car A uses tires for which the coefficient of static
friction is 1.1 on a particular unbanked curve. The
maximum speed at which the car can negotiate this
curve is 25 m/s. Car B uses tires for which the
coefficient of static friction is 0.95 on the same curve.
What is the maximum speed at which car B can
negotiate the curve? Known
v  ac r
vA = 25m/s
µA = 1.1
µB = 0.95
v  gr
Find
vB
Example – comparing speeds
Car A uses tires for which the coefficient of static friction is 1.1
on a particular unbanked curve. The maximum speed at which
the car can negotiate this curve is 25 m/s. Car B uses tires for
which the coefficient of static friction is 0.95 on the same
curve. What is the maximum speed at which car B can
negotiate the curve?
vB   B gr
v A   A gr
Known
vA = 25m/s
µA = 1.1
µB = 0.95
vB   B
Find
vB
vA   A
Example – comparing speeds
Car A uses tires for which the coefficient of static friction is 1.1
on a particular unbanked curve. The maximum speed at which
the car can negotiate this curve is 25 m/s. Car B uses tires for
which the coefficient of static friction is 0.95 on the same
curve. What is the maximum speed at which car B can
negotiate the curve?
vB   B
vA   A
B
vB 
vA
A
vB = 23 m/s
Example – think fuzzy dice
A block is hung by a string from the inside roof of a
van. When the van goes straight ahead at a speed of
28 m/s, the block hangs vertically down. But when
the van maintains this same speed around an
unbanked curve (radius = 150 m), the block swings
toward the outside of the curve. Then the string
makes an angle θ with the vertical. Find θ.
θ
Example problem
A penny is placed 15 cm
from the center of a
turntable. The table has
period of 1.5 s. What is
the mimimum
coefficient of static
friction so that the
penny will rotate with
the turntable and not
slide off?
Example problem - answer
A penny is placed 15 cm
from the center of a
turntable. The table
has period of 1.5 s.
What is the mimimum
coefficient of static
friction so that the
penny will rotate with
the turntable and not
slide off?
Known
r = 15 cm
T = 1.5 s
Find
µsmax
EOC # 23
A ride at a carnival consists of chairs that are swung in a circle by
15.0-m cables attached to a vertical rotating pole, as shown in
the drawing. The total mass of a chair and occupant is 179 kg.
a) Draw a free body diagram for this situation. Note that angle is
from vertical, but the Indian Princess can cope. Oh, and what is
the radius in this problem?
b) Determine the Tension in cable
c) Determine speed of chair
EOC # 23
A ride at a carnival consists of chairs that
are swung in a circle by 15.0-m cables
attached to a vertical rotating pole, as
shown in the drawing. The total mass
of a chair and occupant is 179 kg.
60°
ΣFy = 0, no acceleration in y direction.
ΣFx = mac where ac = mv2/r
Known
L = 15.0 m, r = L sin θ
m = 179 kg
θ = 60° (from vertical)
Find
v, T
A ride at a carnival consists of chairs
that are swung in a circle by 15.0-m
cables attached to a vertical rotating
pole, as shown in the drawing. The
total mass of a chair and occupant is
179 kg.
b. 3508 N
c.
14.8 m/s
60°
Friction in the back seat
A woman is sitting in the middle of the back seat of the
car. The car rounds a 60-m radius curve while
moving at a constant speed of 20 m/s. What is the
minumum coefficient of static friction between her
pants and the car seat so that she doesn’t slide across
the seat?
Draw a fbd and write down Newton’s second law in
component form.
Stone in the tire tread
• A stone has a mass of 6.00 10-3 kg and is
wedged into the tread of an automobile
tire, as the drawing shows. The coefficient
of static friction between the stone and
each side of the tread channel is 0.90.
When the tire surface is rotating at a
maximum speed of 13 m/s, the stone flies
out of the tread. The magnitude FN of the
normal force that each side of the tread
channel exerts on the stone is 1.8 N.
Assume that only static friction supplies
the centripetal force, and determine the
radius r of the tire.
Friction in the back seat
A woman is sitting in the middle
of the back seat of the car. The
car rounds a 60-m radius curve
while moving at a constant
speed of 20 m/s. What is the
minimum coefficient of static
friction between her pants and
the car seat so that she doesn’t
slide across the seat?
r=60m, v = 20m/s
Draw a fbd and write down
Newton’s second law in
component form.
She refused to tell me
her mass, so you have
to solve without it…
Find us
ΣFy = may = 0
ΣFx = mac = mv2/r.
Friction in the back seat
A woman is sitting in the middle
of the back seat of the car. The
car rounds a 60-m radius curve
while moving at a constant
speed of 20 m/s. What is the
minimum coefficient of static
friction between her pants and
the car seat so that she doesn’t
slide across the seat?
Draw a fbd and write down
Newton’s second law in
component form.
r=60m, v = 20m/s
Find us
ΣFy = may = 0
ΣFx = mac = mv2/r.
μs = 0.68