#### Transcript R - IBPhysicsLund

```Topic 2: Mechanics
2.4 Uniform circular motion
2.4.1 Draw a vector diagram to illustrate that
the acceleration of a particle moving with
constant speed in a circle is directed towards
the center of the circle.
2.4.2 Apply the expressions for centripetal
acceleration.
2.4.3 Identify the force producing circular
motion in various situations. Examples include
friction of tires on turn, gravity on
planet/moon, cord tension.
2.4.4 Solve problems involving circular motion.
Topic 2: Mechanics
2.4 Uniform circular motion
On the next pass,
however, Helen failed to
clear the mountains.
r
m
v
What force must be
applied to Helen to
keep her moving in a
circle?
How does it depend
r?
How does it depend
on Helen’s velocity
v?
How does it depend
on Helen’s mass m?
Topic 2: Mechanics
2.4 Uniform circular motion
Draw a vector diagram to illustrate that the
acceleration of a particle moving with constant
speed in a circle is directed towards the center
of the circle.
A particle is said to be in uniform circular
motion if it travels in a circle (or arc) with
constant speed v.
v red
Observe that the velocity vector is always
r blue
tangent to the circle.
y
Note that the magnitude of the
v
velocity vector is NOT changing.
r
Note that the direction of the
velocity vector IS changing.
x
Thus, there is an acceleration,
even though the speed is not
changing!
Topic 2: Mechanics
2.4 Uniform circular motion
Draw a vector diagram to illustrate that the
acceleration of a particle moving with constant
speed in a circle is directed towards the center
of the circle.
In order to find the direction of the
acceleration (a = ∆v/∆t ) we observe two nearby
snapshots of the particle:
v red
The direction of the acceleration is gotten
r blue
from ∆v = v2 – v1 = v2 + (-v1):
v
y 2
v2
v1
-v1
v1
-v1
∆v
∆v
The direction of the acceleration
is toward the center of the circleyou must be able to sketch this.
x
Topic 2: Mechanics
2.4 Uniform circular motion
Apply the expressions for centripetal
acceleration.
Centripetal means center-seeking.
How does centripetal acceleration ac depend on r
and v?
We define the centripetal force Fc:
Fc = mac
centripetal force
Picture yourself as the passenger
in a car that is rounding a left turn:
The sharper the turn, the harder
you and your door push against
each other. (Small r = big Fc.)
 The faster the turn, the harder
you and your door push against
each other. (Big v = big Fc.)
Fc
Topic 2: Mechanics
2.4 Uniform circular motion
Apply the expressions for centripetal
acceleration.
PRACTICE: For each experiment A and B, label the
control, independent, and dependent variables.
manipulated
No change
r
r
r
ac
manipulated
ac
v
ac
no change
ac
v
v
responding F
c
responding
Fc
r
Fc
CONTROL: r
INDEPENDENT: v
A DEPENDENT: Fc and ac
Fc
CONTROL: v
INDEPENDENT: r
B DEPENDENT: Fc and ac
v
Topic 2: Mechanics
2.4 Uniform circular motion
Apply the expressions for centripetal
acceleration.
We know the following things about ac:
If v increases, ac increases.
If r increases, ac decreases.
v
ac =
r
first guess
formula
From dimensional analysis we have
ac =
v
r

m ?
=
s2
m/s
m
?
=
1
s
What can we do to v or r to “fix” the units?
ac = v2/r
ac
v2
=
r
centripetal acceleration

m ? m2/s2 ? m
= 2
=
2
s
m
s
This is the correct one!
Topic 2: Mechanics
2.4 Uniform circular motion
Apply the expressions for centripetal
acceleration.
ac = v2/r
centripetal acceleration
Fc = mac
centripetal force
EXAMPLE: A 730-kg Smart Car negotiates a 30. m
radius turn at 25. m s-1. What is its centripetal
acceleration and force? What force is causing
this acceleration?
SOLUTION:
ac = v2/r = 252/30 = 21 m s-2.
Fc = mac = (730)(21) = 15000 n.
The centripetal force is caused by the friction
force between the tires and the pavement.
Topic 2: Mechanics
2.4 Uniform circular motion
Apply the expressions for centripetal
acceleration.
The period T is the time for one complete
revolution.
One revolution is one circumference C = 2r.
Therefore v = distance / time = 2r/T.
Thus v2 = 42r2/T2 so that
ac = v2/r
= 42r2/T2r
= 42r/T2.
ac = v2/r
ac = 42r/T2
centripetal acceleration
Topic 2: Mechanics
2.4 Uniform circular motion
Apply the expressions for centripetal
acceleration.
ac = 42r/T2
ac = v2/r
centripetal acceleration
EXAMPLE: Albert the 2.50-kg physics cat is being
a radius of 3.00 meters. He takes 5.00 seconds to
complete one fun revolution. What are ac and Fc?
What is the tension in the string?
SOLUTION:
ac = 42r/T2
= 42(3)/(5)2 = 4.74 m s-2.
Albert the
Fc = mac = (2.5)(4.74) = 11.9 n.
Physics
The tension is causing the
Cat
centripetal force, so the tension
is Fc = 11.9 n.
Topic 2: Mechanics
2.4 Uniform circular motion
Solve problems involving circular motion.
EXAMPLE: Dobson is watching a 16-pound bowling
ball being swung around at 50 m/s by Arnold. If
the string is cut at the instant the ball is next
to the ice cream, what will the ball do?
(a)It will follow path A and strike Dobson's ice
cream.
(b)It will fly outward along curve path B.
(c)It will fly tangent to the original circular
path along C.
A
C
B
Topic 2: Mechanics
2.4 Uniform circular motion
Solve problems involving circular motion.
EXAMPLE: A 3.0-kg mass is tied to a string having
a length of 1.5 m, and placed in uniform circular
motion as shown. The string traces out a cone
with a base angle of 60°, with the mass
traveling the base of the cone.
(a) Sketch in the forces acting on the mass.
SOLUTION:
The ONLY two forces acting on the
mass are its weight W and the tension
T
in the string T.
Don’t make the mistake of drawing
Fc into the diagram.
Fc is the resultant of T and W, as the
next questions will illustrate.
W
Topic 2: Mechanics
2.4 Uniform circular motion
Solve problems involving circular motion.
EXAMPLE: A 3.0-kg mass is tied to a string having
a length of 1.5 m, and placed in uniform circular
motion as shown. The string traces out a cone
with a base angle of 60°, with the mass
traveling the base of the cone.
(b) Draw a FBD in the space
provided. Then break down the
tension force in terms of the
unknown tension T.
Ty
T
T
SOLUTION:
Tx = T cos 


= T cos 60° = 0.5T.
Tx
Ty = T sin 
= T sin 60° = 0.87T.
W
W
Topic 2: Mechanics
2.4 Uniform circular motion
Solve problems involving circular motion.
EXAMPLE: A 3.0-kg mass is tied to a string having
a length of 1.5 m, and placed in uniform circular
motion as shown. The string traces out a cone
with a base angle of 60°, with the mass
traveling the base of the cone.
(c) Find the value of the components of the
tension Tx and Ty.
SOLUTION:
Ty
Note that Ty = mg = 4(10) = 40 n.
T
But Ty = 0.87T.

Thus 40 = 0.87T so that T = 46 n.
Tx = 0.5T = 0.5(46) = 23 n.
Tx
W
Topic 2: Mechanics
2.4 Uniform circular motion
Solve problems involving circular motion.
EXAMPLE: A 3.0-kg mass is tied to a string having
a length of 1.5 m, and placed in uniform circular
motion as shown. The string traces out a cone
with a base angle of 60°, with the mass
traveling the base of the cone.
(d) Find the speed of the mass as it
travels in its circular orbit.
SOLUTION:
The center of the UCM is here:
Since r / 1.5 = cos 60°,r = 0.75 m.
Fc = Tx = 23 n.
Thus Fc = mv2/r

23 = 3v2/0.75
r
v = 2.4 m s-1.
Topic 2: Mechanics
2.4 Uniform circular motion
Solve problems involving circular motion.
EXAMPLE: Suppose a 0.500-kg baseball is placed in
a circular orbit around the earth at slightly
higher that the tallest point, Mount Everest
(8850 m). Given that the earth has a radius of RE
= 6400000 m, find the speed of the ball.
SOLUTION:
The ball is traveling in a circle
of radius r = 6408850 m.
Fc is caused by the weight of the
ball so that
Fc = mg = (0.5)(10) = 5 n.
Since Fc = mv2/r we have
5 = (0.5)v2/6408850
v = 8000 m s-1!
Topic 2: Mechanics
2.4 Uniform circular motion
Solve problems involving circular motion.
EXAMPLE: Suppose a 0.500-kg baseball is placed in
a circular orbit around the earth at slightly
higher that the tallest point, Mount Everest
(8850 m). How long will it take the ball to
SOLUTION:
We want to find the period T.
We know that v = 8000 m s-1.
We also know that r = 6408850 m.
Since v = 2r/T we have
T = 2r/v
T = 2(6408850)/8000
T = (5030 s)(1 h / 3600 s)
= 1.40 h.
Topic 2: Mechanics
2.4 Uniform circular motion
Solve problems involving circular motion.
EXAMPLE: Explain how an object can remain in
orbit yet always be falling.
SOLUTION:
Throw the ball at progressively larger
speeds.
In all instances the force of
gravity will draw the ball
toward the center of the earth.
When the ball is finally thrown
at a great enough speed, the
curvature of the ball’s path
will match the curvature of the
earth’s surface.
The ball is effectively falling
around the earth!
Topic 2: Mechanics
2.4 Uniform circular motion
Preparation for future topics.
UCM will be revisited in Topic 4 Oscillations and
waves. We now extend UCM a bit to get ready.
In uniform circular motion the x and y
coordinates of the particle at any instant can be
determined using the following relationships:
x = r cos θ
relationship between rotational
and linear variables
y = r sin θ
We call  the angular position
of the particle.
Take note that  varies with
time.
y
r
θ
x
y
x
Topic 2: Mechanics
2.4 Uniform circular motion
Preparation for future topics.
x = r cos θ
relationship between rotational
and linear variables
y = r sin θ
EXAMPLE: Find the x- and y-components of a
particle whose path follows a circle of radius
3.00 m at the precise instant its angular
position is θ = 135°. Then sketch it into the
grid.
(-2.12 m, 2.12 m ) y
SOLUTION:
r = 3 m and θ = 135°.
x = r cos θ = 3 cos 135° = -2.12 m.
y = r sin θ = 3 sin 135° = +2.12 m.
x
Topic 2: Mechanics
2.4 Uniform circular motion
Preparation for future topics.
x = r cos θ
relationship between rotational
and linear variables
y = r sin θ
EXAMPLE: A particle in uniform circular motion at
a radius of 3.00 m has an x-coordinate of -2.00 m
and a y-coordinate that is negative. Sketch the
particle. Then find the values of y and θ.
SOLUTION:
r2 = x2 + y2 so that
y
32 = 22 + y2 or y = 2.24 m.
x = 2 (adj) and r = 3 (hyp) so
cos θ = adj/hyp = 2/3

θ = cos-1(2/3) = 48.2°.
r
From +x dir, 180° + 48.2° = 228°.
x
Topic 2: Mechanics
2.4 Uniform circular motion
Preparation for future topics.
Just as we defined displacement in linear motion
as ∆x = x2 – x1 we can define angular displacement
in circular motion to be ∆ = 2 – 1.
And just as we defined a velocity as v = ∆x/∆t,
we define angular velocity  = ∆/∆t.
∆ = 2 – 1 ang. displacement angular displacement
and angular velocity
 = ∆/∆t
angular velocity
y
At first glance we would expect
t2
that the units for  = ∆/∆t are
∆
(deg/s) but we would be wrong!
t1
θ2
The next slides will explain.
θ
1
x
Topic 2: Mechanics
2.4 Uniform circular motion
Preparation for future topics.
Some of you may already know the formula for arc
length s.
s = r∆θ
∆θ in radians arc length s
Radians is a more natural way to measure angular
displacement than degrees. Here is how they are
related.
 rad = 180° = 1/2 rev
2 rad = 360° = 1 rev
conversions
Arc length s is measured in meters (m) and tells
us how far we have traveled in an arc.
Angular velocity (or angular speed)  is
Topic 2: Mechanics
2.4 Uniform circular motion
Preparation for future topics.
s = r∆θ
∆θ in radians arc length s
 rad = 180° = 1/2 rev
2 rad = 360° = 1 rev
conversions
EXAMPLE: How far have you traveled if you are on
the outer perimeter of a merry-go-round having a
diameter of 7.00 m and it has turned 135°? Sketch
SOLUTION:
90°
135°
We are looking for arc length s.
45°
∆θ = 135°. But s needs radians.
180°
0°
r = diameter/2 = 7/2 = 3.5 m.
s = r∆θ = (3.5)(2.356) = 8.25 m.
225°
315°
The red arc is 8.25 m long.
270°
Topic 2: Mechanics
2.4 Uniform circular motion
Preparation for future topics.
s = r∆θ
 rad = 180° = 1/2 rev
2 rad = 360° = 1 rev
arc length s
conversions
PRACTICE: How far have you traveled if you are
2.75 m from the center of a merry-go-round having
a diameter of 7.00 m and it has turned 135°?
We are looking for arc length s.
90°
∆θ = 135°. But s needs radians.
135°
45°
But r = 2.75 m so that
0°
s = r∆θ = (2.75)(2.356) = 6.48 m. 180°
FYI
Arc length depends on arc radius.
225°
315°
270°
Topic 2: Mechanics
2.4 Uniform circular motion
Preparation for future topics.
s = r∆θ
∆θ in radians arc length s
Recall that velocity v is just the displacement s
divided by the time. Thus
v = s/∆t
= r∆θ/∆t
= r(∆θ/∆t)
= r
Angular speed  is therefore related to linear
speed v like this:
v = r
relation between v and 
12
Topic 2: Mechanics
2.4 Uniform circular motion
9
3
6
Preparation for future topics.
v = r
relation between v and 
EXAMPLE: The second hand of a clock has a length
of 12.0 cm. What are its angular speed (in rad/s)
and the linear speed, of its tip, and its base?
SOLUTION:
The second hand makes one revolution each 60 s.
Therefore  = 2 rad / 60 s
Angular speed is the same for the base and tip.
v = r = (0.12)(0.105) = 0.013 m s-1 for the tip.
v = r = (0)(0.105) = 0 m s-1 for the base.
Topic 2: Mechanics
2.4 Uniform circular motion
Preparation for future topics.
We define the frequency f of uniform circular
motion as how many revolutions or cycles the
object in UCM executes per unit time.
Since the period T is how many seconds per cycle,
the relationship between T and f is simply
f = 1/T
T = 1/f
relation between T and f
Frequency f is measured in (cycles / second)
which are known as (Hertz).
Since v = 2r/T and v = r, we can derive
2r/T = r
2/T = 
2f = 
 = 2/T
 = 2f relation between , T and f
Topic 2: Mechanics
2.4 Uniform circular motion
Preparation for future topics.
f = 1/T
T = 1/f
relation between T and f
 = 2/T
 = 2f relation between , T and f
EXAMPLE: The wheel of a car rotates once each
0.500 s. Find its period, frequency, and angular
speed.
SOLUTION:
The period T is just 0.500 s.
The frequency f = 1/T = 1/0.5 = 2.00 Hz.
The angular speed can be found in two ways.
METHOD 1:  = 2f = 2(2) = 12.6 rad s-1.
METHOD 2:  = ∆/∆t