Transcript Document

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
on the Helen’s radius
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
swung around your head by a string harness having
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 = 3.0(9.8) = 29.4 n.
T
But Ty = 0.87T.

Thus 29.4n = 0.87T so that T = 33.8 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
return to Everest?
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!