Uniform Circular Motion

1
Uniform Circular Motion

• the motion of an object traveling in a circular
  path
• an object will not travel in a circular path
  naturally
• an object traveling in a circular path must have
  a net force acting upon it!

2
The object wants to travel in a straight line!
That net force has a very specific name
It cannot because a force (friction, tension,
normal, etc.) constantly pulls/pushes toward the
center every split second
3
The net force that produces circular motion is
called Centripetal Force (Fc).
The acceleration vector is considered to be in a
constant direction because it is always towards
the center of the path
v
v
a
a
a
a
v
v
The velocity is tangent to the path- the object
wants to travel in a straight line
4
Centripetal Force (Fc) depends upon the mass,
speed squared and radius of the object
Fc mv2 r
This is a net force, therefore
ac v2 r
5
A ball with mass 0.480 kg moves at a constant
speed. A centripetal force of 23.0 N acts on the
ball, causing it to move in a circle with radius
1.70 m. What is the speed of the ball?
m .480 kg
v Fc? r m
Fc 23.0 N
r 1.70 m
(23.0 N)(1.70 m) .480 kg
v ?
9.03 m/s
6
A rubber stopper of mass 13.0 g is swung at the
end of a cord .85 m long with a period of .65 s.
What is the tension in the cord?
m .013 kg
r .85 m
T .65 s
Fc mv2 r
v ?d ?t
Fc ?
2pr T
(.013 kg)(8.2 m/s)2 .85 m
2(3.14)(.85 m) .65 s
1.0 N
8.2 m/s
7
An object of mass 15.0 g is spun from a .750 m
string so that the centripetal force acting on
the object is 4.50 N. If the object is spun in a
horizontal circle at a height above the floor of
2.00 m and then released, how far away will it
land?
Horizontally
Vertically
m .0150 kg
vo 0
r .750 m
?x ?
a -9.80 m/s2
Fc 4.50 N
?x vx?t
?y -2.00 m
?y 2.00 m
?t ?
?x ?
vx ?
?t ?
8
2?y a
v
?t
Fcr m
2(-2.00 m) -9.80 m/s2
(4.50 N)(.750 m) .0150 kg
.639 s
15.0 m/s
?x vx?t (15.0 m/s)(.639 s) 9.58 m
9
Imagine firing a projectile with a very high
speed from a very high mountaintop
With an increased speed
10
Faster and faster, farther and farther until
At this point, the weight force is the
centripetal force
mg mv2 r
The speed needed to orbit would depend only
upon the acceleration due to gravity and the
radius of the path.
11
Critical Velocity-- The minimum velocity needed
to maintain a circular path when gravity is a
factor.
?the critical velocity only depends upon the
radius of the path and the acceleration due to
gravity!
v rg
this equation is ONLY for finding the minimum
velocity needed to maintain a circular path!
12
1) A car with a mass of 1250 kg rounds a curve
where the coefficient of friction is measured to
be .185. If the radius of the curve is 195 m,
what speed must the car be traveling?
2) A student spins a 15.0 g rubber stopper above
his head from a .750 m string. The tension in
the string is measured to be 8.00 N. He lets the
stopper go and it lands 12.8 m away. How high is
he spinning the stopper above the floor before he
releases it?
3) A student spins a 13.5 g stopper from a 56.0
cm string and times 12 revolutions as taking 6.56
s. What is the tension in the string?
13
How fast would you have to throw a baseball in
order to get it to come all the way around back
to you? Use 6.37 X 106 m for the radius of the
earth, and, of course, ignore air resistance.
A roller coaster car of mass 725 kg enters a loop
with a circumference of 511 m. What minimum
speed must the car maintain in order to
successfully navigate the loop?
14
Newtons Law of Universal Gravitation

• Newton, in studying gravity, discovered that
  gravitation is not limited to planets, but
  applies to all bodies in the universe.
• The force of attraction between any 2 bodies in
  the universe depends directly upon the masses of
  the bodies.
• This force also depends inversely upon the square
  of the distances between the center of the masses.

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This means if the mass (?) of an object changes,
so does its gravity.
This also means that if the distance between the
centers of the objects changes, the gravity will
change.
Therefore, if you move closer to the center of
the earth, you will experience a greater gravity
and you will weigh more!
Conversely, if you move out into space, you will
weigh less!
16
Newtons Law of Universal Gravitation can be
written in equation form
Gm1m2 r2
F
G?Newtons Universal Gravitational Constant
G 6.67 X 10-11 N?m2/kg2

• This is a force due to gravity- another way of
  finding Fw!

17
What is the weight of a 1250 kg object that is in
the payload bay of the shuttle orbiting at 7.25 X
106 m above the surface of the earth?
Fw ?
Gm1m2 r2
F
m 1250 kg
d 7.25 X 106m
r (7.25 X 106 6.37 X 106)m 1.362 X 107m
(6.67X10-11Nm2/kg2)(1250kg)(5.96X1024kg)
F
(1.362 X 107 m)2
2680 N
18
1) What is the gravitational force of attraction
between a 980.0 N man on earth and the moon,
which has a mass of 7.27 X 1022 kg. The center
of the moon is 3.90 X 109 m away from the surface
of the earth.
2) A satellite on the surface of the earth has a
weight of 12, 800 N. When it is in orbit, its
weight is 3200 N. How far above the surface of
the earth is the satellite orbiting?
3) What is the force of attraction between a 90.0
kg boy and his 65.0 kg girlfriend sitting 1.23 m
away?
19
Rotary MotionAngular Kinematics

• Motion about an internal axis
• Describing the motion of a spinning/rotating
  object
• Defining movement in terms of rotations!

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All quantities are linear unless specifically
stated as angular/rotational
All motion begins with understanding displacement
Displacement (?d) distance in a direction
Angular Displacement (?Ø) rotations in a
particular direction
Angular
Quantity
Linear
revolution(rev) radian (rad)
Displacement
meter(m)
radian radius length
1 rev 2p rad 6.28 rad
21
Linear Motion
Angular Motion
v vo a?t
? ?o a?t
1 2
1 2
?d vo?t
a?t2
?ø ?o?t
a?t2
v vvo2 2a?d
? v?o2 2a??
Angular Velocity
rev/s or rad/s
Angular Acceleration
rev/s2 or rad/s2
In the rotary system, all motion is based upon
how much an object has turned! (rev or rad--gt ø)
22
Find the angular displacement (in radians) after
15.0 s for a wheel that accelerates at a constant
rate from rest to 725 rev/min in 10.0 min.
?ø ?(rad)
?ø ?i?t .5a?t12
?t1 15.0 s
.5(.127 rad/s2)(15.0 s)2
?0 0
? 725 rev/min
14.3 rad
?t2 10.0 min
? ?0 ?t2
75.9 rad/s - 0 600 s
a
600 s
725 rev/min(6.28 rad/rev)(1 min/60 s) 75.9
rad/s
.127 rad/s2
23
A car tire is decelerated from 25.0 rev/s to rest
in 20.0 s. If the radius of the tire is 30.0 cm,
how far does the car move down the road in this
time?
?0 25.0 rev/s
?ø ?i?t .5a?t2
? 0
(25.0 rev/s)(20.0 s)
?t 20.0 s
.5(-1.25rev/s2)(20.0 s)2
r 30.0 cm .300 m
a ? ?0 ?t
0 - 25.0 rev/s 20.0 s
?ø ?
-1.25 rev/s2
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This is how many times the wheel turned!
250 rev
How linear distance covered relates to rotational
displacement
?d ( rev)(Circumference) ?d ( rad)(radius
length)
C 2p r
?d (250 rev)(2)(3.14)(.300 m)
471 m
25
1) A car tire makes 65.0 revolutions in slowing
from 35.0 rad/s to 12.0 rad/s. How long did it
take for the car to slow?
2) A car tire of radius 40.0 cm slows to a stop
from an angular speed of 40.0 rad/s in 25.0 s.
How far did the car move down the road during
this time?
3) A bicycle slows down from 8.40 m/s to rest
over a distance of 115 m. If the diameter of the
wheels are 68.0 cm, how many times must they have
turned over that 115 m?
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1. What is the final angular speed of an object
that is rotating at 25.0 rad/s and is accelerated
at 5.00 rad/s2 until it has turned 250.0 rads? 2.
An object is accelerated from rest to 825
rev/min in 7.50 min. How many radians will it
have turned after the first 25.0 s? 3. A
flywheel is accelerated from rest with an
acceleration of 8.40 rad/s2. How long will it
take to reach 20.0 rev/s? 4. A spinning bicycle
wheel is rotating at 2.00 rev/s and comes to rest
in 5.00 s. How many revolutions does it make
during this time?
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5. A car tire of radius .350m is accelerated
from rest to 35.0 rev/s in 10.0s. How far has
the car moved during this time? (Hint-- find the
angular displacement of the tire in radians and
use this and the radius of the tire to find out
how far the car moved. Remember, radian means
radius length.) 6. A flywheel is slowed to
12.0 rev/s in 10.0s. If the wheel lost speed at
the rate of 2.50 rad/s2, how many revolutions did
the wheel make in these 10.0 s?