Transcript Motion in Two Dimensions

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!
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
The net force that produces circular motion is
called Centripetal Force (Fc).
v
a
a
v
a
a
v
v
The acceleration
vector is
considered to be in
a constant direction
because it is always
towards the center
of the path
The velocity is tangent to the path- the object
wants to travel in a straight line
Centripetal Force (Fc) depends upon the mass,
speed squared and radius of the object:
Fc = mv2
r
This is a net force,
therefore:
ac = v 2
r
m = .013 kg
r = .85 m
T = .65 s
Fc = ?
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?
Fc = mv2
r
= (.013 kg)(8.2 m/s)2
.85 m
= 1.0 N
v = ∆d
= 2πr
∆t
T
= 2(3.14)(.85 m)
.65 s
= 8.2 m/s
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?
m = .0150 kg
r = .750 m
Fc = 4.50 N
∆y = 2.00 m
∆x = ?
Horizontally:
∆x = ?
∆x = vx∆t
vx = ?
∆t = ?
Vertically:
vo = 0
a = -9.80 m/s2
∆y = -2.00 m
∆t = ?
∆t =
2∆y
v=
Fcr
a
m
2(-2.00 m)
(4.50 N)(.750 m)
-9.80 m/s2
.0150 kg
= .639 s
= 15.0 m/s
∆x = vx∆t = (15.0 m/s)(.639 s) =
9.58 m
Imagine firing a projectile with a very high speed
from a very high mountaintop:
With an increased speed:
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.
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!
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?
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?
Newton’s 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.
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!
Newton’s Law of Universal Gravitation can be
written in equation form:
F=
Gm1m2
r2
GNewton’s Universal Gravitational Constant
G = 6.67 X 10-11 Nm2/kg2
• This is a force due to gravity- another way of
finding Fw!
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
F
=
m = 1250 kg
r2
d = 7.25 X 106m
r = (7.25 X 106 + 6.37 X 106)m = 1.362 X 107m
F=
(6.67X10-11N•m2/kg2)(1250kg)(5.96X1024kg)
(1.362 X 107 m)2
= 2680 N
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?
Rotary Motion
Angular Kinematics
•Motion about an internal axis
•Describing the motion of a
spinning/rotating object
•Defining movement in terms of
rotations!
All quantities are linear unless specifically stated as
angular/rotational:
All motion begins with understanding displacement:
Displacement (∆d): distance in a direction
Angular Displacement (Ư): rotations in a
particular direction
Angular
Quantity
Linear
revolution(rev)
Displacement
meter(m)
radian (rad)
radian = radius length
1 rev = 2π rad= 6.28 rad
Linear Motion
Angular Motion
v = vo + a∆t
ω = ωo + α∆t
∆d = vo∆t +
1
2
a∆t2
v = √vo2 + 2a∆d
Angular Velocity
Angular Acceleration
∆ø = ωo∆t +
1
2
α∆t2
ω = √ωo2 + 2α∆θ
rev/s or rad/s
rev/s2 or rad/s2
In the rotary system, all motion is based upon how
much an object has turned! (rev or rad--> ø)
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 = ωi∆t + .5α∆t12
∆t1 = 15.0 s
2)(15.0 s)2
=
.5(.127
rad/s
ω0 = 0
ω = 725 rev/min
= 14.3 rad
∆t2 = 10.0 min
ω – ω0 = 75.9 rad/s - 0
α=
= 600 s
∆t2
600 s
725 rev/min(6.28 rad/rev)(1 min/60 s)
= 75.9 rad/s
= .127 rad/s2
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?
ω0 = 25.0 rev/s
ω=0
∆t = 20.0 s
r = 30.0 cm
= .300 m
∆ø = ?
∆ø = ωi∆t + .5α∆t2
=(25.0 rev/s)(20.0 s)
+.5(-1.25rev/s2)(20.0 s)2
a = ω – ω0
∆t
= 0 - 25.0 rev/s
20.0 s
= -1.25 rev/s2
= 250 rev This is how many times the wheel
turned!
How linear distance covered relates to rotational
displacement:
∆d = (# rev)(Circumference)
C = 2π r
∆d = (# rad)(radius length)
∆d = (250 rev)(2)(3.14)(.300 m)
= 471 m
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?