Transcript a R

Circular Motion and Gravitation
Centripetal Acceleration – acceleration towards
the center of a circle.
– a.k.a. Radial Acceleration (aR)
v
Ball rolling in a
straight line (inertia)
v
aR
aR
Same ball, hooked to
a string
v
aR =
2
v
r
If you are on a carousel
at constant speed, are
you experiencing
acceleration?
If you twirl a yo-yo and let go of the string,
what way will it fly?
Period and Frequency
Period (T)
– Time for one complete (360o) revolution
– seconds
Frequency
– Number of revolutions per second
– rev/s or Hertz (Hz)
T=1
f
Formulas
aR = v2
r
v = 2pr
T
T=1
f
v = 2prf
Centripetal Acceleration: Ex. 1
A 150-g ball is twirled at the end of a 0.600 m
string. It makes 2.00 revolutions per
second. Find the period, velocity, and
acceleration.
(0.500 s, 7.54 m/s, 94.8 m/s2)
Centripetal Acceleration: Ex. 2
The moon has a radius with the earth of about
384,000 km and a period of 27.3 days.
a. Calculate the acceleration of the moon
toward the earth.
b. Convert the answer to g’s.
(Ans: 2.72 X 10-3 m/s2, 2.78 X 10-4 g )
Jupiter is about 778 X 106 km from the sun. It
takes 4332.6 days to orbit the sun.
a) Calculate the circumference of Jupiter’s . orbit.
(4.89 X 1012 m)
b) Calculate Jupiter’s period in seconds. (3.74 X
108 s)
c) Calculate Jupiter’s orbital speed. (1.30 X 104
m/s)
d) Calculate Jupiter’s centripetal acceleration
towards the sun. (2.18 X 10-4 m/s2)
Centripetal Force
Centripetal Force – “center seeking” force that
pulls an object in a circular path.
–
–
–
–
Yo-yo
Planets
Merry-go-round
Car rounding a curve?
“Centrifugal Force?”
• Doesn’t exist
• “apparent outward force”
• When you let the string go, the ball will
continue in a straight line path. No new
acceleration involved.
• Water in swinging cup example
Direction water
wants to go
Centripetal
Force of
string
Circular Motion
SF = maR = mv2
r
A 0.150 kg yo-yo is attached to a 0.600 m string
and twirled at 2 revolutions per second.
a) Calculate the circumference of the circle (3.77 m)
b) Calculate the linear speed (7.54 m/s)
c) Calculate the centripetal force in the string (14.2
N)
An electron orbits the nucleus with a radius of 0.5 X
10-10 m. The electron has a mass of 9 X 10-31 kg
and a speed of 2.3 X 106 m/s.
a. Calculate the centripetal force on the electron.
(9.52 X 10-8 N)
b. Calculate the frequency of an electron. (7.32 X
1015 Hz)
c. Convert the velocity to miles/s. (1429 miles/s)
d. What provides the centripetal force?
Circular Motion: Example 2
Thor’s Hammer
(mjolnir) has a mass
of 10 kg and the
handle and loop have
a length of 50 cm. If
he can swing the
hammer at a speed of
3 m/s, what force is
exerted on Thor’s
hands?
(Ans: 180 N)
Can Thor swing his hammer so that it is
perfectly parallel to the ground?
FH
What angle will the hammer take with the
horizontal?
Mass = 10 kg
r = 50 cm
q
mg
v = 3 m/s
What angle will the hammer take with the
horizontal?
Let’s resolve the FR vector into it’s components:
FHx = Fhcosq
q
FHx = mv2/r
FHcosq = mv2/r
mg
SFy = 0 (the hammer is not rising or falling)
0 = FHsinq – mg
Two Equations: Two Unknowns
FHcosq = mv2/r
FHsinq = mg
FH = mv2
rcosq
mv2 sinq = mg
r cosq
sinq = gr
cosq
v2
tan q = gr
v2
q = 28.6o
A 0.0050 kg walnut is swung in a radius of 50.0
cm. The walnut makes 2.50 revolutions per
second.
a) Calculate the linear speed of the walnut. (7.85
m/s)
b) Draw a free-body diagram of the walnut.
c) Calculate the centripetal force needed to keep it
in a circle. (0.616 N)
d) Calculate the force of tension and the angle the
string makes with the horizontal. (0.618 N,
4.5o)
Circular Motion: Example 3
A 0.150 kg ball is swung on a 1.10-m string in a
vertical circle. What minimum speed must it
have at the top of the circle to keep moving in
a circle?
mg
FT
At the top of the circle, both the
weight and the tension in the string
contribute to the centripetal force
SF = FT + mg
v = 3.28 m/s
What is the tension in the cord at the bottom
of the arc if the ball moves at the minimum
speed? (v = 3.28 m/s)
FT = 2.94 N
FT
mg
A rollercoaster vertical loop has a radius of
20.0 m. Assume the coaster train has a
mass of 3,000 kg.
a) Calculate the minimum speed the coaster
needs to have to make the loop. (14.0 m/s)
b) Calculate the normal force the tracks
provide to the train at the bottom of the
curve if the train is travelling at 25.0 m/s.
(123,150 N)
c) Calculate the normal force the tracks
provide at the top of the curve if the train is
travelling at 25.0 m/s. (64, 350 N)
The ferris wheel at Knoebels has a radius of
16.8 m and travels at a speed of 3.52 m/s.
a) Calculate the frequency and period (0.033 Hz,
30 s)
b) Calculate the normal force that the seat
provides to a 56.0 kg rider at the top. (507 N)
c) Calculate the normal force that the seat
provides to a 56.0 kg rider at the bottom. (590
N)
A car travels over a round hill (radius = 50.0 m).
a) Calculate the maximum speed at which the car
can take the hill. (22.0 m/s)
b) Calculate the normal force on a 1000.0 kg car if
it is travelling over the hill at 10.0 m/s. (7.80 X
103 N)
Car Rounding a Turn
• Friction provides centripetal force
• Use (ms). Wheels are turning, not sliding, across
the surface
• Wheel lock = kinetic friction takes over. mk is
always less than ms, so the car is much more
likely to skid.
Car Rounding a Turn: Example 1
A 1000-kg car rounds a curve (r=50 m) at a speed of
14 m/s.
FN
Ffr = Fc
mg
a) Calculate the centripetal force
needed to keep the car on the
road
b) Calculate if the car will skid if
the road is dry and ms = 0.60
c) Calculate if the car will skid if
the road is icy and ms = 0.25
Car Rounding a Turn: Example 2
A 15,000-kg truck can safely round a 150 m curve
at a speed of 20 m/s.
a) Calculate the centripetal force needed to keep the car on the
road (40,000 N )
b) Calculate the coefficient of static friction (0.27)
c) Calculate the maximum speed a 1000 kg Cube car can take
the turn. (20 m/s)
The Rotor
The Rotor at an amusement park has a radius of
7.0 m and makes 30 rev/min.
a) Calculate the speed of the rotor. ( 22.0 m/s)
b) Draw a free body diagram of a person in the
rotor. What causes the FN?
c) Calculate the coefficient of static friction
between the person and the wall. (0.14)
Banked Curves
• Banked to reduce the reliance on friction
• Part of the Normal Force now contributes to
the centripetal force
FC = Ffr + FNsinq
(ideally, we bank the
road so that no friction
is required: Ffr = 0)
Banked Curves: Example 1
A 1000-kg car rounds a 50 m radius turn at 14
m/s. What angle should the road be banked
so that no friction is required?
FN
q mg
q
Now we will simply work with the Normal
Force to find the component that points to
the center of the circle
First consider the y forces.
SFy = FNcosq - mg
Since the car does not
move up or down:
FNcosq
FN
q
SFy = 0
FNsinq
0 = FNcosq – mg
FNcosq = mg
q mg
q
FN = mg/cosq
mv2 = FNsinq
r
mv2 = mgsinq
r
cosq
v2 = gtanq
r
v2 = gtanq
r
v2 = tanq
gr
tan q =
q = 22o
(14 m/s)2
(50 m)(9.8m/s2)
=
0.40
A 2,000-kg Nascar car rounds a 300 m radius
turn at 200 miles/hr.
a. Convert the speed to m/s. (89.4 m/s)
b. What angle should the road be banked so that
no friction is required? (70o)
c. Suppose a track is only banked at 35.0o,
calculate the maximum speed that a car can
take the turn. (45.3 m/s, 101 mph)
d. Looking at the formula for banking angle, how
could a track designer decrease that angle?
Weightlessness
• True weightlessness exists only very far
from planets
• “Apparent weightlessness” can be achieved
on earth
Elevator at Constant Velocity
a= 0
SF = FN – mg
0 = FN – mg
FN = mg
Suppose Chewbacca has a mass
of 102 kg:
FN = mg = (102kg)(9.8m/s2)
FN = 1000 N
FN
mg
a is zero
Elevator Accelerating Upward
a = 4.9 m/s2
SF = FN – mg
ma = FN – mg
FN = ma + mg
FN = m(a + g)
FN=(102kg)(4.9m/s2+9.8 m/s2)
FN = 1500 N
FN
mg
a is upward
Elevator Accelerating Downward
a = 4.9 m/s2
SF = mg - FN
ma = mg - FN
FN = mg - ma
FN = m(g - a)
FN=(102kg)(9.8m/s2 – 4.9 m/s2)
FN = 500 N
FN
mg
a is down
At what acceleration will he feel weightless?
FN = 0
SF = mg - FN
ma = mg -FN
ma = mg - 0
ma = mg
a = 9.8 m/s2
Apparent weightlessness
occurs if a > g
FN
mg
Calculate the apparent weight of a 56.0 kg man in
an elevator if the elevator is:
a) Accelerating upwards at 2.00 m/s2. (661 N)
b) Accelerating downwards at 2.00 m/s2 (436 N)
c) Accelerating downwards at 9.80 m/s2 (0 N)
d) Accelerating sideways at 9.80 m/s2. (549 N)
Other examples of apparent
weightlessness
Even when you are running, you fell weightless
between strides.
Why don’t satellites fall back onto the earth?
• Speed
• They are “falling”
due to the pull of
gravity
• Can feel
“weightless” (just
like in the elevator)
Gravitation
Is gravity caused by the earth’s rotation?
Will a man down here fall
off if the earth stops
rotating?
Gravitation
Newton’s Law of Universal Gravitation
1. Every object in the universe is attracted to every
other object. (based on mass)
2. The force drops off with the distance squared.
(As distance increases, the force of gravity drops
very quickly)
Gravitation: Formula
F= Gm1m2
r2
G = 6.67 X 10-11 N-m2/kg2
m1 = mass of one object
m2 = mass of second object
r = distance from center of objects
Cavendish proves the law in
1798
Gravitation: The Solar System
Everything in the solar system pulls on everything
else.
Sun pulls on Earth
All the other planets
also pull on the
Earth
Some comets/meteors are actually from outside our solar system and
were captured by our sun’s gravity.
Gravitation: Example 1
What is the force of gravity between two 60.0
kg (132 lbs) people who standing 2.00 m
apart?
F= Gm1m2 = (6.67 X 10-11 N-m2/kg2)(60kg)(60kg)
r2
(2.00m)2
F = 6.00 X 10-8 N
Such a force is so small that it is almost impossible to
measure.
Gravitation: Example 2
What is the force of gravity between a 60 kg person
and the earth? Assume the earth has a mass of
5.98 X 1024 kg and a radius of 6,400,000 m
(4,000 miles).
F= Gm1m2 = (6.67 X 10-11 N m2/kg2)(60kg)(5.98 X 1024 kg)
r2
(6,400,00 m)2
F = 584 N
Gravitation: Example 3
A 2000-kg satellite orbits the earth at an altitude of
6380 km (the radius of the earth)above the
earth’s surface. What is the force of gravity on
the satellite?
F= Gm1m2 = (6.67 X 10-11 N m2/kg2)(2000kg)(5.98 X 1024 kg)
r2
(6,380,00 m + 6,380,00 m)2
F = 4900 N
Gravitation: Example 3
What is the net force on the moon when it is at a
right angle with the sun and the earth?
Relevant Data:
MM = 7.35 X 1022 kg
ME = 5.98 X 1024 kg
MS = 1.99 X 1030 kg
rMS = 1.50 X 1011 m
rME = 3.84 X 108 m
Calculate each force separately:
FME
Earth
1020 N
FME = 1.99 X
FMS = 4.34 X 1020 N
FR2 = FME2 + FMS2
q
FMS
FR = 4.77 X 1020 N
tan q = opp = FME
adj
FMS
q = 24.6o
Sun
FR
Calculating the Mass of the Earth
Calculate the mass of the earth knowing that it has a
radius of 6.38 X 106 m. Start using the weight
formula.
Calculating “g”
g = GmE
rE 2
g = (6.67 X 10-11 N-m2/kg2)(5.98 X 1024 kg)
(6.38 X 106 m)2
g = 9.80 m/s2
Calculating “g”: Example 1
Calculate the value of g at the top of Mt.
Everest, 8848 m above the earth’s surface.
g = GmE
r2
g = (6.67 X 10-11 N-m2/kg2)(5.98 X 1024 kg)
(6.38 X 106 m + 8848 m)2
g = 9.77 m/s2
g varies with:
• Altitude
• Location
– Earth is not a perfect sphere
– Different mineral deposits can change density
– “salt domes” are low density salt regions near
petroleum deposits
Objects weigh about 1/6 their weight on Earth
on the Moon. Calculate the mass of the
moon, knowing that the radius of the moon
is 1734 km.
An object weighs 200 N on earth.
a) Calculate the acceleration of gravity on
Mars (3.71 m/s2)
b) Calculate its weight on Mars (75.5 N)
Re = 6370 km
Rm = 3440 km
Me = 5.98 X 1024 kg
Mm = 0.11Me
Three 5.00 kg bowling balls are placed at the
corners of an equilateral triangle whose sides are
1.50 m long. Calculate the magnitude and
direction of the gravitational force on the top
ball.
Four 5.00 kg bowling balls are placed at the
corners of an square whose sides are 1.50 m
long. Calculate the magnitude and direction of
the gravitational force on the lower left ball.
(1.42 X 10-9 N, 45o)
A geosynchronous satellite has a period of one
day. The radius of the Earth is 6380 km and
the mass of the Earth is 5.98 X 1024 kg.
a) Convert the period to seconds (8.64 X 104 s)
b) Calculate the height above the earth that a
geosynchronous satellite must orbit. (Hint: use
mv2/r , and subtract the radius of the earth) (3.59
X7 m)
c) Calculate the speed of the orbit. (3070 m/s)
A satellite orbits with a period of 5.00 hours. The
radius of the Earth is 6380 km and the mass of
the Earth is 5.98 X 1024 kg.
a) Calculate the height of the satellite above the
earth. (8.47 X 106 m)
b) Calculate the speed of the orbit. (5.183X103 m/s)
Kepler’s Laws (1571-1630)
1. The orbit of each planet is an ellipse, with the
sun at one focus
2. Each planet sweeps out equal areas in equal
time
3. T12 = r13
T22 r23
1. The orbit of each planet is an
ellipse, with the sun at one focus
2. Each planet sweeps out equal areas in
equal time
• Suppose the travel time in both cases is
three days.
• Shaded areas are exactly the same area
The Third Law: Example 1
Mars has a year that is about 1.88 earth years. What
is the distance from Mars to the Sun, using the
Earth as a reference (rES = 1.496 X 108 m)
T12 = r13
T22
r23
TM2 = rM3
TE2
rE3
rM3 = TM2rE3
TE2
rM3 = (1.88y)2(1.496 X 108 m)3
(1 y)2
rM3 = 1.18 X 1025 m3
rM = 2.28 X 108 m
Third Law: Example 2
How long is a year on Jupiter if Jupiter is 5.2
times farther from the Sun than the earth?
TJ2 = rJ3
TE 2 r E 3
TJ2 = rJ3 TE2
rE 3
TJ2 = rJ3 TE2 = (5.2)3(1 y)2
rE 3
(1)3
TJ2 = 141 y2
TJ = 11.9 y
Third Law: Example 3
How high should a geosynchronous satellite
be placed above the earth? Assume the
satellite’s period is 1 day, and compare it to
the moon, whose period is 27 days. The
average distance between the earth and the
moon is 384,000 km.
Ts2 = rs3
TM2 rM3
rs3 = rM3 Ts2
TM2
rs3 = rM3 Ts2
TM2
rS3 = rM3
729
=
rM3 (1 day)2
(27 day)2
Take the 3rd root of
both sides
rs = rM
9
The satellite must orbit 1/9 the distance to the
moon. (4.27 X 107 m)
Deriving the Third Law
To derive Kepler’s Law, we will need two
formulas.
F= Gm1mJ
r2
mJ
m1
F=m1v2
r
Gm1mJ = m1v2
r2
r
GmJ = v2
r
GmJ = 4p2r2
r
T2
T2 = 4p2
r3 GmJ
Substitute v=2pr
T
T2 = 4p2
r3 GmJ
T12 = 4p2
r13 GmJ
T12 = T22
r13 r23
We can do this for two
different moons
T22 = 4p2
r23 GmJ
A Useful Form
This form of the equation:
S could be the Sun,
Earth, or other body
with satellites.
T2 = 4p2
r3 GmS
(solve it for ms)
Useful for determining the mass of the central
planet, using only the period and distance of
one of the satellites.
Third Law: Example 4
What is the mass of the sun, knowing that the
earth is 1.496 X 1011 m from the sun.
T2 = 4p2
r3 GmS
mS= 4p2r3
GT2
mS= 4p2(1.496 X 1011 m)3
(6.67 X 10-11 N-m2/kg2) (3.16 X 107 s)2
mS = 2.0 X 1030 kg
Calculate the mass of Neptune if you know
that the period of its moon Galatea is 0.429
days, and the radius is 61,953 km from the
center of Neptune.
(1.02 X 1026 kg)
The Gemini 11 spacecraft sent two astronauts to
a height of 1374 km above the earth’s surface.
The radius of the Earth is 6380 km and the
mass of the Earth is 5.98 X 1024 kg.
a) Calculate the speed of the orbit. (7173 m/s)
b) Calculate the period of the satellite (1.89 hour)
The mass of Mars is 6.40 X 1023 kg.
Calculate the period of its moon Phobos if
Phobos has an orbital radius of 9377 km.
(7.67 h)
Pluto has a radius of 1150 km and a mass of 1.20
X 1022 kg.
a) Calculate the acceleration of gravity on Pluto
(0.605 m/s2)
b) Calculate the weight of a 70.0 kg person on
Pluto. (42.4 N)
c) Calculate the acceleration of gravity on Pluto in
terms of “g’s” (0.0618 g’s)
A student is given the following data and asked to
calculate the mass of Saturn. The data describes
the orbital periods and radii of several of
Saturn’s moons.
Orbital Period, T
Orbital Radius, R
(seconds)
(m)
8.14 X 104
1.85 X 108
1.18 X 105
2.38 X 108
1.63 X 105
2.95 X 108
2.37 X 105
3.77 X 108
Let’s use this equation:
T2 = 4p2
r3
GmS
And rearrange it:
GmS= 1
4p2 r3 T2
Once more:
1 = GmS
T2
4p2 r3
Calculate the following values and graph them.
1
T2
G
4p2 r3
1.60E-10
1.40E-10
1.20E-10
1.00E-10
8.00E-11
6.00E-11
4.00E-11
2.00E-11
0.00E+00
0.00E+00 5.00E-38 1.00E-37 1.50E-37 2.00E-37 2.50E-37 3.00E-37
Calculate the slope of the graph
y
= m
x
1 = mS G
T2
4p2 r3
ms = 5.9 X 1026 kg
+b
The Four Fundamental Forces
1.
2.
3.
4.
Gravity
Electromagnetic
Strong Nuclear Force
Weak Nuclear Force
2. a) 1.52 m/s2, center
b) 38.0 N, center
4.12 m/s
6.14 m/s, no effect
8.a) 3.14 N
b) 9.02 N
10. 27.6 m/s, 0.439 rev/s
12. 9.2 m/s
14. 11 rev/min
16.FT1 = 4p2f2(m1r1 + m2r2)
FT2 = 4p2f2m2r2
18. 0.20
26. 1.62 m/s2
28. 24.5 m/s2
30. 0.91 gsurface
32. 2.0 X 107 m
34. a) 9.8 m/s2
b) 4.3 m/s2
36. 9.6 X 1017 N away from the Sun
38. 2.0 X 1030 kg
40. 3.14 m/s2 upward
42. 5.07 X 103 s (1.41 h), independent of mass
44.a) 58 kg
b) 58 kg
c) 77 kg
d) 39 kg
e) 0
44.a) 58 kg (569N) b) 58 kg(569N)
c) 77 kg (755 N) d) 39 kg (382N)
e) 0
52. RIcarus = 1.62 X 1011 m
54. 5.97 X 1024 kg
56. 3.4 X 1041 kg, 1.7 X 1011 “Suns”
58.REuropa = 6.71 X 105 km
RGanymede = 1.07 X 106 km
RCallisto = 1.88 X 106 km
68.1840 rev/day
70. a) 3000 m b) 5500 N
c) 3900 N
Graphing Centripetal Force A
A 1 kg yo-yo was swung in a circle at a constant speed. The
force on the string was measured as the string was let out
slowly.
Radius (m)
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
Force (N)
10.00
5.00
3.33
2.50
2.00
1.67
1.43
1.25
1.11
1.00
Graphing Centripetal Force B
In a second experiment, the speed was
changed while the length of the string (r)
was kept constant
Speed (m/s)
1.0
2.0
3.0
4.0
5.0
6.0
Force (N)
1.0
4.0
9.0
16
25
36