Transcript Circular Motion, Orbits, and Gravity

Circular Motion, Orbits, and
Gravity
AP Physics B
UNIFORM CIRCULAR MOTION
Angular Position
• Because a particle in circular motion travels in
a circle with a fixed radius r, 𝜃 is equal to the
angular position.
• 𝜃 is positive when the angle is measured
counterclockwise from the positive x-axis.
• 𝜃 is negative when measured clockwise from
the positive x-axis.
Angular Position
• Angular position
measures 𝜃in radians.
• 𝜃 radians =
s
r
• s = arc length
• r = radius
• An angle of 1 rad has an
arc length exactly equal
to the radius r.
• 1 𝑟𝑒𝑣 = 360° = 2𝜋 𝑟𝑎𝑑
Angular Displacement and Velocity
• Just as linear motion 𝑣𝑥 = ∆𝑥/∆𝑡, angular
velocity is
•𝜔 =
∆𝜃
∆𝑡
• Similarly, angular displacement is
• 𝜃𝑓 − 𝜃𝑖 = ∆𝜃 = 𝜔∆𝑡
EX. 1 - Comparing Angular Velocities
• A particle travels one-quarter of the way
around a circle after 5 seconds. What is its
angular velocity?
∆𝜃 (2𝜋 𝑟𝑎𝑑)/4
𝜔=
=
= 0.31 rad /s
∆𝑡
5𝑠
A particle travels one-half of the way around a circle
after 5 seconds. What is its angular velocity?
∆𝜃 (𝜋 𝑟𝑎𝑑)
𝜔=
=
= 0.63 rad /s
∆𝑡
5𝑠
EX. 2 - Kinematics at the roulette
wheel
• A small steel ball rolls counterclockwise
around the inside of a 30.0 cm diameter
roulette wheel. The ball completes exactly 2
rev in 1.20 s.
– What is the ball’s angular velocity?
2(2𝜋 𝑟𝑎𝑑)
𝜔=
= 10.5 rad/s
1.20 𝑠
– What is the ball’s angular position at t = 2.00 s?
∆𝜃 = 𝜔∆𝑡 = 10.5 2.00 = 21.0 rad
Angular Speed
• Angular speed 𝜔 is related to the period T and
the frequency f of the motion.
• 𝜔=
2𝜋
𝑇
or
• 𝜔 = 2𝜋𝑓 , where f is in rev/s or s-1
EX. 3 - Rotations in a car engine
• The crankshaft in your car engine is turning at
3000 rpm. What is the shaft’s angular
velocity?
– Convert rpm to rev/s:
3000
𝑟𝑒𝑣
𝑚𝑖𝑛
×
1 𝑚𝑖𝑛
60 𝑠
= 50 rev/s
– Find angular velocity:
ω = 2𝜋 50 = 314 rad/s
FROM ANGULAR TO LINEAR
Speed
• We know the speed of a particle moving with
frequency f around a circular path of radius r
is 𝑣 = 2𝜋𝑓𝑟
• Combining this with the equation for angular
speed:
𝑣 = 𝜔𝑟
𝜔 must be in units of rad/s.
EX. 4 - Finding the speed at points on a
CD
• The diameter of a CD is 12.0 cm. When the
disc is spinning at its maximum rate of 540
rpm, what is the speed of a point (a) at a
distance of 3.0 cm from the center and (b) at
the outside edge of the disc, 6.0 cm from the
center?
EX. 4 - Finding the speed at points on a
CD
• During one period T, the disc rotates once,
and both points rotate through the same
angle, 2𝜋.
• Thus the angular speed, 𝜔 = 2𝜋/𝑇, is the
same for these two points.
• But as they go around one time, the two
points move different distances; the outer
point B goes around a larger circle.
• Therefore, they have different speeds.
EX. 4 - Finding the speed at points on a
CD
• Convert the frequency of the disc from rpm to
rev/s:
𝑟𝑒𝑣 1 𝑚𝑖𝑛
𝑓 = 540
×
= 9.00 rev/s
𝑚𝑖𝑛
60 𝑠
• Compute angular speed:
𝜔 = 2𝜋 9.00 = 56.5 rad/s
• Compute speed at given points; 𝑟𝑎 = 3.0, 𝑟𝑏 =
6.0:
𝑣𝑎 = 𝜔𝑟 = 56.5 0.030 = 1.7 m/s
𝑣𝑏 = 𝜔𝑟 = 56.5 0.060 = 3.4 m/s
Centripetal Acceleration
𝑣2
𝑎=
= 𝜔2 𝑟
𝑟
• Acceleration depends on speed, but also
distance from the center of the circle.
EX. 5 - Who has the larger
acceleration?
• Two children sit side-by-side on a merry-goround. Which child experiences the larger
acceleration?
– A larger value of r results in a larger acceleration,
so the child further from the center of the circle
moves faster and has a larger acceleration.
Finding the period of a carnival ride
• In the Quasar carnival ride, passengers travel
in a horizontal 5.0-m-radius circle. For safe
operation, the maximum sustained
acceleration that riders may experience is 20
m/s2, approximately twice the free fall
acceleration. What is the period of the ride
when it is being operated at the maximum
acceleration?
EX. 6 - Finding the period of a carnival
ride
• Calculate angular speed from the acceleration:
𝜔=
𝑎
=
𝑟
20
= 2.0 rad/s
5.0
• At this angular speed, the period is:
2𝜋
𝑇=
= 3.1 s
𝜔
DYNAMICS OF UNIFORM CIRCULAR
MOTION
Force and Centripetal Acceleration
𝑚𝑣 2
𝐹𝑛𝑒𝑡 = 𝑚𝑎 =
= 𝑚𝜔2 𝑟
𝑟
• A particle of mass m moving at constant speed
v around a circle of radius r must always have
𝑚𝑣 2
𝑟
a net force of magnitude
= 𝑚𝜔2 𝑟
pointing toward the center of the circle.
• Not a new force. One or more of the forces we
already know is responsible for centripetal.
EX. 7 - Forces on a car
• A car drives through a circularly shaped valley at a
constant speed. At the very bottom of the valley,
is the normal force on the road on the car greater
than, less than, or equal to the car’s weight?
• Acceleration vector points toward the center of
the circle, or upward in this case, so net force also
points upward. Weight force cannot change, so
normal force has to increase and be greater than
the weight.
EX. 8 - Analyzing the motion of a cart
• An father places his 20 kg child on a 5.0 kg cart
to which a 2.0-m-long rope is attached. He
then holds the end of the rope and spins the
cart and child around in a circle, keeping the
rope parallel to the ground. If the tension in
the rope is 100 N, how much time does it take
for the cart to make one rotation?
EX. 8 - Analyzing the motion of a cart
• There is no net force in the y-direction,
perpendicular to the circle, so w and n must
be equal and opposite. There is a net force in
the x-direction, toward the center of the
circle, as there must be to cause centripetal
acceleration, so Newton’s second law is:
𝑚𝑣 2
𝐹𝑛𝑒𝑡 = 𝑇 =
𝑟
EX. 8 - Analyzing the motion of a cart
• We know the mass, the radius of the circle,
and the tension, so we can solve for v:
𝑣=
𝑇𝑟
=
𝑚
(100)(2.0)
= 2.83 m/s
25
• From this, we can compute the period:
2𝜋𝑟
2𝜋 2.0
𝑇=
=
= 4.4 s
𝑣
2.83
EX. 9 - Finding the maximum speed to
turn a corner
• What is the maximum speed with which a
1500 kg car can make a turn around a curve of
radius 20 m on a level road without sliding?
• The only force in the x-direction, toward the
center of the circle, is static friction. Newton’s
second law for this is:
𝑚𝑣 2
𝐹𝑥 = 𝑓𝑠 =
𝑟
EX. 9 - Finding the maximum speed to
turn a corner
• Newton’s second law in the y-direction is:
𝐹𝑦 = 𝑛 − 𝑤 = 0, so 𝑛 = 𝑤 = 𝑚𝑔
• Recall that static friction has a max value of:
𝑓𝑠 = 𝜇𝑠 𝑛 = 𝜇𝑠 𝑚𝑔
• This max on static friction also puts a max value on v. If
the car enters a curve at a speed higher than the max,
the car will slide.
𝑚𝑣 2
𝑓𝑠 =
= 𝜇𝑠 𝑚𝑔
𝑟
• Rearranging:
𝑣 = 𝜇𝑠 𝑔𝑟 = (1.0)(9.8)(20) = 14 m/s
EX. 10 - Finding speed on a banked
turn
• A curve on a racetrack of radius 70 m is
banked at a 15° angle. At what speed can a car
take this curve without assistance from
friction?
• Without friction, the x-component of n is the
only force toward the center of the circle. This
force causes the car to turn the corner.
EX. 10 - Finding speed on a banked
turn
𝑚𝑣 2
𝐹𝑥 = 𝑛 sin 𝜃 =
𝑟
𝐹𝑦 = 𝑛 cos 𝜃 − 𝑤 = 0
• From the y-equation:
𝑤
𝑚𝑔
𝑛=
=
cos 𝜃 cos 𝜃
• Substituting this into the x-equation and solving for v:
𝑚𝑔
𝑚𝑣 2
sin 𝜃 = 𝑚𝑔 tan 𝜃 =
cos 𝜃
𝑟
𝑣=
𝑟𝑔 tan 𝜃 =
(70) tan 15 = 13.56 m/s
APPARENT FORCES IN CIRCULAR
MOTION
Centrifugal Force?
• If you are a passenger in a car that turns a corner
quickly, you may feel “thrown” by some force
against the door.
• In reality, the door keeps you from obeying
Newton’s first law; you want to keep moving
forward, but the door pushes you inward toward
the center of the circle. This force is called
centrifugal force.
• Centrifugal force is not an actual force but only a
“felt” force. Never draw centrifugal force on your
free-body-diagrams or use it in Newton’s laws.
Apparent Weight in Circular Motion
• Imagine a rollercoaster going through a loopthe-loop. Why don’t the people fall out?
• What are the forces acting on a person while
they’re at the bottom of the loop?
• Weight and normal force of the seat pushing
up.
• Recall apparent weight is equal to the force
that supports, so 𝑤𝑎𝑝𝑝 = 𝑛.
Apparent Weight in Circular Motion
• Passengers are moving in a circle, so there
must be a net force directed toward the
center of the circle—or directly above the
head—to provide centripetal acceleration.
• The net force points upward, so it must be the
case that 𝑛 > 𝑤.
• 𝑤𝑎𝑝𝑝 = 𝑛, so apparent weight is greater than
true weight, and passengers feel heavier at
the bottom.
Apparent Weight in Circular Motion
• Newton’s second law for this motion is:
𝑚𝑣 2
𝐹 =𝑛−𝑤 =
𝑟
• So apparent weight is:
𝑚𝑣 2
𝑤𝑎𝑝𝑝 = 𝑛 = 𝑤 +
𝑟
• Therefore 𝑛 > 𝑤.
Apparent Weight in Circular Motion
• What are the forces at the top of the loop?
• Still weight force and normal force, but now
normal force pushes downward because the
seat is above the passenger.
• The passenger is still moving in a circle, so net
force is downward toward the center of the
circle.
Apparent Weight in Circular Motion
• Newton’s second law for this motion is now:
𝑚𝑣 2
𝐹 =𝑛+𝑤 =
𝑟
• So apparent weight is:
𝑚𝑣 2
𝑤𝑎𝑝𝑝 = 𝑛 =
−𝑤
𝑟
• If v is large enough, wapp > w, just as at the
bottom of the track.
Apparent Weight in Circular Motion
𝑚𝑣 2
𝑤𝑎𝑝𝑝 = 𝑛 =
−𝑤
𝑟
• But if the car goes slower, and v decreases, there
comes a point
𝑚𝑣 2
where
𝑟
= 𝑤 and 𝑛 = 0.
• At that point, the seat is not pushing against the
passenger at all.
• The speed for which 𝑛 = 0 is called critical speed,
vc
Critical Speed
𝑟𝑤
𝑟𝑚𝑔
𝑣𝑐 =
=
= 𝑔𝑟
𝑚
𝑚
• What happens if speed is slower than critical
speed?
𝑚𝑣 2
𝑟
• In this case, 𝑛 =
− 𝑤 gives a negative value
for n, but that is impossible, so critical speed is
the slowest possible speed to complete the circle.
• If v < vc the passenger cannot turn the full loop
but will fall from the car as a projectile.
Ex. 11 - How slow can you go?
• A motorcyclist in the Globe of Death rides in a
2.2 m radius vertical loop. To keep control of
the bike, the rider wants the normal force on
his tires at the top of the loop to equal or
exceed his and the bike’s combined weight.
What is the minimum speed at which the rider
can take the loop?
How slow can you go?
𝑚𝑣 2
𝐹 =𝑤+𝑛 =
𝑟
• Minimum acceptable speed is when n = w, so:
𝑚𝑣 2
2𝑤 = 2𝑚𝑔 =
𝑟
• Solving for speed, we find
𝑣 = 2𝑔𝑟 = 2(9.8)(2.2) = 6.6 m/s
CIRCULAR ORBITS AND
WEIGHTLESSNESS
Orbital Motion
• Projectile motion assumes the earth is flat and that g is
everywhere straight down.
• Imagine launching a projectile with speed v from an
extremely high tower. As v increases in magnitude, it seems
to the projectile that the ground is curving out from
beneath it.
• If v is large enough, the curve of the projectile’s trajectory
and the curve of the earth are parallel. In this case, the
projectile is “falling” but never getting closer to the ground.
• Projectile returns to the point which it was launched, at the
same speed which it was launched.
• Such a closed trajectory is called an orbit.
Weightlessness in Orbit
• Orbiting objects are in free-fall, so astronauts
and their space crafts are in free-fall as well.
This produces a feeling of weightlessness.
• Weightlessness in space is no different from
the weightlessness in a free-falling elevator.
• This weightlessness does not occur from an
absence of weight or an absence of gravity,
but from everything in orbit falling together.
NEWTON’S LAW OF GRAVITY
Newton’s Law of Gravity
• Newton believed his laws of motion to be
universal, so gravity must be present everywhere.
• He proposed the idea that the earth’s force of
gravity decreases with the square of the distance
from the earth.
• So, the force that attracts the moon to the earth
(and the planets to the sun) is identical to the
force that attracts falling objects to the surface.
• In other words, gravitation is a universal force
between all objects in the universe.
Gravity Obeys an Inverse-Square Law
• Every object in the universe attracts every other object with a force
that:
– Is inversely proportional to the square of the distance between the
objects.
– Is directly proportional to the product of the masses of the two
objects.
•
•
•
If two objects with masses m1 and m2 are a distance r apart, the
objects exert attractive forces on each other of magnitude:
𝐺𝑚1 𝑚2
𝐹1 𝑜𝑛 2 = 𝐹2 𝑜𝑛 1 =
𝑟2
The forces are directed along the line joining the two objects.
The constant G is called the gravitational constant and is equal to:
𝐺 = 6.67 × 10−11 N·(m/kg)2
Ex. 12 – Varying gravitational force
• The gravitational force between two giant lead
spheres is 0.010 N when the centers of the
spheres are 20 m apart. What is the distance
between their centers when the gravitational
force between them is 0.160 N?
Ex. 12 – Varying gravitational force
• We can solve this problem without knowing the
masses of the two spheres by considering the ratio of
forces and distances.
• Gravity is an inverse-square relationship; the force is
related to the inverse square of the distance.
0.160
• The force increases by a factor of
= 16, so the
0.010
distance must decrease by a factor of 16 = 4.
• The distance is
20
thus
4
= 5.0 m.
Inverse-square relationships
• Two quantities have an inverse-square
relationship if y is inversely-proportional to the
square of x:
𝐴
𝑦= 2
𝑥
• If you double x, you decrease y by a factor of 4.
• If you halve x, you increase y by a factor of 4.
• Generally, if x increases by a factor of C, y
decreases by a factor of C2, and vice versa.
Ex. 13 – Gravitational force between
two people
• You are seated in your physics class next to
another student 0.60 m away. Estimate the
magnitude of the gravitational force between
you. Assume that you each have a mass of 65 kg.
𝐺𝑚𝑦𝑜𝑢 𝑚𝑜𝑡ℎ𝑒𝑟 𝑠𝑡𝑢𝑑𝑒𝑛𝑡
𝐹 𝑦𝑜𝑢 𝑜𝑛(𝑜𝑡ℎ𝑒𝑟 𝑠𝑡𝑢𝑑𝑒𝑛𝑡) =
𝑟2
(6.67 × 10−11 )(65)(65)
-7 N
7.8
x
10
=
(0.60)2
Gravitational force of the earth on a
person
• What is the magnitude of the gravitational
force of the earth on a 60 kg person? The
earth has a mass 5.98 x 1024 kg and radius
6.37 x 106 m.
𝐺𝑀𝑒 𝑚
𝐹𝑒𝑎𝑟𝑡ℎ 𝑜𝑛 𝑝𝑒𝑟𝑠𝑜𝑛 =
𝑅𝑒 2
(6.67 × 10−11 )(5.98 × 1024 )(60)
590 N
=
6.37 × 106 2
Gravity on Other Worlds
• On the surface of the earth, 𝑤 = 𝑚𝑔. The same is true on
other planets, but the value of g changes:
𝑤 = 𝑚𝑔𝑝𝑙𝑎𝑛𝑒𝑡
• The weight of objects comes from the gravitational force
exerted by the planet on those objects, so also true is:
𝐺𝑀𝑝𝑙𝑎𝑛𝑒𝑡 𝑚𝑜𝑏𝑗𝑒𝑐𝑡
𝐹𝑝𝑙𝑎𝑛𝑒𝑡 𝑜𝑛 𝑜𝑏𝑗𝑒𝑐𝑡 =
𝑅𝑝𝑙𝑎𝑛𝑒𝑡 2
• Because these two equations are expressions for the same
force, we can set the equal to each other and find that:
𝐺𝑀𝑝𝑙𝑎𝑛𝑒𝑡
𝑔𝑝𝑙𝑎𝑛𝑒𝑡 =
𝑅𝑝𝑙𝑎𝑛𝑒𝑡 2
Ex. 14 - Gravity on Saturn
• Saturn, at 5.68 x 1026 kg,has nearly 100 times
the mass of the earth. It is also much larger,
with a raidus of 5.85 x 107 m. What is the
value of g on the surface of Saturn?
𝐺𝑀𝑆𝑎𝑡𝑢𝑟𝑛
𝑔𝑆𝑎𝑡𝑢𝑟𝑛 =
𝑅𝑆𝑎𝑡𝑢𝑟𝑛 2
(6.67 × 10−11 )(5.68 × 1026 )
2
11.1
m/s
=
5.85 × 107 2
GRAVITY AND ORBITS
Gravity and Orbits
• Planets of the solar system orbit the sun
because the sun’s gravitational pull causes the
centripetal acceleration of circular motion.
Mercury experiences the largest acceleration,
while Neptune experiences the smallest.
• When a small body of mass m orbits a much
larger body of mass M, the smaller body is
called a satellite.
Gravity and Orbits
• Newton’s second law tells us that:
𝐹𝑀 𝑜𝑛 𝑚 = 𝑚𝑎
• Where FM on m is the gravitational force of the large body on the
satellite and a is the satellite’s acceleration.
• FM on m is given by Newton’s law of gravitation, and because it’s
moving in a circular orbit, the satellite’s acceleration is its
centripetal acceleration. Therefore:
𝐺𝑀𝑚
𝑚𝑣 2
𝐹𝑀 𝑜𝑛 𝑚 = 2 = 𝑚𝑎 =
𝑟
𝑟
• Solving for v, we find that the speed of a satellite in a circular orbit
is:
𝑣=
𝐺𝑀
𝑟
Gravity and Orbits
• A satellite must have this specific speed in
order to maintain a circular orbit of radius r
about the larger mass M. If the velocity differs
from this value, the orbit will become elliptical
rather than circular.
• Notice that the orbital speed does not depend
on the satellite’s mass m.
Gravity and Orbits
• For a planet orbiting the sun, the period T is the time to
complete one full orbit. The relationship among speed,
radius, and period is the same as for any circular motion,
2𝜋𝑟
𝑣=
𝑇
• Combining this with the value of v for a circular orbit gives:
𝐺𝑀 2𝜋𝑟
=
𝑟
𝑇
• If we square both sides and rearrange, we find the period
of a satellite:
2
4𝜋
𝑇2 =
𝑟3
𝐺𝑀
Locating a geostationary satellite
• Communication satellites appear to “hover”
over one point on the earth’s equator. A
satellite that appears to remain stationary as
the earth rotates is said to be in a
geostationary orbit. What is the radius of the
orbit of such a satellite?
Locating a geostationary satellite
• We’ll rearrange the equation for the period of a satellite
to solve for radius:
2
4𝜋
𝑇2 =
𝑟3
𝐺𝑀
1
𝑇2 3
𝐺𝑀𝑒
𝑟=
4𝜋 2
(6.67 × 10−11 )(5.98 × 1024 )(8.64 ×
4𝜋 2
1
104 )2 3
= 4.22 x 107