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## PowerPoint Slideshow about 'Circular Motion' - elom

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Linear speed, v

How far you go in a certain amount of time

Miles per hour, meters per second

Rotational (angular) speed,

How many times you go around in a certain amount of time

Revolutions per minute, rotations per hour, radians per second

Linear speed, v

How far you go in a certain amount of time

Miles per hour, meters per second

Rotational (angular) speed,

How many times you go around in a certain amount of time

Revolutions per minute, rotations per hour, radians per second

Which horse has a larger linear speed on a merry go round, one on the outside or one on the inside?

Outside.

Which horse has a greater rotational speed?

Neither, all the horses complete the circle in the same amount of time.

The number of revolutions per second is called the frequency, f.

Frequency is measured in Hertz, Hz.

The time it takes to go all the way around once is called the period, T.

Frequency is related to period by

f = 1 / T

How do you find the linear velocity if it is not directly provided?

Velocity = distance / time

In circular motion, the distance traveled is all around the circle… the circumference.

The circumference = 2pr

So…

v = 2pr / T

Where r is the radius of the circle and T is the period of time to go around once.

Uniform Circular Motion, UCM: moving in a circle with a constant speed.

Question: Is there a constant velocity when an object moves in a circle with a constant speed?

No, the direction changes, therefore the velocity changes.

If the velocity changed, the object is actually ACCELERATING even while moving at the same speed.

Suppose an object was moving in a straight line with some velocity, v.

According to Newton’s 1st Law of Motion, “An object in motion continues that motion unless a net external force acts on it”.

If you want the object to move in a circle, some force must push or pull it towards the center of the circle.

A force that pushes or pulls an object towards the center of a circle is called a centripetal force

Centripetal means “center seeking”

According to Newton’s 2nd Law, SF = ma, If there is a centripetal force, there must be a centripetal acceleration.

ac = v2 / r

Where r is the radius of the circle and v is the velocity of the object.

Centripetal force

Since SF= ma, the net centripetal force is given by

Lots of forces can help in pushing or pulling an object towards (or away from) the center of a circle. Sometimes it takes more than one force to get an object to move in uniform circular motion.

Centripetal force is the NET force pointing along the radius of a circle, NOT a new kind of force. If an object moves in a circle (or an arc), there must be at least one force that is acting toward the center of the circle.

In a free body diagram, NEVER label an individual force as “Centripetal force”!!

When can these forces contribute to a net centripetal force acting on an object that is moving in a circular pathway?

Gravity?

Moon revolving around the Earth

Tension?

Twirling a pail at the end of a string

Friction?

Cars rounding a curve.

Air Resistance (“Lift”)?

Airplane or birds flying in a circle.

Normal?

Riders in a carnival ride

What happens if the string breaks? Which way will the ball move?

The ball will continue to move in a straight line path that is “tangent” to the circle.

Tension in a string as a centripetal force

A student twirls a rock around and around in a horizontal circle at the end of the string.

The only force that contributes to a NET centripetal force is the tension in the string.

Fnet centripetal = Tension = m(v2/r)

Example

A boy twirls a ½ kg rock in a horizontal circle on the end of a 1.6 meter long string. If the velocity of the rock was 4 m/s, what is the Tension in the string?

m = ½ kg

r = 1.6 m

v = 4 m/s

The only centripetal force is Tension.

Fnet centripetal = Tension = m (v2/ r)

T = ½ 42 / 1.6

T = 5 N

Example

How fast was the ½ kg rock moving if the Tension was 10 N and the string was 1.6 m long?

m = ½ kg

r = 1.6 m

T = 10 N

T = m(v2/ r)

Tr/m = v2

10 x 1.6 / .5 = v2

v = 5.7 m/s

For training, astronauts are required to ride in a special centrifuge to simulate the great accelerations they will experience during lift-off. If the radius is 10 meters, what velocity must the astronaut move to not exceed the acceleration of free-fall?

a = v2 /r = 9.8 m/s2

v2 = 10 m x 9.8 m/s2

Friction along a surface as a centripetal force

A 1500 kg race car goes around a curve at 45 m/s. If the radius of the curve is 100 m, how much friction is require to keep the car on the track? What is m, the coefficient of friction?

m = 1500 kg

v = 45 m/s

r = 100 m

The centripetal force is friction.

f = m(v2/r)

f = 1500 (452/ 100)

f = 30375 N

f = mN

m= f / N but what is N?

N = mg = 15000 N

m = 30375 N / 15000 N

m = 2.02

The Normal force

In some cases the normal force can contribute to the net centripetal force.

For example, on the carnival ride where the riders stand against the walls of the circular room and the floor drops out! And yet, the rider does not slide down!

Draw the free-body diagram!

What keeps them from sliding down?

The wall pushes against the rider toward the center of the circle.

N = m (v2/ r)

Also, if he is not sliding down, we know

f = mg and therefore…..mN = mg

We can combine those two equations

(divide one by the other!)

f

N

mg

Vertical loops

Twirling a rock at the end of a string in a vertical loop.

At the top of the loop, both the Tension and the weight point towards the center of the circle!

SF = T + mg = mv2/r

At the bottom of the loop, the Tension points toward the center, the weight away from the center:

SF = T – mg = mv2/r

What about an object on a vertical track?

At the top of the track, both the Normal force (the track pushing against the ball) and the weight point down toward the center of the circle, therefore, they are both positive:

SF = N + mg = mv2/r

At the bottom of the track, the Normal force points toward the center and the weight points away from the center:

SF = N – mg = mv2/r

Loop the Loop

What is the minimum speed that a rider must be moving at in order to complete a loop the loop of radius 12 m?

At the top of the loop, both the Normal force and weight point towards the center of the circle, so

SFcentripetal = N + mg = mv2 / r

However, at the minimum required speed, contact is lost for a moment at the top of the loop, so that…

The Normal force goes all the way to ZERO.

The weight is the only centripetal force when the rider is moving at the minimum required speed.

mg = mv2/r

g = v2/r

v2 = rg

v2 = 12 x 10

v = 10.95 m/s

“Artificial Gravity”

Occupants of a space station feel weightless because they lack a support (Normal) force pushing up against their feet. By spinning the station as just the right speed, they will experience a “simulated gravity” when the Normal force of the floor pushing up on their feet becomes a centripetal force. The closer their centripetal acceleration, v2/r is to g, the acceleration due to Earth’s gravity, the more they feel the sensation of normal weight.

Artificial Gravity Example…

A circular rotating spacestation has a radius of 40 m. What linear velocity, v, must be maintained along the outer edge, to maintain a sense of “normal” gravity?

We want the centripetal acceleration, v2/r (due to the rotation) to be the same as the acceleration due to gravity on Earth- 9.8 m/s2

v2/r = 9.8 m.s2

v2 = 9.8 x 40

v = 19.8 m/s

Newton’s Universal Law of Gravitation states that every particle in the universe exerts an attractive force on every other particle.

Where “G” is the “universal gravitational constant” G = 6.67 x 10-11

What happens to the Force if one of the masses is doubled?

1. It will now be F x ?

What happens to the Force if both of the masses were doubled?

2. It will now be F x ?

What happens to the Force if one of the masses is doubled and then other one is halved?

3. It will now be F x ?

This is an “inverse square” law, since the

Force is proportional to the inverse of the distance squared.

Example:

At twice the distance, the gravitational force between two objects would be less. How much less?

Two objects are separated by some distance, d. How would the gravitational force differ if the distance was tripled?

1/9 the original force

What if the distance was 4d?

1/16 the original force

4. the distance was 5d?

5. The distance was 10d?

½ d?

6. At ½ d, the force would by F x ??

Example: Two masses of 5 kg and 9 kg are separated by 1.5 m. What is the gravitational force they exert on each other?

How do you enter all those numbers in your calculator? Use your exponent button (EE) for “G”!! Do NOT type in “ x 10^ ”

6.67E-11*5*9÷1.52 =

F = 1.3 x 10-9 N

- Enter 1.3 (one decimal place)
- Enter the exponent, -9

G = 6.67 x 10-11

What is the gravitational force between a 600 kg mass and a 850 kg mass if they are 0.4 meters apart?

- Enter the number with 1 decimal place
- Enter the exponent

G = 6.67 x 10-11

Example: Two masses of 3 x 103 kg and 1.8 x 1015 kg are separated by

d = 1.4 x 1021m. What is the gravitational force they exert on each other?

How do you enter all those numbers in your calculator? Use your exponent button!!

6.67E-11*3E3*1.8E15÷1.4E212 =

- Enter the number with 1 decimal place.
- Enter the exponent

G = 6.67 x 10-11

If the gravitational force between a 95 kg mass and a 120 kg mass is 4 x 10-4N, how far apart are they?

What’s the shortcut to get d2 out of the denominator?

Trade places with F!! And don’t forget to take the square root!

13. Enter the value of “d”

G = 6.67 x 10-11

NET Gravitational Force

Two masses pull on the central mass.

How would you get the NET gravitational force?

Subtract the two forces.

NET Gravitational Force

Two masses pull on the left mass.

How would you get the NET gravitational force?

Add the two forces. (Be careful about your distances!)

NET Gravitational Force

Two masses pull on the mass at the origin.

How would you get the NET gravitational force?

Pythagorize the two forces.

2nd tan for angle.

Cavendish and “G”, the gravitational constant

Henry Cavendish, a British scientist, first devised an experiment to determine “G” in 1797.

He suspended two small known masses from a “torsion wire” of which he knew the strength. These two small masses were gravitationally attracted to two large known masses, which caused the wire to twist until the torsion force was balanced by the gravitational force. Because he knew the strength of the torsion force, he also knew the strength of the gravitational force.

With known masses, known Force, and known distance, the only “unknown” left was G!

* You need to know who first determined “G”,

Finding “g”

Weight is the gravitational force a planet exerts.

Weight = Gravitational Force

mg = G

“g”, the acceleration due to gravity can be found by canceling an “m”.

The distance, d, is measured from the center of the planet to the location of interest. (often, the radius)

The acceleration due to gravity, “g”, is also called the “gravitational field strength”.

planet

How large is “g” on the planet Venus, which has a mass of 4.87 x 1024 kg and has a radius of 6,050,000 meters?

6.67E -11 x 4.87 E24 ÷ 6,050,0002 =

g = 8.87 m/s2

Example: An asteroid of radius 500 m has a mass of 6.5 x 1013 kg. What is the gravitational field strength at its surface?

6.67E -11 x 6.5 E13 ÷ 5002 =

g = 0.0173 m/s2

How much would a 60 kg astronaut WEIGH on this asteroid?

W = mg

W = 60 kg x 0.0173 m/s2

W = 1.04 N

Ptolemy, 83 AD

Ptolemy (also geocentric universe) presented his astronomical models in convenient tables, which could be used to compute the future or past position of the planets, the Sun, and Moon, the rising and setting of the stars, and eclipses of the Sun and Moon. His model showed the planets turning in small circles as they orbited the Earth!

The tables actually produced fairly good predictions, but his model and his geocentric universe was….. WRONG!

Ptolemy was also the first to use latitude and longitude lines.

Copernicus 1473heliocentric universe

“sun-centered” universe

Although others before him had proposed that the planets orbit the sun rather than the Earth, Copernicus was the first to publish mathematical evidence

Tycho Brahe

- 1546
- Built “The Castle of the Stars”
- Had an accident in a duel
- Died an unusual death…

1. The Law of Orbits: All planets have elliptical orbits with the sun at one focus.

Planets’ orbits are only slightly elliptical

Comets have highly elliptical orbits

2. The Law of Areas: A line that connects a planet to the sun sweeps out equal areas in equal times.

(Planets move faster when they are closer to the sun.)

Perigee- closest distance

Apogee- most distance location

3. The Law of Periods: The square of the period of a planet is proportional to the cube of its average orbital radius. (this is easy IF you measure using some different units than we usually use)

(earth years) T2 = r3 (AU)

An AU is an “astronomical unit” and is the distance from the Sun to the Earth.

Kepler’s Third Law in fundamental units of seconds and meters (it’s a LOT more complicated…

We know that the centripetal force for satellites in circular orbit is the gravitational force.

Solve this for v of a satellite in CIRCULAR ORBIT.

We also know the velocity for any circular motion.

Set those two values for v equal and solve for the period T.

Cancelling and rearranging give us Kepler’s Third Law with the standard units of meters and seconds.

This law will work for circular orbits around ANY body as long as you know the mass of that body.

X cubed

But… let’s stick to the easier units for now…Venus is located 0.72 AU’s from the Sun. How many years does it take Venus to orbit the sun?

T2 = a3

0.61 years

cube root

If it takes an asteroid 5 Earth years to orbit the sun, how far is the asteroid from the sun?

T2 = a3

2.92 AU

Galileo 1564

- Made the first telescope (not really, but he gets credited with it)
- Observed the rings of Saturn and some of the moons of Jupiter

Sir Isaac Newton 1642

- Said that gravity was not just an Earthly force, but a Universal force.
- Wrote the Universal Law of Gravitation

Gravity is not really a “force” at all!

The effects of gravity are just the results of the distortion of the “fabric of space-time” or the “space-time continuum”

Albert Einstein 1879THE FABRIC OF SPACE-TIME

According to Einstein, SPACE-TIME is rather like a stretchy fabric.

Any mass will distort the fabric of space-time, like a bowling ball sitting on a rubber sheet.

Einstein proposed that the effects of GRAVITY were only the results of the distortion of the fabric of space-time.

These “indentations” around masses are called gravity wells.

If you release an object near a gravity well, it will accelerate down into the well.

If the object was given some velocity tangent to the circumference of the well, it will either spiral down into the well or began to orbit the well.

“Newton’s Cannon”

Newton proposed a “thought” experiment. If you fire a rocket horizontally from the top of a very high mountain, gravity will pull it towards the center of the Earth and it will go a certain distance before it hits the ground.

“Newton’s Cannon”

The faster the rocket is launched, the further it will go as it falls until it hits the ground.

“Newton’s Cannon”

If it is launched fast enough, the pathway of the rocket as it falls will exactly match the curvature of the Earth. The satellite will continue to fall and fall and fall, but it will never fall to the ground. It goes into orbit about the Earth!

“Newton’s Cannon”

The horizontal speed of the orbiting rocket must be very high and it must maintain that speed or it will fall into the Earth. Satellites in orbit are always FALLING… and FALLING… and FALLING…

“Newton’s Cannon”

This is why the astronauts appear to be weightless. There is in fact plenty of gravity at the elevation of the space shuttle and space station, but since they are always falling, they appear to be in zero gravity- weightless.

Satellite Motion

The gravitational force provides the centripetal force for an orbiting satellite in circular orbit.

For any object moving in a circle the velocity is given by

v = circumference / time = 2pr / T

Where T is the period of the motion – the time to go around once.

Therefore, for a satellite in circular orbit:

= 2pr / T

If a rocket is launched VERTICALLY from the surface of a planet, is it true that what goes up must come down?

If we throw a ball into the air, it reaches some highest point and then gravity pulls it back down.

If we throw it faster, it will go higher before it comes back down.

However, if we throw it fast enough, it can escape the gravitational pull of the Earth and keep moving upward.

The speed at which that will occur is called the ESCAPE SPEED. The escape speed for a planet is given by

ESCAPE SPEED =

Geosynchronous Satellites

A geosynchronous satellite is one whose orbital period is the same as Earth’s rotational period.

So, as Earth rotates once every 24 hours, the satellite orbits the Earth once every 24 hours.

This means that when the orbit lies entirely over the equator, the satellite remains stationary relative to the Earth's surface and the antennae’s do not have to “track” it continually. These satellites are used for communications, and intelligence!

How many satellites are in Earth’s orbit?

One natural satellite

Over 8000 artificial satellites!

There are two main processes constantly going on in the super massive stars:

nuclear fusion (which tends to blow the star's hydrogen outward from the star's center) and

Gravitation (which tends to pull all hydrogen back in the direction it had come).

These two processes balance one another until all the star's hydrogen is exhausted, allowing gravitation to take over.

Once gravitation dominates, the star becomes unstable and starts to collapse.

Once a super massive star starts to collapse, it does not stop, and the star (and ultimately its atoms) will cave inward upon itself, resulting in the formation of a black hole (Hewitt 186).

When a star has exhausted its fuel supply, gravitational forces crush the star to one of three possible outcomes:

1) The star shrinks and stabilizes into a white dwarf.

2) The star crunches into a neutron star.

3) The star collapses to a black hole.

A star less than 1.4 times the mass of the sun will become a white dwarf. A star between 1.4 and 3 times the mass of the sun will become a neutron star. It's only those stars greater than 3 times the mass of the sun that become black holes upon collapse.

How do you “see” a black hole when it can’t be seen??

When a star collapses and changes into a black hole, the strength of its gravitational field still remains the same as it had been before the collapse. Therefore the planets in orbit would not be affected. The planets would continue in their orbits as usual and would not be drawn into the black hole. Because black holes do not give off any light, the planets would appear to be orbiting around nothing. There is reason to believe that the planets could just be orbiting about a star that is too faint to be seen, but there is an equal chance that a black hole could be present (Hewitt 187).

Because the gravity of a black hole is so intense, dust particles from nearby stars and dust clouds are pulled into the black hole. As the dust particles speed and heat up, they emit x-rays. Objects that emit x-rays can be detected by x-ray telescopes outside of the Earth's atmosphere (Miller).

Black holes can also be detected through a technique called gravity lensing. Gravity lensing occurs when a massive object, in this case a black hole, passes between a star and the Earth. The black hole acts as a lens when its gravity bends the star's light rays and focuses them on the Earth. From an observer's point of view on the Earth, the star would appears to brighten or to be distorted or to be in a different part of the sky.

The event horizon is the boundary around a black hole where gravity has become so strong that nothing- not even light- can escape. The escape velocity at the event horizon = c. The escape velocity inside the event horizon > c, therefore, escape is impossible. The event horizon is the point of no return.

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