pg 551 Uniform Circular Motion When an object is moving in a circle and its speed the magnitude of its velocity is constant it is said to be moving with uniform circular motion The direction of the objects velocity is always tangent to the circle hence at the exhibition when going around ID: 787186
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Slide1
11.2 Uniform Circular Motion
pg 551
Slide2Uniform Circular Motion
When an object is moving in a circle and its speed (the magnitude of its velocity) is constant, it is said to be moving with uniform circular motion.
The direction of the objects velocity is always tangent to the circle (hence at the exhibition, when going around in the crazy swirly ride, your body wants to fly out of the ride.
When you drive around and around the roundabout, your body (and car) also want to fly out the side
Centripetal Acceleration – the centre directed acceleration of a body moving continuously along a circular path.
Slide3Slide4Slide5If a comet has a radius of 20 meters, and a satellite orbits it every 3 minutes, what velocity is the satellite orbiting at?
What would be the centripetal force?
Slide6Centripetal Acceleration formula is
a
c
= V
2
/r
ac = centripetal acceleration in m/s2
V= velocity (or magnitude) in m/s
R = radius of the circle (m)
Slide7Centripetal Force
I
s the force that causes centripetal acceleration
http://www.youtube.com/watch?v=GM_2Ceh05Gs&feature=relmfu
Formula is
Fc = mv
2
/ r
Fc = centripetal force (N)
M = mass in kgV = velocity in m/sR = radius of circular path (m)Fc = FnFn = Ff (if at a constant speed)
Slide8A yo yo has a mass of 225 g. The full length of the string is 1.2 m. You decide to see you slowly you can swing it in a vertical circile while keeping the string fully extended, even when the yo yo is at the top of its swing.
a) Calculate the minimum speed at which you can swing th yo yo while keeping it on a circular path
b)At the speed in a, find the tension in the string when the yo yo is at the side and at the bottom of its swing.
a)
To swing the yo yo to keep it on a circular path, you have to swing it just above the force of gravity. Therefore Fg=Fc
Fg = Fc
mg = mv
2
/ rSolve for vv
2 = grV = 3.4 m/s
Slide9A yo yo has a mass of 225 g. The full length of the string is 1.2 m. You decide to see you slowly you can swing it in a vertical circile while keeping the string fully extended, even when the yo yo is at the top of its swing.
·Calculate the minimum speed at which you can swing th yo yo while keeping it on a circular path
·At the speed in a, find the tension in the string when the yo yo is at the side and at the bottom of its swing.
b) At the speed in a, find the tension in the string
at the side
at the bottom Fc
At the side
Fc = Ft
Ft = mv2 /r Ft = 2.2 Fg
At the bottomFc = Fc + FgFt = mv2 / r + fgFt = .2kg(3.431)2 / 1.2 m +.2*9.81
Ft = 4.1 N At the top subtract Fg Fc = .238 N
Slide10With your partner, decide with what minimum velocity you would need to spin one of yourselves with a string
a) .2 meters longb) 10 meters long
Then determine with what the tension in the string would be at the
top
bottom
right and left side
Slide11A string used to make a pendulum has a breaking strenght of 14 N and a length of 1.2 m. A 2.00 kg mass is used as a bob and set in motion.
a) if the bob moves with a speed of 3.0 m/s at the bottom of the swing, will the swing break?
b) What is the max speed the bob can go so the string does not break (at the bottom of the swing)
c) What is the max speed the bob can go so the string does not break (at the top of the swing)
Slide12A string used to make a pendulum has a breaking strenght of 12 N and a length of .8 m. A 1.00 kg mass is used as a bob and set in motion.
a) if the bob moves with a speed of 1.0 m/s at the bottom of the swing, will the swing break?
b) What is the max speed the bob can go so the string does not break (at the bottom of the swing)
c) What is the max speed the bob can go so the string does not break (at the top of the swing)
Slide13A car with a mass of 2135 kg is rounding a roundabout on a level road. If the radius of the curvature of the road is 52 m and the coefficient of friction between the tires and the road is .70, what is the maximum speed at which the car can make the curve without skidding off the road?
Fc = mv2 / r
Fc = Fn = Ff (if at a constant speed)
Fn = mv2 / r
uFn = mv2 / r
.7(2135)(9.8)= 2135(v 2 ) /52
v = 19m/s
To remain on the roundabout the car must not go faster than 19m/s.
Slide14Determine the mass of a car of someone in the class. How fast can you go around the roundabout in Charlottetown if the radius of the roundabout is 20 meters and the friction between the tires and the road is .65?
Slide15Centripetal Force vs Centrifugal Force
Every force needs an equal and
Opposite force…hence
Centripetal force = centrifugal force
Fictitious forces, like centrifugal
Forces, are needed to explain
motion in non inertial frames of
reference .
http://www.youtube.com/watch?v=fmf2nPuv2OI
http://www.youtube.com/watch?v=hZOekFFSoWI
Slide16In a celtic field event called the hammer throw, a 12 kg ball is whirled in a circle of radius 2.0m with a frequency of 1.5Hz.
a) What is the velocity when it is released?
b) What is the centripetal force?
Slide17Slide18Chapter 12 Universal Gravitation pg 572-594
5 classes
Bas
ically, universal gravitation states that all objects
are attracted toward eachother…causing gravity. The lighter object is always attracted toward the heavier object. The distance between these objects decreases the strength of gravity.
Slide19Kepler, was the guy who discovered three key concepts when discussing the motion of planets
Kepler's 3 Laws
·Planets move in elliptical orbits, with the Sun at one focus of the ellipse
·An imaginary line between the Sun and a planet sweeps out equal areas in equal time intervals.
·The quotient of the square of the period of a planets revolution around the sun and the cube of the average distance from the sun is constant and the same for all planets.
The area of the shaded regions is the same. Therefore a
planet has to move faster on the left side of the sun than
the right
In the case with Earth, the left side corresponds to
our winter and the right to our summer. Notice that the months in the year are longer in May-Aug and shorterIn Nov through Janht
tp://www.youtube.com/watch?v=NiWK5z7z_Oc&feature=related
Slide20K
epler's Third Law : The quotient of the square of the period of a planets revolution around the sun and the cube of the average distance from the sun is constant and the same for all planets.
This constant is the formula
r
3
=k
T
2
r is the radius and T is period (time it takes to go one full rotation)k is the constant
If you are comparing two planets, than r3= r3=kT2 T2http://www.youtube.com/watch?v=WTJa8DXlerchttp://www.youtube.com/watch?v=NG18fObqMV4&feature=related
Slide21The orbital period and radius for Jupiter is 4332.62 days and 5.2 AU respectively. Calculate Saturn's radius if the orbital period for Saturn is 10 759.2 days.
r3=
r
3
=k
T
2
T
2
Slide22Newton’s Universal Law of Gravitation
F = force of gravity
M1 = first mass (the bigger mass)
M2 = second mass
G = universal gravitational constant
R = distance between the centre of the masses
G = 6.67+10
-11
Slide23A 65 kg astronaut is walking on the surface of the moon, which has a mean radius of 1.74*10
3 km and a mass of 7.35*10
22
kg. What is the weight of the astronaut? B/c I’ve forgotten what gravity is on the moon, we can’t use
F =mg rather we use
Fg =
G (m1m2)
r2 Fg =6.67 *10-11
* (7.35* 1022)(65) (1.74*106
)2 Fg = 105 N (this is the weight of the astronaut)Now, determine how much gravity is on the moonF = mg105 = 65(g)1.61=g
Slide24Now that we've done the moon, a
65 kg astronaut is walking on the surface of the Saturn, which has a mean radius of 6*104 km and a mass of 5.6851 × 1026
kg. What is the weight of the astronaut?
Slide25Looking at your partner, determine your mass and distance between each other.
Now, calculate the a)force between you
b)gravity between you (to calculate gravity, use only one person's mass)
Slide26Page 580
Questions # 1-6use appendix B pg 955
Slide27Notice Newton’s Law of Universal Gravitation looks a lot like the centripetal force equation
Fc = mv
2
/ r
The two forces are equal !
Therefore
mv
2 = G (m1m2) r r2
If I were to solve for velocityV = √(Gm) /r
Slide28The general formula that combines Kepler’s third law and Newton’s Law of Universal Gravitation, is
m
v
2
=
G (m1m2)
r r
2Which can be rearranged intoR3 = GmT
2 4π 2When do I use which???…it depends upon what variables I have.To utilize this law,
Slide29Find the mass of the sun, using Earth’s orbital radius and period of revolution (use page 955 to help you)
G = 6.67* 10
-11
T = 365.25 days = 3.1558* 10
7
seconds
R = 1.49 * 1011 Using the formula, what are we looking for….mR3
= GmT2 4π 2M =
R3 *4π 2 T2 * GFill in your numbers andM = 1.97* 10 30 kgNow use Venus and Mercury to find the mass of the sun. (use page 955 to help you)
Slide30The period of the moons orbit is 27.3 day. If the mean radius of moons orbit is 3.84 * 10
8 m, calculate the mass of Earth?M
=
R
3
*4π
2
T2 * G
Slide31How fast is a satellite orbiting Earth if it is 200km from the surface of the Earth?
mv
2
=
G (m1m2)
r r
2
V = √(Gm) /r V = √(6.67* 10-11 )(5.98*10
24 )/200000 metersV =
Slide32R
3 =
Gm
T
2
4π
2
can be flipped for easy switching toT2 = 4π 2R
3 GmRemember, all your units have to be inseconds and meters !Page 586 Questions 9-14
Slide3312.2 Planetary and Satellite Motion Page 587
As Earth is a circle, if you can maintain a strong enough speed, you should be able to stay in the atmosphere forever. The reasoning, is that if you can match your decent (downwards), to match the curvature of the earth (4.9 meters every 8 km), you can stay in the air permanently.
Formula for Period
.
Slide34F
ormula’s
R
3
=
Gm
T
2
4π
2orFc = Universal law of gravitationmv2 =
G (m1m2) r r2Which m cancels out? (the m that is rotating around)V = √(Gm) rorT = 2π √ R3/Gm
orT = 1/f Formula for radius, if you do not use one of the formula’s from up aboveR = vT 2π
Slide351) At satellite orbits earth at an altitude of 200 km. What is its orbital speed?
2) W
hat is the satellite's period?
3)If you want a geosyncrohnise orbit, what would be your radius ?
Slide361) At satellite orbits earth at an altitude of 200 km. What is its orbital speed?
V =
√(Gm)
r
2) What
is the satellite's period?
T =
2
π √ R3/Gm3)If you want a geosyncrohnise orbit, what would be your radius ?
4.2*107 meters - Earths radius = 3.5*107 meters
Slide37The problem on Earth is air friction. If though, you go high enough in the air (outer layers of atmosphere), the friction caused by air decreases enough that a geostationary orbit can be achieved
If you go farther out still, you can go at a speed that is identical to earths rotation, resulting in a geosynchronous orbit.
A satellite that is in geosynchronous orbit is still falling toward earth…its just falling at the same speed that Earth is moving away…meaning it never catches up or slows down.
Slide38The planet Uranus has an orbital radius around the Sun of about 2.3 * 10
8 m. What is its period & orbital speed?
T = ?
G = 6.67* 10
-11
M of Uranus = 9.98* 10
24
kgRadius of Uranus = 6.38 * 106Mass of Sun =1.99*10
30Formula for velocityV = √(Gm) /rT = 2π √
R3/Gm
Slide39Review Questions
1)Two masses 4.0 kg and 8.0 kg, are located 2 m apart. What is the gravitational attraction force between them.
2)
What is the mass of an object which experiences a pull of 10 N at the Earth's surface?
3)At what height above the earth's surface would an objects weight be one half the value at the surface?
Slide40Review Questions
Two masses 4.0 kg and 8.0 kg, are located 2 m apart. What is the gravitational attraction force between them.Fg =
G (m1m2)
r
2
Wha
t is the mass of an object which experiences a pull of 10 N at the Earth's surface?
10g = g(m1m2) r2
At what height above the earth's surface would an objects weight be one half the value at the surface?Fg = G (m1m2) Fg =
6.67*10-11 (5.98*1024)(90) r2 (6.38*106)2Fg = 881= weight 440 = 6.67*10-11
(5.98*1024)(90) r2 r = 9 * 106 metersSubtract this new r from the radius of earth and you get 2.6*106
meters
Slide41At what position between the Earth and the moon would a spaceship experience no net force?
This is where the centripetal force = the force of gravitymv^2/r = GMm m/r^2
Slide42Microgravity
Ev
en though astronauts in the space ship are floating, they still experience gravity. In fact, gravity is still fairly strong (Earth’s gravity is strong enough to keep the moon in orbit)
Hence apparent weighlessness experienced by astronauts is called microgravity. There still is gravity...and they are still falling towards the center of the Earth…but because they are so far away and Earth is moving so fast…there falling never quite catches up to Earth’s movement.
If I were to drop an apple in the spaceshuttle, it would be in a free fall towards earth…but it would appear to float
Slide43Page 591
Questions 15-17