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What is Gravity?
• An attractive force between all masses.
• No one knows why it acts.
• What’s is measured in?
• Newtons.
• Fg = mg
Is gravity an action/reaction pair?
• The earth pulls on you – do you pull on the earth?
• Yes.
• Same force?
• Yes.
• Equal in magnitude and opposite in direction?
• Yes.
Why do objects fall down?
• Do they?
• What does “down” mean?
• Define “down” relative to Earth.
• Define “fall.”
• More accurate to say that masses on earth are pulled in a path toward the center of the earth.
What is the Universal Gravitational Force?
• Newton’s Universal Law of Gravitation says that every mass in the Universe exerts a gravitational force on every other mass.
• This is how a solar system is born.
• No one had proof until astronauts went to the moon.
How is the Universal Force of Gravity calculated?
• Given two objects, m1 and m2, separated by a distance r (or d), then
• Fg = Gm1m2/d2 where G = 6.7 x 10-11
units for G?
• Nm2/kg2
g
• But we have two equations:
• Fg = Gm1m2/d2 and Fg = mg
• Well….. Try this calculation:
• Gm1/d2 where G = 6.7 x 10-11Nm2/kg2
• M1 = mass of Earth = 6 x 10 24 kg,
• d = radius of Earth = 6.38 x 106 m
• What does it equal???
What is weight?
• Each planet has a little g!
• It’s specific to its mass and radius.
• Weight is mg
Hw #20
• M1 = 45 kg
• Fg=m1g = (45kg)(9.8 m/s2) = 441 N
• Fnet = -Fg +Fair = -441 + 250N = -191 N
• Fnet = ma
• a= Fnet/m = -191n/45 kg = -4.24 m/s2
•
Hw #21
• m = 115 kg
• Fg on earth = (115kg)(9.8m/s2) = 1127N
• Mass is constant everywhere
• Weight = 0 if it is not near any other masses.
HW #22
• Fg1= 11000N- m1=Fg1/9.8 = 1122kg
• Fg2 = 3400 N - m2=Fg2/9.8 = 347kg
• r = 12 m
• Fg = Gm1m2/d2
• Fg = 1.8 x 10-7 N
HW #23• mA = 363 kg mB = 517 kg mC = 154 kg
• rab = .5 rac = .75 rbc = .25• Net force on A = FBA + FCA
• = Gmamb/(rab)2 + Gmamc/(rac)2 = 5.7x 10-5N• Net Force on B = FAB + FCB
• = -Gmamb/(rab)2 + Gmbmc/(rbc)2 = 3.5x 10-5N toward c
• Net force on C = FAC + FCB
• = 9.2x 10-5N toward A
HW #24
• Use Pythagorean theorem• F sun on moon = Gmsmm/rsm2
• = 4.3 x 1020N• F earth on moon = Gmemm/rem2
• = 2 x 1020N
• Net Force on moon = 4.7 x 1020N
HW #25
• Acceleration due to gravity on Mars =
• Gmmars/rmars2
• 3.75 m/s2
• Fg = mgmars = (65kg)(3.75 m/s2)
• = 243.7 N
Put our thinking caps on…
• How can we explain the motion of objects moving in a circle, compared to objects moving in a straight line?
• What equations can we use to solve problems and interpret the motion of objects moving in a circle?
• Our old friends the Kinematics equations and Newton’s Laws!
Circular Motion Definitions
• 2) Rotation – when an object turns around an internal axis.
• Internal axis is located within the body
Definitions continued
• 3) Revolution – when an object turns around an external axis.
• The Earth’s revolutionary period is 1 year.
Definitions continued
• 4) Uniform Circular motion – describes an object travelling in a circle at a constant (uniform) speed. NOT constant velocity!
Definitions continued• 5) Linear Speed (also called Tangential
speed) – distance travelled per time.
• 6) Rotational speed (also called angular speed) - # of rotations per time.
Definitions continued• 7) Centripetal Force –
• Center seeking
• Force necessary to keep an object in uniform circular motion
• 8) Centrifugal Force –
• center fleeing
• Fake!
• Fictitious!
Definitions continued
• 9) Centripetal Acceleration – acceleration that points in to the center of the circle.
• It is a change in direction
To move in a circle, a force must point in toward the center.
• What supplies that force?
• No new forces!
• It is gravity, friction, tension or a normal force.
• How does the object accelerate?
• In a direction in toward the center.
• Is an object in circular motion in equilibrium?
• Good way to get ketchup out of a bottle?
Handout Examples
• Rubber stopper attached to a string.
• m=0.013 kg r = length of string = 0.93m
• Distance = 2πr
• T = period = 1.18s
• A) speed = distance/time = 2πr/T = 4.95 m/s
• B) ac = v2/r = (4.95 m/s)2 /0.93m= 26.4 m/s2
• C) Fc = mac = mv2/r = 0.34N
• directed radially inward
Jupiter orbiting the Sun
• What supplies the centripetal force? • Fg
• What is the magnitude of this force?
• First let’s use Fc = mv2/r
• Convert T = 3.74 x 108s
• Then v = 2πr/T = 13,070 m/s
• Fc = mv2/r = 3.95 x 1022 N
Jupiter continued:
• Now let’s try by calculating Fg:
• Fg = Gm1m2/d2 where G = 6.67 x 10-11
• Fg = (6.67x 10-11)(2 x 1030)(1.8 x 1026)
• (7.78 x 1011m)2
• Fg = 3.97 x 1022 N
Try some problem solving.
• #1. v = 2πr/T
• Given v and r
• T = 2πr/v
• T = 160 (rounded from 162.8s)
• #2. v = 21m/s, r = 0.053m
• T = time for one revolution
• T = 2πr/v = 0.016s
• #3. T = 118 s, v = 17 m/s
• r = vT/2 π
• r= 320m
• #5. v = 98.8 m/s
• ac = 3.00g = 3.00 x 9.8 m/s2
• ac = v2/r r = v2/ac =
• r = 332 m
Windshield Wiper
• #7. T = 4(.28s)
• r = 0.76m
• ac = v2/r ac = (2πr/T)2/r
• v = 2π (.76)/4(.28) = 4.26 m/s
• ac = (4.26m/s)2/0.76
• ac = 23.9 m/s2
Helicopter blades
• #9.
• R = 6.7 m r = 3m
• acR = (v2/R) = (2πR/T)2(1/R)
• acr (v2/r) (2πr/T)2(1/r)
• R2/R = R/r = 6.67m/3m = 2.2
• r2/r
• #12. fold = mv2/r fnew = 1/3(fold)
• mvnew2/r = 1/3 (mvold2/r)
• m and r don’t change, so cancel them out
• Then…
• vnew2 = 1/3(vold2)
• vnew = square root of vold2/3
• vnew = 12.12 m/s
Rotor Ride• Do we need a sign that
says you need to be a certain mass to ride this?
• Vertical: f = mg
• Horizontal: Fn = mv2/r
• Since f also = μFn, then
• μ(mv2/r) = mg
• μ(v2/r) = g
• Mass cancels!
So the answer is NO!
Homework Ch5 #19 • Swing Ride• Angle is 65 to the y axis, Length = 12m, m=220 kg• Find Tension, T, and linear speed, v.
• Horizontal: Tsin65 = mv2/r
• Vertical: Tcos65 = mg
• T = mg/cos65 = 5101 N
• v2 = (Tsin65)( r) / m
• r = 12sin65 = 10.9 m
• v2 = (5101 N)sin(65)(10.9 m) / 220kg=229
• v = 15.1 m/s
• The force equation for the y direction is
• FN cos θ - mg= 0
• FN cos θ= mg
• FN = mg/cos θ
• Then looking at the x force equation:
• FN sin θ= mv2/r
• FN sin θ = mg sin θ/cos θ = mv2/r
• tan θ= mv2/r / mg = v2/gr
• θ= tan-1[v2/(gr)]• This is the banking angle!
Quiz Topics
• Gravitational Interactions
• Big G vs. little g
• Three sphere in a line – finding Force of gravity on one.
• Find centripetal acceleration, v, T, & force
• Flat bend, rotor ride, banked curves
• Vocabulary
Satellite Homework• #27. Satellite over Jupiter• Placed 6 x 105m above surface, given
mass of Jupiter, find v.
• So r = radius of Jupiter + height over surface
• r = 7.2 x 107 m• V2 = GMJ/r• v2 = 1.76 x 109 m/s• v = 4.195 x 104 m/s
• #28. Find r so astronauts on space station weigh half of their weight on Earth, if v = 35.8 m/s.
• Fn = Fc = mv2/r
• Fn = (1/2) mg
• (1/2) mg = mv2/r
• r = 2v2/g
• r = 262 m
• #29. Satellites circling unknown planet. Sat1 has v1 = 1.7 x 104 m/s, and r = 5.25 x 106 m. Sat2 has r = 8.6 x 106 m. Find v for Sat2.
• V2 = GMPLANET/r
• So MPLANETG = constant = v2r
• V12r1 = v2
2r2
• V2 = 1.33 x 104 m/s
Ball on a string: At the bottom:
• T – mg = ma = mv2/r
• Max speed at the bottom means
• T = mg + mv2/r
• Max T
Ball on a string: At the Top
• T + mg = ma = mv2/r
• T = mv2/r – mg
• Min speed at the top, occurs if T = 0, then
• Vmin = √rg
Minimum Velocity
• Occurs always at the top!
• Fn + mg = mv2/r
• Vmin occurs if Fn goes to zero for an instant.
• So Fn + mg = mv2/r becomes
• mg = mv2/r
• Or v = √(rg)
Ch. 5: Homework
• #37. Plane flies over the top of a vertical circle, passengers experience weightlessness. Given v = 215 m/s, what’s r?
• At the top, -Fn + mg = mv2/r• Fn = 0 N, so -Fn + mg = mv2/r becomes• mg = mv2/r• Or r = v2/g• r = 4717 m
#38. Fighter Pilot
• Dives to ground at 230 m/s. • http://physics.bu.edu/~duffy/semester1/c8_vertical.html
• Find r so Fn = 3mg.
• At bottom of circle:
• Fn – mg = mv2/r
• 3mg – mg = mv2/r
• 2g = v2/r
• r = v2/ 2g = 2699 m
#39. Ball spinning on string.
• Given r, Tmax,and m, find v at top and bottom.
• Speed at the bottom: Max T at bottom,
• T – mg = ma = mv2/r
• v2= (T – mg)(r/m) = 23.1
• V = 4.8 m/s
#39. continued
• Speed at Top:
• NOTE – It doesn’t ask for min velocity
• Assume max T at top too
• T + mg = ma = mv2/r
• v2= (T + mg)(r/m) = 32.9
• v = 5.7m/s at top
#40. Downhill Skiier• Bottom of the curve, Fn – mg = mv2/r
• Fn = mg + mv2/r, given r, v, m, Fn=?
• Fn = 1306 N
#41. Demolition Ball
• Given m, r, and v, and motion is at the bottom of swing, find Tension in cable.
• T – mg = mv2/r
• T = mg + mv2/r
• T = 2.87 x 104 N
#42. Roller-Blader• Given r, find v.
• At the top of the hill,
• -Fn + mg = mv2/r
• Vmin occurs if Fn goes to zero for an instant.
• -Fn + mg = mv2/r • mg = mv2/r
• v2 = rg
• v = 14 m/s