Gravity and Circular Motion. First, The story of Gravity… What do we think we know?

Gravity and Circular Motion

First,

• The story of Gravity…

• What do we think we know?

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

What is weight?

• The pull you experience due to your gravitational interaction with a planet.

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

Force and Weight

• Does not depend upon

• The same

• Mass

• Weight

• Weight

• Weight

Gravitational Interactions

Hw #19

• Fg = Gm1m2/d2

• where G = 6.7 x 10-11Nm2/kg2

• F= 8.7 x 10-12 N

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

Moving in Circles

What happens to YOU when you move in a circle?

What in our physical experience seems to move “naturally” in a circle? (By itself.)

Is it true?

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

• 1) Axis – a straight line around which rotation takes place.

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

Definitions

• 10) Period of uniform circular motion – Time required to complete one revolution.

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

Crate on Flat bed Truck

• #11. Force is supplied by friction

• f = Fc = mv2/r

• f = 426 N

• #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

Just you and the wall!

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!

Swing Ride

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

Banked Curves

A banked section of Highway #427 in Toronto. 

• 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

Vertical Circular Motion

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

#43. Spinning stone 2 ways.• Vertical equation:

• Given Tmax = 1.1 TH

• At bottom

• Tmax – mg = mv2/r

• 1.1(mv2/r) – mg = mv2/r

• 1.1 (mv2/r) - mv2/r = mg

• 0.1(mv2/r) = mg

• v2= (gr)/0.1

• V= 9.6 m/s

Given r= 0.95m

Tmax = 1.1 TH

Horizontal eq:

TH = mv2/r

Find v (same for both)