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. F g = 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 F g = Gm 1 m 2 /d 2 where G = 6.7 x units for G? Nm 2 /kg 2

What is weight? The pull you experience due to your gravitational interaction with a planet.

g But we have two equations: F g = Gm 1 m 2 /d 2 and F g = mg Well….. Try this calculation: Gm 1 /d 2 where G = 6.7 x Nm 2 /kg 2 M1 = mass of Earth = 6 x kg, d = radius of Earth = 6.38 x 10 6 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

Gravitational Interactions

Hw #19 F g = Gm 1 m 2 /d 2 where G = 6.7 x Nm 2 /kg 2 F= 8.7 x N

Hw #20 M 1 = 45 kg F g =m 1 g = (45kg)(9.8 m/s 2 ) = 441 N F net = -F g +F air = N = -191 N F net = ma a= F net /m = -191n/45 kg = m/s 2

Hw #21 m = 115 kg F g on earth = (115kg)(9.8m/s 2 ) = 1127N Mass is constant everywhere Weight = 0 if it is not near any other masses.

HW #22 F g1 = 11000N-  m 1 =F g1 /9.8 = 1122kg F g2 = 3400 N -  m 2 =F g2 /9.8 = 347kg r = 12 m F g = Gm 1 m 2 /d 2 F g = 1.8 x N

HW #23 m A = 363 kg m B = 517 kg m C = 154 kg r ab =.5 r ac =.75 r bc =.25 Net force on A = F BA + F CA = Gm a m b /(r ab ) 2 + Gm a m c /(r ac ) 2 = 5.7x N Net Force on B = F AB + F CB = -Gm a m b /(r ab ) 2 + Gm b m c /(r bc ) 2 = 3.5x N toward c Net force on C = F AC + F CB = 9.2x N toward A

HW #24 Use Pythagorean theorem F sun on moon = Gm s m m /r sm 2 = 4.3 x N F earth on moon = Gm e m m /r em 2 = 2 x N Net Force on moon = 4.7 x N

HW #25 Acceleration due to gravity on Mars = Gm mars /r mars m/s 2 Fg = mg mars = (65kg)(3.75 m/s 2 ) = 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) a c = v 2 /r = (4.95 m/s) 2 /0.93m= 26.4 m/s 2 C) F c = ma c = mv 2 /r = 0.34N directed radially inward

Jupiter orbiting the Sun What supplies the centripetal force? F g What is the magnitude of this force? First let’s use F c = mv 2 /r Convert  T = 3.74 x 10 8 s Then v = 2πr/T = 13,070 m/s F c = mv 2 /r = 3.95 x N

Jupiter continued: Now let’s try by calculating F g : F g = Gm 1 m 2 /d 2 where G = 6.67 x F g = (6.67x )(2 x )(1.8 x ) (7.78 x m) 2 F g = 3.97 x 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 a c = 3.00g = 3.00 x 9.8 m/s 2 a c = v 2 /r  r = v 2 /a c = r = 332 m

Windshield Wiper #7. T = 4(.28s) r = 0.76m a c = v 2 /r  a c = (2πr/T) 2 /r v = 2π (.76)/4(.28) = 4.26 m/s a c = (4.26m/s) 2 /0.76 a c = 23.9 m/s 2

Helicopter blades #9. R = 6.7 m r = 3m a cR = (v 2 /R) = (2πR/T) 2 (1/R) a cr (v 2 /r) (2πr/T) 2 (1/r) R 2 /R = R/r = 6.67m/3m = 2.2 r 2 /r

Crate on Flat bed Truck #11. Force is supplied by friction f = F c = mv 2 /r f = 426 N

#12. f old = mv 2 /r f new = 1/3(f old ) mv new 2 /r = 1/3 (mv old 2 /r) m and r don’t change, so cancel them out Then… v new 2 = 1/3(v old 2 ) v new = square root of v old 2 /3 v new = 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 = mv 2 /r Since f also = μFn, then μ(mv 2 /r) = mg μ(v 2 /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 = mv 2 /r Vertical: Tcos65 = mg T = mg/cos65 = 5101 N v 2 = (Tsin65)( r) / m r = 12sin65 = 10.9 m v 2 = (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 F N cos θ - mg= 0 F N cos θ= mg F N = mg/cos θ Then looking at the x force equation: F N sin θ= mv 2 /r F N sin θ = mg sin θ/cos θ = mv 2 /r tan θ= mv 2 /r / mg = v 2 /gr θ= tan-1[v 2 /(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 10 5 m above surface, given mass of Jupiter, find v. So r = radius of Jupiter + height over surface r = 7.2 x 10 7 m V 2 = GM J /r v 2 = 1.76 x 10 9 m/s v = x 10 4 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 = mv 2 /r Fn = (1/2) mg (1/2) mg = mv 2 /r r = 2v 2 /g r = 262 m

#29. Satellites circling unknown planet. Sat1 has v1 = 1.7 x 10 4 m/s, and r = 5.25 x 10 6 m. Sat2 has r = 8.6 x 10 6 m. Find v for Sat2. V 2 = GM PLANET /r So M PLANET G = constant = v 2 r V 1 2 r 1 = v 2 2 r 2 V 2 = 1.33 x 10 4 m/s

Vertical Circular Motion

Ball on a string: At the bottom: T – mg = ma = mv 2 /r Max speed at the bottom means T = mg + mv 2 /r Max T

Ball on a string: At the Top T + mg = ma = mv 2 /r T = mv 2 /r – mg Min speed at the top, occurs if T = 0, then Vmin = √rg

Minimum Velocity Occurs always at the top! Fn + mg = mv 2 /r Vmin occurs if Fn goes to zero for an instant. So Fn + mg = mv 2 /r becomes mg = mv 2 /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 = mv 2 /r Fn = 0 N, so -Fn + mg = mv 2 /r becomes mg = mv 2 /r Or r = v 2 /g r = 4717 m

#38. Fighter Pilot Dives to ground at 230 m/s. Find r so Fn = 3mg. At bottom of circle: Fn – mg = mv 2 /r 3mg – mg = mv 2 /r 2g = v 2 /r r = v 2 / 2g = 2699 m

#39. Ball spinning on string. Given r, T max,and m, find v at top and bottom. Speed at the bottom: Max T at bottom, T – mg = ma = mv 2 /r v 2 = (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 = mv 2 /r v 2 = (T + mg)(r/m) = 32.9 v = 5.7m/s at top

#40. Downhill Skiier Bottom of the curve, Fn – mg = mv 2 /r Fn = mg + mv 2 /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 = mv 2 /r T = mg + mv 2 /r T = 2.87 x 10 4 N

#42. Roller-Blader Given r, find v. At the top of the hill, -Fn + mg = mv 2 /r V min occurs if Fn goes to zero for an instant. -Fn + mg = mv 2 /r  mg = mv 2 /r v 2 = rg v = 14 m/s

#43. Spinning stone 2 ways. Vertical equation: Given T max = 1.1 T H At bottom T max – mg = mv 2 /r 1.1(mv 2 /r) – mg = mv 2 /r 1.1 (mv 2 /r) - mv 2 /r = mg 0.1(mv 2 /r) = mg v 2 = (gr)/0.1 V= 9.6 m/s Given r= 0.95m T max = 1.1 T H Horizontal eq: T H = mv 2 /r Find v (same for both)