PHY 151 Lecture 6 6 3 Extending Particle

PHY 151: Lecture 6 • 6. 3 Extending Particle in Uniform Circular Motion Model (Continued) • 6. 4 Nonuniform Circular Motion • 6. 5 Motion in Accelerated Frames • 6. 6 Motion in Presence of Resistive Forces

PHY 151: Lecture 6 Circular Motion Other Applications of Newton’s Laws 6. 3 Extending Particle in Uniform Circular Motion Model (Continued)

Gravitron - 1 • Circular room with a person pressed against the wall • What is ms to keep the person from sliding down? • Radius is 3 meters • Rotating at 20 revolutions/minute

Gravitron - 2 • v = 20 rpm = 20 x 2 pr/60 m/s • v = 20 rpm =2. 09(3) = 6. 28 m/s • Friction force = weight • ms. N = mg • Normal force is centripetal force • Fc = N = mv 2/r • msmv 2/r = mg • ms = gr/v 2 = (9. 8)(3)/6. 282 = 0. 76

Satellites in Uniform Circular Motion

Centripetal Force Satellites in Orbit - 1 • Satellite of mass m is in orbit around earth • Gravitational pull of earth provides centripetal force ØFc = GMem/r 2 = mv 2/r ØGMe/r = v 2 Øv = sqrt(GMe/r) • Orbital velocity for a given radius is independent of mass of satellite

Centripetal Force Satellites in Orbit - 2 • • • v = sqrt(GMe/r) (from prior slide) v = 2 pr/T (definition of period) 2 pr/T = sqrt(GMe/r) T 2 = 4 p 2 r 3/Gme r 3 = T 2 Gme/4 p 2 Relationships between period, T, and orbital radius r

Low Earth Orbit - 1 • • • Velocity of a satellite in low earth orbit Me = 5. 98 x 1024 kg re = 6. 37 x 106 m Height of satellite = 120 miles = 0. 19 x 106 m rorbit = 6. 56 x 106 m G = 6. 67 x 10 -11 nm 2/kg 2 v = sqrt(GMe/r) v = sqrt(6. 67 x 10 -11 x 5. 98 x 1024/ 6. 56 x 106) v = 7797 m/s = 17, 446 mi/hour

Low Earth Orbit - 2 • • • Period of satellite in low earth orbit T 2 = 4 p 2 r 3/Gme re = 6. 38 x 106 m Height of satellite = 120 miles = 0. 19 x 106 m rorbit = 6. 56 x 106 m me = 5. 98 x 1024 kg T 2 = 4 p 2(6. 38 x 106)3/[6. 673 x 10 -11 x 5. 98 x 1024] T 2 = 27941796 s 2 T = 5286 s = 88 minutes

Geosynchronous Orbit - 1 • A satellite in geosynchronous orbit stays above a specific place on the earth • For this to occur, the period of the orbit must be 24 hours = 86400 s • r 3 = T 2 Gme/4 p 2 • r 3 = (86400)2 x [6. 673 x 10 -11 x 5. 98 x 1024] /4 p 2 • r 3 = 7. 55 x 1022 m 3 • r = 4. 23 x 107 m = 26, 268 miles • This is distance from center of the earth

Geosynchronous Orbit - 2 • • • r = 4. 23 x 107 m = 26, 268 miles This is distance from center of the earth Radius of earth is 6. 38 x 106 m Distance from surface is 4. 23 x 107 – 0. 64 x 107 = 3. 59 x 107 m 22, 312 miles

Satellites in Orbit – Example 3 • • Earth orbits the sun as a satellite. What is sun’s mass? Distance from earth to sun, rse = 1. 5 x 1011 m Period of earth around sun Ø T = 365. 24 days x 24 x 60 = 3. 156 x 107 s • • d = vt v = d/t = 2 prse /T GMsun/rse = v 2 = 4 p 2 rse 2/T 2 Msun = 4 p 2 rse 3/GT 2 Msun = 4 p 2(1. 5 x 1011)3/6. 67 x 10 -11/(3. 156 x 107)2 Msun = 2 x 1030 kg Msun/Mearth = 2 x 1030 / 6 x 1024 = 330, 000

Satellites in Orbit – Example 4 • • • Moon is satellite of Earth Io is satellite of Jupiter What is ratio of Jupiter’s mass to earth’s mass Period, Tmoon , of the moon is 27 days Period, Tio, of Io is 1. 5 days Radii of the moon and Io are approximately the same • Mearth = 4 p 2 rme 3/GTmoon 2 • Mjupiter = 4 p 2 rij 3/GTio 2 • Mjupiter/Mearth = Tmoon 2/Tio 2 = (27/1. 5)2 = 324

Satellites in Orbit – Example 5 • • • Earth and Jupiter are satellites of the sun What is ratio of Jupiter’s period to earth’s period 1 AU = 1. 5 x 1011 m, distance from earth to sun Earth’s distance from sun is 1 AU (astronomical unit) Jupiter’s distance from sun is 5. 2 AU Msun = 4 p 2 rse 3/GTearth 2 = 4 p 2 rse 3/GMsun Tjupiter 2 = 4 p 2 rsj 3/GMsun Tjupiter/Tearth = sqrt(rsj 3/rse 3) = sqrt(5. 23/13) = 11. 86

Stars Orbiting Edge of Galaxy - 1 • Mass of galaxy is 2 x 1041 kg • This is the mass of visible stars and dust • A star orbits the galaxy at 2. 7 x 1020 m from the center of the galaxy • This is close to the edge of the galaxy • Assume 90% or more of the galactic mass is inside the star’s orbit • Find a formula for the orbital velocity of star’s at the edge of the galaxy

Stars Orbiting Edge of Galaxy - 2

Graph

Actual Data Rotation curve of a typical spiral galaxy: predicted (A) and observed (B). The discrepancy between the curves is attributed to dark matter.

PHY 151: Lecture 6 Circular Motion Other Applications of Newton’s Laws 6. 4 Nonuniform Circular Motion

Non-Uniform Circular Motion • The acceleration and force have tangential components • produces the centripetal acceleration • produces the tangential acceleration • The total force is •

Vertical Circle with Non-Uniform Speed • The gravitational force exerts a tangential force on the object – Look at the components of Fg • Model the sphere as a particle under a net force and moving in a circular path – Not uniform circular motion • The tension at any point can be found

Top and Bottom of Circle • The tension at the bottom is a maximum • • The tension at the top is a minimum • If Ttop = 0, then

PHY 151: Lecture 6 Circular Motion Other Applications of Newton’s Laws 6. 5 Motion in Accelerated Frames

Motion in Accelerated Frames • A fictitious force results from an accelerated frame of reference – The fictitious force is due to observations made in an accelerated frame – A fictitious force appears to act on an object in the same way as a real force, but you cannot identify a second object for the fictitious force • Remember that real forces are always interactions between two objects – Simple fictitious forces appear to act in the direction opposite that of the acceleration of the non-inertial frame

“Centrifugal” Force • From the frame of the passenger (b), a force appears to push her toward the door • From the frame of the Earth, the car applies a leftward force on the passenger • The outward force is often called a centrifugal force – It is a fictitious force due to the centripetal acceleration associated with the car’s change in direction • In actuality, friction supplies the force to allow the passenger to move with the car. – If the frictional force is not large enough, the passenger continues on her initial path according to Newton’s First Law

“Coriolis Force” • This is an apparent force caused by changing the radial position of an object in a rotating coordinate system • The result of the rotation is the curved path of the thrown ball • From the catcher’s point of view, a sideways force caused the ball to follow a curved path

Fictitious Forces, examples • Although fictitious forces are not real forces, they can have real effects • Examples: – Objects in the car do slide – You feel pushed to the outside of a rotating platform – The Coriolis force is responsible for the rotation of weather systems, including hurricanes, and ocean currents

Fictitious Forces in Linear Systems • • The inertial observer models the sphere as a particle under a net force in the horizontal direction and a particle in equilibrium in the vertical direction The non-inertial observer models the sphere as a particle in equilibrium in both directions The inertial observer (a) at rest sees • The non-inertial observer (b) sees • These are equivalent if Ffictiitous = ma •

PHY 151: Lecture 6 Circular Motion Other Applications of Newton’s Laws 6. 6 Motion in the Presence of Resistive Forces

Motion with Resistive Forces • Motion can be through a medium – Either a liquid or a gas • The medium exerts a resistive force, , on an object moving through the medium • The magnitude of depends on the medium • The direction of is opposite the direction of motion of the object relative to the medium – This direction may or may not be in the direction opposite the object’s velocity according to the observer • nearly always increases with increasing speed

Motion with Resistive Forces, cont. • The magnitude of can depend on the speed in complex ways • We will discuss only two: – is proportional to v • Good approximation for slow motions or small objects – is proportional to v 2 • Good approximation for large objects

Resistive Force Proportional To Speed • The resistive force can be expressed as • b depends on the property of the medium, and on the shape and dimensions of the object • The negative sign indicates is in the opposite direction to

Resistive Force Proportional To Speed, Example • Assume a small sphere of mass m is released from rest in a liquid • Forces acting on it are: – Resistive force – Gravitational force • Analyzing the motion results in

Resistive Force Proportional To Speed, Example, cont. • Initially, v = 0 and dv/dt = g • As t increases, R increases and a decreases • The acceleration approaches 0 when R ® mg • At this point, v approaches the terminal speed of the object

Terminal Speed • To find the terminal speed, let a = 0 • Solving the differential equation gives • t is the time constant and • t = m/b

Resistive Force Proportional To v 2 • For objects moving at high speeds through air, the resistive force is approximately equal to the square of the speed. • R = ½ Dr. Av 2 – D is a dimensionless empirical quantity called the drag coefficient – r is the density of air – A is the cross-sectional area of the object – v is the speed of the object

Resistive Force Proportional To v 2, example • Analysis of an object falling through air accounting for air resistance

Resistive Force Proportional To v 2, Terminal Speed • The terminal speed will occur when the acceleration goes to zero • Solving the previous equation gives

Some Terminal Speeds

Example: Skysurfer • Step from plane – Initial velocity is 0 – Gravity causes downward acceleration – Downward speed increases, but so does upward resistive force • Eventually, downward force of gravity equals upward resistive force – Traveling at terminal speed

Skysurfer, cont. • Open parachute – Some time after reaching terminal speed, the parachute is opened – Produces a drastic increase in the upward resistive force – Net force, and acceleration, are now upward • The downward velocity decreases – Eventually a new, smaller, terminal speed is reached

Example: Coffee Filters • A series of coffee filters is dropped and terminal speeds are measured • The time constant is small – Coffee filters reach terminal speed quickly • Parameters – meach = 1. 64 g – Stacked so that front-facing surface area does not increase • Model – Treat the filter as a particle in equilibrium

Coffee Filters, cont. • Data obtained from experiment: • At the terminal speed, the upward resistive force balances the downward gravitational force • R = mg

Coffee Filters, Graphical Analysis • Graph of resistive force and terminal speed does not produce a straight line • The resistive force is not proportional to the object’s speed

Coffee Filters, Graphical Analysis 2 • Graph of resistive force and terminal speed squared does produce a straight line • The resistive force is proportional to the square of the object’s speed

Resistive Force on a Baseball – Example • • The object is moving horizontally through the air The resistive force causes the ball to slow down Gravity causes its trajectory to curve downward The ball can be modeled as a particle under a net force – Consider one instant of time, so not concerned about the acceleration • Analyze to find D and R