Physics: Principles with Applications, 6th edition Lecture PowerPoints Chapter 5 Physics: Principles with Applications, 6th edition Giancoli © 2005 Pearson Prentice Hall This work is protected by United States copyright laws and is provided solely for the use of instructors in teaching their courses and assessing student learning. Dissemination or sale of any part of this work (including on the World Wide Web) will destroy the integrity of the work and is not permitted. The work and materials from it should never be made available to students except by instructors using the accompanying text in their classes. All recipients of this work are expected to abide by these restrictions and to honor the intended pedagogical purposes and the needs of other instructors who rely on these materials.

Circular Motion Gravitation

5-1 Kinematics of Uniform Circular Motion Uniform circular motion: motion in a circle of constant radius at constant speed Instantaneous velocity is always tangent to circle.

5-1 Kinematics of Uniform Circular Motion This acceleration is called the centripetal, or radial, acceleration, and it points towards the center of the circle.

5-2 Dynamics of Uniform Circular Motion For an object to be in uniform circular motion, there must be a net force acting on it. We already know the acceleration, so can immediately write the force: (5-1)

5-2 Dynamics of Uniform Circular Motion We can see that the force must be inward by thinking about a ball on a string:

5-2 Dynamics of Uniform Circular Motion Net forces that can cause centripetal acceleration Friction of tires on road Tension of cord Gravity keeping object in orbit Source: physicsclassroom.com

5-2 Dynamics of Uniform Circular Motion Net forces that can cause centripetal acceleration Normal force Multimedia: Roller Coaster G forces Source: physicsclassroom.com

5-2 Dynamics of Uniform Circular Motion There is no centrifugal force pointing outward; what happens is that the natural tendency of the object to move in a straight line must be overcome. If the centripetal force vanishes, the object flies off tangent to the circle.

5-1 Kinematics of Uniform Circular Motion Frequency (f): Number of revolutions per second Period (T): Time it takes for one revolution T = 1 / f Frequency: units are s-1 Period: units are s

5-2 Dynamics of Uniform Circular Motion Formulas for Circular Motion Where R is radius of circle, T is period, v is velocity, m is mass Average speed = 2πR / T Acceleration = v2 / R Substitute 2πR / T for v Acceleration = (2πR/T)2 / R = 4π2R / T2 ; velocity not required Fnet r= mar Substitute mv2/R for ar Fnetr = mv2 / R Substitute 4π2R / T2 for ar Fnet r= m4π2R / T2 ; velocity not required

Circular Motion Examples

Circular Motion Examples Force on revolving ball (horizontal)

Circular Motion Examples Force on revolving ball (vertical) Calculate FT at bottom of arc At bottom, mg is downward, FT2 upward FT2 - mg = m ((v2)2 / r) FT2 = m ((v2)2 / r) + mg = 0.150 kg (6.56 m/s2) / 1.10 m + (0.150 kg)(9.8 m/s2) = 7.34 N (continued)

Circular Motion Examples Force on revolving ball (vertical) 0.150 kg ball at end of 1.10 m cord swung vertically Calculate maximum speed at top of arc so ball doesn’t fall At top, forces are mg and FT1 downward FT + mg = m ((v1)2 / r) v = √gr = √(9.80 m/s2 1.10m) v = 3.28 m/s

Circular Motion Examples Tetherball Acceleration is toward pole, not top So, net force is toward pole FTy – mg = 0 FTx is centripetal acceleration, towards center

Circular Motion Examples Centripetal acceleration = aR = v2 / r ; / 9.8 m/s2 to find ‘g’s Particle revolves in circle of circumference = 2πr = 2π(0.0600m) = 0.377 m /revolution 5 x 104 rpm x 60 s / min = 833 rev / s T = 1 / (833 rev / s) = 1.2 x 10-3 s / rev v = 2πr / T = 3.14 x 102 m/s aR = v2 / r = (3.14 x 102 m/s)2 /.0600 m = 1.64 x 106 m/s2 ‘g’s = 1.64 x 106 m/s2 / 9.8 m/s2 = 1.67 x 105

5-3 Highway Curves, Banked and Unbanked When a car goes around a curve, there must be a net force towards the center of the circle of which the curve is an arc. If the road is flat, that force is supplied by friction.

5-3 Highway Curves, Banked and Unbanked If the frictional force is insufficient, the car will tend to move more nearly in a straight line, as the skid marks show.

Circular Motion Examples Forces are mg, FN and horizontal FF FN = mg = 9800 N Net horizontal force to keep car moving in circle (continued)

Circular Motion Examples Skidding on a curve Dry pavement Maximum static friction of tires (μS = 0.60) 5900N > 3900N so car stays on road Icy road Maximum static friction of tires (μS = 0.250) 2500N < 3900N so car skids off of road

Circular Motion Examples Car moving along horizontal circle, so aR is horizontal. FN sin(θ) = mv2 / r FN cos(θ) - mg = 0 (no vert. motion) FN = mg / cos(θ) [FN > mg] Substitute for FN in first equation sin(θ) (mg / cos(θ)) = mv2 / r or mg x tan(θ) = mv2 / r tan(θ) = v2 / rg = (14 m/s)2 / (50m)(9.8 m/s2) = 0.40 Θ = 22°

5-6 Newton’s Law of Universal Gravitation If the force of gravity is being exerted on objects on Earth, what is the origin of that force? Newton’s realization was that the force must come from the Earth. He further realized that this force must be what keeps the Moon in its orbit.

5-6 Newton’s Law of Universal Gravitation The gravitational force on you is one-half of a Third Law pair: the Earth exerts a downward force on you, and you exert an upward force on the Earth. When there is such a disparity in masses, the reaction force is undetectable, but for bodies more equal in mass it can be significant.

5-6 Newton’s Law of Universal Gravitation Therefore, the gravitational force must be proportional to both masses. By observing planetary orbits, Newton also concluded that the gravitational force must decrease as the inverse of the square of the distance between the masses. In its final form, the Law of Universal Gravitation reads: where (5-4)

5-6 Newton’s Law of Universal Gravitation The magnitude of the gravitational constant G can be measured in the laboratory. This is the Cavendish experiment.

5-7 Gravity Near the Earth’s Surface; Geophysical Applications Now we can relate the gravitational constant to the local acceleration of gravity. We know that, on the surface of the Earth: Solving for g gives: Now, knowing g and the radius of the Earth, the mass of the Earth can be calculated: (5-5)

5-7 Gravity Near the Earth’s Surface; Geophysical Applications The acceleration due to gravity varies over the Earth’s surface due to altitude, local geology, and the shape of the Earth, which is not quite spherical.

5-8 Satellites and “Weightlessness” Satellites are routinely put into orbit around the Earth. The tangential speed must be high enough so that the satellite does not return to Earth, but not so high that it escapes Earth’s gravity altogether. Velocity at which satellite escapes from Earth’s gravity called escape velocity

Escape Velocities Source: http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html

5-8 Satellites and “Weightlessness” The satellite is kept in orbit by its speed – it is continually falling, but the Earth curves from underneath it.

5-8 Satellites and “Weightlessness” Objects in orbit are said to experience weightlessness. They do have a gravitational force acting on them, though! The satellite and all its contents are in free fall, so there is no normal force. This is what leads to the experience of weightlessness.

5-8 Satellites and “Weightlessness” More properly, this effect is called apparent weightlessness, because the gravitational force still exists. It can be experienced on Earth as well, but only briefly:

Newton’s Law of Universal Gravitation Examples Since satellite at r=2, FG is ¼ as great FG = ¼ mg = ¼ (2000 kg) (9.80 m/s2) = 4900 N

Newton’s Law of Universal Gravitation Examples a) Period of satellite must be 1 day Only force on satellite is gravity, so G mSat x mE / r2 = mSat v2 / r ; unknowns are v and r But, period of satellite = period of rotation of Earth = 24 h 24 h x 3600 s / 1 h = 86,400 s Speed of satellite v = 2πr / T (same speed at Earth) Substitute: G x mE / r2 = (2πr)2 / r T2 r = 4.23 x 107 m = 42,300 km Radius of Earth = 6380 km, so satellite orbits at 36,000 km

Newton’s Law of Universal Gravitation Examples Geosynchronous satellite b) v = √ G mE = √ (6.67x10-11 x Nm2/kg2) (5.98 x 1024 kg) / 4.23 x 107 m = 3070 m/s

Newton’s Law of Universal Gravitation Examples

5-9 Kepler’s Laws and Newton's Synthesis Kepler's First and Second Laws (FYI) Orbits of planets are ellipses with Sun at one focus Planets sweep out equal areas in equal times (velocity not constant)

5-9 Kepler’s Laws and Newton's Synthesis Deriving Kepler's Third Law Since the centripetal force equals the force of gravity for an orbiting body M is mass of Sun, m is mass of planet We know the formulas for centripetal acceleration and velocity so Substituting for ac in original formula divide out m (If the units of R are AU (astronomical units) and the units of time are Earth years, then R3 = T2)

5-9 Kepler’s Laws and Newton's Synthesis Kepler’s 3rd Law: The ratio of the square of a planet’s orbital period is proportional to the cube of its mean distance from the Sun.

Gravitational field Gravitational field exists for every object that has mass Field is responsible for gravitational force on objects Field is force per unit mass Magnitude of Earth’s gravitational field at any point above Earth’s surface is G ME / r2 At Earth’s surface, gravitation field = g

Gravitational field Since gravity is an attractive only force, field lines point toward center of Earth (center of mass)

Summary An object moving in a circle at constant speed is in uniform circular motion. It has a centripetal acceleration There is a centripetal force given by The centripetal force may be provided by friction, gravity, tension, the normal force, or others. Period: Time required for one revolution Frequency: Revolutions per second

Summary of Chapter 5 Newton’s law of universal gravitation: Satellites are able to stay in Earth orbit because of their large tangential speed. Objects in freefall are said to be weightless because there is no normal force, but there is still gravitational force

Summary of Chapter 5 Kepler’s Third Law states that the square of the period of revolution is proportional to the cube of the mean distance from the Sun

Summary of Chapter 5 Multimedia – Circular Motion