Rotational Motion and the Law of Gravity 1 Linear motion is described by position, velocity, and acceleration. Circular motion repeats itself in circles around the axis of rotation Ex. Planets in orbit, children on a merry-go-round, spinning a basketball Rotational quantities Rotational motion: when an object spins Axis of rotation: line about which the rotation occurs Point on an object that rotates has circular motion around that axis o Can t use linear quantities because we are considering the rotational motion of the extended object instead of just the linear motion of a particle Circular motion = change in angular position o All points on a rigid rotating object (except points on the axis) move through the same angle during any time interval Angular quantities Angular displacement (θ) describes how much an object has rotated Change in angular position Change in arc length/distance from axis Θ = s Or, s = rθ (only works r in radians) Measured in radians (rad) o Unit has no dimension (placeholder only) When s = r, Θ = 1 rad Conversion factors: Θ = s = 2πr = 2 π rad r r Full circle: s = 2πr Reference line: defines Θ = 0 at some time (t) = 0. analogous to the origin in the linear coordinate system. Begins at axis of rotation. Can be any point, but must be used consistently thereafter. 1 revolution (rev) = 2π rad = 360 Θ (rad) = π Θ (degree) 180 Clockwise = negative Counterclockwise = positive Sample problem A While riding on a carousel that is rotating clockwise, a child travels through an arc length of 11.5 m. If the child s angular displacement is 165, what is the radius of the carousel?
Rotational Motion and the Law of Gravity 2 Angular velocity (ω) Rate at which an object rotates or revolves Also called rotational velocity Amount an object turns in a specific time period Ratio of angular displacement to time interval ω avg = ΔΘ Unit: rad/s = s -1 Δt o Unit may also be revolutions/time period (RPM) Convert revolutions to radians Radian is more convenient unit to calculate angular velocity than degrees Ratio of two lengths Magnitude of angular velocity = angular speed (not a vector) Instantaneous angular velocity ω = lim ΔΘ Δt 0 Δt Sample problem B A child at an ice cream parlor spins on a stool. The child turns counterclockwise with an average angular speed of 4.0 rad/s. In what time interval will the child s feet have an angular displacement of 8.0π rad? Angular acceleration (α) = change in angular velocity α avg = Δω = ω f - ω i Unit: rad/s 2 = s -2 counterclockwise = +, clockwise = Δt Δt Every point on a rigid rotating object has the same angular velocity and angular acceleration. Instantaneous angular acceleration α = lim Δω Δt 0 Δt Practice problem C A car s tire rotates at an initial angular speed of 21.5 rad/s. The driver accelerates, and after 3.5 s, the tire s angular speed is 28.0 rad/s. What is the tire s average angular acceleration during the 3.5 s time interval? Kinematic equations for constant angular acceleration Rotational motion ω f = ω i + α t Linear motion v f = v i + a t θ = ω i t + ½α( t) 2 x = v i t + ½a( t) 2 ω f 2 = ω i 2 + 2α( θ) θ = ½(ω i + ω f ) t v f 2 = v i 2 + 2a( x) x = ½(v i + v f ) t ω represents the instantaneous angular speed of the rotating object instead of the average angular speed.
Rotational Motion and the Law of Gravity 3 Practice problem D The wheel on an upside-down bicycle moves through 11.0 rad in 2.0 s. What is the wheel s angular acceleration if its initial angular speed is 2.0 rad/s? Tangential speed There is a relationship between angular speed and angular acceleration and linear speed and linear acceleration of a point on the rotating object. A wheel rolling along the ground has both a linear speed and an angular speed Horses on a carousel rotating about its center: Same angular speed Same angular acceleration Different tangential speed depending upon their distance from the axis of rotation Angular speed and linear speed are related by the circumference of the wheel. Circumference is the distance around the circle and depends upon the radius of the circle Circumference = 2πr A point at the edge of a wheel moves 1 circumference in each turn of the circle Linear speed (v) is the circumference divided by the time it takes to make 1 turn Angular speed (ω) = 2π t Linear speed = angular speed x radius v = ωr o ω is instantaneous angular speed, not average angular speed because the time is very short o ω must be in rad/s Tangential speed: instantaneous linear speed of any point on a rotating object For horses on a carousel = speed along a line drawn tangent to its circular path. Angular speed is the same for all points on the wheel Linear speed on a point on a wheel depends on how far the point is from the center of rotation Outer object covers more distance in the same time, so its speed is greater! Point at center (r=0) doesn t move at all. Tangential speed of an object is its speed along a line drawn tangent to its circular path. Time for a demonstration!
Rotational Motion and the Law of Gravity 4 Practice problem E The radius of a CD in a computer is 0.0600 m. If a microbe riding on the disc s rim has a tangential speed of 1.88 m/s, what is the disc s angular speed? Another practice problem Two children are spinning around on a merry-go-round. Joe is standing 4 meters from the axis of rotation and Sue is standing 2 meters from the axis. Calculate each child s linear speed when the angular speed = 1 rad/s. Back on the carousel If the carousel speeds up, the horses on it experience an angular acceleration. There will also be a linear acceleration tangent to the circular path, and it is called tangential acceleration. Practice problem F A spinning ride at a carnival has an angular acceleration of 0.50 rad/s 2. How far from the center is a rider who has a tangential acceleration of 3.3 m/s 2? Tangential = radius x angular acceleration acceleration v t = rω t t a t = rα Instantaneous angular acceleration, and unit must be in rad/s 2.
Centripetal Acceleration Rotational Motion and the Law of Gravity 5 A Ferris wheel moving at constant speed. Acceleration? Yes, change in direction Centripetal acceleration: always directed toward the center of a circular path; center-seeking. a c = v t 2 r a c = rω 2 Units: m/s 2 a = v f v i t f - t i Acceleration and velocity are both vectors Acceleration is due to its change in direction, change in magnitude of velocity, or both. Practice problem 7A (back in the textbook!) A test car moves at a constant speed around a circular track. If the car is 48.2 m from the track s center and has a centripetal acceleration of 8.05 m/s 2, what is the car s tangential speed? Note: Centripetal acceleration is due to changes in direction. Tangential acceleration is due to a change in speed. A car on a circular track has centripetal acceleration because it is moving in a circle. When the car speeds up to pass another car, it has tangential acceleration because of its change in speed. Total acceleration Any object moving in a circle has both angular acceleration and centripetal acceleration. When both components exist simultaneously, tangential acceleration is tangent to the circular path and centripetal acceleration points to the center of the path. These two components are perpendicular to each other, so total acceleration can be found by using Pythagorean theorem. With tangential acceleration, tangential A total = a 2 2 c + a t speed is changing. Not uniform circular motion. Direction can be found by using tan -1.
Centripetal Force Rotational Motion and the Law of Gravity 6 Ball of mass, m, tied to a string of length r. The ball moves with constant speed, so it has centripetal acceleration. a c = v t 2 r The ball s inertia wants to travel in a straight path, but the string exerts a force on the ball, making it travel in a circular path. Forces on the ball: Force exerted by string horizontal and vertical Gravitational force F c = ma c F c = mv t 2 r Centripetal force: name given to any object in uniform circular motion. Forces that provide centripetal force: Friction ex. Car wheels and the road around a curve Gravity keeps the moon in orbit Gravitational force is directed downward. The vertical component of the string s force counteracts the force of gravity upward. Net force is the horizontal force. The net force is directed toward the center of the object s circular path, so it is the centripetal force. Practice problem 7B (book) A pilot is flying a small plane at 56.6 m/s in a circular path with a radius of 188.5 m. The centripetal force needed to maintain the plane s circular motion is 1.89 x 10 4 N. What is the plane s mass? The tangent is determined by the object s location along its circular path. Centripetal force acts at right angles to an object s circular motion, and therefore, changes the object s direction. If the force disappears, the object no longer moves in a circle, but instead moves along its tangent line.
Rotational Motion and the Law of Gravity 7 Centrifugal force object is forced to the outside while spinning. Not a true force. Ex: a car exits to the right at a high rate of speed and encounters a sharp turn to the left. The passenger slides right and hits the right door. Centrifugal force? Newton s 1 st law law of inertia Not enough friction to hold the passenger stationary Slides right, which was toward the original path (straight) Newton s Law of Universal Gravitation Earth and many planets in our solar system travel in nearly circular orbits around the sun. A centripetal force keeps the planets in orbit. Newton realized that the same force that caused an apple to fall to the ground is the same centripetal force that keeps planets in orbit gravitational force. Orbiting objects are in free-fall Newton s law of universal gravitation: F g = Gm 1 m 2 r 2 Gravitational = Constant x mass1 x mass2 Force (distance between masses) 2 G = 6.673 x 10-11 N m 2 kg 2 [inverse square law] Suppose a cannon were fired from on top of a mountain Path of cannonball is a parabola Horizontal distance increases as the initial speed increases If shot at the right speed, the cannonball would continue falling while the Earth curved out from under the it o Cannonball would orbit the Earth o Gravitational force between the cannonball and the Earth is a centripetal force that keeps the cannonball in orbit Newton s 3 rd law also applies F Em = F me Also centripetal forces Earth and the moon orbit around the center of mass of the Earth-moon system.
Rotational Motion and the Law of Gravity 8 Newton s 2 nd law F = ma, so a = F m The apple falls to the earth, the earth accelerates up to meet the apple! But, acceleration is very small because earth s mass is so large; acceleration is virtually negligible. Practice problem C (textbook) Find the distance between a 0.300 kg billiard ball and a 0.400 kg billiard ball if the magnitude of the gravitational force between them is 8.92 x 10-11 N. Shortcomings to Law of Universal Gravitation. Not accurate for: Very small objects Objects moving at or close to the speed of light Weight changes with location F w = F g = mg (mass x gravitational field strength) Your weight depends on where you are on Earth As the distance from the center of the Earth increases, the value of g decreases, so your weight decreases Gravitational mass = inertial mass Newton s 2 nd law F = ma m = inertial mass that resists acceleration Newton s Law of Universal Gravitation: F g = Gm 1 m 2 r 2 m = gravitational mass relates how objects attract one another How do we know this is true? Acceleration of objects in free-fall is always the same confirms that the two types of masses are equal. Gravity is a field force Masses create a gravitational field in space A gravitational force is an interaction between a mass and the gravitational field created by other masses Ex. Raising an object above the Earth increases its potential energy. The GPE is stored in the gravitational field. Gravitational field strength: g = F g m Gravitational field is a vector that points in the direction of the gravitational force. g = gravitational field strength and free-fall acceleration Free-fall acceleration (a g ) always equals gravitational field strength for that location [equivalent but not the same!] Spring scale shows gravitational field strength, but not acceleration the object is at rest. Gravitational field strength decreases rapidly as the distance from Earth increases. o Intuitively obvious since gravity is an inverse-square law.
Rotational Motion and the Law of Gravity 9 Motion in Space Scientific thought prior to the mid-16 th century: Earth is at the center of the universe All orbits (around Earth) are perfect circles Problem: Didn t account for all the observations of planetary motion. Ptolemy (2 nd century CE) Earth is center of universe, but planets travel in small circles (epicycles) while simultaneously traveling in larger circular orbits Copernicus (1543) Earth and other planets orbit the sun in perfect circles o Not first to theorize a suncentered universe Aristarchus first theorized it 1700 years before Copernicus. o Most scientists did not accept his theory. Kepler s 3 laws describe planetary motion 1. Each planet travels in an elliptical orbit around the sun, and the sun is at one of the focal points. 2. An imaginary line drawn from the sun to any planet sweeps out equal areas in equal time intervals. 3. The square of a planet s orbital period (T 2 ) is proportional to the cube of the average distance (r 3 ) between the planet and the sun, or T 2 r 3 Kepler worked with astronomer Tycho Brahe and used his data for the orbit of Mars. Data only fit if the orbit is an ellipse rather than a circle, with the sun at one focal point of the ellipse. Kepler s 2 nd law shows an imaginary line from the sun to any planet sweeps out equal areas in equal times Kepler s law also applies to satellites orbiting the Earth, including our moon. r is the distance between the orbiting satellite and Earth Proportionality constant depends on the mass of the central object For the times to be equal, the planet must travel faster when they are closer to the sun. 3 rd law relates orbital periods and distances of one planet to those of another planet. o Period (T) = time it takes a planet to finish 1 full o revolution Radius (r) = mean distance between the planet and the sun o T 2 1 3 = r 1 2 T 2 3 r 2
Rotational Motion and the Law of Gravity 10 Kepler and Newton s Law of Gravitation Newton used Kepler s laws to support his law of gravitation Force of gravity is inversely proportional to the distance squared, the resulting orbit must be an ellipse or a circle Law of gravitation can be used to derive Kepler s 3 rd law Sample problem 7D The color-enhanced image of Venus was compiled from data taken by Magellan, the first planetary spacecraft to be launched from a space shuttle. During the spacecraft s 5 th orbit around Venus, Magellan traveled at a mean altitude of 361 km. If the orbit had been circular, what would Magellan s period and speed have been? Kepler s 3 rd law describes orbital period T 2 r 3 Proportionality constant between these 2 variables is 4π 2 /Gm o m = mass of the object being orbited o T 2 = 4π 2 x r 3 Gm o Take the square root of T and you have the period of any object that is in a circular orbit. o The speed of an object in a circular orbit depends on the same factors as the period o Most planets (except Mercury) have a nearly circular orbit T = 2π r 3 /Gm orbital period = 2π [sq. root radius 3 ] constant x mass) v t = Gm/r orbital speed = [sq.root constant x mass ] radius m in both equations is the mass of the central object that is being orbited. Mass of the object that is in orbit does not affect its speed or period Mean radius (r) is the distance between the centers of the 2 bodies o For an artificial satellite orbiting Earth, r is equal to Earth s mean radius plus the satellite s distance from the Earth s surface. Weight and weightlessness Bathroom scales measure the amount of force exerted on it when you step on it You exert the downward force, the scale exerts an upward normal force Riding in an elevator and measuring your weight o At rest = your weight o o Descending = normal force is less At free-fall, F N = 0 apparent weightlessness
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