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## Circular Motion, Orbits, and Gravity

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**Circular Motion, Orbits, and Gravity**AP Physics B**Angular Position**• Because a particle in circular motion travels in a circle with a fixed radius r, is equal to the angular position. • is positive when the angle is measured counterclockwise from the positive x-axis. • is negative when measured clockwise from the positive x-axis.**Angular Position**• Angular position measures in radians. • s = arc length • r = radius • An angle of 1 rad has an arc length exactly equal to the radius r.**Angular Displacement and Velocity**• Just as linear motion , angular velocity is • Similarly, angular displacement is**EX. 1 - Comparing Angular Velocities**• A particle travels one-quarter of the way around a circle after 5 seconds. What is its angular velocity? A particle travels one-half of the way around a circle after 5 seconds. What is its angular velocity? • rad /s • rad /s**EX. 2 - Kinematics at the roulette wheel**• A small steel ball rolls counterclockwise around the inside of a 30.0 cm diameter roulette wheel. The ball completes exactly 2 rev in 1.20 s. • What is the ball’s angular velocity? • What is the ball’s angular position at t = 2.00 s? • rad/s • rad**Angular Speed**• Angular speed is related to the period T and the frequency f of the motion. or • , where f is in rev/s or s-1**EX. 3 - Rotations in a car engine**• The crankshaft in your car engine is turning at 3000 rpm. What is the shaft’s angular velocity? • Convert rpm to rev/s: rev/s • Find angular velocity: rad/s**Speed**• We know the speed of a particle moving with frequency f around a circular path of radius r is • Combining this with the equation for angular speed: must be in units of rad/s.**EX. 4 - Finding the speed at points on a CD**• The diameter of a CD is 12.0 cm. When the disc is spinning at its maximum rate of 540 rpm, what is the speed of a point (a) at a distance of 3.0 cm from the center and (b) at the outside edge of the disc, 6.0 cm from the center?**EX. 4 - Finding the speed at points on a CD**• During one period T, the disc rotates once, and both points rotate through the same angle, • Thus the angular speed, , is the same for these two points. • But as they go around one time, the two points move different distances; the outer point B goes around a larger circle. • Therefore, they have different speeds.**EX. 4 - Finding the speed at points on a CD**• Convert the frequency of the disc from rpm to rev/s: • Compute angular speed: • Compute speed at given points; , : 9.00 rev/s 56.5 rad/s 1.7 m/s 3.4 m/s**Centripetal Acceleration**• Acceleration depends on speed, but also distance from the center of the circle.**EX. 5 - Who has the larger acceleration?**• Two children sit side-by-side on a merry-go-round. Which child experiences the larger acceleration? • A larger value of r results in a larger acceleration, so the child further from the center of the circle moves faster and has a larger acceleration.**Finding the period of a carnival ride**• In the Quasar carnival ride, passengers travel in a horizontal 5.0-m-radius circle. For safe operation, the maximum sustained acceleration that riders may experience is 20 m/s2, approximately twice the free fall acceleration. What is the period of the ride when it is being operated at the maximum acceleration?**EX. 6 - Finding the period of a carnival ride**• Calculate angular speed from the acceleration: • At this angular speed, the period is: 2.0 rad/s 3.1 s**Force and Centripetal Acceleration**• A particle of mass m moving at constant speed v around a circle of radius r must always have a net force of magnitude pointing toward the center of the circle. • Not a new force. One or more of the forces we already know is responsible for centripetal.**EX. 7 - Forces on a car**• A car drives through a circularly shaped valley at a constant speed. At the very bottom of the valley, is the normal force on the road on the car greater than, less than, or equal to the car’s weight? • Acceleration vector points toward the center of the circle, or upward in this case, so net force also points upward. Weight force cannot change, so normal force has to increase and be greater than the weight.**EX. 8 - Analyzing the motion of a cart**• An father places his 20 kg child on a 5.0 kg cart to which a 2.0-m-long rope is attached. He then holds the end of the rope and spins the cart and child around in a circle, keeping the rope parallel to the ground. If the tension in the rope is 100 N, how much time does it take for the cart to make one rotation?**EX. 8 - Analyzing the motion of a cart**• There is no net force in the y-direction, perpendicular to the circle, so w and n must be equal and opposite. There is a net force in the x-direction, toward the center of the circle, as there must be to cause centripetal acceleration, so Newton’s second law is:**EX. 8 - Analyzing the motion of a cart**• We know the mass, the radius of the circle, and the tension, so we can solve for v: • From this, we can compute the period: 2.83 m/s 4.4 s**EX. 9 - Finding the maximum speed to turn a corner**• What is the maximum speed with which a 1500 kg car can make a turn around a curve of radius 20 m on a level road without sliding? • The only force in the x-direction, toward the center of the circle, is static friction. Newton’s second law for this is:**EX. 9 - Finding the maximum speed to turn a corner**• Newton’s second law in the y-direction is: , so • Recall that static friction has a max value of: • This max on static friction also puts a max value on v. If the car enters a curve at a speed higher than the max, the car will slide. • Rearranging: 14 m/s**EX. 10 - Finding speed on a banked turn**• A curve on a racetrack of radius 70 m is banked at a 15° angle. At what speed can a car take this curve without assistance from friction? • Without friction, the x-component of n is the only force toward the center of the circle. This force causes the car to turn the corner.**EX. 10 - Finding speed on a banked turn**• From the y-equation: • Substituting this into the x-equation and solving for v: 13.56 m/s**Centrifugal Force?**• If you are a passenger in a car that turns a corner quickly, you may feel “thrown” by some force against the door. • In reality, the door keeps you from obeying Newton’s first law; you want to keep moving forward, but the door pushes you inward toward the center of the circle. This force is called centrifugal force. • Centrifugal force is not an actual force but only a “felt” force. Never draw centrifugal force on your free-body-diagrams or use it in Newton’s laws.**Apparent Weight in Circular Motion**• Imagine a rollercoaster going through a loop-the-loop. Why don’t the people fall out? • What are the forces acting on a person while they’re at the bottom of the loop? • Weight and normal force of the seat pushing up. • Recall apparent weight is equal to the force that supports, so .**Apparent Weight in Circular Motion**• Passengers are moving in a circle, so there must be a net force directed toward the center of the circle—or directly above the head—to provide centripetal acceleration. • The net force points upward, so it must be the case that . • , so apparent weight is greater than true weight, and passengers feel heavier at the bottom.**Apparent Weight in Circular Motion**• Newton’s second law for this motion is: • So apparent weight is: • Therefore .**Apparent Weight in Circular Motion**• What are the forces at the top of the loop? • Still weight force and normal force, but now normal force pushes downward because the seat is above the passenger. • The passenger is still moving in a circle, so net force is downward toward the center of the circle.**Apparent Weight in Circular Motion**• Newton’s second law for this motion is now: • So apparent weight is: • If v is large enough, wapp > w, just as at the bottom of the track.**Apparent Weight in Circular Motion**• But if the car goes slower, and v decreases, there comes a point where and . • At that point, the seat is not pushing against the passenger at all. • The speed for which is called critical speed, vc**Critical Speed**• What happens if speed is slower than critical speed? • In this case, gives a negative value for n, but that is impossible, so critical speed is the slowest possible speed to complete the circle. • If v < vcthe passenger cannot turn the full loop but will fall from the car as a projectile.**Ex. 11 - How slow can you go?**• A motorcyclist in the Globe of Death rides in a 2.2 m radius vertical loop. To keep control of the bike, the rider wants the normal force on his tires at the top of the loop to equal or exceed his and the bike’s combined weight. What is the minimum speed at which the rider can take the loop?**How slow can you go?**• Minimum acceptable speed is when n = w, so: • Solving for speed, we find 6.6 m/s**Orbital Motion**• Projectile motion assumes the earth is flat and that g is everywhere straight down. • Imagine launching a projectile with speed v from an extremely high tower. As v increases in magnitude, it seems to the projectile that the ground is curving out from beneath it. • If v is large enough, the curve of the projectile’s trajectory and the curve of the earth are parallel. In this case, the projectile is “falling” but never getting closer to the ground. • Projectile returns to the point which it was launched, at the same speed which it was launched. • Such a closed trajectory is called an orbit.**Weightlessness in Orbit**• Orbiting objects are in free-fall, so astronauts and their space crafts are in free-fall as well. This produces a feeling of weightlessness. • Weightlessness in space is no different from the weightlessness in a free-falling elevator. • This weightlessness does not occur from an absence of weight or an absence of gravity, but from everything in orbit falling together.**Newton’s Law of Gravity**• Newton believed his laws of motion to be universal, so gravity must be present everywhere. • He proposed the idea that the earth’s force of gravity decreases with the square of the distance from the earth. • So, the force that attracts the moon to the earth (and the planets to the sun) is identical to the force that attracts falling objects to the surface. • In other words, gravitation is a universal force between all objects in the universe.**Gravity Obeys an Inverse-Square Law**• Every object in the universe attracts every other object with a force that: • Is inversely proportional to the square of the distance between the objects. • Is directly proportional to the product of the masses of the two objects. • If two objects with masses m1 and m2are a distance r apart, the objects exert attractive forces on each other of magnitude: • The forces are directed along the line joining the two objects. • The constant G is called the gravitational constant and is equal to:**Ex. 12 – Varying gravitational force**• The gravitational force between two giant lead spheres is 0.010 N when the centers of the spheres are 20 m apart. What is the distance between their centers when the gravitational force between them is 0.160 N?**Ex. 12 – Varying gravitational force**• We can solve this problem without knowing the masses of the two spheres by considering the ratio of forces and distances. • Gravity is an inverse-square relationship; the force is related to the inverse square of the distance. • The force increases by a factor of , so the distance must decrease by a factor of • The distance is thus m.**Inverse-square relationships**• Two quantities have an inverse-square relationship if y is inversely-proportional to the square of x: • If you double x, you decrease y by a factor of 4. • If you halve x, you increase y by a factor of 4. • Generally, if x increases by a factor of C, y decreases by a factor of C2, and vice versa.**Ex. 13 – Gravitational force between two people**• You are seated in your physics class next to another student 0.60 m away. Estimate the magnitude of the gravitational force between you. Assume that you each have a mass of 65 kg. 7.8 x 10-7 N**Gravitational force of the earth on a person**• What is the magnitude of the gravitational force of the earth on a 60 kg person? The earth has a mass 5.98 x 1024 kg and radius 6.37 x 106 m. 590 N