Circular Motion and Gravitation Chapter 7 Circular Motion and Gravitation

Centripetal Acceleration An object traveling in a circle, even though it moves with a constant speed, will have an acceleration The centripetal acceleration is due to the change in the direction of the velocity

Centripetal Acceleration, cont. Centripetal refers to “center-seeking” The direction of the velocity changes The acceleration is directed toward the center of the circle of motion

Centripetal Acceleration, cont. a = Δv (eq. I) Δt By similar triangles Δv = Δs v r therefore Δv = Δs v r Sub into eq. I a = Δs v = v2 r Δt r Since Δs = v

Centripetal Acceleration and Centripetal Force The centripetal acceleration can also be related to the angular velocity ac = vt2 Fc = mac r Therefore, Fc = mvt2

Forces Causing Centripetal Acceleration Newton’s Second Law says that the centripetal acceleration is accompanied by a force F = maC F stands for any force that keeps an object following a circular path Tension in a string Gravity Force of friction

Level Curves Friction is the force that produces the centripetal acceleration Can find the frictional force, µ, v Fc= mac = mv2 r Fc =f !!!!!! f = μn = μmg = mv2 Therefore, Example: The coefficient of friction between The tires of a car and the road is 0.6. What is the smallest circle that the car will be able to Negotiate at 60mph (26.8 m/s)?

Vertical Circle Consider the forces at the top of the circle If car just makes it over the top then the forces exerted by track are zero. Ftrack = 0 The minimum force at the top of the circle to continue circular motion is mg Therefore mg = mv2 r v gr

Newton’s Law of Universal Gravitation (1687) Every particle in the Universe attracts every other particle with a force that is directly proportional to the product of the masses and inversely proportional to the square of the distance between them.

Law of Gravitation, cont. G is the constant of universal gravitational G = 6.673 x 10-11 N m²/kg² This is an example of an inverse square law Example: find the magnitude and direction of the gravitational force between the sun and the earth.

Escape Speed (see p. 206 in text) If an object is launched from the earth with a high enough velocity it will escape the earth’s gravity. The escape speed is the speed needed for an object to soar off into space and not return For the earth, vesc is about 11.2 km/s Note, v is independent of the mass of the object

Satellite Law Motion GMm = mv2 r2 r r = GM v2 Assuming a circular orbit is a good approximation Fg = Fc F GMm = mv2 r2 r r = GM v2

Kepler’s Laws Based on observations made by Tycho Brahe Newton later demonstrated that these laws were consequences of the gravitational force between any two objects together with Newton’s laws of motion

Kepler’s Laws: All planets move in elliptical orbits with the Sun at one of the focal points. A line drawn from the Sun to any planet sweeps out equal areas in equal time intervals. The square of the orbital period of any planet is proportional to cube of the average distance from the Sun to the planet.

Kepler’s First Law All planets move in elliptical orbits with the Sun at one focus. Any object bound to another by an inverse square law will move in an elliptical path Second focus is empty

Kepler’s Second Law A line drawn from the Sun to any planet will sweep out equal areas in equal times Area from A to B and C to D are the same

5.5 Satellites in Circular Orbits GM

Kepler’s Third Law The square of the orbital period of any planet is proportional to cube of the average distance from the Sun to the planet. T = circumference of orbit orbital speed For orbit around the Sun, KS = 2.97x10-19 s2/m3 K is independent of the mass of the planet K = 42 GM Example: A planet is in orbit 109 meters from the center of the sun. Calculate its orbital period and velocity. Ms=1.991 x 1030 kg

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

Gravitational Force Chapter 7 Orbiting objects are in free fall. Section 2 Newton’s Law of Universal Gravitation Chapter 7 Gravitational Force Orbiting objects are in free fall. To see how this idea is true, we can use a thought experiment that Newton developed. Consider a cannon sitting on a high mountaintop. Each successive cannonball has a greater initial speed, so the horizontal distance that the ball travels increases. If the initial speed is great enough, the curvature of Earth will cause the cannonball to continue falling without ever landing.

Torque The force can be resolved into its x- and y-components The x-component, F cos Φ, produces 0 torque The y-component, F sin Φ, produces a non-zero torque

Mechanical Equilibrium In this case, the First Condition of Equilibrium is satisfied The Second Condition is not satisfied Both forces would produce clockwise rotations

Kepler’s Third Law The square of the orbital period of any planet is proportional to cube of the average distance from the Sun to the planet. T = circumference of orbit orbital speed For orbit around the Sun, KS = 2.97x10-19 s2/m3 K is independent of the mass of the planet K = 42 GM Example: A planet is in orbit 109 meters from the center of the sun. Calculate its orbital period and velocity. Ms=1.991 x 1030 kg

Simple Machines Chapter 7 Section 4 Torque and Simple Machines Chapter 7 Simple Machines A machine is any device that transmits or modifies force, usually by changing the force applied to an object. All machines are combinations or modifications of six fundamental types of machines, called simple machines. These six simple machines are the lever, pulley, inclined plane, wheel and axle, wedge, and screw.

Simple Machines, continued Section 4 Torque and Simple Machines Chapter 7 Simple Machines, continued Because the purpose of a simple machine is to change the direction or magnitude of an input force, a useful way of characterizing a simple machine is to compare the output and input force. This ratio is called mechanical advantage. If friction is disregarded, mechanical advantage can also be expressed in terms of input and output distance.

Simple Machines, continued Section 4 Torque and Simple Machines Chapter 7 Simple Machines, continued The diagrams show two examples of a trunk being loaded onto a truck. In the first example, a force (F1) of 360 N moves the trunk through a distance (d1) of 1.0 m. This requires 360 N•m of work. In the second example, a lesser force (F2) of only 120 N would be needed (ignoring friction), but the trunk must be pushed a greater distance (d2) of 3.0 m. This also requires 360 N•m of work.

Simple Machines, continued Section 4 Torque and Simple Machines Chapter 7 Simple Machines, continued The simple machines we have considered so far are ideal, frictionless machines. Real machines, however, are not frictionless. Some of the input energy is dissipated as sound or heat. The efficiency of a machine is the ratio of useful work output to work input. The efficiency of an ideal (frictionless) machine is 1, or 100 percent. The efficiency of real machines is always less than 1.

Horizontal Circle Find ac of yoyo The horizontal component of the tension causes the centripetal acceleration

5-9 Kepler’s Laws and Newton's Synthesis Kepler’s laws describe planetary motion. The orbit of each planet is an ellipse, with the Sun at one focus.