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Chapter 7 Review.

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Presentation on theme: "Chapter 7 Review."— Presentation transcript:

1 Chapter 7 Review

2 Centripetal Acceleration
The equation is ar = v2 / r where ar is centripetal acceleration, V is velocity in m/s and r is radius of the circle. The equation is used when orbiting an object, turning a corner or riding a rollercoaster. It’s important to remember that velocity is tangential whereas acceleration is to the center of the circle. This is how Weightlessness can happen, falling toward the center, but going fast enough to miss the horizon.

3 Centripetal Acceleration (Ex 1)
A rock tied to a string is traveling at a constant speed of 4 m / s in a circle of radius 1.5 m. Calculate the magnitude of the centripetal acceleration of the rock. 1 ac = v2 / r v2 = 16 m / s v = 4 m / s / r = 10.67 r = 1.5 m ac = m/s2 The equation Given information Squaring velocity Dividing by radius

4 Centripetal Acceleration (Ex 2)
A car is moving with a constant velocity around a circular path. If the radius of the circular path is 48.2 m and the centripetal acceleration is 8.05 m/s2 , what is the tangential speed of the car? 2 ac = v2 / r = v2 8.05 = v2 / = v The equation Substitution Multiply by 48.2 Take the ½ power

5 Centripetal Force The equation for centripetal force is fc = m v2 / r where m is mass, v is velocity in m/s, r is radius and fc is centripetal force. Centripetal force is the force needed to keep an object in orbit, such as tension on a string.

6 Centripetal Force (Ex 1)
An aeroplane of mass kg travels a horizontal loop of radius 10 m at the rate of 1000 km/hr. Calculate the centripetal force.3 fc = m v2 / r fc = * 108 (1000 * 1000) / 3600 fc = ( ) / 10 The equation Converting to m/s Substitution Simplification

7 Centripetal Force (Ex 2)
A 1.3 m long fishing line rated as "10 lb test" that can stand a force of 10 lb (44.48 N) is attached to a rock of mass 0.5 kg. Calculate the maximum speed at which the rock can be rotated without breaking the line.4 fc = m v2 / r = v2 44.8 = .5 v2 / v = 10.8 m/s 58.24 = .5 v2 The equation Substitution Multiply by 1.3 divide by .5 take the ½ power

8 Newton Newton’s law of universal gravitation states that the force of gravity between two objects is proportional to the sum of the masses divided by distance squared. This means that two more massive objects attract each other more than two less massive objects and even more so at closer distances. This explains why the earth holds us down, and not the sun.

9 Newton (Cont.) Newton's equation is fg = (G m1 m2 ) / d2 . fg is the force of gravity, m is mass, d is distance and G is a constant. G = 6.67 * This can be used to find the gravity of things such as planets, and solar systems.

10 Newton (Ex 1) Calculate the gravitational attraction between a person of mass 60 kg and a building of mass 10,000 kg when the person is 5 m from the building.5 fg = (G m1 m2) / d2 fg = fg = (G ) / 52 fg = / 52 The equation Substitution Multiply the numerator Divide by 25

11 Newton (Ex 2) The Earth has a mass of 5.98 x 1024 kg, the Moon has a mass of 7.34 x 1022 kg, and the distance from the center of the Earth to the center of the Moon is 3.8 x 105 km. Calculate the gravitational attractive force between the Earth and the Moon.6 fg = (G m1 m2) / d fg = * 1026 fg = (G 5.98 * * 1022) / (3.8 * 105)2 fg = * 1035 / * 1011

12 Kepler’s First Keplers laws have to deal with planetary orbits. The first is that planets orbit in ellipses. Ellipses are shapes in which any point’s sum of distance from the two foci are equal.

13 Kepler’s Second Kepler’s second law states that the area covered in an orbit are always equal in equal amounts of time. Objects orbit faster when closer to each other, and slower when farther away. The distance apart counteracts the speed, however and the area of all the sections are always equal.

14 Kepler’s Third Keplers third, unlike the other two, states that the sum of the square of the period over the cube of the average distance is always the same. This is true of all satellites, be it man made, or otherwise.

15 Orbital speed The equation is v = (G M / r)½. V is velocity, G is the gravitational constant (6.67 * 10-11), M is the mass of the body in the center, and r is radius. This is used to find the speed of a satellite orbiting around a center object.

16 Orbital Speed (Ex) A 600-kg satellite is in a circular orbit about Earth at a height above Earth equal to Earth’s mean radius. Find the satellite’s orbital speed.7 v = (G M / r)½ v = (G * 1024 / * 107)½ v = (3.13 * 107)½ v = 5595 m/s

17 Orbital period The equation for orbital period is T = 2πr / v where t is time, r is radius and v is velocity. It’s the circumference divided by speed, or distance over velocity.

18 Orbital period (Ex) A 600-kg satellite is in a circular orbit about Earth at a height above Earth equal to Earth’s mean radius. Find the orbital period.8 T = 2πr / v T = 2π * 107 / 5595 T =14037 s

19 Force and Torque The difference between force and torque is that torque has radius in its equation. This means that no matter the force you apply, if the radius from the source of the force to the fulcrum is 0, you get no torque.

20 Sources 1 2 3 4 5 6 7 8

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