1. Acceleration is equal to Δv/Δt. Velocity is a vector and there are two ways a vector can be changed: by changing magnitude or by changing direction.

2. An acceleration due to a change in direction is called a centripetal acceleration.

3. The equation is: a c = v t 2 /r.

5. v i and v f are equal and can be called v. Also, ∆d = v ∆t. Therefore, ∆v/ v i = ∆d/r becomes: ∆v/v = v∆t/r

6. ∆v/v = v∆t/r becomes: ∆v/∆t = v 2 /r

7. ∆v/∆t = v 2 /r ∆v/∆t = a SO, a = v 2 /r

8. A test car moves at 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 its tangential speed?

9. The direction of centripetal acceleration is toward the center of the circle.

10. An object can have a centripetal acceleration and a tangential acceleration. These two accelerations are perpendicular to each other.

11. When both these components exist simultaneously, the total acceleration can be found using the Pythagorean theorem.

12. A Hot Wheels‘ car moves around a curve in the track. The curve has a radius of 0.20 m, the instantaneous velocity is 1 m/s, and friction is causing a tangential acceleration of -0.5 m/s2. What is the aC and what is the total acceleration?

13. A force is required to accelerate an object: F = ma. A centripetal force is required to cause a centripetal acceleration. F c = ma c

14. F c = ma c F c = mv 2 /r F c is measured in newtons.

15. A pilot is flying a small plane at 30.0 m/s in a circular path with a radius of 100.0 m. If a force of 635 N is needed to maintain the pilot’s circular motion, what is the pilot’s mass?

16. The centripetal force needed to pull an object into a curved path is directed toward the center of the circle formed by the path.

17. Inertia is often misrepresented as a force: centrifugal force. Centrifugal force is a fictitious force.

18. Gravitational force is what pulls orbiting objects into a elliptical path.

19. The magnitude of this force can be calculated from Newton’s Law of Universal Gravitation.

20. The force of attraction between two objects is directly proportional to the product of the masses of the objects and inversely proportional to the square of the distance between their centers of mass.

21. F = Gm 1 m 2 \d 2 F is force of gravitation m 1 & m 2 are masses d is distance G is the universal gravitational constant

22. Henry Cavendish measured G in 1797. G = 6.67 x 10 -11 Nm 2 /kg 2

23. Find the distance between a 0.300 kg billiard ball and a 0.400 kg billiard ball if the magnitude of the gravitational force is 8.92 x 10 -11 N.

24. We can use this law to find the mass of the earth: F w = Gm e m p /d 2

25. F w = Gm e m p /d 2 divide both sides by m p F w /m p = Gm e /d 2

26. F w /m p = Gm e /d 2 F w = m p g, or F w /m p = g, so g = Gm e /d 2

27. G & m e are constant,so g is determined only by d.

28. g = Gm e /d 2, solve for me, m e = gd 2 /G

29. m e = gd 2 /G g = 9.81 m/s 2 d = 6.37 x 10 6 m G = 6.67 x 10 -11 Nm 2 /kg 2 m e = 5.96 x 10 24 kg

30. Weight is the measure of the force of attraction between the object and the Earth.