1. Chapter 7: Rotational Motion and the Law of Gravity Angular Speed & Acceleration  A unit of angular measure: radian y x P r  s = r  where s,r in m, and  in rad(ian) length of the arc from the x-axis s: A complete circle: s = 2  r 360 o = 2  rad 57.3 o = 1 rad

2.  Angular displacement and velocity Momentum and Impulse y x P at t f r ff ii P at t i Angular displacement:  f  i in a time interval  t = t f – t i Average angular velocity: rad/s Instantaneous angular velocity: rad/s  counter) clockwise rotation

3.  Angular acceleration Momentum and Impulse y x P at t f r ff ii P at t i Average angular acceleration: Instantaneous angular acceleration: rad/s 2

4.  An example of a rigid body Momentum and Impulse The distance of any two points in a rigid object does not change when the body is even in motion. When a rigid object rotates about a fixed axis, every portion of the object has the same angular speed and the same angular acceleration.

5.  One-to-one correspondence between linear and angular quantities Rotational Motion under Constant Angular Acceleration Similarity between  av and v av Similar derivation used for linear quantities can be used for angular quantities.

6.  An example Rotational Motion under Constant Angular Acceleration Example 7.2 : A rotating wheel A wheel rotates with a constant angular acceleration of 3.50 rad/s 2. If the angular speed of the wheel is 2.00 rad/s at t i =0, (a) through what angle does the wheel rotate between t=0 and t=2.00 s? (b) What is the angular speed of the wheel at t=2.00 s?

7.   and v Relations between Angular and Linear Quantities Consider an object rotating about the z-axis and a point P on it. tangential speed tangent to circle The tangential speed of a point on a rotating object equals the distance of that point from the axis of rotation multiplied by the angular speed.

8.   and a Relations between Angular and Linear Quantities Consider an object rotating about the z-axis and a point P on it. tangential acceleration tangent to circle The tangential acceleration of a point on a rotating object equals the distance of that point from the axis of rotation multiplied by the angular acceleration.

9.  Acceleration at a constant speed Centripetal Acceleration Consider a car moving in a circular path with constant linear speed v. Even though the magnitude of v i and v f are the same,  v can be non-zero if their directions are different. This leads to non-zero acceleration called centripetal acceleration.

10.  Centripetal acceleration Centripetal Acceleration Consider a car moving in a circular path with constant linear speed v. Triangle OAB and the triangle in Fig. (b) are similar. Total acceleration

11.  Vector nature of angular quantities Centripetal Acceleration Angular quantities are vector and their directions are defined as:  points into the page  points out of the page

12. Centripetal Acceleration  Forces causing centripetal acceleration An object can have a centripetal acceleration only if some external force acts on it. An example is a ball whirling in a circle at the end of a string. In this case the tension in the string is the force that creates the centripetal force. Net centripetal force F c is the sum of the radial components of all forces acting on a given object. T=F c A net force causing a centripetal acceleration acts toward the center of the circular path. If it vanishes, the object would immediately leave its circular path and move along a straight line tangent to the circle.

13. Centripetal Acceleration  Examples Example 7.7 : Buckle up for safety v=13.4 m/s r=50.0 m Find the minimum coefficient of static friction  s between tires and roadway to keep the car from sliding.

14. Centripetal Acceleration  Examples Example 7.8 : Daytona International Speedway (a) Find the necessary centripetal acceleration on the banked curve so that the car will not slip due to the inclination (neglect friction). q=31.0 o r=316 m y-component (vertical) : x-component (horizontal) : (b) Find the speed of the car.

15. Centripetal Acceleration  Examples Example 7.9 : Riding the tracks (a) Find the speed at the top. R=10.0 m (b) Find the speed at the bottom.

16. Centripetal Acceleration  Examples Example 7.9 : Riding the tracks (cont’d) (a)Find the normal force on a passenger at the bottom if R=10.0 m n does not depend on R!

17. Newtonian Gravitation  Law of universal gravitation Using accumulated data on the motions of the Moon and planets, and his first law, Newton deduced the existence of the gravitational force that is responsible for the movement of the Moon and planets and this force acts between any two objects. If two particles with mass m 1 and m 2 are separated by a distance r, then a gravitational force acts along a line joining them with magnitude constant of universal gravity Newton’s 2 nd law

18. Newtonian Gravitation  Law of universal gravitation (cont’d) The gravitational force exerted by a uniform sphere on a particle outside the sphere is the same as the force exerted if the entire mass of the sphere were concentrated at its center. This is a result from Gauss’s law and stems from the fact that the gravitational force is inversely proportional to square of the distance between two particles. The expression F=mg is valid only near the surface of Earth and can be derived from Newton’s law of universal gravitation.

19. Newtonian Gravitation  Gravitational potential energy revisited Gravitational potential energy near Earth (approximation) General form of gravitational potential energy due to Earth radius of Earthmass of Earth This is a special case where the zero level for potential energy is at an infinite distance from the center of Earth. The gravitational potential energy associated with an object is nothing more than the negative of the work done by the force of gravity in moving the object.

20. Newtonian Gravitation  Gravitational potential energy revisited (cont’d) Derivation of gravitational potential energy near Earth

21. Newtonian Gravitation  Escape speed If an object is projected upward from Earth’s surface with a large enough speed, it can soar off into space and never return. This speed is called Earth’s escape speed v esc. The initial mechanical energy of the object-Earth system is: If we neglect air resistance and assume that the initial speed is large enough to allow the object to reach infinity with a speed of zero, this value of v i is the escape speed v esc. 4.3 km/s for Mercury 11.2 for Earth 2.3 for Moon 60.0 for Jupiter

22. Newtonian Gravitation  Examples Example 7.10 : Billiards m 1,2,3 =0.300 kg (a)Find the net gravitational force on the cue ball.

23. Newtonian Gravitation  Examples Example 7.10 : Billiards (cont’d) m 1,2,3 =0.300 kg (b) Find the components of the force of m 2 on m 3.

24. Newtonian Gravitation  Examples Example 7.12 : A near-Earth asteroid m 1,2,3 =0.300 kg An asteroid with mass m=1.00x10 9 kg comes from infinity, and falls toward Earth. Find the change in potential energy when it reaches a point 4.00x10 8 m from Earth. Find the work done by gravity. r i =0. (b) Find the speed of the asteroid when it reaches r f =4.00x10 8 m. (c) Find the work needed to reduce the speed by half.