1. Unit 4, 5, and 6 Review

2. Newton’s Law of Universal Gravitation is based on two points First: As distance between two objects increases, gravitational force decreases Distance increases Force decreases Distance decreases Force increases

3. Second: Gravitational attraction or force on a body depends on its MASS. A large mass and a large mass = Large gravitational force A large mass and a small mass = Large gravitational force A small mass and a small mass = Small gravitational force

4. Combining 1 and 2 produces the formula for Newton’s Law of Universal Gravitation FForce of gravitational attractionN GUniversal Gravitational Constant6.67 x 10 -11 Nm 2 /kg 2 m1m1 Mass of object 1kg m2m2 Mass of object 2kg d (r)Distance between the centers of m 1 and m 2 m

5. In the drawing below, two small masses are separated by 20.0 cm. They attract each other with a force of 10.0 N. a)When they are 10.0 cm apart, these masses will attract each other with what force? b)When they are 5.0 cm apart, these masses will attract each other with what force? c)When they are 40.0 cm apart, these masses will attract each other with what force?

6. Calculate the gravitational force of attraction between two 55 kg people standing 0.25 m apart.

7. What is the relationship between the “Gravitational Force” (F) in Newton’s Law of Universal Gravitation and “Weight” (W) ? Both are measured in Newtons (N), therefore both are forces. Most importantly both measure the same thing…the force of gravity on an object. Therefore…

8. Torque Torque is how we describe a force that causes an object to rotate, such a twisting force. The symbol for torque is the Greek symbol tau (  ) The ability to rotate an object depends not only on the force applied (F), but also on how far that force is applied from the point of rotation, or pivot point (O). The distance between the point where the force is applied to the object, and the point about which the object rotates is called the lever arm. Thus the torque, or twisting force is given by the relation

9. Torque Torque = (Magnitude of the force) X (Lever arm distance)  = FdsinƟ The units of torque are derived from the equation, Nm Force: F unit N (Newton) Lever arm distance: dunit m (meter) Ɵ angle that force is ……………………………………… applied to lever arm

10. Torque Torque is a pseudo-vector, its direction is described by how it would make the object rotate; clockwise, or counter- clockwise. Torque that causes a clockwise rotation is considered negative Torque that causes a counter-clockwise rotation is considered positive

11. The torque required to loosen a nut on a wheel has a magnitude of 90 Nm and the force exerted by the technician is 260 N. How far from the nut must the technician apply the force?

12. Torque Torque Producing Forces in Equilibrium How can two people of different weights balance on a see saw? THE TORQUES MUST BE EQUAL!!!!  1 =  2 F 1 d 1 = F 2 d 2 (m 1 g) d 1 = (m 2 g) d 2

13. Two kids are on a 14m see saw. If kid A weighs 600N and sits 3.5m from the center, and kid B weighs 700N, where should kid B sit so that they are perfectly balanced?  1 =  2 F 1 d 1 = F 2 d 2 (400N) 3m = (600N) d 400N 600N 3m 7m d? 7m

14. 14 Uniform circular motion motion of an object in a circle with a constant speed With a constant change in direction *Is the velocity constant?

15. 15 a c = v² r Centripetal Acceleration Because the direction of the motion is changing, the VELOCITY IS CHANGING, then by definition there must be an ACCELERATION. The acceleration CAN’T be in the direction of motion because the object would speed up and the MOTION WOULDN’T BE UNIFORM.

16. 16 A car moving at 40 m/s goes around a curve with a radius of 60 m. What is the car’s acceleration? a c = v² = (40m/s) 2 r 60m a c = 27 m/s 2

17. 17 Applying Newton’s 2 nd Law: to produce any acceleration requires a net force Centripetal Force a net force is needed to produce centripetal acceleration. This net force is CENTRIPETAL FORCE, which is also directed INWARD toward the center of the circle. Centripetal force is the magnitude of the force required to maintain uniform circular motion.

18. 18 What is the centripetal force on the car in the previous example, if the mass of the car is 700 kg? F c = mv 2 = (700kg)(40m/s) 2 r 60m F c = 18667N

20. 20 Direction of Centripetal Force, Acceleration and Velocity Without a centripetal force, an object in motion continues along a straight-line path.

22. 22 Direction of Centripetal Force, Acceleration and Velocity

23. 23 Relationship Between Variables of Uniform Circular Motion Suppose two identical objects go around in horizontal circles of identical diameter but one object goes around the circle twice as fast as the other. The force required to keep the faster object on the circular path is A.the same as B.one fourth of C.half of D.twice E.four times the force required to keep the slower object on the path. The answer is E. As the velocity increases the centripetal force required to maintain the circle increases as the square of the speed.

24. 24 Relationship Between Variables of Uniform Circular Motion Suppose two identical objects go around in horizontal circles with the same speed. The diameter of one circle is half of the diameter of the other. The force required to keep the object on the smaller circular path is A.the same as B.one fourth of C.half of D.twice E.four times the force required to keep the object on the larger path. The answer is D. The centripetal force needed to maintain the circular motion of an object is inversely proportional to the radius of the circle. Everybody knows that it is harder to navigate a sharp turn than a wide turn.

25. 25 Relationship Between Variables of Uniform Circular Motion Suppose two identical objects go around in horizontal circles of identical diameter and speed but one object has twice the mass of the other. The force required to keep the more massive object on the circular path is A.the same as B.one fourth of C.half of D.twice E.four times Answer: D.The mass is directly proportional to centripetal force.

26. 26 Tension Can Yield a Centripetal Acceleration: If the person doubles the speed of the airplane, what happens to the tension in the cable? Doubling the speed, quadruples the force (i.e. tension) required to keep the plane in uniform circular motion.

27. 27 Centripetal Force: Question Smaller radius: larger force required to keep it in uniform circular motion. A car travels at a constant speed around two curves. Where is the car most likely to skid? Why?

28. 28 A 615 kg race car completes 1.00 lap in 14.3 sec around a circular track with a radius of 50.0 m. What is the car’s T, v, a, F (on tires)? V = (2)(π)(50.0) 14.3 = 22.0 m/s a = (22.0 m/s)² 50.0 = 9.68 m/s² F (on tires) = (615 kg)(22.0 m/s)² 50.0 m = 5,950 N

29. Chapter 6 Linear Momentum Momentum is defined as mass times velocity. Momentum is represented by the symbol p, and is a vector quantity. p = mv momentum = mass  velocity As mass increase momentum increases As velocity increase momentum increases.

30. Day 1: Practice Problem 1.When comparing the momentum of two moving objects, which of the following is correct? a)The object with the higher velocity will have less momentum if the masses are equal. b)The more massive object will have less momentum if its velocity is greater. c)The less massive object will have less momentum if the velocities are the same. d)The more massive object will have less momentum if the velocities are the same. 2.What is the momentum of a 60 kg child running at 3 m/s?

31. Day 2: Practice Answers : 1. C 2. p = mv = (60.0 kg) (3.0 m/s) = 180 kg * m/s