Chapter 6 – Circular Motion and Other applications on Newton’s Laws

1. Chapter 6 – Circular Motion and Other applications on Newton’s Laws • Uniform circular motion. • Drag forces and terminal speed. • Other Applications of Newton’s Laws

2. Uniform Circular Motion • A force, Fr, is directed toward the center of the circle • This force is associated with an acceleration, ac • Applying Newton’s Second Law along the radial direction gives

3. Uniform Circular Motion • A force causing a centripetal acceleration acts toward the center of the circle • It causes a change in the direction of the velocity vector • If the force vanishes, the object would move in a straight-line path tangent to the circle

4. Centripetal Force • The force causing the centripetal acceleration is sometimes called the centripetal force • This is not a new force, it is a new role for a force • It is a force acting in the role of a force that causes a circular motion

5. Conical Pendulum • The object is in equilibrium in the vertical direction and undergoes uniform circular motion in the horizontal direction (1) Tcosθ = mg (2) Tsinθ = mac= mv2/r

6. Conical Pendulum • Dividing (2) by (1) and using sinθ/cosθ = tanθ we eliminate T and find: • v is independent of m Using r = Lsinθ

7. Consider a conical pendulum with an 80.0-kg bob on a 10.0-m wire making an angle of 5.00 with the vertical. Determine (a) the horizontal and vertical components of the force exerted by the wire on the pendulum and (b) the radial acceleration of the bob.

8. A 4.00-kg object is attached to a vertical rod by two strings, as in Figure. The object rotates in a horizontal circle at constant speed 6.00 m/s. Find the tension in (a) the upper string and (b) the lower string.

9. How fast can it spin? • The speed at which the object moves depends on the mass of the object and the tension in the cord. The centripetal force is supplied by the tension: This shows that v increases with T and decreases with larger m.

10. Horizontal (Flat) Curve • The force of static friction supplies the centripetal force • The maximum speed at which the car can negotiate the curve is • Note, this does not depend on the mass of the car Fr

11. Banked Curve • These are designed with friction equaling zero • There is a component of the normal force that supplies the centripetal force (1), and component of the normal force that supplies the gravitational force (2). • Dividing (1) by (2) gives:

12. Loop-the-Loop • A pilot in a jet aircraft executes a loop-the loop. This is an example of a vertical circle • At the bottom of the loop, (b), the upward constant force, that keeps the pilot moving in a circular path at a constant speed, is greater than its weight, because the normal and gravitational forces act in opposite direction:

13. Loop-the-Loop • At the top of the circle, (c), the force exerted on the object is less than its weight, because both the gravitational force and the normal force, ntop, exerted on the pilot by the seat act in same direction:

14. A 0.400-kg object is swung in a vertical circular path on a string 0.500 m long. If its speed is 4.00 m/s at the top of the circle, what is the tension in the string there?

15. Non-Uniform Circular Motion • When the force acting on a particle moving in a circular path has a tangential component , the particles speed changes. • The acceleration has a tangential components • Fr produces the centripetal acceleration • Ftproduces the tangential acceleration. • The total force is the vector sum or the radial force and tangential force:

16. Vertical Circle with Non-Uniform Speed • The gravitational force exerts a tangential force on the object • Look at the components of Fg • The gravitational force resolve into a tangential component mg sinθ and a radial component mg cosθ.

17. Vertical Circle with Non-Uniform Speed • Applying II NL to the tangential and radial directions: The tension at any point can be found:

18. Tarzan (m = 85.0 kg) tries to cross a river by swinging from a vine. The vine is 10.0 m long, and his speed at the bottom of the swing (as he just clears the water) will be 8.00 m/s. Tarzan doesn't know that the vine has a breaking strength of 1 000 N. Does he make it safely across the river?

19. Motion in accelerated Frames When Newton’s laws of motion was introduced in Chapter 5, we emphasized that they are valid only for inertial frames of reference. In this section, we will analyze the noninertial frames, that is, on that is accelerating. Example: Let’s consider a hockey puck on a table in a moving train. The train moving with a constant velocity represents an inertial frame. The puck at rest remains at rest, and Newton’s I low is obeyed.

20. Motion in accelerated Frames The accelerating train is not an inertial frame. For the observer on the train, there appears to be no visible force on the puck, but it will accelerate from rest toward the back of the train, as the train start to accelerate. The Newton’s I law is violated. The observer on the accelerating train, if he applied the N II law to the puck, might conclude that a force has acted on the puck to cause it to accelerate. We call an apparent force such as this a fictitious force.

21. Motion in Accelerated Frames • A fictitious force results from an accelerated frame of reference • A fictitious force appears to act on an object in the same way as a real force, but you cannot identify a second object for the fictitious force

22. “Centrifugal” Force • From the frame of the passenger (b), a force appears to push her toward the door • From the frame of the Earth, the car applies a leftward force on the passenger • The outward force is often called a centrifugal force • It is a fictitious force due to the acceleration associated with the car’s change in direction

23. If the coefficient of static friction between your coffee cup and the horizontal dashboard of your car is μs = 0.800, how fast can you drive on a horizontal roadway around a right turn of radius 30.0 m before the cup starts to slide? If you go too fast, in what direction will the cup slide relative to the dashboard?

24. A person stands on a scale in an elevator. As the elevator starts, the scale has a constant reading of 591 N. As the elevator later stops, the scale reading is 391 N. Assume the magnitude of the acceleration is the same during starting and stopping, and determine (a) the weight of the person, (b) the person's mass, and (c) the acceleration of the elevator.

25. “Coriolis Force” • This is an apparent force caused by changing the radial position of an object in a rotating coordinate system • The result of the rotation is the curved path of the ball

26. The Earth rotates about its axis with a period of 24.0 h. Imagine that the rotational speed can be increased. If an object at the equator is to have zero apparent weight, (a) what must the new period be? (b) By what factor would the speed of the object be increased when the planet is rotating at the higher speed? Note that the apparent weight of the object becomes zero when the normal force exerted on it is zero.

27. Fictitious Forces, examples • Although fictitious forces are not real forces, they can have real effects • Examples: • Objects in the car do slide • You feel pushed to the outside of a rotating platform • The Coriolis force is responsible for the rotation of weather systems and ocean currents

28. Fictitious Forces in Linear Systems • The inertial observer (a) sees • The noninertial observer (b) sees

29. Fictitious Forces in a Rotating System • According to the inertial observer (a), the tension is the centripetal force • The noninertial observer (b) sees

30. Motion with Resistive Forces • Motion can be through a medium • Either a liquid or a gas • The medium exerts a resistive force, R, on an object moving through the medium • The magnitude of R depends on the medium • The direction of R is opposite the direction of motion of the object relative to the medium • R nearly always increases with increasing speed

31. Motion with Resistive Forces • The magnitude of R can depend on the speed in complex ways • We will discuss only two • R is proportional to v • Good approximation for slow motions or small objects • R is proportional to v2 • Good approximation for large objects

32. R Proportional To v • The resistive force can be expressed as R = - b v • b depends on the property of the medium, and on the shape and dimensions of the object • The negative sign indicates R is in the opposite direction to v

33. R Proportional to v, Example • Analyzing the motion results in

34. R Proportional to v • Initially, v = 0 and dv/dt = g • As tincreases, R increases and adecreases • The acceleration approaches 0 when R→mg • At this point, vapproaches the terminal speed of the object

35. Terminal Speed • To find the terminal speed, let a = 0 • Solving the differential equation gives • t is the time constant and t = m/b

36. R Proportional to v2 • For objects moving at high speeds through air, the resistive force is approximately equal to the square of the speed R = ½ DrAv2 • D is a dimensionless empirical quantity that called the drag coefficient -r is the density of air • A is the cross-sectional area of the object • v is the speed of the object

37. R Proportional to v2 • Analysis of an object falling through air accounting for air resistance

38. R Proportional to v2, Terminal Speed • The terminal speed will occur when the acceleration goes to zero • Solving the equation gives

39. Some Terminal Speeds

40. Process for Problem-Solving • Analytical Method • The process used so far involves the identification of well-behaved functional expressions generated from algebraic manipulation or techniques of calculus

41. Analytical Method • Apply the method using this procedure: • Sum all the forces acting on the particle to find the net force, SF. • Use this net force to determine the acceleration from the relationship a =SF/m • Use this acceleration to determine the velocity from the relationship dv/dt = a • Use this velocity to determine the position from the relationship dx/dt = v

42. Analytic Method, Example • Applying the procedure: Fg = may = - mg ay = -g and dvy/dt = -g vy(t) = vyi – gt y(t) = yi + vyi t – ½ gt2

43. Numerical Modeling • In many cases, the analytic method is not sufficient for solving “real” problems • Numerical modeling can be used in place of the analytic method for these more complicated situations • The Euler method is one of the simplest numerical modeling techniques

44. Euler Method • In the Euler Method, derivatives are approximated as ratios of finite differences • Dtis assumed to be very small, such that the change in acceleration during the time interval is also very small

45. Equations for the Euler Method

46. Euler Method • It is convenient to set up the numerical solution to this kind of problem by numbering the steps and entering the calculations into a table • Many small increments can be taken, and accurate results can be obtained by a computer

47. Euler Method Set Up

48. Euler Method Final • One advantage of the method is that the dynamics are not obscured • The relationships among acceleration, force, velocity and position are clearly shown • The time interval must be small • The method is completely reliable for infinitesimally small time increments • For practical reasons a finite increment must be chosen • A time increment can be chosen based on the initial conditions and used throughout the problem • In certain cases, the time increment may need to be changed within the problem

49. Accuracy of the Euler Method • The size of the time increment influences the accuracy of the results • It is difficult to determine the accuracy of the result without knowing the analytical solution • One method of determining the accuracy of the numerical solution is to repeat the solution with a smaller time increment and compare the results • If the results agree, the results are correct to the precision of the number of significant figures of agreement

50. Euler Method, Numerical Example