Introduction to Mechanics Non-uniform Circular Motion Introducing Energy

Transcription

1 Introduction to Mechanics Non-uniform Circular Motion Introducing Energy Lana Sheridan De Anza College Nov 20, 2017

2 Last time applying the idea of centripetal force banked turns

3 Overview non-uniform circular motion and tangential acceleration energy and work

4 of the test tube a linthe Non-uniform axis of rotation, Circular Motion 60 g A particle can speed up or slow down while following a circular arc. It it does this it must have a component of its acceleration along the direction of its velocity. y that are many thouevices referred to as on g. Even in the relolved in a centrifuge st tube have a mass ottom of the tube is a total a t a cp x r decrease its speed. O tial to its path that cular to its path, a! cp, FIGURE 6 15 A particle moving in a ated in Figure a! circular path with tangential acceleration cp. We will 1 Figure explore from Walker, In this Physics. case, the particle s speed is

5 Non-uniform Circular Motion = 9560 g y ons that are many thouct, devices referred to as illion g. Even in the relinvolved in a centrifuge he test tube have a mass he bottom of the tube is a total a t a cp ase or decrease its speed. gential to its path that endicular to its path, a! cp, lustrated in Figure and a! cp. We will explore O FIGURE 6 15 A particle moving in a circular path with tangential acceleration In this case, the particle s speed is increasing at the rate given by a t. The centripetal acceleration a cp is toward the center of the circle and changes the direction of the velocity. The tangential acceleration a t is tangent to the circle and causes a change of speed. x

6 Problems Radial and Tangential Accelerations cm inate cenbe at e the te of mall dge). arch ntal, Figin a away. ond, lera- 41. A train slows down as it rounds a sharp horizontal M turn, going from 90.0 km/h to 50.0 km/h in the 15.0 s it takes to round the bend. The radius of the curve is 150 m. Compute the acceleration at the moment the train speed reaches 50.0 km/h. Assume the train continues to slow down at this time at the same rate. 42. A ball swings counterclockwise in a vertical circle at the end of a rope 1.50 m long. When the ball is 36.9 past the lowest point on its way up, its total acceleration is i^ j^2 m/s 2. For that instant, (a) sketch a vector diagram showing the components of its acceleration, (b) determine the magnitude of its radial acceleration, and (c) determine the speed and velocity of the ball. 43. (a) Can a particle moving with instantaneous speed 3.00 m/s on a path with radius of curvature 2.00 m have an acceleration of magnitude 6.00 m/s 2? (b) Can 1 Pageit 93, have Serway an & acceleration Jewett. of magnitude 4.00 m/s 2? In

7 Problems Radial and Tangential Accelerations cm inate cenbe at e the 41. A train slows down as it rounds a sharp horizontal M turn, going from 90.0 km/h to 50.0 km/h in the 15.0 s it takes to round the bend. The radius of the curve is 150 m. Compute the acceleration at the moment the train speed reaches 50.0 Chapter km/h. 4 Assume 159 the train continues to slow down at this time at the same rate. te of 42. A ball swings counterclockwise in a vertical circle at mall both Sketch: its speed the end and of a rope 1.50 m long. When the ball is 36.9 dge). vector will past be the lowest point on its way up, its total acceleration ial arch and radial is i^ j^2 m/s 2. For that instant, (a) sketch a The ntal, tangential vector diagram showing the components of its acceleration, (b) determine the magnitude of its radial acceleration, and (c) determine the speed and velocity of from Figin the changing hile a the radial away. the ball. from the radius of ond, 43. (a) Can a particle moving with instantaneous speed speed. lera m/s on a path with radius of curvature 2.00 m ed units from have km/h an acceleration of magnitude 6.00 m/s 2? (b) Can 1 Pageit 93, have Serway an & acceleration Jewett. ANS. FIG. of magnitude P m/s 2? In

8 Radial and Tangential Accelerations Problems cm inate cenbe at e the te of mall dge). arch ntal, Figin a away. ond, lera- 41. A train slows down as it rounds a sharp horizontal M turn, going from 90.0 km/h to 50.0 km/h in the 15.0 s it takes to round the bend. The radius of the curve is 150 m. Compute the acceleration at the moment the train speed reaches 50.0 km/h. Assume the train continues to slow down at this time at the same rate. 42. A ball swings counterclockwise in a vertical circle at the end of a rope 1.50 m long. When the ball is 36.9 past the lowest point on its way up, its total acceleration is i^ j^2 m/s 2. For that instant, (a) sketch a vector diagram showing the components of its acceleration, (b) determine the magnitude of its radial acceleration, and (c) determine the speed and velocity of the ball. 43. (a) Can a particle moving with instantaneous speed 3.00 m/s on a path with radius of curvature 2.00 m 1 Pagehave 93, Serway an acceleration & Jewett. of magnitude 6.00 m/s 2? (b) Can

9 Radial and Tangential Accelerations Problems cm inate cenbe at e the te of mall dge). 41. A train slows down as it rounds a sharp horizontal M turn, going from 90.0 km/h to 50.0 km/h in the 15.0 s it takes to round the bend. The radius of the curve is 150 m. Compute the acceleration at the moment the train speed reaches 50.0 km/h. Assume the train continues to slow down at this time at the same rate. arch ntal, Figin a away. ond, lera- 42. A ball swings counterclockwise ( ) in a vertical circle at the end of a rope 1.50 m 90.0 v i = 90.0 km/h = long. m/s When = 25.0 the m/s ball is 36.9 past the lowest point on its 3.6way up, its total acceleration is i^ j^2 m/s ( 2. For ) that instant, (a) sketch a 50.0 vector v f = diagram 50.0 km/h showing = the components m/s = 13.9of m/s its acceleration, (b) determine the 3.6 magnitude of its radial acceleration, and (c) determine the speed and velocity of the ball. 43. (a) Can a particle moving with instantaneous speed 3.00 m/s on a path with radius of curvature 2.00 m 1 Pagehave 93, Serway an acceleration & Jewett. of magnitude 6.00 m/s 2? (b) Can

10 Radial and Tangential Accelerations Problems cm inate cenbe at e the te of mall dge). 41. A train slows down as it rounds a sharp horizontal M turn, going from 90.0 km/h to 50.0 km/h in the 15.0 s it takes to round the bend. The radius of the curve is 150 m. Compute the acceleration at the moment the train speed reaches 50.0 km/h. Assume the train continues to slow down at this time at the same rate. arch ntal, Figin a away. ond, lera- 42. A ball swings counterclockwise ( ) in a vertical circle at the end of a rope 1.50 m 90.0 v i = 90.0 km/h = long. m/s When = 25.0 the m/s ball is 36.9 past the lowest point on its 3.6way up, its total acceleration is i^ j^2 m/s ( 2. For ) that instant, (a) sketch a 50.0 vector v f = diagram 50.0 km/h showing = the components m/s = 13.9of m/s its acceleration, (b) determine the 3.6 magnitude of its radial acceleration, and (c) determine the speed and velocity of Tangential the ball. accel. corresponds to changing speed: a t,avg = v t Centripetal 43. (a) Can accel. a corresponds particle moving to changing with instantaneous direction: a speed cp = v 2 r 3.00 m/s on a path with radius of curvature 2.00 m 1 Pagehave 93, Serway an acceleration & Jewett. of magnitude 6.00 m/s 2? (b) Can

11 Radial and Tangential Accelerations Problems cm inate cenbe at e the te of mall dge). arch ntal, Figin a away. ond, lera- 41. A train slows down as it rounds a sharp horizontal M turn, going from 90.0 km/h to 50.0 km/h in the 15.0 s it takes to round the bend. The radius of the curve is 150 m. Compute the acceleration at the moment the train speed reaches 50.0 km/h. Assume the train continues to slow down at this time at the same rate. 42. A ball swings counterclockwise in a vertical circle at the end of a rope 1.50 m long. When the ball is 36.9 past the lowest point on its way up, its total acceleration is i^ j^2 m/s 2. For that instant, (a) sketch a vector diagram showing the components of its acceleration, (b) determine the magnitude of its radial acceleration, and (c) determine the speed and velocity of the ball. 43. (a) Can a particle moving with instantaneous speed 1 Page , Serway m/s on & Jewett. a path with radius of curvature 2.00 m

12 Radial a circular andloop Tangential of radius Accelerations 20.0 m. same speed, 13.0 m/s at the top, l acceleration of the riders at the n the normal force at the top in ed in part (c) and on the advanrop-shaped loops. One end of a cord is fixed and a small kg object is attached to the other end, where it swings in a section of a vertical circle of radius 2.00 m as shown. When θ = 20.0, the speed of the object is 8.00 m/s. At this instant, find (a) the tension in the string, (b) the tangential and radial components of acceleration. fixed and a small attached to the ings in a section radius 2.00 m as. When u , ject is 8.00 m/s. (a) the tension e tangential and of acceleration, u S v Figure P Serway & Jewett, 9th ed., ch 6, prob 18.

13 Energy Energy is a difficult concept to define, but it is very important for physics. Energy can take many different forms. Knowing the amount of energy a system has can tell us about what states or configurations we can find the system in.

14 Energy Energy is a difficult concept to define, but it is very important for physics. Energy can take many different forms. Knowing the amount of energy a system has can tell us about what states or configurations we can find the system in. One way that energy is often described is that it represents the ability of a system to do work. We need to know what work is!

15 Work Work is an amount of energy. The amount of work done on an object depends on the applied force and the displacement of the object as the force acts.

16 Work we do work. The greater the force, the greater the work; the grea tance, the greater the work. These simple ideas form the basis for ou of work.! Work is an To amount be specific, of energy. suppose we push a box with a constant force a Figure 7 1. If we move the box in the direction of F! F, through a displace work W we have done is Fd: The amount of work done on an object depends on the applied force and Definition the displacement of Work, W, of When thea object Constant as Force the force Is in the acts. Direction of Displac W = Fd If the force is in the same direction as the displacement, SI unit: newton-meter 1N # m2 = joule, J W = Fd rce in hrough ase, t are in on the F d F

17 Work W = Fd Units of Work? They have a special name: Joules, symbol J. 1 joule = 1 J = 1 Nm Work is not a vector. Work is a scalar.

18 Let us imagine The work done by the agent on the gravitational force book Earth system is rest through a ver What is the work done in mgy lifting f mgy a 3.0-kg i. book 0.50 m? to our discussion o Example y f y i S r Physics S F app Physics S mg Figure 7.15 An external agent lifts a book slowly from a height y i to a height y f. appear as an incre the work and is at the kinetic energy Because the en the work-kinetic e appear as some fo book, we could rel (and therefore, the that was done in l tem had the potenti allowed to fall. Th is released potenti only be associated The amount of po the system. Movin may change the co

19 Let us imagine The work done by the agent on the gravitational force book Earth system is rest through a ver What is the work done in mgy lifting f mgy a 3.0-kg i. book 0.50 m? to our discussion o Example y f y i S r Physics S F app Physics appear as an incre the work and is at the kinetic energy Because the en the work-kinetic e appear as some fo book, we could rel (and therefore, the that was done in l tem had the potenti allowed to fall. Th is released potenti Figure 7.15 An external agent To lift a book withlifts constant a book slowly velocity, from requires a height Fy i app = only mg be associated to a height y f. The amount of po the system. Movin may change the co S mg

20 gest that when Work, we must cone of the force. nt represents a N through the rtant. Moving ving it 2 cm if What if the force is not in the direction of the displacement? system) undert force of magt. done by the hammer on the nail identifies the nail as the system, and the force from the hammer represents the influence from the environment. r orce on the itude Dr of Figure 7.2 An object undergoes We need to extend our definition of work. u, where u is u S F S a displacement D r S under the action of a constant force F S. (7.1) Work done by a

21 Work F is zero; W = Fd cos 90 = 0. This result leads naturally to an alternative way to think about the expression W = Fd cos u. In Figure 7 3 we show the displacement and the force for the suitcase gle p at. The of e by F F cos d F F cos For a constant applied force, Work is defined as: W = F d = Fd cos θ

22 Work In this expression: W = F d = Fd cos θ we use something called the dot product of two vectors A and B: A B = AB cos θ

23 d u Units of Work Thus, whenever we calculate work, we must be careful about its sign and assume it to be positive. Work can be positive or negative! the e- nerro gles F F F d d d 90 < < 90 = ± < < 270 W = Fd cos θ > 0 (a) (b) (c) W = Fd cos θ = 0 W = Fd cos θ < 0 positive work zero work negative work For work done on a system: Positive energy is transferred to the system. Negative energy is transferred from the system.

24 Work done by 180 individual forces Chapter 7 Energy of a System S F is the only force that does work on the block in this situation. S n mg S u r S S F have done a co tion, however, the chair, but does not move the situation d Also notice is zero when th application. Th Figure 7.3, the the gravitation dicular to the tion of D S r. The sign of done by the ap is in the same If there are several Figure forces7.3 acting An on object a system, is dis- each one can have an associated work. placed on a frictionless, horizon-

25 Work done by individual forces 180 Chapter 7 Energy of a System S F is the only force that does work on the block in this situation. S n mg S u have done a considerable tion, however, you have do the chair, but you do not does not move through a the situation depicted in F Also notice from Equa is zero when the force ap application. That is, if u 5 Figure 7.3, the work done the gravitational force on dicular to the displaceme tion of D S r. The sign of the work al done by the applied force is in the same direction a the work done by the app Figure 7.3 An object is displaced on a frictionless, horizon- surface. W mg The = normal 0 Wforce F = S n Fd cos θ W n = 0tal and the gravitational force mg S do of that force is upward, i In other words, we can no work ask on what the object. is the work done on the system by application. When the pro each force separately. placement, W is negative. Pitfall Prevention 7.3 gravitational force on the r S S F

26 Net Work The net work is the sum of all the individual works. W net = i W i where W i = F i d cos θ is the work done by the force F i. If the system can be modeled as a particle (the only case we consider in this course): W net = F net d cos θ assuming the net force is constant.

27 Question the car as it travels a distance d along the road. Picture the Problem Because f is the angle the slope makes with the horizontal, it is also the angle between was shown in Figure It follows that the angle between and the displacement is N! d! mg! d! d! mg! u u = 180. A car coasts the angle down between a hill that and makes is u = 90, anand angle the angle φ between to the F horizontal.! air and is F air N d N = 90 d mg Strategy For each force we calculate the work using where is the angle between ment d! W = Fd cos u, u. The total work is the sum of the work done by each of the three forces. The work done by the weight (mg force) is (A) positive Solution (B) negative (C) zero 1. We start with the work done by the normal force, N!. From the figure we see that u = 90 for this force: 2. For the force of air resistance, u = 180 : (D) cannot be determined W N = Nd cos u W air = F air d cos 3. For gravity the angle u is u = 90 - f, as indicated in the W mg = mgd cos figure. Recall that cos190 - f2 = sin f (see Appendix A):

28 Question the car as it travels a distance d along the road. Picture the Problem Because f is the angle the slope makes with the horizontal, it is also the angle between was shown in Figure It follows that the angle between and the displacement is N! d! mg! d! d! mg! u u = 180. A car coasts the angle down between a hill that and makes is u = 90, anand angle the angle φ between to the F horizontal.! air and is F air N d N = 90 d mg Strategy For each force we calculate the work using where is the angle between ment d! W = Fd cos u, u. The total work is the sum of the work done by each of the three forces. The work done by the normal force, N, is (A) positive Solution (B) negative (C) zero 1. We start with the work done by the normal force, N!. From the figure we see that u = 90 for this force: 2. For the force of air resistance, u = 180 : (D) cannot be determined W N = Nd cos u W air = F air d cos 3. For gravity the angle u is u = 90 - f, as indicated in the W mg = mgd cos figure. Recall that cos190 - f2 = sin f (see Appendix A):

29 Question the car as it travels a distance d along the road. Picture the Problem Because f is the angle the slope makes with the horizontal, it is also the angle between was shown in Figure It follows that the angle between and the displacement is N! d! mg! d! d! mg! u u = 180. A car coasts the angle down between a hill that and makes is u = 90, anand angle the angle φ between to the F horizontal.! air and is F air N d N = 90 d mg Strategy For each force we calculate the work using where is the angle between ment d! W = Fd cos u, u. The total work is the sum of the work done by each of the three forces. The work done by the air resistance (F air force) is (A) positive Solution (B) negative (C) zero 1. We start with the work done by the normal force, N!. From the figure we see that u = 90 for this force: 2. For the force of air resistance, u = 180 : (D) cannot be determined W N = Nd cos u W air = F air d cos 3. For gravity the angle u is u = 90 - f, as indicated in the W mg = mgd cos figure. Recall that cos190 - f2 = sin f (see Appendix A):

30 Question the car as it travels a distance d along the road. Picture the Problem Because f is the angle the slope makes with the horizontal, it is also the angle between was shown in Figure It follows that the angle between and the displacement is N! d! mg! d! d! mg! u u = 180. A car coasts the angle down between a hill that and makes is u = 90, anand angle the angle φ between to the F horizontal.! air and is F air N d N = 90 d mg Strategy For each force we calculate the work using where is the angle between ment d! W = Fd cos u, u. The total work is the sum of the work done by each of the three forces. The net (or total) work done by all forces on the car is (A) positive Solution (B) negative (C) zero 1. We start with the work done by the normal force, N!. From the figure we see that u = 90 for this force: 2. For the force of air resistance, u = 180 : (D) cannot be determined W N = Nd cos u W air = F air d cos 3. For gravity the angle u is u = 90 - f, as indicated in the W mg = mgd cos figure. Recall that cos190 - f2 = sin f (see Appendix A):

31 Summary non-uniform circ. motion and tangential acceleration energy and work Homework Walker Physics: Ch 6, onward from page 177. Problems: 29, 105, 109, 110 Ch 7, onward from page 210. Questions: 1, 3, 5; Problems: 5, 7