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Motion in Two Dimensions: Projectile Motion Circular Motion Angular Speed Simple Harmonic Motion Torque Center of Mass.

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Presentation on theme: "Motion in Two Dimensions: Projectile Motion Circular Motion Angular Speed Simple Harmonic Motion Torque Center of Mass."— Presentation transcript:

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2 Motion in Two Dimensions: Projectile Motion Circular Motion Angular Speed Simple Harmonic Motion Torque Center of Mass

3 Projectile Motion A red marble is dropped off a cliff at the same time a black one is shot horizontally. At any point in time the marbles are at the same height, i.e., they’re falling down at the same rate, and they hit the ground at the same time. Gravity doesn’t care that the black ball is moving sideways; it pulls it downward just the same. Since gravity can’t affect horiz. motion, the black particle continues at a constant rate. With every unit of time, the marbles’ vertical speed increases, but their horiz. speed remains the same (ignoring air resistance). continued on next slide

4 Projectile Motion continued on next slide Gravity’s downward pull is independent of horiz. motion. So, the vertical acceleration of each marble is - g (for the whole trip), and the sideways acceleration of each is zero. (Gravity can’t pull sideways). Whatever horiz. velocity the black one had when shot is a constant throughout its trip. Only its vertical velocity changes. (A vertical force like gravity can only produce vertical acceleration.) 9.8 m/s 2

5 Projectile Motion (cont.) continued on next slide t = 0 t = 1 t = 2 t = 3 t = 4 v y = 1 v y = 2 v y = 3 v y = 4 v y = 0 If after one unit of time the marbles have one unit of speed downward, then after two units of time they have two units of speed downward, etc. This follows directly from v f = v 0 + a t. Since v 0 = 0, downward speed is proportional to time. Note: The vectors shown are vertical components of velocity. The shot marble has a horizontal component too (not shown); the dropped one doesn’t.

6 Since the shot black marble experiences no horiz. forces (ignoring air), it undergoes no horiz. acceleration. Therefore, its horiz. velocity, doesn’t change. So, the horiz. vector has a constant magnitude, but the vertical vector gets longer. The resultant (the net velocity vector in blue) gets longer and points more downward with time. When t = 0, v = v x for the shot marble. v = v y for the dropped marble for the whole trip. Projectile Motion (cont.) continued on next slide v x = v vxvx vxvx vxvx vxvx v v v v y t = 0 t = 1 t = 2 t = 3 t = 4

7 Projectile Motion (cont.) The trajectory of any projectile is parabolic. (We’ll prove this later.) If its initial velocity vector is horizontal, as with the black marble, the launch site is at the vertex of the parabola. The velocity vector at any point in time is tangent to the parabolic trajectory. Moreover, velocity vectors are always tangent to the trajectory of any moving object, regardless of its shape. v v v continued on next slide

8 x = 1 Projectile Motion (cont.)  y = 1  y = 3  y = 5  y = 7 x = 2 x = 3 x = 4 continued on next slide The vertical displacements over consecutive units of time show the familiar ratio of odd numbers that we’ve seen before with uniform acceleration. Measured from the starting point, the vertical displacements would be 1, 4, 9, 16, etc., (perfect squares), but the horiz. displacements form a linear sequence since there is no acceleration in that direction. t = 0 t = 1 t = 2 t = 3 t = 4

9 Projectile Example A rifle is held perfectly horizontally 1.5 m over level ground. At the instant the trigger is pulled, a second bullet is dropped from the tip of the barrel. The muzzle velocity of the gun is 80 m/s. 1. Which bullet hits the ground first? answer: 2. How fast is each bullet moving after 0.3 s ? answer: dropped bullet after 0.3 s fired bullet after 0.3 s 80 m/s They hit at same time. vyvy vyvy Use v f = v 0 + a t and use vertical info only: v 0 = 0, a = -9.8 m/s 2, and t = 0.3 s. We get v y in the pic for each bullet is -2.94 m/s. Using the Pythagorean theorem for the fired bullet we get 80.054 m/s in a direction tangent to its path. continued on next slide

10 Projectile Example (cont.) 80 m/s 1.5 m 3. How far away does the fired bullet land (its range)? answer: The first step is to find the its hang time. This is the same hang time as the dropped bullet. Use  y = v 0 t + 0.5 a t 2 with only vertical data: -1.5 = (0) t + (0.5) (-9.8) t 2. So, t = 0.5533 s. The whole time the bullet is falling it’s also moving to the left at a constant 80 m/s. Since horizontally v is constant, we use d = v t with only horiz. info: d = (80 m/s) (0.5533 s) = 44.26 m. Note: When a = 0,  x = v 0 t + 0.5 a t 2 breaks down to d = v t.

11 Now let’s find range of a projectile fired with speed v 0 at an angle . Step 1: Split the initial velocity vector into components. Projectiles Fired at an Angle v 0 cos  v 0 sin  v0v0  continued on next slide

12 Step 2: Find hang time. Use  y = v 0 t + ½ a t 2 with only vertical data: v 0 cos  v 0 sin  v0v0   y = ( v 0 sin  ) t + ½ (-g)t 2 Over level ground,  y = 0. Divide through by t : 0 = v 0 sin  - 4.9 t, and t = ( v 0 sin  ) / 4.9 Note: If we had shot the projectile from a 100 m cliff,  y would be -100 m. continued on next slide Projectiles Fired at an Angle (cont.)

13 v 0 cos  Step 3: Now that we know how long it’s in the air, we know how long it travels horizontally. (The projectile’s vertical and horizontal movements are completely independent.) Use  x = v 0 t + ½ a t 2 again, this time with only horizontal data: v 0 sin  v0v0   x = (v 0 cos  ) t + ½ (0) t 2 = (v 0 cos  ) t This is the same as saying: horiz. distance = horiz. speed  time In other words, d = v t continued on next slide Projectiles Fired at an Angle (cont.)

14 Picklemobile Example A stuntman drives a picklemobile off a 350 m cliff going 70 mph. The angle of elevation of the cliff is 21 . He’s hoping to make it across a 261 m wide river and land on a ledge 82 m high. Does he make it ? 70 mph 350 m 82 m Well, the first thing we have to do is convert the initial velocity into m/s: 70 mi h 1609 m mi h 3600 s = 31.2861 m/s 261 m continued on next slide 21 

15 Picklemobile Example (cont.) 31.2861 m/s 350 m We resolve the initial velocity into components. 21  11.2119 m/s 29.2081 m/s Then we find the picklemobile’s hang time (which is the same as if it had been shot straight up at about 11.2 m/s), with  y = 82 m - 350 m = -268 m. -268 = 11.2119 t - 4.9 t 2 4.9 t 2 - 11.2119 t - 268 = 0 t = -6.3394 s or 8.6276 s (using quadratic formula or computer) continued on next slide 82 m continued on next slide 261 m

16 Picklemobile Example (cont.) 29.2081 m/s continued on next slide We want the positive answer for t. The interpretation of the negative answer is that if the pickle car had been launched from the height of the ledge, it would have taken about 6.3 s to reach the edge of the cliff. Anyway, for 8.62757 s the pickle mobile is in the air and traveling to the right at about 29 m/s. Therefore, its range is (29.2081 m/s) (8.6276 s)  252 m < 261 m. Alas, the poor picklemobile doesn’t make it. 82 m

17 Picklemobile Example (cont.) parabolic trajectory 350 m What max height does the pickle mobile attain? 11.2119 m/s It attains the same max height as if it had been shot up at about 11.2 m/s. Since its vertical velocity is zero at its high pt., we have 0 2 - (11.2119) 2 = 2(-9.8)  y. So,  y = 6.41 m. Add 350 m and the max height is 356.41 m. continued on next slide 82 m

18 83.5805 m/s 29.2081 m/s Picklemobile Example (cont.) What is the impact velocity of the pickle mobile (the velocity upon splash down)? 29.2081 m/s 11.2119 m/s The horiz. component is the same at landing as it was on liftoff. We must find the final vertical velocity: v f 2 - (11.2119) 2 = 2(-9.8) (-350). So, v f = -83.5805 m/s. 350 m The Pythag. theorem gives us the magnitude of the resultant.  = tan -1 (83.5805 / 88.5371) = 70.74 . Thus the impact velocity is about 88.5 m/s at 71  below the horizontal.  88.5371 m/s

19 Parabolic Proof A projectile is shot with speed v 0 at an angle . Its vertical position is given by y = (v 0 sin  ) t + ½ (-g) t 2. Here y is the dependent quantity, and t is the independent quantity. Everything else is a constant. The projectile’s horizontal position is given by x = (v 0 cos  ) t. Only x and t are variables, and t = x / (v 0 cos  ). Let’s substitute this for t in the equation for y : y = (v 0 sin  ) t + ½ (-g) t 2 y = (v 0 sin  ) [ x / (v 0 cos  ) ] - ½ g [ x / (v 0 cos  ) ] 2 y = (tan  ) x - g 2 v 0 2 cos 2  x 2x 2 The coefficients of x and x 2 are constants. Since the leading coef. is negative, this is the equation of a parabola opening down.

20 Symmetry and Velocity   The projectile’s speed is the same at points directly across the parabola (at the same vertical position). The angle is the same too, but with opposite orientation. Horizontal speeds are the same throughout the trajectory. Vertical speeds are the same only at points of equal height.  The vert. comp. shrinks then grows in opposite direction at a const. rate (- g ). The resultant velocity vector’s orientation and magni- tude changes, but is always tangent. The horiz. comp. doesn’t change. At the peak, the horiz. comp. equals the resultant velocity vector.

21 Symmetry and Time t = 0 t = 10 t = 20 t = 15 t = 5 t = 3 t = 17 Over level ground, the time at the peak is half the hang time. Notice the symmetry of times at equal heights relative to the 10 unit mark. The projectile has covered half its range when it has peaked, but only over level ground. Note: near the peak the object moves more slowly than when lower to the ground. It rises 3/4 of its max height in only 1/2 of its rising time. (See if you can prove this for an arbitrary launch velocity.)

22 Max height & hang time depend only on initial vertical velocity Each initial velocity vector below has the a different magnitude (speed) but each object will spend the same time in the air and reach the same max height. This is because each vector has the same vertical component. The projectiles will have different ranges, however. The greater the horizontal component of initial velocity, the greater the range.

23 Max Range at 45  Over level ground at a constant launch speed, what angle maximizes the range, R ? First consider some extremes: When  = 0, R = 0, since the object is on the ground from the moment it’s launched. When  = 90 , the object goes straight up and lands right on the launch site, so R = 0 again. The best angle is 45 , smack dab between the extremes. 45  Here all launch speeds are the same; only the angle varies. 38  76  proof on next slide

24 Range Formula & Max Range at 45  First find the time. Note that  y = 0, since the projectile starts and stops at ground level (no change).  y = v 0 t + ½ a t 2. So, 0 = (v 0 sin  ) t - ½ g t 2 Since the ground is level we divide through by t giving us t = 2 v 0 sin  / g. Then, R = ( v 0 cos  ) t = (v 0 cos  ) (2 v 0 sin  / g) = 2 v 0 2 sin  cos  / g. By the trig identity sin 2  = 2 sin  cos , we get R = v 0 2 sin 2  / g. Since v 0 and g are fixed, R is at a max when sin 2  is at a max. When the angle, 2 , is 90 , the sine function is at its maximum of 1. Therefore,  = 45 .  v 0 cos  v 0 sin  v0v0

25 Max Range when  y  0 When fired from a cliff, or from below ground, a projectile doesn’t attain its max range at 45 . 45  is only the best angle when a projectile is fired over level ground. When fired from a cliff, a projectile attains max range with a launch angle less than 45  (see next slide). When fired from below ground, a projectile attains max range with a launch angle greater than 45 .

26 Range when fired from cliff 45  < 45  If ground were level, the 45  launch would win. Because the < 45  parabola is flatter it eventually overtakes 45  parabola. Launch speeds are the same.

27 Ranges at complementary launch angles 15  75  40  50  An object fired at angle  will have the same range as when it’s fired at the same speed at an angle 90  - . Reason: R = 2v 0 2 sin  cos  / g, and the sine of an angle is the cosine of its complement (and vice versa). For example, R at 40  is 2v 0 2 sin 40  cos 40  / g = 2v 0 2 cos 50  sin 50  / g = R at 50 .

28 Monkey in a Tree Monkey in a tree web site Here’s a classic physics problem: You want to shoot a banana at a monkey up in a tree. Knowing that the monkey will get scared and let go of the branch the instant he hears the sound of the banana gun, how should you aim: a little above, a little below, or right at him?

29 Monkey in a tree explanation The reason you should aim right at the monkey even though the monkey lets go right when you pull the trigger is because both the monkey and the banana are in the air for the same amount of time before the collision. So, with respect to where they would have been with no gravity, they fall the same distance. banana path without gravity path with gravity “gravity-free” monkey monkey w/ gravity

30 Homerun Example From home plate to the center field wall at a ball park is 130 m. When a batter hits a long drive the ball leaves his bat 1 m off the ground with a velocity of 40 m/s at 28  above the horizontal. The center field wall is 2.6 m high. Does he hit a homerun? Let’s first check the range to see if it even has a chance: R = v 0 2 sin 2  / g = 40 2 sin 56° / 9.8 = 135.35 m. So, if it can clear the wall, it will make it. We need to determine its vertical position when its horizontal position is 130 m. If it’s 1.6 m or more, it’s a homer. Let’s first find the time when the ball is 130 m away (horizontally) from the point where it was walloped. 40 m/s 130 m 2.6 m 1 m } 28  continued on next slide

31 Homerun (cont.) t = d / v = (130 m) / (40 cos 28  ) = 3.68085 s. Let’s see how high up it is at this time:  y = (40 sin 28  ) (3.68085) - 4.9 (3.68085) 2 = 2.73 m, which is 3.73 m above the ground, out of the reach of a leaping outfielder. Therefore, it’s a homer! In real life the batter wouldn’t be so fortunate. His hit would only have gone a fraction of the distance with air resistance. What is barely a homer in a vacuum is a mere blooper in air. Homerun balls in the major league would go about 700 or 800 ft in a vacuum! This means balls would routinely be hit out over the stadium into the parking lot.

32 wheeled wooden frame that rolls backwards during launch Trebuchet Example A trebuchet launches a 180 kg lead sphere at the wall of a medieval castle 120 m away. The projectile impacts the wall 23 m up. Its high point in the air occurs 2/3 of the way to the wall. The trebuchet takes 0.6 s to launch the sphere and releases it at a height of 5 m. Find the launch velocity of the sphere and the average force the trebuchet exerts on it. Trebuchet lead sphere tree trunk swinging counterweight axle rope continued on next slide

33 Trebuchet (cont.) (0, 5) (120, 23) impact (80, h )  v cos  v sin  v The high pt. occurs horizontally at a distance of half of what its range would be over level ground, which is 2 / 3 of 120 m. continued on next slide So, 80 = R / 2 = v 2 sin 2  / 2g  v 2 = 1568 / sin 2 

34 Trebuchet (cont.) Let t = time at impact. Horizontally: 120 = (v cos  ) t. Vertically: 18 = (v sin  ) t - 4.9 t 2. Now substitute for t : 18 = (v sin  ) [120 / (v cos  ) ] - 4.9 [120 / (v cos  ) ] 2  18 = 120 tan  - 4.9 14400 v 2 cos 2  From the last slide, v 2 = 1568 / sin 2 . Substitute for v 2 into the equation above: 18 = 120 tan  - 4.9 14400 1568 sin 2  cos 2   18 = 120 tan  - 45 sin 2  cos 2   6 = 40 tan  - 15 sin 2  cos 2  continued on next slide

35 Trebuchet (cont.) Now we have an equation with just one variable, but it’s not an easy one to solve. We can solve for  by graphing the equation below and looking for a y -intercept (in radian mode). y = 40 tan  - 15 sin 2  cos 2  - 6 Domain: [ 0,  / 2 ] Domain: [ 0.52, 0.55 ]  = 0.5404 radians 180   radians = 30.9626  continued on next slide

36 Trebuchet (cont.) Now let’s substitute this value for  into v 2 = 1568 / sin 2 . This gives us v = 42.1557 m/s. Using the fact that the trebuchet pushes on the lead sphere for 0.6 s, we can find the sphere’s acceleration, (since it starts from rest and we now know v f for the launch phase). v f = v 0 + a t a avg = (42.1557 - 0) / (0.6) = 70.2595 m/s 2 So, F avg = m a avg = 180 (70.2595) = 12 646.71 N Because the force (and therefore the acceleration) is not constant, what we’ve calculated is the average force and acceleration.

37 Circular Motion Suppose you drive a go cart in a perfect circle at a constant speed. Even though your speed isn’t changing, you are accelerating. This is because acceleration is the rate of change of velocity (not speed), and your velocity is changing because your direction is changing! Remember, a velocity vector is always tangent to the path of motion. v v v

38 Tangential vs. Centripetal Acceleration 10 m/s 15 m/s So how do we calculate the centripetal acceleration ? ? ? Stay tuned! 18 m/s start finish Suppose now you drive your go cart faster and faster in a circle. Now your velocity vector changes in both magnitude and direction. If you go from start to finish in 4 s, your average tangential acceleration is: a t = (18 m/s - 10 m/s) / 4 s = 2 m/s 2 So you’re speeding up at a rate of 2 m/s per second. This is the rate at which your velocity changes tangentially. But what about the rate at which your velocity changes radially, due to its changing direction? This is your centripetal (or radial) acceleration.

39 Centripetal Acceleration v 0 = v r r  v f = v Let’s find a formula for centripetal acceleration by considering uniform circular motion. By the definition of acceleration, a = (v f - v 0 ) / t. We are subtracting vectors here, not speeds, otherwise a would be zero. ( v 0 and v f have the same magnitudes.) The smaller t is, the smaller  will be, and the more the blue sector will approximate a triangle. The blue “triangle” has sides r, r, and v t (from d = v t ). The vector triangle has sides v, v, and | v f - v 0 |. The two triangles are similar (side-angle-side similarity). vfvf v0v0  v f - v 0  r r v tv t continued on next slide

40 Centripetal Acceleration (cont.) v 0 = v r r  v f = v By similar triangles, v v  | v f - v 0 |  r r v tv t v r = v t v t So, multiplying both sides above by v, we have | v f - v 0 | t = r v 2v 2 ac =ac = Unit check: (m/s) 2 m = m2 / s2m2 / s2 m m s 2 =

41 Centripetal acceleration vector always points toward center of circle. v acac atat v acac atat moving counterclockwise; speeding up moving counterclockwise; slowing down “Centripetal” means “center-seeking.” The magnitude of a c depends on both v and r. However, regardless of speed or tangential acceleration, a c always points toward the center. That is, a c is always radial (along the radius).

42 Resultant Acceleration a c atat a The overall acceleration is the vector sum of the centripetal acceleration and the tangential acceleration. That is, a = a c + a t This is true regardless of the direction of motion. It holds true even when an object is not moving in a perfect circle. Note: The equation above does not include v. Vectors of different quantities cannot be added! moving counterclockwise while speeding up or moving clockwise while slowing down

43 v acac acac Non-circular paths acac R2R2 R1R1 P 1 v P 2 Here we have an object moving along the brown path at a constant speed (a t = 0). a c changes, though, since the radius of curvature changes. At P 1 the path is approximated by the large green circle, at P 2 by the smaller orange one. The smaller r is, the bigger a c is. r v 2v 2 ac =ac =

44 Friction and a c a c 20 m/s 60 m You’re cruising at a constant 20 m/s on a winding highway. The radius of curvature where you are is 60 m. Your centripetal acceleration is: a c = (20 m/s) 2 / (60 m) = 6.67 m/s 2 The force that causes this acceleration is friction, which is why it’s hard to turn on ice. Friction, in this case, is the “centripetal force.” The sharper the turn or the greater your speed, the greater the frictional force must be. overhead view continued on next slide

45 Friction and a c (cont.) a c v r overhead view f Since you’re not speeding up, f is the net force, so F net = f =  s N =  s mg = m a. We use  s because you’re not sliding (or even moving) radially. Thus,  s mg = m a. Mass cancels out, showing that your centripetal acceleration doesn’t depend on how heavy your vehicle is. Solving for a we have a =  s g. In the diagram N is pointing out of the slide. Also, a = a c in this case since a t = 0. m

46 Centripetal Force, F c From F = m a, we get F c = m a c = mv 2 / r. F c = mv 2 r If a body is turning, look at all forces acting on it, and find the net force. The component of the net force that acts toward the center of curvature (perpendicular to the body’s motion) is the centripetal force. The component that acts parallel to its motion (forward or backwards) is the tangential component of the net force.

47 Forces that can provide a centripetal force Friction, as in the turning car example Tension, as in a rock whirling around while attached to a string, or the tension in the chains on a swing at the park.* Normal Force, as in a “round-up ride” at an amusement park (that spins & the floor drops out), or the component of normal force on a car on a banked track that acts toward the center.* Gravity: The force of gravity between the Earth and sun keeps the Earth moving in a nearly circular orbit. Any force directed toward your center of curvature, such as an applied force. * Picture on upcoming slides

48  T mg cos  Simple Pendulum m mg  mg sin  Two forces act on a swinging pendulum, tension and weight. Tension acts radially. We break the weight vector into a radial component (green) and tangential one (violet). Blue is bigger than green, otherwise there would be no net centripetal force, and the mass couldn’t turn. F c = T - mg cos  = m a c = mv 2 / L. F t = mg sin  = m a t F t is the force that speeds the mass up or slows it down. The weight is a constant, but since  changes, so do the weight’s components. mg cos  is greatest when the mass is at its low point. This is also where it’s moving its fastest. For these two reasons T is the greatest when the mass is at the low point. string of length L

49 A pendulum’s period is the same for big arcs as it is for little arcs, so long as the angle through which it swings isn’t real large. (The average speed, therefore, is greater when the arc is bigger, because it must cover a bigger distance in the same time.) The period is independent of the mass. The period depends only on the pendulum’s length. The period = T = 2  (proven in advanced physics) This formula gives us a way to measure the acceleration due to gravity (which varies slightly with location) by measuring the period and length of a pendulum. Don’t confuse the symbol T, which is used for both period and tension. Facts about the Simple Pendulum g L

50 Conical Pendulum  T mg acac  T acac m m Like a tether ball, the mass hangs from a rope and sweeps out a circular path, and the rope a cone.  is a constant. The vertical component of T balances mg. The horiz. comp. of T is the centripetal force. v v continued on next slide

51 Conical Pendulum (cont.)  T mg T sin  m  T cos  All vectors shown are forces. v points into or out of the slide, depending on the direction in which the mass moves. a c is parallel to T sin , which serves as the centripetal force. T cos  = mg T sin  = mv 2 / r Dividing equations : tan  = v 2 / rg See it in action

52 Loop-the-loop in a Plane mg N top A plane flies in a vertical circle, so it’s upside down at the high point. Its speed is constant, but because of its nonlinear motion, the pilot must experience centripetal acceleration. This a c is provided by a combination of her weight and her normal force. mg is constant; N is not. N is the force her pilot’s chair exerts on her body. v v continued on next slide N bot

53 Loop-the-loop (cont.) mg N bot mg N top v v Top: Normal force and weight team up to provide centripetal force: N top + mg = mv 2 / r. If the pilot were sitting on a scale, it would say she’s very light. Bottom: Weight works against normal force, so N must be bigger down here to provide the same centripetal force: N bot - mg = mv 2 / r. ( F c has a constant magnitude since m, v, and r are constants.) Here a scale would say that the pilot is very heavy. continued on next slide r

54 Loop-the-loop (cont.) We’ve been discussing the pilot, but what force causes the plane to turn? Answer: The normal force (force on pilot due to seat) changes throughout the loop. This case is similar to the simple pendulum (the only difference being that speed is constant here). Part of the weight opposes N, and the net radial force is the centripetal force: N - mg cos  = m a c = mv 2 / r mg N r   mg cos  mg sin  The air provides the centripetal force on plane

55 Angular Speed,  Linear speed is how fast you move, measured as distance per unit time. Angular speed is how fast you turn, measured as an angle per unit time. The symbol for angular speed is the small Greek letter omega, , which looks like a curvy “w”. Units for angular speed include: degrees per second; radians per second; and rpm (revolutions per minute). 3 m 110  A B Suppose an object moves steadily from A to B along the circle in 5 s. Then  = 110  / 5 s = 22  / s. The distance it covers is (110  / 360  )(2  )(3) = 5.7596 m. So its linear speed is v = (5.7596 m) / (5 s) = 1.1519 m / s.

56 Arc length: s = r  r  s If  is in radians, then the arc length, s, is  times r. This follows directly from the definition of a radian. One radian is the angle made when the radius of a circle is wrapped along the circle. 1 radian r r When the arc length is as long as the radius, the angle subtended is one radian. (A radian is really dimensionless, since it’s found by dividing a length by a length.)

57 v = r  M  C C´C´ The Three Stooges go to the park. Moe and Larry are on a merry-go- round ride of radius r that Larry is pushing counterclockwise, running at a speed v. In a time t, Moe goes from M to M´ and Curly goes from C to C´. Moe is twice as far from the center as Curly. The distance Moe travels is r , where  is in radians. Curly’s distance is ½ r . Both stooges sweep out the same angle in the same time, so each has the same angular speed. However, since Moe travels twice as far, his linear speed in twice as great. M´M´ v v / 2v / 2 s = r  stst r  t = =  t r v = r 

58 Ferris Wheel Problem Schmedrick is working as a miniature ferris wheel operator (radius 2.1 m). He gets a little overzealous and cranks it up to 75 rpm. His little brother Poindexter flies out at point P, when he is 35  from the low point. At the low point the wheel is 1 m off the ground. A 1.5 m high wall is 27 m from the low point of the wheel. Does Poindexter clear the wall? P Strategy outlined on next slide

59 Ferris Wheel Problem-Solving Strategy 1. Based on his angular speed and the radius, calculate Poindexter’s linear speed. 2. Break his launch velocity down into vertical and horizontal components. 3. Use trig to find the height of his launch. 4. Use trig to find the horizontal distance from the launch site to the wall. 5. Calculate the time it takes him to go that far horizontally. 6. Calculate height at that time. 7. Draw your conclusion. 16.4934 m/s v x = 13.5106 m/s v y0 = 9.4602 m/s 1.3798 m 25.7955 m 1.9093 s 1.5798 m He just makes it by about 8 cm !

60 Springs x m m = mass of weight hanger and weight. (We’re assuming the mass of the spring itself is negligible.) x = the amount the spring stretches due to some force, in this case the weight mg. The amount of stretch ( x ) depends on how much force is applied to the spring ( mg ) and the stiffness of the spring. continued on next slide

61 Hooke’s Law: F = - k x x m Hooke’s Law says that the stretch is proportional to the force the spring exerts, F, which is the reaction pair to the force causing the stretch. Here F is mg up, since the spring is pulling up on the weight hanger with a force of mg. The negative sign is there because as the spring is stretched downward, the force it exerts on the weight is upward. The constant of proportionality is the spring constant, k. The bigger k is, the stiffer the spring, and the harder it is to stretch it. continued on next slide

62 Spring Constant, k 100 N (by hand) 5 cm Suppose that to stretch a certain spring 5 cm, you grab it and pull with 100 N of force. From Hooke’s Law, we get k = (100 N) / (5 cm) = 20 N / cm. This means for every cm you want to stretch it, you must apply another 20 N. Note: k is always positive. In the formula F = - k x, F is to the right (the direction the spring pulls on your hand), and x is to the left (the direction in which the spring was elongated). The minus sign accounts for this difference in direction. If you applied different forces and plotted the spring forces against the corresponding stretches they produced (an F vs. x graph), what would graph look like? answer: a line with a slope of - k

63 Simple Harmonic Motion A mass bobbing up and down on a spring undergoes simple harmonic motion (SHM). SHM occurs whenever a body’s position as a function of time is of the form y = A cos [ b (t - c) ] + d. You’ll see this in trig class too. y = vertical position (our dependent variable) t = time (our independent variable) | A | = amplitude (maximum distance from equilibrium) d = vertical displacement of the equilibrium from your reference point (point from which you measure) c = phase shift (time from start of experiment until the mass reaches its first maximum displacement). c = 0 if experiment starts with spring fully stretched or compressed. The period of the mass is the amount of time it takes to bob up and down once. b is used to determine the period. The period, T, is given by T = 2  / b. 2  is the “normal” period of the sine & cosine functions (see next slide). m equilibrium pt.

64 Sine & Cosine Graphs: y = A [ b (t - c) ] + d period = 2  } amp = 1 y = cos t A = 1 b = 1 c = 0 d = 0 T = 2  / 1 = 2  y = 1 cos[ 1(t - 0) ] + 0 -Pi Pi period = 2  y = sin t A = 1 b = 1 c = 0 d = 0 T = 2  / 1 = 2  y = 1 sin[ 1(t - 0) ] + 0 } amp = 1 y = cos t begins at a max; y = sin t begins at equilibrium. Their periods and amplitudes are the same. Their graphs are congruent, differing only by a 90  horizontal shift. sin cos

65 Cosine graph: Amplitude y = A cos [ b (t - c) ] + d y = cos t -Pi Pi y = 2 cos t y = -1.5 cos t The leading coefficient determines the amplitude, which equals | A |. If A is negative, the graph is reflected about the time axis. For the red graph, our mass is pushed up, compressing the spring 2 m and is released at time zero. In the green graph, it’s pulled down 1.5 m and released at time zero. Note: all graphs have the same period or 2 , and none has a phase (horiz.) shift or a vertical shift. A = 1 m A = 2 m A = 1.5 m

66 Cosine graph: Period y = A cos [ b (t - c) ] + d y = cos t -Pi Pi 22 2 / 32 / 3 44 b = 1 s -1 y = cos 3 t y = cos 0.5 t The bigger b is, our “scrunch factor,” the more scrunched up the graph gets. The red graph completes 3 cycles in same time the blue graph does one, and red’s period is 3 times shorter. So, the mass in the red graph goes up and down 3 times more often. The green graph completes half as many cycles as blue and its mass has twice the period. Note: b ’s units cancel t ’s. T = 2  / b b = ½ s -1 b = 3 s -1

67 Cosine graph: vertical displacement y = cos t -Pi Pi d = 0 d = 1.5 d = -0.5 y = cos t + 1.5 y = cos t - 0.5 Adding or subtracting outside of the cosine function shifts the graph vertically. The dotted lines are lines of symmetry. For our mass on a spring, this is where it’s in equilibrium (where the spring is neither stretched nor compressed). For the red graph, the mass is in equilibrium 1.5 meters above the point from which we chose to make our measurements. y = A cos [ b (t - c) ] + d

68 Cosine graph: Phase Shift y = A cos [ b (t - c) ] + d y = cos t y = cos (t + 1.2) y = cos (t - 0.9) } c = 0.9 s } c = -1.2 s c = 0 Adding or subtracting inside the cosine function shifts the graph horizontally. (Adding shifts it left, subtracting right.) Like the mass in the blue graph, the red mass was compressed 1 m and then released, but the clock was started 1.2 s after it had reached max compression. The green mass was also compressed 1 m and released, but the clock was started 0.9 s before the mass reached max compression.

69 SHM / Spring Problem part 1 Schmedrick wants to go bungee jumping but can’t afford it, so he takes a dive off a bridge 60 ft over a ranging river with a giant slinky connecting him to the bridge. After the leap he begins bobbing up and down. While bobbing, he’s moving the fastest when he’s 27 ft above the river. He gets as high as 39 ft above the river. It takes him 5 s to go from his high point to his low point. Schmedrick weighs 105 lb, and the natural length of the slinky is 20 ft. 1. What’s the spring constant of the slinky? The point at which Schmedrick is moving the fastest while bobbing is the equilibrium point. ( v = 0 at the high and low points and v is a max half way in between.) This happens at 27 ft above the river, which is 60 - 27 = 33 ft from the bridge. So, if Schmed were just hanging from the slinky, his weight would stretch it 33 - 20 = 13 ft. Thus, k = (105 lb) / (13 ft) = 8.08 lb / ft. This means every foot of stretch requires about an additional 8 pounds of force.

70 2. If the clock starts when he jumps off the bridge, write Schmedrick’s position as a function of time using the river as a reference point. Ignoring air resistance, a mass on a spring undergoes SHM (derived in advanced physics). Hence we can use y = A cos [ b (t - c) ] + d. Our task, then, is to find A, b, c, and d. d = 27 ft since the equilibrium pt, Eq, is 27 ft above our reference position. The high pt, H, is given to be 39 ft. This is 12 ft above Eq, so A = 12 ft, and the low pt, L, is 12 ft below Eq. Since it takes 5 s for him to go from H to L, his period is 10 s. So, T = 2  / b  b = 2  / (10 s) = 0.628 s -1. If Schmedrick had been at H or L at t = 0, c would be zero. But the clock starts early (when y = 60 ft), so let’s figure out how long it takes him to fall to H (assume free fall). SHM / Spring Problem part 2 continued on next slide river y = 0 bridge y = 60 ft y = 27 ft Eq L y = 15 ft y = 39 ft H

71 SHM / Spring Problem part 2 (cont.) If we’d like to continue working in feet, we can’t use 9.8 m/s 2 for g. Instead we use its equivalent: 32 ft / s 2. He falls 21 ft to H. Thus, -21 = (0) t + 0.5 (-32) t 2  t = 1.146 s, and this is our c value. (This is only an approximation, since the spring begins stretching after he falls 20 ft.) Putting it all together, we have: river y = 0 bridge y = 60 ft y = 27 ft Eq L y = 15 ft y = 39 ft H y(t) = 12 cos [ 0.628 (t - 1.146) ] + 27 where t is the time in seconds from the instant he jumps, and y is his height above the water in feet. Note: This formula only works for t > 1.146 s, since that’s when SHM begins. Check by plugging in some values into the function (using radian mode in the calculator): y(1.146) = 12 cos [0] + 27 = 12(1) + 27 = 39 ft. (H) y(6.146) = 12 cos [0.628 (5) ] + 27 = 12 (-1) + 27 = 15 ft. (L) So he’s at L 5 s after he’s at H.

72 SHM / Spring Problem parts 3 & 4 3. How high up is Schmedrick 14 s into the experiment? 4. At what times is he 30 ft above the river? y (14) = 12 cos [0.628 (14 - 1.146) ] + 27 = 24.4 ft. (When we calculated b, we used 2  rather than 360  for the “normal” period of the cosine function, so we must put our calculators in radian mode.) He can only be at one position at a particular time, but he can be at a particular position at many different times. (This is the nature of a function.) Thus, 30 = 12 cos [ 0.628 ( t - 1.146) ] + 27, and we must solve for t. First isolate the cosine function by subtracting 27 and dividing by 12: 0.25 = cos [0.628 ( t - 1.146) ]. Think of the quantity in the brackets as an angle. We want to know what angles have a cosine of 0.25. One angle is cos -1 (0.25) = 1.318 (radians). But this is only one of an infinite number of angles whose cosine is 0.25. continued on next slide

73 SHM / Spring Problem part 4 (cont.) One angle in which cos  = 0.25 is 1.318 radians  75.5  (in Q I). Another occurs at 4.965  284.5  (in Q IV). To get to P 1 or P 2 from the origin, you must go 0.25 units to the right. Thus, both x y  cos  left & right (green) sin  up & down (brown) tan  slope (of red) On a unit circle (radius 1, center at origin): P1P1 P2P2 angles have the same cosine of 0.25. Negative angles and angles over 2  (or 360  ) that terminate at P 1 or P 2 will also have a cosine of 0.25. These angles are 1.318 + 2 n  and 4. 965 + 2 n , where n = 0,  1,  2,  3, …. Ex: when n = 5, cos[1.318 + 2(5)  ] = 0.25. Here we’ve gone around the circle 5 times and stopped at P 1. continued on next slide

74 SHM / Spring Problem part 4 (cont.) We are in the process of solving 0.25 = cos [0.628 ( t - 1.146) ]. Our “angle” is 0.628 ( t - 1.146). Therefore, t = 1.146 + (1.318 + 2 n  ) / 0.628 or t = 1.146 + (4.965 + 2 n  ) / 0.628 0.628 ( t - 1.146) = 1.318 + 2 n  or 0.628 ( t - 1.146) = 4. 965 + 2 n  We only want t values > 1.146 s (when SHM begins). Substituting in for n, we get t = 3.245 s, 13.250 s, 23.255 s, … or t = 9.052 s, 19.057 s, 29.062 s, … These are all the times when Schmedrick is 30 ft above the water. The top row lists times when he’s on his way down. The bottom row lists times when he’s bouncing back up. The difference between consecutive times on a given row is 10 s (neglecting some slight compounded rounding error), which is the period of his motion. That is, every 10 s he is at the same position moving in the same direction.

75 SHM / Spring Problem part 5 We’ve been neglecting air resistance throughout this problem. How would Schmedrick’s position vary with time in real life? It’s not exactly SHM any more, but it’s similar. He will still bob up and down relative to the same equilibrium point. It would take him a little longer to fall to the high point for two reason. One, drag from the air slows him down. Two, after falling 20 ft (natural length of the slinky) the slinky begins to stretch out and exerts an upward force on him, slowing him down. This would change our c value. The most interesting change, however, is the amplitude. In real life, the amplitude itself is a function of time, diminishing with each bounce due to air resistance and friction in the slinky. This is damped periodic motion. Let’s checked out some graphs of Schmedrick’s motion, damped (real life) and undamped (idealized simple harmonic motion). continued on next slide

76 SHM / Spring Problem part 5 (cont.) undamped damped In real life drag forces would decay his amplitude until he’s hanging at the equilibrium point 27 ft above the water, barely moving. The amplitude decays exponentially, meaning that a certain percentage of it is lost with each bounce. The percentage lost depends on the viscosity of the fluid in which the mass is moving. The damping would be more severe in water than in air.

77 Things that undergo SHM Mass bobbing vertically from a spring w/ no air resistance Mass on a frictionless floor attached to a spring on the wall Simple pendulum or person on a swing where max displacement is not excessive The shadow on a wall of an object moving in a vertical circle at constant speed with light shining on it from the side A person who jumps into a tunnel that goes clear through the center of the Earth all the way to the opposite side All of these situations have common features: periodic motion with constant period an equilibrium point half way between its high and low (or leftmost and rightmost) points where its moving its fastest a constant amplitude that is half of its total span (from high to low) a “restoring force” that always acts in the direction of equilibrium and is a max at the extremes (max distance from equil.)

78 Torque Acme Nut Buster 9/16 F r Informally speaking, torque is the measure of a force’s turning power. Normally it’s impossible to loosen a nut or bolt by hand. A wrench doesn’t make you any stronger, but it increases your turning power, because the farther a force is applied from the axis of rotation, the greater the torque it produces. So long as F and r are perpendicular, the magnitude of the torque is given by  = r F, where r is the magnitude of the displacement vector from the center of rotation to the point of application of the force. Technically, torque is a vector quantity, but we’ll mainly deal with its magnitude.

79 Torque when force is at an angle Acme Nut Buster 9/16 F r  F cos  F sin  When F is not perpendicular to r, we can split F into components. The component parallel to r does nothing in the way of turning the bolt, so it produces no torque. The perpendicular component does produce torque. Therefore, our general formula for the magnitude of torque is:  = r F sin 

80 Torque as a vector F r  Torque is a vector quantity that can be defined relative to any point of interest by the cross product:  = r  F where r is the displacement vector from the point of interest to the point of application of the force. As you learned in the last unit, the magnitude of r  F is r F sin , where  is the angle between r and F. Use the right-hand rule to find the direction of . In the wrench example,  points out toward you, perpendicular to both r and F. The torque vector does not point in the direction of motion, as a velocity vector does. Just as a net force produces acceleration (a change in velocity), a net torque produces angular acceleration (a change in angular speed).

81 See-saw Example part 1 175 lb 7 ft5 ft6 ft 35 lb How much do I weigh? A man, a turtle, and a bike rider are positioned on a giant see-saw as shown. If the see-saw is in equilibrium, how much must the man weigh? Answer: The torque that would produce clockwise rotation (from the weight of the bike and rider) must cancel out the torques that would produce counterclockwise rotation (from the man and turtle). If the man’s weight is W, then (7 ft) W + (5 ft) (35 lb) = (6 ft) (175 lb) (7 ft) W = 875 ft·lb W = 125 lb fulcrum

82 See-saw Example part 2 175 lb 7 ft5 ft6 ft 35 lb The see-saw is in equilibrium at this instant, but which side will go down after a short time? Answer: Because the biker is heavier and, presumably, moving faster than the turtle, the torque he produces will diminish much more quickly than the torque the turtle produces. Therefore, the turtle’s side of the see-saw will go down. Note: A foot-pound is a unit of torque (a force times a distance). The SI unit for torque is the Newton-meter (N m). Also note that the angle  in the formula  = r F sin  is 90° for each person/animal since each weight vector is  to its corresponding displacement vector from the fulcrum.

83 Equilibrium 175 lb 35 lb 125 lb N = 335 lb N is the normal force--the force on the see-saw due to the fulcrum. Its magnitude is the sum of the weights of the 3 see-saw participants. If this were not the case, the see-saw would be accelerating either up or down. So, equilibrium means no net force and no net torque. Note that weights on each side don’t have to be equal. The distance each weight lies from the fulcrum must be taken into account.

84 Center of Mass Schmedrick decides to build his own see-saw, but of course he screws up, and instead of putting the fulcrum in the middle, he puts it only 1 / 3 of the way from the right end. The beam of the see-saw has a mass of 50 kg. If the beam is uniform (equal density throughout), then we pretend all of the weight of the beam acts right at the center of the beam. This point is called the center of mass because it’s the point where the beam would balance if the fulcrum were beneath it. Schmed figures, rather than cutting some of the left end off, he could attach a weight at the right end. In order to attain a balance, the center of mass of the weight-beam system must be shifted to the right, so that it is right above the fulcrum. How much weight should he add? c. m. for beam continued on next slide weight

85 c. m. for beam Center or Mass (cont.) x = 0 x = L / 2 x = L To calculate the center of mass, we set up a coordinate system. (The reference point doesn’t matter.) For our purposes, all 50 kg of the the beam is right at the its center of mass (at L / 2). Mathematically, the center of mass for a system of masses is given by adding up all the products of mass and position and then dividing by the total mass: x = 2 L / 3 m x cm = 50 (L / 2) + m (L ) 50 + m = 2 L / 3 The L’s cancel out and we have: 25 + m 50 + m 2 3 = 75 + 3 m = 100 + 2 m So, m = 25 kg, and its weight is (25 kg) (9.8 m/s 2 ) = 245 N

86 Center of mass is independent of coordinate system c. m. for beam x = L x = L / 2 x = 0 x = L / 3 m x cm = 50 (L / 2) + m (0 ) 50 + m = L / 3 Let’s show we get the same answer using different coordinate systems: one with zero at the right end where positive is to the left (red); and one with zero at the fulcrum where positive is to the right (black). In each equation we set the center of mass equal to the fulcrum’s position in that coordinate system: x = 0 x = -L / 6 x = L / 3 x = -2L / 3 x cm = 50 (-L / 6) + m (L / 3) 50 + m = 0 In each case m comes out to be 25 kg, as it was on the last slide. (Work the algebra out for yourself.) This would be true no matter what coordinate system we chose.

87 Torque Method c. m. for beam m Same problem, another method: This time we’ll find m using torque. In order for the beam to balance where the fulcrum is, the torque (force  distance) must cancel out on each side of the fulcrum. The entire weight of the beam acting at its center of mass is equivalent to its weight being spread out along its length. 50 g (L / 6) = m g (L / 3) Both g and L cancel out and m = 25 kg, just as it did on previous slides. (Since the weight of the beam acts at half the distance as the weight of the green mass, the green mass must have half the weight.) L / 6 L / 3 50 g m gm g


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