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Chapter 4 – Kinematics (cont.). Recap:. Dropping a Package. a) quickly lag behind the plane while falling b) remain vertically under the plane while falling c) move ahead of the plane while falling d) not fall at all.
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Dropping a Package a) quickly lag behind the plane while falling b) remain vertically under the plane while falling c) move ahead of the plane while falling d) not fall at all You drop a package from a plane flying at constant speed in a straight line. Without air resistance, the package will:
Follow-up: what would happen if air resistance is present? Dropping a Package a) quickly lag behind the plane while falling b) remain vertically under the plane while falling c) move ahead of the plane while falling d)not fall at all You drop a package from a plane flying at constant speed in a straight line. Without air resistance, the package will: Both the plane and the package have the samehorizontalvelocity at the moment of release. They will maintain this velocity in the x-direction, so they stay aligned.
a b A battleship simultaneously fires two shells at two enemy submarines. The shells are launched with the same magnitude of initialvelocity. If the shells follow the trajectories shown, which submarine gets hit first? c) both at the same time
Follow-up: Did you need to know that they had the same initial speed? a b A battleship simultaneously fires two shells at two enemy submarines. The shells are launched with the same magnitude of initialvelocity (i.e., speed). If the shells follow the trajectories shown, which submarine gets hit first? The flight time is fixed by the motion in the y-direction. The higher an object goes, the longer it stays in flight. The shell hitting submarine #2 goes less high, therefore it stays in flight for less time than the other shell. Thus, submarine#2 is hit first. c) both at the same time
Range: the horizontal distance a projectile travels As before, use (y = 0 at landing) and Eliminate t and solve for x when y=0
Sin(2θ) θ Range is maximum at 45o sin(2 x 45o) = sin(90o) = 1
Range Gun a) 10 degrees b) 22.5 degrees c) 60 degrees d) 75 degrees e) one would also need the launch velocity of the range gun to know If the range gun launches a ball 6 meters with a launch angle of 45 degrees, at which of these angles should a ball be launched to land in a bucket at 3 meters?
Range Gun a) 10 degrees b) 22.5 degrees c) 60 degrees d) 75 degrees e) one would also need the launch velocity of the range gun to know If the range gun launches a ball 6 meters with a launch angle of 45 degrees, at which of these angles should a ball be launched to land in a bucket at 3 meters? The range is proportional to sin2θ, so to travel half the distance, the ball would need to be launched with sin2θ = 0.5. sin(75o) = 0.5, so θ=75o Note: there are two angles that would work [sin(30o) = 0.5 also]. How are the two solutions different?
Dropping the Ball III a) just after it is launched b) at the highest point in its flight c) just before it hits the ground d) halfway between the ground and the highest point e) speed is always constant A projectile is launched from the ground at an angle of 30°. At what point in its trajectory does this projectile have the least speed?
Dropping the Ball III a) just after it is launched b) at the highest point in its flight c) just before it hits the ground d) halfway between the ground and the highest point e) speed is always constant A projectile is launched from the ground at an angle of 30º. At what point in its trajectory does this projectile have the least speed? The speed is smallest at the highest point of its flight path because they-component of thevelocity is zero.
h a b c d) all have the same hang time Punts I Which of the three punts has the longest hangtime?
Follow-up: Which one had the greater initial speed? h a b c d) all have the same hang time Punts I Which of thethree puntshas thelongest hangtime? The time in the air is determined by the vertical motion! Because all of the punts reach the same height, they all stay in the air for the same time.
•• On a hot summer day, a young girl swings on a rope above the local swimming hole. When she lets go of the rope her initial velocity is 2.25 m/s at an angle of 35.0° above the horizontal. If she is in flight for 0.616 s, how high above the water was she when she let go of the rope?
Time to hit the water: t=0.616 Initial velocity: 2.25 m/s at 35o above the horizontal y = (v0 sinθ) t - 1/2 g t2 y = 2.25 m/s * sin(35o) * (0.616 s) - 1/2 (9.8 m/s2) (0.616 s)2 = - 1.07 m At time t= 0.616 s, the girl is 1.07 m below her starting position, so her initial position was 1.07m above the water.
In a friendly game of handball, you hit the ball essentially at ground level and send it toward the wall with a speed of 14 m/s at an angle of 34° above the horizontal. (a) How long does it take for the ball to reach the wall if it is 3.8 m away? (b) How high is the ball when it hits the wall?
In a friendly game of handball, you hit the ball essentially at ground level and send it toward the wall with a speed of 14 m/s at an angle of 34° above the horizontal. (a) How long does it take for the ball to reach the wall if it is 3.8 m away? (b) How high is the ball when it hits the wall? d = 3.8 m v0 = 14 m/s θ = 34o v0x = (14 m/s) cos(34o) = 11.6 m/s v0y = (14 m/s) sin(34o) = 7.83 m/s (a) d = v0x t t = d/v0x = (3.8m) / (11.6 m/s) = 0.33 s (b) h = v0y t - 1/2 g t2 = (7.83 m/s) (0.33s) -1/2 (9.8 m/s2) (0.33 s)2 = 2.0 m
d = 3.8 m vy = v0y - g t = 7.83 m/s - (9.8m/s2) (0.33s) = 4.6 m/s v0 = 14 m/s θ = 34o v0x = (14 m/s) cos(34o) = 11.6 m/s vx = v0x v0y = (14 m/s) sin(34o) = 7.83 m/s t = d/v0x = (3.8m) / (11.6 m/s) = 0.33 s In a friendly game of handball, you hit the ball essentially at ground level and send it toward the wall (3.8 m away) with a speed of 14 m/s at an angle of 34° above the horizontal. (c) what are the magnitude and direction of the ball's velocity when it strikes the wall? (d) Has the ball reached the highest point of its trajectory at this time? Explain. The vertical component of velocity is still positive, that is, the ball is still going up. So the ball has not yet reached its highest point.
Kinematics: Assumptions, Definitions and Logical Conclusions What have we done so far? • Defined displacement, velocity, acceleration (also position, distance, speed)... • Defined scalers (like speed) and vectors (like velocity) • Laid out assumptions about free-fall • noticed that 2-dimensional motion is really just two, simultaneous, 1-dimensional motions. Used this to shoot a monkey, range out a small cannon, etc. This wasn’t physics. This was preparing the language needed to talk about physics.
Chapter 5 Newton’s Laws of Motion (sections 5.1-5.4)
Newton’s Laws How can we consistently and generally describe the way objects move and interact?
Newton in a 1702 portrait by Godfrey Kneller Newton in a 1689 portrait by Godfrey Kneller Isaac Newton 1643-1727
I do not know what I may appear to the world, but to myself I seem to have been only like a boy playing on the sea-shore, and diverting myself in now and then finding a smoother pebble or a prettier shell than ordinary, whilst the great ocean of truth lay all undiscovered before me. Nature and nature's laws lay hid in night; God said "Let Newton be" and all was light. Newton’s epitaph Alexander Pope (English poet, 1688-1744) from a memoir by Newton
Force Force: push or pull Force is a vector – it has magnitude and direction
Mass Mass is the measure of how hard it is to change an object’s velocity. Mass can also be thought of as a measure of the quantity of matter in an object.
Newton’s First Law of Motion If you stop pushing an object, does it stop moving? Only if there is friction! In the absence of any net external force, an object at rest will remain at rest. In the absence of any net external force a moving object will keep moving at a constant speed in a straight line. This is also known as the Law of Inertia.
Newton’s First Law a) more than its weight b) equal to its weight c) less than its weight but more than zero d) depends on the speed of the puck e) zero A hockey puck slides on ice at constant velocity. What is the net force acting on the puck?
Newton’s First Law a) more than its weight b) equal to its weight c) less than its weight but more than zero d) depends on the speed of the puck e) zero A hockey puck slides on ice at constant velocity. What is the net force acting on the puck? The puck is moving at a constant velocity, and therefore it is not accelerating. Thus, there must be no net force acting on the puck. Follow-up: Are there any forces acting on the puck? What are they?
Newton’s First Law a) a net force acted on it b) no, or insufficient, net force acted on it c) it remained at rest d) it did not move, but only seemed to e) gravity briefly stopped acting on it You put your book on the bus seat next to you. When the bus stops suddenly, the book slides forward off the seat. Why?
Newton’s First Law a) a net force acted on it b) no, or insufficient, net force acted on it c) it remained at rest d) it did not move, but only seemed to e) gravity briefly stopped acting on it You put your book on the bus seat next to you. When the bus stops suddenly, the book slides forward off the seat. Why? The book was initially moving forward (because it was on a moving bus). When the bus stopped, the book continued moving forward, which was its initial state ofmotion, and therefore it slid forward off the seat.
Calibrating force Two equal weights exert twice the force of one; this can be used for calibration of a spring:
Experiment: Acceleration vs Force Now that we have a calibrated spring, we can do more experiments. Acceleration is proportional to force:
Experiment: Acceleration vs Mass Acceleration is inversely proportional to mass:
Newton’s Second Law of Motion Acceleration is proportional to force: Acceleration is inversely proportional to mass: Combining these two observations gives Or, more familiarly,
Newton’s Second Law of Motion An object may have several forces acting on it; the acceleration is due to the net force: SI unit for force Newton is defined using this equation as: 1 N is the force required to give a mass of 1 kg an acceleration of 1 m/s2
The weight of an object is the force acting on it due to gravity Weight: W = Fg = ma = mg vertically downwards Since , the weight of an object in Newtons is approximately 10 x its mass in kg adult human 70 kg 700 N ~ 160 lbs. Force of Gravity WHERE? There is no “conversion” from kg to pounds! (Unless you specify what planet you are assuming)
In order to change the velocity of an object – magnitude or direction – a net force is required. (I) (II) Newton’s First and Second Laws What about the bus... From the perspective of someone who didn’t know they were on the bus?
In order to change the velocity of an object – magnitude or direction – a net force is required. (I) (II) Inertial Reference Frames Newton’s First and Second Laws do not work in an accelerating frame of reference An inertial reference frame is one in which the first and second laws are true. Accelerating reference frames are not inertial. Was the bus an inertial reference frame? Is the earth an inertial reference frame? No, but acceleration due to earth’s rotation around Its axis (0.034 m/s2), and due to earth’s rotation around sun (smaller) are negligible compared to g; so approximately yes.
Analyzing the forces in a system • Free-body diagrams: • A free-body diagram shows every force acting on an object. • Sketch the forces • Isolate the object of interest • Choose a convenient coordinate system • Resolve the forces into components • Apply Newton’s second law to each coordinate direction
Free-body Diagram Example of a free-body diagram:
Newton’s First Law a) there is a net force but the book has too much inertia b) there are no forces acting on it at all c) it does move, but too slowly to be seen d) there is no net force on the book e) there is a net force, but the book is too heavy to move A book is lying at rest on a table. The book will remain there at rest because:
Newton’s First Law a) there is a net force but the book has too much inertia b) there are no forces acting on it at all c) it does move, but too slowly to be seen d) there is no net force on the book e) there is a net force, but the book is too heavy to move A book is lying at rest on a table. The book will remain there at rest because: There are forces acting on the book, but the only forces acting are in the y-direction. Gravity acts downward, but the table exerts an upward force that is equally strong, so the two forces cancel, leaving no net force.
Newton’s Third Law of Motion Forces always come in pairs, acting on different objects: If object 1 exerts a force on object 2, then object 2 exerts a force – on object 1. These forces are called action-reaction pairs.
Newton’s 3rd: F12 = - F21 action-reaction pairs are equal and opposite, but they act on different bodies Action-reaction pair? a) Yes b) No
Newton’s Third Law of Motion Although the forces are the same, the accelerations will not be unless the objects have the same mass. Q: When skydiving, do you exert a force on the earth? Does the earth accelerate towards you? Is the magnitude of the acceleration of the earth the same as the magnitude of your acceleration?