www.lsmsa.edu teacher Robert Dalling's Physics Lectures (see also www.ushumans.net)
Trajectories
In this chapter we will discuss the motion of thrown objects, spinning objects, and objects that are moving relative to another object that is itself moving. The student might like to investigate the mutual motions of crowds of people or of groups of birds or fish. How are all of the group members able to move as a single entity? Hear are some musical trajectories. Here are some studies of motion in computer animations and cartoons. Dance and ballet use motion to communicate. Here is an animation set to music and another that uses motion-capture and equation-based models. Birds, bees, stotting gazelles, and other mammals use motion to communicate. After we understand the basics of motion then we see that no matter how complicated a particular system seems to be, it consists of the same basic physics: speed, acceleration, mass, force, and energy.
We have all tossed objects through the air, resulting in two- or three-dimensional projectile motion. If an object moves at a slow speed and or if the motion of the tossed object lasts only a few seconds, then air friction has too little time to have much of an effect. When we use rulers and stop watches to examine the path of a thrown object more closely, we uncover unexpected aspects of its motion. Surprisingly, its horizontal and vertical motions are independent of each other. The horizontal and vertical components of its position, velocity, and acceleration are x and y, vx and vy, and ax = 0 and ay = g. We find that the projectile moves at a constant horizontal speed described by the equation x = xo + vxt, while also falling downward under the constant vertical acceleration of gravity, which is described by the equation y = yo + vyot + ½at2, where we usually have a = -g. Rolling a ball off a table top results in this sort of motion.
When you drop an object within a moving car, the object falls straight to the floor right under your hand. From the point of view of an observer sitting on the sidewalk, that object undergoes horizontal motion at a constant speed while falling with a constant acceleration–just as was the case for our frictionless projectile. When Apollo astronaut Alan Shepard hit a golf ball on the moon, where there is no air or air friction, the flying ball underwent motion with constant horizontal speed and constant vertical acceleration.
We have all been traveling in a car at 100 km/hr (60 mph) while we tossed a coin straight up into the air and found that the object falls straight back down into our hand. It does not smash through the back window at 100 km/hr and land in the road behind the car. Since your past experience has taught you this, do you believe that a rock dropped from the mast of a moving ship will fall straight to the deck below the mast–that is, it will not land in the water behind the moving ship? If you toss a coin while riding in an airplane moving at 1000 km/hr (600 mph), will the coin crash into the back of the aircraft at 1,000 km/hr? If an object is launched straight upwards from the back of a moving truck or car, the object will later land in the same spot in the vehicle from which it was launched. Astronauts travel inside a space shuttle moving at 20,000 mph and can step out of the ship without being left behind. If any of these tossed objects are in mid-air when the car, boat, airplane, or shuttle accelerates, then the back of the vehicle will move forward and can meet the mid-air object. Galileo used the term inertia to describe the property of objects in motion to remain in motion and for objects at rest to remain at rest. While driving in a car and making a turn, the stuff on our dash board wants to keep moving forward and so the side of the turning car collides with it. In a similar situation, Mr. Schlangen of Two Harbors High School demonstrates a Hewitt Inertia Hat. When we study force in the coming chapters, we will see that objects continue moving at a constant speed–including a speed of zero–until acted upon by an outside force. That is, if you push or pull on an object with a force, then its speed will change. If no force acts, then the speed of an object will not change. If there were no air friction, an object dropped from an airplane would continue in motion directly under the airplane just as does the object dropped under your hand while moving in the car.
For trajectories lasting more than a few seconds, air friction will slow the horizontal motion but calculus is needed to describe such motions. For example, if an object undergoes the one-dimensional motion of falling while experiencing a frictional force proportional to the square of its speed, then the path of that object through time follows a hyperbolic cosine function. Its speed rises from zero to a maximum value called the terminal speed. The terminal speed of an object depends on its density and area. Jack Ord calculates the terminal speed of a tennis ball, a baseball, and a baseball-sized iron sphere to be 9.38, 34.7, and 117 m/s. The typical values for a box-sized objects is 200 mph or 90 m/s. In two or more dimensions, a computer is used to calculate and display the motion. In this course, we want to understand the physical meaning of speed and acceleration and such: we’ll leave friction and its mathematics for later study.
Example:
A ball is launched horizontally at a speed of 4 m/s off a table top that is 1.3 meters above the ground. Draw a picture, show the coordinate system, height h, x, and y. Put the origin at the ground under the edge of the table.
O ->
______
| | ^
| | h
For the x-motion:
xo = 0
x = ? = xo + vx t
vx = 4 m/s = constant
For the y-motion:
yo = h = 1.3 m
y = 0
vyo = 0
vy = ?
a = -9.8 m/s2 (pointing in what we’ve chosen to be the negative y direction)
t = ?
How long does it take for the ball to reach the ground? While solving projectile problems, we use the fact that time ‘t’ is the same in both the x and y motions. In this problem, we get the time from the y-motion using y = yo + vyot + ½at2 or
t = sqrt( - 2yo/a ) = sqrt[ -2 ( 1.3 / -9.8 ) = 0.51 s.
How far will it travel horizontally before hitting the ground?
Now we can find x = xo + vx t = 0 + (4 m/s)( 0.51 s) = 2.04 m.
What is vy just before the ball hits the ground?
vy = vyo + at = 0 + ( -9.8 m/s2 )( 0.51 s) = -5 m/s.
What is the velocity vector just before the ball hits the ground?
V = vx i + vy j = 4 i - 5 j.
What is the magnitude of the velocity just before the ball hits the ground?
|V| = sqrt( vx 2 + vy 2 ) = sqrt[ 4 2 (-5)2 ] = 6.4 m/s.
What angle does the velocity vector make with the positive-x axis?
θ = tan-1( vy / vx ) = tan-1( -5 / 4 ) = -51°, which is 51° below the x-axis.
The range increases when an object is instead launched from increasing heights or with increasing speeds. Here is the trajectory of a skateboarder doing an Ollie off a step.
If the ball is launched off the table at a speed of 4 m/s and an angle of 30° above the positive x-axis, then vxo = v cos θ = 4 cos( 30 ) = 3.46 m/s and vyo = v sin θ = 4 sin( 30 ) = 2 m/s. This is an example of projectile motion. Goalfinder has an animation of projectile motion.
Notice that the horizontal speed is unchanged throughout the trajectory. This is more than 99% true when rolling a marble off a table. It is less accurate when larger speeds and flight-times occur. While solving these low speed problems, try to remove the inconsequential air friction from your mind.
Student:
Try launching projectiles with the PHET animation.
Example:
A ball is batted from a position 1.0 meters above the ground and initially movies at an angle θ = 45° above the horizontal direction. With what speed must the ball begin if it is to pass over a fence that is 2.5 meters high and a distance D = 110 meters away? Let’s place the origin of the coordinate system at the batter’s feet such that positive x and y are upward and toward the fence.
O
/\ | ^
+ ------------------------| h
D
For the x-motion:
xo = 0
x = 110 m and x = xo + vx t
vx = | Vo | cos θ = constant
For the y-motion:
yo = 1.0 m
y = 2.5 m
vyo = | Vo | sin θ
vy = ?
a = -9.8 m/s2 (pointing in what we’ve chosen to be the negative y direction)
t = ?
We use the x motion to determine t. From x = xo + vx t we have
t = ( x - xo ) / vx = x / vx
We put this value of t into the equation y = yo + vyot + ½at2 to get
y = yo + vyox/vx + ½a(x/vx)2.
Now substitute vx = | Vo | cos θ and vyo = | Vo | sin θ to get
y = yo + ( Vo sin θ )( x / Vo cos θ ) + ( ½ )( -g )( x / Vo cos θ )2,
where Vo = | Vo |. This simplifies to
y = yo + x tan θ - ½gx2 / ( Vo cos θ )2.
We can now solve for the single unknown
1 / Vo2 = ( yo - y + x tan θ )( 2 cos2 θ ) / ( gx2 ),
so that
Vo = 33 m/s.
CQ: What is Vo for a taller person having yo = 1.3 m? (Still 33 m/s.)
Example.
If wind blows horizontally at constant speed across a baseball field, the air friction on a batted ball will then create a constant horizontal force. This force will result in a constant horizontal acceleration of the ball that will be combined with the constant vertical acceleration due to gravity. If the wind blows directly toward the batter who hits a ball directly into the wind, then we have y = yo + vyot + ½ayt2 and x = xo + vxot + ½axt2 when using a coordinate system placed on the ground at the location of the batter, with positive-x toward the field. The y-acceleration is ay = -9.8 m/s2. If the x-acceleration is also ax = -9.8 m/s2 and the batter hits the ball upward at an angle of 45°, do you believe that the trajectory of the ball will follow a straight line at that angle, slow to a stop, and then return along the same line back to its place of origin? Plot a few trajectories for varying values and directions of ax when the batted ball has an initial speed of 50 m/s upward at an angle of 65° from the ground. If the wind is toward the batter, the ball might land behind the batter.
Example:
Analyze the trajectory of an animal, person, football, skier, skateboarder, motorcycle, bike, skateboarder, monster truck, or of a thrown or kicked football, basketball, or baseball. Estimate heights and lengths of objects and then estimate trajectory parameters x, y, v, t, and maximum height and such. Estimate the effects of air friction. For example, if you search the web at http://video.search.yahoo.com or www.google.com for videos involving “long jumper” you’ll get results like this.
Student:
Run the Phet Motion: 2-D Motion simulation and then write a paragraph explaining what it does, what you have learned from it, and how it helps you understand this physical phenomenon. Include numerical examples.
Student:
Run one of the ActivPhysics Projectile Motion units and then write a paragraph explaining what it does, what you have learned from it, and how it helps you understand this physical phenomenon. Include numerical examples.
Example:
L. Gladney of the University of Penn teaches the course “Mechanics for the Health Sciences” and discusses the work, power, and trajectory of a jumping flea. He includes an animation of the jumping flea.
Here is another video of a jumping insect.
Arkive.org has movies of leaping dolphins. Use the body dimensions given in their website to estimate the initial speed, maximum height, and range for the motion. Here is a primate. Another primate. A jumping rat. Jumping spiders.
Suppose that a projectile is launched at speed vo at an angle θ from the ground, rises, stops momentarily, and then falls back to the ground. We next find an expression for the horizontal distance traversed by such a projectile. To obtain this relation, we put the origin of the coordinate system at the ground-level launch point. At the top of its trajectory, the vertical speed of the object will be zero momentarily. We use this fact to find the time at which the projectile reaches its maximum.
From vy = vyo - gt = 0 we get t = vyo / g.
Since the object takes this same amount of time to fall back to the ground, the total time-of-flight is twice this time, or t = 2vyo / g. In that time, the particle will travel a horizontal distance given by x = xo + vxo t = 0 + vxo t = vxo t. Using the total travel time, this becomes
x = 2 vxo vyo / g.
Since vxo = vo cos θ and vyo = vo sin θ, we get
x = 2 vo2 cos θ sin θ.
Since 2 cos θ sin θ.= sin 2θ, the final result for the range R = x of a projectile which begins and ends its motion at the same level, is
R = vo2 sin( 2θ ) / g.
This range is maximum when sin( 2θ ) = 1, which happens when 2θ = 90° or θ = 45°. It also occurs that the range is identical for pairs of angles differing from 45 by n degrees–at 45 + n. The pair of angles are spaced equally above and below 45°. For example, the two angles 25° and 65° can be written as 45° + 20° since each differ from 45° by 20°. Two projectiles will have the same range if launched with the same initial speed but one having an launch angle of 25° and the other a launch angle of 65°. These facts can be seen in a range animation.
Suppose we throw an object at a mark on the wall. To hit the mark, we aim above it because we know that gravity will cause the object to fall a bit while it is moving. Suppose that the total travel time is one second. During that time, the object is going to fall a distance y = ½gt2 = 5m. To hit the target, we must aim 5 m above it. Another way to hit the target is to aim straight at it and instead let the target fall at the instant we throw the object. Suppose your brother is sitting in a tree and you want to throw him a coin. If the thrown coin is going to take one second to reach him, you can either aim the coin 5m above him, or have him drop out of the tree at the instant you throw the coin. In both cases, the coin will hit him. This is more complicated when air friction is included.
Suppose we horizontally shoot a bb at a speed of 50 m/s toward a target 50 meters away. The bb will take a time t = x/v = ( 50 m ) / ( 50 m/s ) = 1 second to reach the target. We just decided that in 1 second, which is the time needed for the bb to travel to the target, the bb will fall 5 meters and hit below the target. Suppose we instead use a rifle that shoots a bullet at a speed of 300 m/s. The bullet crosses the same distance in a time t = x/v = ( 50 m ) / (300 m/s ) = 0.17 seconds and so falls a distance y = ½gt2 = 0.014m.
Suppose now we horizontally shoot both the rifle and the bb at the same time from a height of 5 meters. We know that the bb will hit the ground 50 meters away after having traveled for one second. The bullet also takes one second to fall 5 meters to the ground, but during this one second, it will travel a distance of x = vt = ( 300 m/s )( 1 s ) = 300 meters. In the one second needed to fall 5 meters, the bb will travel 50 meters while the bullet travels 300 meters, and both strike the ground at the same time. The rifle bullet just travels farther down range compared to the bb before hitting the ground.
Suppose now we horizontally shoot the rifle and drop a bullet at the same time? Both bullets will hit the ground at the same instant, though the fired bullet will have traversed a horizontal distance.
Suppose we shoot a cannon at an angle of 45° above the ground. If it weren't for gravity the cannonball would move away from the cannon in a straight line. But we just saw that the cannonball will fall 5 meters below this line after one second, 20 meters after 2 seconds, and 45 meters after 3 seconds. The cannonball follows a curved trajectory before returning to the ground.
Now suppose we shoot the cannon parallel to the ground. Since the Earth's surface is curved, the surface sort of drops away from a tangential by 5 vertical meters for every 8000 horizontal meters (as Paul Hewitt explains). We just saw that an object falls 5 meters in 1 second. For the trajectory of the fired canon ball to match the Earth's curvature, which drops one meter for each 8000 meters along the surface, the cannonball must have a speed of 8000 m/s. If the cannon is launched at a speed of 8000 meters per second then its trajectory will be a curved path that sort of fall around the Earth. The canon ball will then be a satellite. This is the same way that the Moon falls around the Earth and that the Earth falls around the Sun. Inertia wants straight line motion but gravity keeps the changing the direction of the satellite’s motion.
The above examples show that nature often surprises us when we look more closely.
Everyday we see objects moving in a circle and we see objects that are spinning. We see spinning wheels, toy tops, dancers, skaters, footballs, bowling balls, carousels and other amusement park rides, and playground equipment such as a merry-go-round. The spinning Earth is undergoing circular motion, as are the other planets and their moons. We undergo circular motion as we ride in a turning car or airplane.
The circumference of a circle of radius r is c = 2πr. The time taken for an object to make one complete revolution is referred to as the period T of its motion. The number of times per second that a complete cycle is made is the frequency f = 1 / T. If an object is moving through one complete circle in T seconds, then its tangential speed is v = Δx / Δt = 2πr/T = 2πrf. This is referred to as a tangential velocity because it lies along a line tangential to the object.
CQ: An object rotates once every 20 seconds. What is the period of its motion? (T = 20 s) What is the frequency of its motion? (F = 0.05 s)
We know from geometry that arc length s = θr. For a spinning object such as a disk, we then have Δs = r Δθ, which is a curving distance along the circumference of the spinning disk. A point located on the rim of the spinning disk moves at a speed v and wants to traverse a straight-line distance Δx = v Δt. For tiny increments, the curving and straight-line distances are equal, giving Δs = Δx or r Δθ = v Δt. We can rearrange this as Δθ = v Δt / r. The velocity vectors form a triangle of sides Δv, v2, and v1. In this triangle, the angle Δθ between v2 and v1 is Δθ = Δv/v because an arc length here is also found from s = θr. Equating the two incremental angles gives Δθ = v Δt / r = Δv/v. This can be rearranged as Δv / Δt = v2 / r, which is the acceleration of the object. This means that whenever an object is moving in a circle, we know just from geometry that the magnitude of its acceleration is ac = v2 / r. This is called centripetal acceleration. It points radially inward toward the center of the circle.
Example.
Children are riding on the outer edge of a merry go round (as shown in the video clip) that has a 1.5 meter radius and is spinning once every 2 seconds. What is the centripetal acceleration?
The speed of a child is v = Δx / Δt = 2πr/T = ( 2π )( 1.5 m ) / ( 2 s )= 4.7 m/s and their centripetal acceleration is ac = v2 / r = ( 4.7 m/s )2 / 1.5 m = 14.8 m/s2, which is 1.5g. In the video, we see that this acceleration makes a thrown ball take a curved trajectory.
Notice that the outer edge of a spinning disk of radius r is moving at a higher tangential speed and is experiencing a higher centripetal acceleration than is a point half way to the edge. A point near the center of a spinning disk completes one small loop in the same amount of time in which a point further out makes a larger loop. The tangential speed is proportional to r.
We see a tangential speed that is proportional to r in various circumstances. You may have noticed at a stop light that has two left-hand turning lanes, that the car in the outer lane has farther to travel than does the car in the inner turning lane. Sometime a line of ice skater or roller skaters join hands and swing in a circle as a unit, with the inner person not moving at all except to spin where he or she stands. The outer person has farther to travel than the barely moving, inner person. While two bikes–one with larger wheels than has the other–travel side-by-side, the smaller wheel makes a greater number of revolutions per minute than dos the larger wheel. Bicycles became usable contraptions around the year 1870, just before they were regrettably replaced by automobiles. For a bike that has no gears, its maximum speed is determined by the radius of its wheels. That is the reason some bikes had large wheels. (Hester Amstel. Since the U.S. no longer has sidewalks, what do you say we close every other street to cars and instead use them for foot and bike traffic and outlaw cars from compact, central downtown areas?). The spinning disc within cd players have a constant tangential speed so that the grooves move past the reading laser at constant speed. They spin more slowly as the laser moves from the inner loop, which is the first track or song, to the outer loop, which holds the last track or song. Check this at home on a cd player that has a window allowing you to see the spinning disc (spelled with a 'c').
Motion is always relative to something else–and in turn, the “else” might be in motion relative to yet another object. For example, while two persons sitting in a moving car ‘A’ have little relative motion between them, they have some motion relative to the car ‘B’ that they are passing, and a lot of motion past a person ‘C’ standing near the road. The speed of ‘A’ relative to ‘C’ is equal to the speed of ‘A’ relative to ‘B’ plus the speed of ‘B’ relative to ‘C’.
Let's add velocity vectors. For example, think of walking down the aisle of a bus that is moving forward down the street at 2 mph, which is the a slow walking speed.. A person can walk 2 mph down the bus aisle, moving either toward the front or the back of the forward moving bus. A person on the curb sees the bus-walker either standing still or going 4 mph depending on whether or not the bus-walker is moving walking opposite to the direction in which the bus is traveling. The sum produces, either
--------> + --------> = -----------------> or
--------> + <------- = 0.
Same as a boat moving either with or against the 5-mph river current. (Visit here, choose "Kinematics" then “River and Boat.”) If the boat can go 20 mph in the water, then relative to the shore the boat will be moving either 20 + 5 = 25 mph or 20 - 5 = 15 mph. Note that if the boat has a limiting water speed of 20 mph then it could not travel upstream in a 30 mph river. Think of boating upstream in rapids. By the way, what makes rapids? A vertical drop in the river bed–just like the inclined plane. Rapids are tilted water falls.
While adding velocity vectors, pilots of small airplanes prefer to take off into the wind to attain the needed relative speed. A small airplane might need a 50 mph wind over its wings before it can lift off. For a runway pointing with or against the wind, such airplanes take off into the wind to add speeds rather than going against the wind. In a 10 mph wind, the plane can either drive 40 mph along the ground in the direction of the wind or 60 mph along the ground in the opposite direction of the wind. To achieve flight, The Wright Flyer needed a 35 mph wind over its wing.
A boat moves with speed Vboat/water relative to still water. When the water is also moving at the speed Vwater/ground relative to the ground (or a person standing on the shore), then the speed Vboat/ground of the boat relative to the ground is determined by the vector sum
Vboat/ground = Vboat/water + Vwater/ground.
Notice that “water” can be sort of cancelled from the numerators and denominators in the subscripts of the summed vectors. The same vector sum occurs for an airplane moving relative to air that is moving relative to the ground. The sum also describes the speed of a bird walking across the back of a walking hippopotamus.
CQ: Give another example of relative speeds.
Example:
A boat moves diagonally across a river.
Vwater/ground = 5 m/s Vboat/water = 5 m/s at angle θ =60° relative to the direction of current flow.
------------------>..........•
------------------>.......•
------------------>....•.......θ
------------------>.• .............
What is Vboat/ground?
Put the origin at the lower left with positive x toward the right and positive y upwards so that Vwater/ground = 5i m/s is entirely in the +x direction. The components of Vboat/water are
x: Vboat/water cos(θ) = 5cos( 60 ) = 2.5 m/s
y: Vboat/water sin(θ) = 5sin( 60 ) = 4.3 m/s.
These are added to the components of Vwater/ground to obtain the components of Vboat/ground, which are
x: 5 m/s + 2.5 m/s = 7.5 m/s
y: 0 m/s + 4.3 m/s = 4.3 m/s.
The magnitude of Vboat/ground is then sqrt[ ( 7.5 m/s )2 + ( 4.3 m/s )2 ] = 8.6 m/s. From the positive-x axis, the vector Vboat/ground makes an angle tan-1( 4.3 / 7.5 ) = 30°.
The current is carrying the boat along with it, so the boat has an increased speed relative to the ground.
CQ: A bird runs at a speed of 5 km/h left to right across the back of hippo that is running straight west at 5 km/h. What is the speed Vbird/ground of the bird relative to the ground? We are adding a 5 m/s due west speed to a 5 m/s due north speed. Hewitt reminds us that the magnitude of this speed can be found from hypotenuse of a 45° triangle that has sides in the ratio 1:1:√2. In this case we have 5:5:(√2 )( 5 ). So the speed of the bird relative to the ground is
( √2 )( 5 m/s ) = ( 1.41 )( 5 m/s ) = 7 m/s.
Example:
Relative to the boat, this water skier undergoes an oscillatory motion like a pendulum. At some point, calculate the velocity of the skier relative to the ground. Calculate projectile parameters when the skier goes over the jump.
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