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PHYSICS CHAPTER 7
The curved arch of thrown or launched objects such as baseballs, basketballs,
golf balls, and missiles illustrates motion you have probably seen many times.
Mathematically, the curve is a parabola. Newton's laws of motion are
all that are needed to understand how objects like water drops and golf balls
move.
7.1 PROJECTILE MOTION
Objects that are thrown like baseballs are projectiles. The path they follow is called a trajectory.
The motion of a projectile is described in terms of its position, velocity, and acceleration. These
are all vector quantities, and we have seen that the perpendicular components of vector
quantities are independent. How does this fact let us analyze the motion?
Independence of Motion in Two Dimensions
Let's start with a simple trajectory. Figure 7-1 shows a photograph of two falling golf balls, taken
with a strobe light that flashed 30 times each second.
One ball was launched horizontally at 2.0 m/s, the other ball was simply dropped. Notice that the
thrown ball moves the same distance to the right during each time interval. That is, its horizontal
velocity is constant. The dropped ball also has a constant horizontal velocity, equal to zero. In
each case, the force acting on the ball has no horizontal component, so there is no horizontal
acceleration. (DRAD)
Disregarding air drag
The golf ball photograph shows a remarkable property of motion. Look carefully at the height of
the balls. At each flash, the vertical position of the dropped ball is the same as that of the thrown
ball. This means that the vertical motion of the thrown ball and the dropped ball are the same. In
addition, the change in vertical position between two successive flashes is the same for the two
balls. This means that their average vertical velocities during these intervals also must be equal.
The spaces between images grow larger because the force of gravity accelerates both balls
downward. The photograph shows that the horizontal motion of the thrown ball does not affect
its vertical motion at all. Bullet and shell from a gun
Suppose that Kyle slowly passes Adam in his car as he stands by the side of a road, Figure 72. Jami throws an apple out the window of the car at Tito.
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Imagine that Adam can watch the apple only. What path will Adam see it take? Remember that
the apple has the same horizontal velocity as the car. From Adam’s frame of reference, or point
of view, the path will be curved like that of the launched golf ball. Will Jami see the same path?
Jami and the apple both move with the same horizontal velocity. Thus, Jami will see a straight
vertical path like that of the dropped golf ball. The trajectory as seen by the two observers will be
different, but both observers will agree that it takes the same time for the apple to hit the ground.
Thus, the shape of the trajectory and the horizontal motion depend on the viewpoint, or frame of
reference, of the observer. The vertical motion does not.
The independence of vertical and horizontal motion and our motion equations can be used to
determine the position of thrown objects. If we call the horizontal displacement dx and the initial
horizontal velocity vx then, at time t,
dx = vxt from d=vt
and vxf = vxi same as vf = vi
The equations for an object falling with constant acceleration, g, describe the vertical motion. If
dy is the vertical displacement, the initial vertical velocity of the object is v y. At time t, the
vertical displacement is
dy = vyt + ½gt2 from d = vit + ½ gt2
also, vyf = vy + gt from vf = vi + gt
If an object is not thrown or projected vertically, meaning it is dropped, pushed, or thrown only
with horizontal velocity (vx) its vertical velocity initial (vy) is 0, when vy = 0 t = the square root
of 2dy/g
Using these equations, we can analyze the motion of projectiles. The simplicity of these
calculations is due, in part, to the fact that for the golf ball, the force due to air resistance is
negligible. For an object such as a feather these equations could not be used because air
resistance would play an important factor in determining their motion.
Objects Launched Horizontally
We'll first look at the motion of objects that have an initial vertical velocity of zero. vy = 0
Solve: Megan throws a stone horizontally at +15.0m/s from the edge of a vertical cliff that is 44.0
meters high A. how long does it take the stone to hit the ground below? B. How far from the
bottom of the cliff does the stone strike the ground? C. Sketch the trajectory of the stone.
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Do practice problems 7-1
Objects Launched at an Angle
Figure 7-4a shows what happens when a golf ball bounces off a hard surface. (Tin Cup) The
vertical and horizontal velocity components during one of the bounces are shown in Figure 74b. The horizontal component, vx, is constant because very little force is exerted in the
horizontal direction. Gravity, however, acts on the ball, giving it a downward acceleration. Thus,
the vertical velocity component is large and positive at the beginning, decreases to zero at the
top of the path, and then increases in the negative direction as the projectile falls. When the
object returns to its launch position, the vertical speed is the same as it was at launch, but its
direction is reversed. The range, R, is the horizontal distance from the point of bounce until the
projectile returns to the surface height.
PROBLEM SOLVING STRATEGY
When solving a projectile motion problem, first determine the horizontal and vertical components
of the initial velocity. Vx= vi cos Ө Vy= vi sin Ө Then the parts of the problem involving either
component can be solved separately. The symmetry of the trajectory can be used when the
launching and landing locations are at equal height. Rising and falling times are equal, as are
the horizontal distances moved in each half of the trajectory. We will ignore air resistance in
projectile motion problems.
When objects are thrown or launched at an angle and land at the same height vy ≠ 0, but dy is
equal to 0. So
dy = vyt + ½gt2 so 0 = vyt + ½gt2 so -2 vyt = gt2 so
t = -2 vy/g
Solve: In the strobe photo of the flight of the ball in figure 7-5 below the initial velocity of the ball
was 4.47 m/s at an angle of 66.00 the ball left and landed at the same height. A. How long was
the ball in the air? B. How high did the ball go during its flight? C. How far did the ball travel?
PROBLEM SOLVING STRATEGY
Always check a problem to see if your answers are reasonable. In this case, you can use the
photo and the knowledge that there were 30 flashes per second. The flight time was calculated
to be 0.833 s. At 30 flashes/second this would be 25 flashes. You can count 25 flash intervals
during this bounce, so the answer is fine. We don't know the photo scale, but the calculated
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maximum height is 0.85 m and the range is 1.5 m. The ratio of these two is 0.57/1. On the
photo, the ratio of height to range is 57 mm/100 mm = 0.57/1, also in very good agreement.
The equation below can be used to determine the horizontal distance (range) of a
projectile from the velocity given at an angle with the horizontal. Using this equation, you
could determine the maximum range a projectile would travel if its initial velocity was
12.5m/s when  equals 45°, 75°, 15°, 60°, 30°, 10°, and 80°.
A sketch of the paths would look like the diagram shown here. The symmetry of the
motion is a very interesting concept.
45 – 12.52 * sin 2 45 = 156 m
In baseball and most other games, the player does not throw a ball so that it goes farthest, but
so that it goes a certain distance in the least time. Consider a throw from third to first base. If
thrown at 38 m/s, the smallest angle it could be thrown to go the 37 m is 7° above the horizontal.
It would take 0.97 s without air resistance. The drag caused by the air means the thrower must
increase the angle to 9°, and makes the ball take about 1.1s.
Use example problem 7-2 to help you solve practice problems 7-2
Do concept review 7-1
7.2 PERIODIC MOTION
Projectile motion is two dimensional, but the motion does not repeat. Projectiles move along
their trajectories only once. On the other hand, if you swing a toy on a string, such as a yo-yo,
horizontally around your head, its motion repeats itself at regular intervals. That is, its motion is
periodic. An object bouncing on a spring or the pendulum of a clock is also periodic. Although
different terms are used to describe this type of motion, Newton's laws can still be used to
understand it.
Circular Motion
Your body can't sense constant velocity, but it is a sensitive accelerometer. Your stomach can
detect the acceleration of an elevator or of an airplane in turbulence rough weather.
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Pilots say those drops are typically no more than 50 feet -- not the hundreds of feet many
passengers perceive. They also emphasize that avoiding turbulence is mostly a matter of
comfort, not safety.
Aircraft manufacturers have been collecting data since the 1970s to determine the maximum
stress that planes experience in turbulence, and they then design aircraft to withstand one and a
half times that. In fact, Boeing's test of the wing of a 777; using cables, show that the wing is
bent upward about 24 feet at the tip before it breaks.
Systems to detect and respond to turbulence have also improved, including the technology that
automatically adjusts to lateral gusts of wind. And Boeing's 787 aircraft will have a new vertical
gust suppression system to minimize the stomach-churning sensation of the plane suddenly
dropping midair
It can also feel acceleration on an amusement park ride, such as the Round up below, that
spins you around to make the ride thrilling and exciting. Except when the ride is starting or
stopping, the speed is constant, but the velocity is changing because its direction is changing.
The riders are moving with uniform circular motion.
A vector diagram like Figure one below can help you find the acceleration of an object moving
in a circle at constant speed.
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Points A and B are two successive positions of an object moving with uniform circular motion.
The radius of the circle is r. The radius and speed are constant, so the velocity is always
perpendicular to the radius. That is, the velocity is tangent to the circle. The vector v 1 represents
the instantaneous velocity of the object at A. The vector v2 represents the instantaneous velocity
of the object at B. Note that the two vectors have the same length, but their directions are
different.
Acceleration is the change in velocity, ∆v, divided by the time interval, ∆t. The difference
between the two vectors, v1 and v2, is ∆v. That is, ∆v = v2 – v1. You know how to find a vector
sum, but not a difference. We can convert the equation into a sum by adding v1 to both sides of
this equation to get v2 = v1 + ∆v. In Chapter 6, you found a vector sum by placing two vectors
head to tail. Consider Figure two above. The size of ∆v can be found by using a ratio of the
sides of similar triangles ABC and DEF. The distance traveled by the object between point A
and point B is the arc AB. If we choose A and B so they are close together, then arc AB is
approximately equal to chord AB. Thus,
∆v/v = chord AB/r
But the distance traveled in the time interval ∆t is just v∆t. Thus, we obtain
∆v/v = v∆t/r
Since acceleration is given by a = ∆v/∆t, we find a = ∆v/∆t is equal to v2/r
The direction of ∆v can be seen in Figure 7-9b. As positions A and B become very close
together, ∆v becomes perpendicular to the two velocity vectors. That is, it is in the direction of
the radius, pointing toward the center of the circle. Newton originated the word centripetal for a
quantity that always points toward the center of the circle. Centripetal means, "center seeking."
For this reason, we call the acceleration just defined centripetal acceleration and write
ac = ∆v/∆t = v2/r
Centripetal acceleration always points toward the center of the circle. It is directly proportional to
the square of the speed and inversely proportional to the radius of the circle.
It is often difficult to measure the speed of an object, but easier to measure the time needed to
make a complete revolution, or the period, T. During this time, an object travels a distance equal
to the circumference of the circle, 2πr. The speed of the object is then v = d/t and d = 2πr and t
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= (T period) so
v = 2πr/T period of an object also = T=1/frequency
This expression for v can be substituted into the equation for centripetal acceleration to yield
(V)2
ac = (2πr/T)2 / r = 4π2r / T2 so ac = 4π2r / T2
Newton's second law tells us that an object does not accelerate unless a net force acts on it.
The acceleration is in the same direction as the force. What exerts the force that gives a body its
centripetal acceleration? In the case of the amusement park ride, the drum exerts a force
against the backs of the riders. The force that causes the centripetal acceleration of a model
plane flying in a circle is the force of the string on the plane. The moon orbits Earth with nearly
uniform circular motion. Earth's gravity is the force that causes this acceleration. Because the
force, as well as the acceleration, is toward the center of the circle, these forces are often called
centripetal forces. To understand centripetal acceleration, it is very important to identify the
source of the force. For a body in uniform circular motion, Newton's second law can be written in
the form
Fnet = Fc which = mac and ac = 4π2r / T2 so Fc = m(4π2r / T2)
In what direction does an object fly if the force giving it centripetal acceleration suddenly
disappears? Remember that with no net force, an object moves in a straight line at constant
speed. Consider an Olympic hammer thrower. The "hammer" is a heavy ball on a chain that the
thrower whirls in a circle. As he whirls it at greater and greater speeds, his arms and the chain
supply the force that gives it centripetal acceleration. At the correct time, he lets go of the chain.
The hammer flies off in a straight-line tangent to the circle. After its release, only the
gravitational force acts on it. Water in a bucket!
Use example problem 7- 3 (on 7-3 page) to help you solve practice problems 7-3
Changing Circular Motion: Torque
As we have seen, if an object moves with constant circular motion its acceleration is always
perpendicular to its velocity. In this case, the acceleration can change only the direction of the
velocity, not its magnitude. How then can you start or stop circular motion? You have to apply
another force on the object, one that has a component parallel to its velocity. But, changing the
speed of circular motion is slightly more complicated.
Consider how you close a door, Figure 7-12.
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How do you make the door rotate rapidly about its axis of rotation, its hinges? First, you exert a
large force. Second, you must exert the force as far from the hinges as possible. Pushing on the
hinges does not start it rotating. Pushing halfway to the edge does some good, but pushing on
the knob or the far edge gets the door rotating most rapidly. Third, you push perpendicular to the
door, not toward the hinges. We combine the information about distance and direction into one
concept, the lever arm. The lever arm, d, Figure 7-13,
is defined as the perpendicular distance from the axis of rotation to a line along which the force
acts. The product of the force and the lever arm is called torque. The greater the torque applied,
the greater the change in rotational motion. Thus, torque plays the role of force for rotational
motion.
Torques can stop, start, or change the direction of rotation. To stop the door from closing, or to
open it, you exert a force in the opposite direction. Another example of torque is the seesaw. If a
seesaw is balanced there is no net torque. If it is not rotating, it remains non-rotating. How would
Vido and Ryan balance a seesaw? Each person exerts a torque of the same magnitude, but in
the opposite direction. Vido would have to sit closer to the axis of rotation, the pivot. Each torque
is the product of the force, the person's weight, mg, and the lever arm, d, the distance to the
pivot. That is, the seesaw is balanced if m1gd1 = m2gd2.
Simple Harmonic Motion
A playground swing moves back and forth over the same path. A vibrating guitar string, a
pendulum, and a metal block on a spring are other examples of back and forth or vibrational
motion.
In each case, the object has an equilibrium position. Whenever the object is pulled away from
its equilibrium position, a force in the system pulls it back toward equilibrium. The force might be
gravity, as in the case of the swing or pendulum. It could be the stretch of a guitar string or a
spring. If the restoring force varies linearly with the displacement, the motion that results is
called simple harmonic motion.
The motion of objects with SHM can be described by two quantities, period and amplitude.
The period, T, is the time needed to repeat one complete cycle of motion. The amplitude of the
motion is the maximum distance the object moves from its equilibrium position.
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Let's look more closely at the motion of a metal block on a spring. The spring exerts a
restoring force, a force that acts to oppose the force that stretched or compressed it. The force
increases linearly with the amount the spring is stretched or compressed. An object that exerts a
force in this way obeys a relationship called Hooke's law. Because the force dependence is
linear, the movement of an object suspended on a spring is an example of simple harmonic
motion.
When the block is at the equilibrium point, as shown in Figure 7-15a,
the weight of the block is balanced by the force of the spring, F1. Suppose you start the motion
as shown in Figure 7-15b by pulling the block down just a few centimeters and letting go. At
first, the restoring force of the spring is more than the weight. There is a net upward force, so
the block is accelerated upward. As the block rises, the spring is stretched less and less and the
force it exerts gradually decreases. Thus, the net force decreases and the acceleration is less.
When the block returns to its equilibrium position, the net force is again zero. Why doesn't the
block stop? According to Newton's laws, a force is needed to change velocity. With no net force,
the block continues with its upward motion. When the block is above the equilibrium position,
the weight is larger than the force exerted by the spring. There is a net downward force and
resulting downward acceleration. The speed of the block decreases. When the speed is zero,
the block is as high as it will go. Even though the speed is zero, there is a net downward force,
and the block accelerates down. When it again reaches the equilibrium position, there is no
force, but the downward motion continues. When the block returns to its lowest location, one
period, T, has elapsed, and the cycle of motion starts over. For a small displacement, the period
depends on the mass of the block and stiffness of the spring. It does not depend on the
amplitude of the motion.
Figure 7-16 shows graphs of the position, velocity, and acceleration of the block. At the
vertical line, the block is at its largest distance from equilibrium. Its velocity is zero; it is
momentarily at rest. Its acceleration is most negative, and thus the force pulling it back down to
equilibrium is largest. Notice also that the speed is largest when the acceleration is zero, which
occurs when the block is at the equilibrium position.
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The swing of a pendulum also demonstrates SHM. A simple pendulum consists of an object (the
bob) suspended by a string of length, l. The bob swings back and forth, as shown in Figure 717.
The force on the bob when it is pulled away from the vertical is also shown. The gravitational
force, the weight, W, is resolved into two components. FII is along the direction of the string. The
component F┴ is at right angles to the direction of the string. When the bob is pulled to the right,
F┴, is to the left, and vice versa. That is, the component of the weight, F┴, is a restoring force.
Further, for small angles (Ө less than 150 ) the magnitude of F┴, is proportional to the
displacement of the bob. Thus, small-displacement pendulum motion is an example of SHM.
The period of the simple pendulum of length, l, is given by the equation
____
T = 2 π√ l /g manipulate T2 = 4 π2 l/g
Notice that for small-angle displacements, the period depends only on the length of the
pendulum, not its mass or amplitude. By measuring the length and period of a pendulum, you
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can find the magnitude of the local value of the gravitational acceleration, │g│
The frequency of a pendulum is the number of complete cycles of motion in one second. It can
be found from the period using f = 1/T
Use example problem 7-4 to help you solve practice problems 7-4
The amplitude of any vibrating object can be greatly increased by applying small external
forces at specific regular intervals of time. This effect is called mechanical resonance. You were
probably first introduced to resonance when you learned to "pump" a playground swing. You
found that you could increase the amplitude of the swing if your friend applied forces to it at just
the right times. The time interval between the applied forces had to equal the period of the
swing. Other familiar examples of mechanical resonance include rocking a car to free it from a
snow bank and jumping up and down on a trampoline or diving board. Rubbing rack under
chair. Resonances can also cause damage. To avoid resonance, Roman soldiers do not march
across bridges in cadence. The rhythm of their steps could otherwise create large oscillations
caused by their steps resonating with the natural period of the bridge. Audiences in theater balconies who jump up and down in time with the music have caused damage to the balcony when
the jumping frequency matched the natural vibration frequency of the balcony.
Our hallway - Resonances have caused the collapse of bridges (Tacoma Washington)
Do Concept Review 7-2.
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