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MOTION IN TWO DIMENSIONS
projectile
any object thrown or otherwise projected into the air
trajectory
the parabolic path of a projectile
In projectile motion, the horizontal and the vertical components of the motion are treated
separately. A projectile moves both horizontally and vertically. Its horizontal motion is constant.
Its vertical motion is affected by the acceleration due to gravity. The only variable shared by both
types of motion is time. Every point on the trajectory is the vector sum of the horizontal and the
vertical components of the velocity.
See if you can score a "hole-in-one."
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An object projected horizontally (projected perfectly parallel to the surface) will reach the ground
in the same time as an object dropped vertically. Since speed at any point in a trajectory is the
vector sum of the horizontal and vertical velocity components at that point, the projected object
will have a greater speed when it strikes.
The maximum range for a given initial velocity is obtained when the angle of projection is 45.
Equations that are used to describe the horizontal and vertical motion:

horizontal motion (constant velocity):
d=vt
where v is the horizontal velocity component and d is the horizontal distance (range)

vertical motion (acceleration):
d = vit + ½ a t2
vf = v i + a t
vf2 = vi2 + 2 a d
where d is the vertical distance, vi is the initial vertical velocity, and vf is the final vertical velocity
Speed
The speed of an object at any point on the trajectory can be found by calculating the horizontal and
vertical velocity components at that point. The speed is the vector sum of the components at that
point. This is a very common "trick" that arises on the AP Physics B exam. You are asked to
calculate speed, acceleration, etc. It is very easy to forget that it is the resultant of the horizontal
and the vertical components of speed, acceleration, etc. For example, assume you are asked to
calculate the speed of an object which rolls off a vertical cliff with horizontal velocity v. You must
calculate the final vertical velocity (vf)of the object the instant before the object strikes. The speed
is the square root of the sum of v2 and (vf2.
Directions
It is very important to consistently define directions in projectile motion. If the acceleration due to
gravity is defined to be negative, then all velocities in the "down" direction are also negative. If
displacement is measured from the ground up, it is positive. If it is measured from the "top" down,
it is negative.
AP assignment of an origin
In AP problems, it helps to assign an "origin." For example, an object is thrown upward from the
top of a cliff with speed v and at an angle . Call this point the origin. It is convenient to set the
origin where t=0. The initial vertical velocity will thus be positive. The height of the cliff would
be negative because it is measured from your origin downward in the -y direction.
Advanced calculations
The general equation of motion, d = vit + ½ a t2, can be easily used to calculate vertical
displacement and/or time at any point in a trajectory. First, find the initial vertical velocity
component.




If horizontal distance is known: Use the horizontal velocity component to calculate time
in the air to that point in the trajectory. Substitute that value for time in d = vit + ½ a t2
and solve for vertical displacement, knowing the initial vertical velocity component.
If vertical displacement is known (remember to assign your origin system and use the
appropriate positive and negative signs for vertical displacement and initial vertical
velocity): Use d = vit + ½ a t2 to solve for time to that point in the trajectory knowing the
initial vertical velocity component.
Calculate the total time the object is in the air by using d = vit + ½ a t2 and the initial
vertical velocity component. Remember, to set your origin and use the appropriate
positive and negative signs for the initial vertical velocity component and for vertical
displacement.
Suppose an object is launched from the ground with speed v and at an angle . You can
use d = vit + ½ a t2 and the initial vertical velocity component to find the total time in the
air. Remember, when it hits the ground, the vertical displacement would be zero.
We will be working two types of problems.

The first type of problem involves an object that is thrown or projected at the top (or
apex) of its trajectory (at point B in the picture). At this point, the vertical velocity is zero.
Since it starts at this point, we let vi = zero. It accelerates vertically downward at a rate of
-9.8 m/s2. We usually use the first acceleration formula listed to find the time the
projectile is in the air. The time that it travels vertically is the same as it travels
horizontally. We can use this time to determine its horizontal range. Its vertical
displacement is given by distance BD and it is negative. Its horizontal range is given by
distance DC.
vertical: vi = v sin 
horizontal: vi = v cos 
In this case,  = 0. Thus, cos  = 1 and sin  = 0 which makes the initial vertical velocity
component equal to zero and the horizontal velocity component equal the the horizontally
projected velocity.
Advanced calculations
Consider this problem: An object rolls down an incline and then off the edge of a cliff. Motion
information is given allowing one to calculate to speed and angle with which the object leaves the
edge of the cliff. Find the initial vertical and horizontal velocity components at the edge of the
cliff. Remember, set your origin at the point t=0. The object is going down, so the sign of the
initial vertical velocity component is negative. If the height of the cliff is known, the time to fall
can be calculated using d = vit + ½ a t2, the initial vertical velocity component, and the
acceleration due to gravity. Once time to fall is known, the horizontal velocity component can be
used to calculate the distance the object strikes from the base of the cliff. One can also calculate
the vertical velocity component the instant before it strikes the ground. It can be used to determine
the speed and angle with which the object strikes the ground (remember, it is the vector sum of the
horizontal and vertical velocity components at that point).

The second type of problem involes an object that is projected from the ground (point A) and
whose path is parabolic. At this point, the speed at which it is projected, v, is the vector sum of its
horizontal and its vertical components. At point A, its initial vertical velocity component and its horizontal
velocity component are given by:
vertical: vi = v sin 
horizontal: vi = v cos 
At pointB, its vertical velocity component is zero. The acceleration at all points is -9.8 m/s2. At
point B, the object is accelerating (even though its vertical velocity is zero), because its direction is
changing. At all points in the trajectory, the horizontal velocity component remains the same (in
the absence of air friction). We will work these problems by finding total time in the air using the
second acceleration formula. vf at point C is numerically equal to vi at point A but of the opposite
sign. Remember, velocity is a vector quantity! Once we have total time, we can use d = v t to find
the horizontal range, knowing that this uses the horizontal velocity componet. We can use the last
acceleration formula to predict the maximum vertical displacement (working from points B and
D).
A formula can be derived for the horizontal range:
Range = (v2 sin 2) / g
where v is the projectile velocity,  is the projectile angle, and g is the acceleration due to gravity.
Advanced calculations
The general equation of motion, d = vit + ½ a t2, can be easily used to calculate vertical
displacement and/or time at any point in a trajectory. You can also calculate the vertical velocity
component at any point knowing the initial vertical velocity component and the time to that point.
Remember to set your origin and use the appropriate positive and negative signs. You can then use
vf = vi + at to find the vertical velocity component at that point.
http://jersey.uoregon.edu/vlab/newCannon/NewCannon_plugin.html
A virtual lab in which the user can control the angle of the cannon. Cursor interrogation allows
maximum vertical height and total horizontal range to be determined. A target is provided for
amusement.
Projetile Motion applet
http://www.phy.ntnu.edu.tw/java/projectile/projectile.html
These two types of problems can also be solved graphically using parametric equations on the
graphing calculator. If you are interested in learning how to do this, it is explained in the link
below.
Graphing Calculator Solution for Two Dimensional Motion Problems
Your TI-83 needs to be in MODE=Degree, Par, Simul. This allows you to graph the horizontal motion (x)
as a function of time (t) and to graph the vertical motion (y) as a function of time (t). Remember the physics
of the problem! The horizontal motion is independent of the vertical motion and the only variable that they
share is time.
Write an equation of motion for the horizontal motion of the object. Its horizontal distance(d) at any time
(t)is a function of its horizontal velocity (v).
d = v cos  t Its vertical displacement (d) at any time (t) is a function of its vertical velocity (v) and
the acceleration.
d = vi sin cos  t + 1/2 a t2
In our first type of problem, the object is launched from the apex of the trajectory. At this point, 
= 0 and the cos  = 1 and sin  = 0.
Let's work an example of the first type of problem on the graphing calculator.
An object is projected from the top of a cliff 40 m with a horizontal velocity of 20 m/s. How long
does it take to strike the ground? How far does the object travel horizontally?
Write the equation of horizontal motion. Enter it into "X1T ="
X1T=20 T
the horizontal velocity is 20 m/s. Write the equation of vertical motion. Enter it into "Y1T ="
Y1T = 0 + 1/2(-9.8)T2 + 40
it's initial vertical velocity is zero, the acceleration is that of gravity, and +40 is its initial vertical
displacement.
Graph the functions. View the graph. Set Window as needed. Appropriate window settings for this
problem are:
Tmin=0
Tmax=3
Tstep=0.1
Xmin=0
Xmax=60
Xscl=1
Ymin=0
Ymax=50
Yscl=1
Use the trace key to determine the maximum time is 2.86 sec and the horizontal distance is 57.14
m. You can use the "zoom in" function to get an exact answer.
Our second type of problem differs in that the object is projected from the horizontal and has a
value for the angle . Also, the initial vertical displacement is zero.
Let's work an example.
An object is projected with a velocity of 20 m/s at an angle of 37X1T ="
X1T=20 cos 37 T
the horizontal velocity is 20 cos 37 Write the equation of vertical motion. Enter it into "Y1T ="
"Y1T = 20 sin 37 + 1/2(-9.8)T2
it's initial vertical velocity is 20 sin 37, the acceleration is that of gravity, and the initial vertical
displacement is zero.
Graph the functions. View the graph. Set Window as needed. Appropriate window settings for this
problem are:
Tmin=0
Tmax=3
Tstep=0.1
Xmin=0
Xmax=50
Xscl=1
Ymin=0
Ymax=10
Yscl=1
Use the trace key to determine the maximum time is 2.86 sec, the horizontal distance is 39.29 m,
and the maximum vertical displacement is 7.40 m. You can use the "zoom in" function to get an
exact answer. You can determine the vertical velocity at any time t by using the dy/dt function
found under 2nd calc. Enter twice for the answer.
One of the classic questions in physics is this: What should a monkey in a tree do when a gun
pointed at him fires -- jump down or stay where it is? A graphing calculator solution to this
problem is found at Monkey and Hunter Problem
http://www.cyberclassrooms.net/~pschweiger/monkeygun.html
An interactive "shoot the monkey" activity"
http://www.explorelearning.com/index.cfm?method=cResource.dspDetail&ResourceID=42
AP Multiple Choice Questions -- Motion in Two Dimensions
1.
2.
3.
4.
Be prepared to perform simple calculations. For example, an object is rolled off a table a
height h above the ground. The object has initial horizontal speed v. Calculate how long
the object is in the air. Calculate how far it lands from the edge of the table. Predict how
time in the air and distance it lands from the edge of the table would change if the
horizontal velocity of the object were increased or decreased.
Be prepared to perform simple calculations. For example, and projectile is launched with
velocity v from the ground at an angle . Points are labeled on its trajectory, one being at
its apex. Be able to identify the acceleration vector for each point. Be able to compare the
speeds at each point. Be able to compare the velocities at each point. Be able to identify
graphs of its vertical velocity, its horizontal velocity, and its acceleration for each point in
its trajectory.
Important - know that the horizontal velocity is constant for each point in the trajectory.
Know that the vertical velocity is accelerating. At the apex of the trajectory, the object
has a vertical velocity of zero. It still has acceleration (gravity) and horizontal velocity.
The speed of the object at any point is the vector sum of the horizontal and vertical
velocities at that point. Remember -- going up, vertical velocity is positive and
acceleration is negative; going down, vertical velocity is negative and acceleration is
negative.
Remember -- acceleration involves a change in speed and/or direction.
AP Free Response Questions -- Motion in Two Dimensions
1.
2.
3.
4.
5.
These questions occur over and over in free response questions! They all involve an
object that is some height h above the ground. Something has given it horizontal velocity
v at this point. Its initial vertical velocity is zero. You will be asked to describe its motion
as it falls to the ground. Remember -- describe its horizontal motion (it moves with
constant horizontal speed v); describe its vertical motion (it accelerates downward at -9.8
m/s/s); and, describe the trajectory that you see as the vector sum of these two motions.
Be able to recognize this type of problem which is asked over and over. They all involve
an object that is some height h above the ground. Something has given it horizontal
velocity v at this point. Its initial vertical velocity is zero. You have to calculate the time
it takes for the object to reach the ground. You have to calculate how far it lands
(horizontally). You have to calculate its speed when it hits the floor (remember, it is the
vector sum of its final vertical velocity and its horizontal veloctiy).
Be able to recognize an Atwood machine or its variation. Two objects with different
masses are hung by a cord across a frictionless pulley (in its variation, one object is on a
table and the other hangs off the edge or the object is on an incline and the other object
hangs off its edge). Calculate the acceleration of the system. You have to calculate the
time it takes for the object to reach the ground. You have to calculate how far it lands
(horizontally). You have to calculate its speed when it hits the floor (remember, it is the
vector sum of its final vertical velocity and its horizontal veloctiy).
Trajectory problems and their variations (common variation is to launch the object and
have it clear a fence, etc.). Be able to calculate how high the object goes and how long it
is in the air. Be able to calculate its height at a specific point above the ground.
Be able to draw horizontal velocity vs time, vertical velocity vs time, and acceleration vs
time graphs for projectile motion.
SIMPLE HARMONIC MOTION
simple harmonic motion
periodic motion where the unbalanced force varies directly with the displacement from the
equilibrium point; this motion is described by the period, the frequency, and the amplitude of the
motion
period (T)
the time in seconds needed to complete one cycle of motion
amplitude
the distance from the equilibrium point to the point of greatest displacement
frequency
the number of vibrations in a time interval; its SI unit is hertz (Hz)
I like to use the Greek letter, , as the symbol for frequency. This can be confusing to students
since its appearance is similar to the letter v. It can also be represented using the symbol f
1 Hz = 1 sec-1
T = 1/ or T = 1/f
 = 1/T or f = 1/T
characteristics of a simple pendulum:
1.
2.
3.
4.
period is independent of mass
period is directly proportional to the square root of its length
period is indirectly proportional to the square root of the acceleration due to gravity
period is independent of amplitude if the arc is less than 10
Where l is the length of the pendulum and g is the acceleration due to gravity at that point.
For a pendulum, speed is zero and acceleration is a maximum at the point of maximum
displacement (point A). For a pendulum, speed is a maximum and acceleration is zero at the
equilibrium point (point B).
This Pendulum Applet
(http://www.sciencejoywagon.com/physicszone/lesson/otherpub/wfendt/pendulum.htm)shows the
variation of displacement, velocity, acceleration, and force in a pendulum's swing. You can choose
whichever graph display that you would like.
The same thing as above, but for a Spring
Pendulum(http://www.sciencejoywagon.com/physicszone/lesson/otherpub/wfendt/springpendulu
m.htm).
An Advanced Look at Simple Harmonic Oscillation
A block is attached to a spring. The block is pulled to position A, stretching the spring from its
equilibrium position x=0. This amount of stretch represents the displacement or the amplitude of
the oscillating block on the spring. At this point (maximum displacement or amplitude of the
oscillation), the speed of the block is zero (v=0) and the spring experiences a restoring force F. An
external force must be applied to stretch the spring. This force is given by Hooke's Law (F =-kx,
where k is the spring constant and x is the displacement). (see image 1) When the block is
released, the spring exerts a force (given by Hooke's Law) that restores the spring/block to the
equilibrium position. As the block moves closer to the equilibrium position, the speed increases
and the force decreases. The speed is its maximum and the restoring force is zero at the
equilibrium position (see image 2). As the block continues to move past its equilibrium position,
the force acting on it tends to slow it down until the speed is again zero at the point of maximum
displacement (see image 3).
Work must be done to stretch the spring. Potential energy is stored in a stretched or compressed
spring. This elastic potential energy is given by
PE = 1/2 kx2
The total mechanical energy of an oscillating spring/block system is the sum of the kinetic and
potential energies at that point. The total mechanical energy is equal to the elastic potential energy
at maximum displacement.
E = 1/2 mv2 + 1/2 kx2
The total mechanical energy is proportional to the amplitude. If the spring is stretched twice as far,
the force is twice as great. Since the energy is proportional to the square of the amplitude,
stretching it twice as far quadruples the energy. Since the force is twice as great, the acceleration is
twice as great.
The period of a simple harmonic oscillator is dependent upon the spring constant and the mass that
is oscillating.
AP Multiple Choice Questions on Pendulums and Springs in Simple Harmonic Motion
1.
2.
3.
4.
5.
6.
Be able to predict how the period of a pendulum would change if its length were doubled.
Be able to perform simple calculations to predict the period of a pendulum.
Be able to predict what would happen to the period of a pendulum if the mass of its bob
were doubled.
Be able to predict what would happen to the period of a mass on a spring in SHM if the
mass were doubled.
Be able to predict what the period and the amplitude for an object on a spring in SHM
would be for each point in its oscillation. In other words, what would they be after 1/4 the
period has elapsed? 1/2 the period? 3/4 the period?
Be able to identify when the acceleration and/or the velocity is the greatest (the least) for
an object in SHM.
AP Free Response Questions on Pendulums and Springs in Simple Harmonic Motion
1.
Not commonly asked unless it is combined with energy concepts. This will be discussed
in the next chapter.
UNIFORM CIRCULAR MOTION
acceleration involves a change in speed and/or direction; it is caused by an unbalanced force
in circular motion, the object moves at constant speed but is accelerating because its direction is
constantly changing
Uniform Circular Motion
An object moving in a circle of radius r with constant speed v has an acceleration whose direction
is toward the center of the circle and whose magnitude is a R = v 2/r. Acceleration depends upon
speed and radius. The greater the speed, the faster the velocity changes direction; the larger the
radius, the less rapidly the velocity changes direction. Since the acceleration is directed toward the
center of the circle, the net force must be directed toward the center of the circle too. The net force
must be applied by other objects.
centripetal acceleration:
where r is radius and v is velocity
centripetal force:
Newspaper Article About Sky Diving
Accident
You can use your graphing calculator to determine how the magnitude of the centripetal force
varies the speed with which the object is swung in the horizontal circle, the mass of the object, or
the radius of the horizontal circle.
Remember: when a mass moves in a horizontal circle, it completes one revolution in its period. It
goes a distance of one revolution or the circumference of the circle, given by 2r. Since it is
moving at constant speed in a horizontal circle, the speed can be found by
v = d/t = (2r)/T
where T is the period
Usually, one takes frequency data. In other words, one counts revolutions for a given time interval.
Remember, the period is the inverse of the frequency.
Graphing Calculator Tips - Periodic Motion
Use the graphing calculator to study how the centripetal force varies with changes in the speed with which
the object is swung in a horizontal circle, the mass of the object, or the radius of the horizontal circle.
1.
How the magnitude of the centripetal force varies with speed:
In the formula for centripetal force, the force is dependent upon the speed of the object.
The independent variable is the speed v; the dependent variable is the force F.
The formula can be written for the graphing calculator and entered into "y1="
y = (m * x2) / r
Substitute numerical values for the radius r and the mass m. Graph the function. Use the trace key
to determine the value of the force (y) for various values of the speed (x).
2.
How the magnitude of the centripetal force varies with mass:
In this case, the independent variable is the mass (m).
The formula can be written for the graphing calculator and entered into "y1="
y = (x * v2) / r
Substitute numerical values for the radius r and the speed v. Graph the function. Use the trace key
to determine the value of the force (y) for various values of the mass (x).
3.
How the magnitude of the centripetal force varies with radius:
In this case, the independent variable is the radius (r).
The formula can be written for the graphing calculator and entered into "y1="
y = (m * v2) / x
Substitute numerical values for the speed v and the mass m. Graph the function. Use the trace key
to determine the value of the force (y) for various values of the radius (x).
Horizontal Circles The image below shows the forces acting on a ball attached to a cord swung in
a perfectly horizontal circle. There are two forces acting on the ball, the tension in the string and
the weight of the ball. In reality, the ball's weight makes it impossible to swing the ball in a
perfectly horizontal circle. We will assume that the ball's weight is small enough that we can
ignore it. The tension (labeled T in the image) is the unbalanced center-seeking force and provides
the centripetal force on the ball.
Fc = mv2/r = T
Horizontal Circles The image below shows the forces acting on a ball attached to a cord swung in
a horizontal circle. The cord is attached to a pole at angle  to the pole. There are two forces
(shown in red) acting on the ball, the tension in the string and the weight of the ball. The diagram
to the right shows the x and y components of the tension. The y-component of the tension is equal
in magnitude and opposite in direction to the weight of the ball. The x-component of the tension is
the unbalanced center-seeking force which provides the centripetal force on the ball. If q is
measured between the cord and the pole, (A handy trig identity to remember is that tan=sin
/cos)
T cos = mg
T sin = Fc = mv2/r
Vertical Circles The image shows a picture of a ball attached to a cord with tension T swung in a
vertical circle. We will look at three points on the circle, labeled A, B, and C. There are always
two forces acting on the ball, the tension in the cord T and the weight mg. At the top of the circle
(point A), both the weight and the tension act down. Their sum (T + mg) is the unbalanced force
which is the source of the centripetal force. The minimum speed that the ball must have for the
ball to just clear the top of the circle occurs when T=0 (thus, mg = mv,sup>2/r). At the bottom of
the circle (point C), the tension is acting up and the weight is acting down. The unbalanced force
which is the center-seeking centripetal force is T - mg. At point B, only the tension is centerseeking (the weight is acting perpendicular to the tension) and is the sole source of the centripetal
force.
At point A, T + mg = Fc = mv2/r
At point C, T - mg = Fc = mv2/r
At point B, T = Fc = mv2/r
A common example of centripetal acceleration is a car rounding a curve. If the road surface is flat,
friction is the source of the centripetal force. If the curve is banked, a component of the normal
force provides the centripetal force. For a given banking angle, there would be one speed for
which no friction is required.
flat surface, Fc = Ff = FN
Banking Angle
tan  = v2/rg
AP FORCE EXAMPLES

Horizontal Circles
A string is attached to mass m and swung in a horizontal circle of radius r. There is tension T in
the string. The angle  is measured from the vertical (where the string is held).
The free-body diagram showing the forces acting on the mass is shown below. Notice, there is no
centripetal force indicated. It is not a separate force.
The vector diagram showing the force components is shown below. The tension has a vertical
component which offsets the weight and a horizontal component. Notice the horizontal component
is the force that is center-seeking, or the centripetal force. Also, notice the horizontal component is
unbalanced.
You can solve for the speed of the mass, knowing the angle and the radius:
1.
2.
3.
4.
T cos  = mg
T sin  = mv2/r
Divide the second equation by the first equation.
Knowing that tan  = sin  / cos , you get
tan  = v2/rg

Vertical Circle
A mass m is twirled in a vertical circle or radius r. The mass is held at the length of a string with
tension T.
The free-body diagrams showing the forces acting on the mass are shown below for each point.
Notice, there is no centripetal force indicated. It is not a separate force.
Point A
Point B
Point C
The vector diagrams showing the force components at each point are shown below. At point A:
the tension and the weight act in the same direction. Their sum represents the centripetal force. At
point B: the tension acts toward the center of the circle and the weight acts down. Their sum
represents the centripetal force. At point C: only the tension acts toward the center of the circle.
The weight has no effect.
Point A
Point B
Point C
mv2/r = mg + T
mv2/r = T - mg
mv2/r = T
If the mass is being twirled at constant minimum speed, the tension at the top of the vertical
circle is zero. At point A, only the weight provides the centripetal force. The net force acting on
the mass is therefore equal to mg. At point B, the difference between the tension and the weight
provides the centripetal force. It also must equal mg. At point C, the tension provides the
centripetal force. It also must equal mg.
At point A: mv2/r = mg
At point B: mv2/r = 2 mg
At point C: mv2/r = mg

Flat Curve:
The drawing below shows the free body diagram for an object of mass m traveling around a flat
curve of radius r at constant speed v. Friction is keeping the car on the curve.
The drawing below shows the vector diagram for the object. Notice, the frictional force is centerseeking. Also, notice that the normal force is equal to the weight of the object.
The frictional force can thus be set equal to the centripetal force yielding:
mv2/r = mg

Banked Curve:
The drawing below shows the free body diagram for an object of mass m traveling around a
banked curve of radius r at constant speed v. A component of the normal force is keeping the car
on the curve.
The drawing below shows the vector diagram for an object of mass m traveling around a banked
curve of radius r at constant speed v.
The horizontal component of the normal force provides the centripetal force. This can be set equal
to the centripetal force. The vertical component of the normal force offsets the weight. These two
expressions can be solved yielding:
mv2/r = N sin 
mg = N cos 
tan  = v2/rg

Static Friction:
The drawing below shows the forces acting on an object being spun in a horizontal surface. Static
frictional forces keep the object in place.
The vector diagram below shows the forces acting on an object being spun in a horizontal surface.
Static frictional forces offset the weight. The normal force provides the centerseeking force.
The static frictional force can be set equal to the weight of the object. The normal force can be set
equal to the centripetal force. These two expressions can be combined yeilding:
mg = mg
N = mv2/r
 = rg/v2

Gravitational Force:
The speed of an object in orbit around another object is independant of the mass of the first object.
Its speed only depends upon the distance it is from the center of mass of the second object.
Consider an object of mass m in orbit at a radial distance r from object of mass M.
mv2/r = GmM/r2
AP Multiple Choice Questions on Centripetal Motion
1.
2.
3.
4.
5.
Be able to predict how the speed of an object twirled in a horizontal circle would change
if the radius were doubled or halved. If the force supplying the centripetal force were
doubled or halved.
Be able to perform simple ratio calculations in which you set the Universal Law of
Gravitation equal to mg. For example, this could be used to determine the acceleration
due to gravity on Planet X comparing everything to earth data. It also could be used to
find the weight of an object (mg) on Planet X comparing everything to earth data.
Be able to identify the direction of the force providing the centripetal force for an object
in circular motion. For example, the frictional force acts toward the center for a car
rounding a flat curve.
Be able to perform simple calculations at the bottom (or at the top) of the swing of an
object swung in a vertical circle. You should be able to calculate an expression for the
centripetal force, knowing weight and tension. Once you know this, you could also
determine centripetal acceleration.
Know that the work done on an object in circular motion in one period (or one
revolution) is zero.
AP Free Response Questions on Centripetal Motion
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6.
This is frequently asked!
Remember how to calculate speed when an object moves in a circle. Remember, it goes a
distance of one circumference in a time of one period. Its speed is constant.
You may be given a problem in which an object on a string is twirled in a horizontal
circle. The object makes an angle theta with respect to the vertical. Represent all forces
using a FBD. Calculate the tension in the string (remember: sin /cos  = tan ). They
also combine motion concepts with this type of problem by asking you to describe the
motion of the object when the string breaks. You will also be asked to calculate how long
it takes the object to hit the ground, hwo far it travels horizontally, and its speed the
instant before it strikes.
You may be given a problem in which an object on a string is twirled in a vertical circle.
Represent all forces using FBDs at the top of the swing, the bottom of the swing, and a
radial distance above the bottom. Express the tension in the string at each one of these
points. Calculate the minimum speed needed for the object to clear the top of the swing.
They also combine motion concepts with this type of problem by asking you to describe
the motion of the object when the string breaks at each point. You will also be asked to
calculate how long it takes the object to hit the ground, hwo far it travels horizontally,
and its speed the instant before it strikes.
An object can roll down a frictionless arc of a circle onto a plane. Calculate the speed of
the object at the bottom of the arc. Calculate the centripetal force acting on the object
while on the arc. They also combine motion with this type of problem.
Another common type of problems combines centripetal motion with electrostatics and
magnetism concepts. These will be described in the spring.
Kepler's Laws of Planetary MotionThis is not asked on the AP B test.
 Kepler's First Law: The path of each planet about the sun is an ellipse with the sun at one focus.
 Kepler's Second Law: Each planet moves so that an imaginary line drawn from the sun to the
planet sweeps out equal areas in equal periods of time.
 Kepler's Third Law: The ratio of the squares of the periods (the time needed for one revolution
about the sun) of any two planets revolving about the sun is equal to the ratio of the cubes of their
mean distances from the sun.
Relative Velocity
This is not asked on the AP B test.
How are measurements made in different reference frames related to one another?
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2.
Velocities along a straight line:
o Simple addition or subtraction is used.
o Example: Car A moving at 20 m/s passes Car B moving at 15 m/s. The velocity
of the Car A relative to Car B is 20 m/s - 15 m/s = 5 m/s.
o Example: Car A moving at 20 m/s approaches Car B moving at 15 m/s in the
opposite direction. Observers in either car will measure a speed of 20 m/s + 15
m/s = 35 m/s relative to their position for the approaching car.
When determining relative velocities, we will draw a diagram and use a labeling process
in which each velocity is labeled by two subscripts. The first subscript refers to the object
and the second subscript refers to the reference frame in which it has this velocity.
o Example: A boat heads at angle  across a river. The boat has velocity v
measured with respect to the shore (This is the velocity the boat would have if
the water were still). Using our labeling convention, we will call the boat's
velocity vbs. A current is flowing perpendicular to the boat. The velocity of the
current with repsect to the shore is vws. To find the velocity of the boat with
respect to the shore, we perform the necessary vector addition. (This would be
the velocity that is perpendicular to the shore.)
vbs = vbw + vws
o
Note: using our labeling method, the two outer subscripts for the sum of the vectors on
the right are the same as the subscripts of the velocity vector on the left.
o
AP Objectives - MOTION IN TWO DIMENSIONS
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Students should know how to deal with displacement and velocity vectors so they can:
1. Relate velocity, displacement, and time for motion with constant velocity.
2. Calculate the componet of a vector along a specified axis, or resolve a vector into
components along two specified mutaually perpendicular axes.
3. Add vectors in order to find the net displacement of a particle that undergoes successive
straight-line displacements.
4. Subtract displacement vectors in order to find the location of one particle relative to
another, or calculate the average velocity of a particle.
5. Add or subtract velocity vectors in order to calculate the velocity change or average
acceleration of a particle, or the velocity of one particle relative to another.
Students should understand the motion of projectiles in a uniform gravitational field so they can:
1. Write down expressions for the horizontal and vertical components of velocity and
position as functions of time, and sketch or identify graphs of these components.
2. Use these expressions in analyzing the motion of a projectile that is projected above level
ground with a specified initial velocity.
AP Objectives - Circular Motion & Gravitation
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Students should understand the uniform circular motion of a particle so they can:
1. Relate the radius of the circle and the speed or rate of revolution of the particle to the
magnitude of the centripetal acceleration.
2. Describe the direction of the particle's velocity and acceleration at any instant during the
motion.
3. Determine the components of the velocity and acceleration vectors at any instant, and
sketch or identify graphs of these quantities.
Students should know Newton's Law of Universal Gravitation so they can:
1. Determine the force that one spherically symmetrical mass exerts on another.
2. Determine the strength of the gravitational field at a specified point outside a spherically
symmetrical mass.
Students should understand the motion of a body in orbit under the influence of gravitational
forces so they can:
1. Recognize that the motion for a circular orbit does not depend on the body's mass,
describe qualitatively how the velocity, period of revolution, and centripetal acceleration
depend upon the radius of the orbit, and derive expressions for the velocity and period of
revolution in such an orbit.
2. Apply for a general orbit conservation of angular momentum to determine the velocity
and radial distance at any point in the orbit.
3. Apply for a general orbit angular momentum conservation and energy conservation to
relate the speeds of a body at the two extremes of an elliptic orbit.
Motion in Two Dimensions Homework
Motion in Two Dimensions Homework
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A steel projectile is shot horizontally at 20 m/s from the top of a 40 m tower. How far from the
base of the tower does the projectile hit the ground? Ans: 57.14 m
A projectile is fired from a cannon at a speed of 301 m/s and at an angle of 3. How long does it
take the projectile to reach its highest point? How far does it go horizontally? Ans: 1.61 sec;
964.89m
A bomber releases a bomb at a height of 50 m above the surface. The bomber is flying at a
constant horizontal speed of 88.9 m/s. How long does it take the bomb to fall to the surface? How
far away is the point of impact? Ans: 3.19 sec; 283.59 m
A golfer launches a ball with an initial speed of 30 m/s at an angle of 34 with respect to the
ground. At what time does the golf ball reach its maximum height? What is this height? What is
the total time in the air? Ans: 1.71 sec; 14.37m; 3.42 sec; 85.06 m
A diver running at 3.6 m/s dives out horizontally from the edge of a vertical cliff and reaches the
water below 20 sec later. How high was the cliff and how far from the base did the diver hit the
water? Ans: 1960 m; 72 m
A football is kicked at an angle of 37 with a speed of 20 m/s. What is its maximum height, the
time it takes to hit the ground, and how far away it lands? Ans: 7.4 m; 2.46 sec; 39.29 m
A package in an airplane moving horizontally at 150 m/s is dropped when the altitude is 490 m.
How long does it take the package to reach the ground? How far horizontally from the spot over
which it was dropped does the package land? Ans: 10 sec; 1500 m
An arrow is fired with a horizontal speed of 89 m/s directly at a target 60 m away. When it is fired,
the arrow is 1 m above the ground. How far short of the target is it when it strikes the ground?
Ans: 20 m short
9.
A bullet is fired horizontally from a height of 78.4 m and hits the ground 2400 m away. With what
velocity does the bullet leave the gun? Ans: 600 m/s
10. A bullet is fired from a gun at a 30 angle to the horizontal with a muzzle velocity of 600 m/s.
What is the horizontal range of the gun? Ans: 31,812 m
Periodic Motion Homework
Periodic Motion
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Convert the following speeds to ones in m/s: An object moves in a circle of 2 m radius at 25 rpm,
at 8 rpm, and at 10 rpm. Ans: 5.24 m/s; 1.68 m/s; 2.09 m/s
A pendulum is 2 m long and has a period of 2.9 sec. What is the acceleration due to gravity at this
location? Ans: 9.4 m/s2
A pendulum located on earth has a period of 1 sec. What is its length? Ans: 0.25 m
A pendulum 0.25 m long has a frequency of 0.91 Hz. What is the period of a pendulum in the
same location if it has a length of 0.10 m? Ans: 0.696 sec
Find the acceleration due to gravity on Planet Rocket. A pendulum on this planet has a length of
2.5 hakeems and a period of 6 drexlers. Ans: 2.74 hakeems/drexlers
A mass moves in a circular path at a velocity of 2 ms and with a centripetal acceleration of 4.3
m/s2. What is the radius of the mass’ path? Ans: 0.93 m
A 25 g ball is swung at the end of a string in a horizontal circular path at a speed of 5 ms. If the
length of the string is 2 m, what centripetal force does the string exert on the ball? Ans: 0.3125 N
How fast must a 6 N stone be twirled in a circle of 3 m radius in order to have a centripetal
acceleration of 2 m/s2? Ans: 2.45 m/s
An object weighing 14 N is swung at the end of a cord 0.8 m long. The object moves at a speed of
2 rev/sec. What is the centripetal force on the moving object? Ans: 181 N
Motion in Two Dimensions Sample Problems
Two Dimensional Motion
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An object is thrown horizontally at 15 m/s from the top of a 44 m building. How long does it take
for it to strike? How far from the building's base does it land?
A missle is dropped from a bomber, giving it an initial horizontal velocity of 8 m/s. The bomber is
122.5 m above the ground. What is the missile's range?
A plane flying horizontally at 220 m/s drops a tank when it is 490 m above the ground. How long
does it take to fall? What is its range?
A rock is dropped from a cliff 90 m high. What is its velocity the instant before it strikes the
ground?
A missile is launched at a speed of 25 m/s at an angle of 35 to the ground. What is its range? How
high did it go vertically?
Another is launched at 39.2 m/s and 30.
Another is launched at 67 m/s and 20.
Another is launched at 60 m/s and 40.
A missile is traveling at 7 m/s horizontally and 9.6 m/s vertically when it leaves the launchpad.
What is its range and how long is it in the air?
6.
Grease pops onto your hand from a frying pan 1 m away from you (measured horizontally). If it
were in the air 0.50 sec, what was the initial velocity (resultant) of the grease? Assume optimum
angle.
Centripetal Force and Acceleration
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4.
A moving ball is spun in a circle with a diameter of 1 m at a speed of 3 m/s. What is its centripetal
acceleration?
A 250 g mass is spun in a horizontal circle. It is held at the end of a 1 m length of string. If it is
spun at 15 m/s, what force is applied?
A ball weighing 5 N is attached to a 1 m string and swung in a horizontal circle above one's head
at the rate of 5 rev/sec (5 rps). What centripetal force is required?
Convert 10 rpm (rev/min) into m/s for a horizontal circle of 50 cm radius.
Pendulums
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What is the period of a pendulum (located on earth) that is 70 cm long?
It takes 230 sec for a pendulum to reach its starting point 100 times after its release. What is its
period?
A pendulum bob is suspended on a string 75 cm long. It has a frequency of 0.57 sec -1. What would
the period of a pendulum be that is located at the same location and is 50 cm long?
A pendulum on earth has a length of 35 cm. What is its period?
What is the acceleration due to gravity on planet X? A 1 m long pendulum has a period of 2 sec on
this planet.
AP Motion in Two Dimensions Sample Problems
AP Vectors Sample Problems
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Find the resultant of the following vectors: 1 N, E; 3 N, N; 5 N, 143; and 6 N 233. Ans: 6.71 N,
169.5
A 400 N weight hangs from two cords (F1 and F2) attached to the ceiling. F1 is directed at a 37 to
the horizontal and is pointed west. F2 is directed at a 53 angle to the horizontal and is pointed
east. What is the magnitude of each force? Ans: 240 N, 320 N
An airplane flies at 180 km/h at a 30 N of E. A wind blows south at 50 km/h. What is the velocity
of the airplane with respect to the ground? Ans: 161 km/h, 14.4°
A boat’s speed in still water is 5.55 m/s. If the boat is to travel directly across a river whose
current has speed 3.60 m/s, at what upstream angle must the boat head? Ans: 40.4
The same boat now heads directly across the river. What is the velocity of the boat with respect to
the shore? If the river is 300 m wide, how long will it take for the boat to cross the river? How far
downstream will it land? Ans: 6.6 m/s, 33; 54 sec; 195 m
AP Two Dimensional Motion
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2.
A projectile is fired with an initial speed of 75.2 m/s, 34.5. Determine the maximum height
reached by the projectile, the total time in the air, the total horizontal distance covered by the
projectile, and the velocity of the projectile 1.5 sec after firing. Ans: 92.6 m; 8.69 sec; 539 m; 68
m/s, 24.2
A ball is thrown with a velocity of 30 m/s, 37. It is 15 m from a wall when thrown. It leaves the
pitcher’s hand 2 m above the ground. At what height above the ground will it hit the wall? Ans:
11.43 m
3.
A rescue plane is flying at a constant elevation of 1200 m with a speed of 430 km/h toward a point
directly over a person struggling in the water. At what angle should the pilot release the rescue
capsule if is to strike at the person? Ans: 57
4.
A car is parked on a cliff overlooking the ocean on an incline that makes an angle of 24 below the
horizontal. The negligent driver leaves the car in neutral, and the emergency brakes are defective.
The car rolls from rest down the incline with a constant acceleration of 4 m/s/s and travels 50 m to
the edge of the cliff. The cliff is 30 m above the ocean. How long is the car in the air? How far
from the base of the cliff does it land? With what speed does it hit? Ans: 1.78 sec; 32.52 m; 31.43
m/s
Mass M rests on a smooth surface. It is attached by a massless rope to mass m. Mass M is a
distance 2 h from the edge of the surface. Mass m is a distance h from the horizontal surface above
which it hangs. In words, describe the motion of mass m. In words, describe the motion of mass
M. What is the acceleration of the system in terms of M, m, g, and h? How long does it take mass
m to fall to the surface? What is the velocity of mass M when mass m hits the surface? Predict the
impact point of mass M. Ans: mg/(M + m); [2h(M + m)/mg]1/2; [2mgh/(M + m)]1/2; 2h [M/(M +
m)]1/2
5.
AP Dynamics of Circular Motion Sample Problems
AP Dynamics of Uniform Circular Motion Class Problems
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A 150 g ball is twirled at the end of a string 60 cm long. The ball makes 2 revolutions every 12
seconds. What is the centripetal acceleration of the ball? What is the centripetal force exerted?
Ans: 0.66 m/s2; 0.10 N
A pendulum bob with mass of 1.5 kg whirls around in a horizontal circle at constant speed v at the
end of a 1.7 m cord. The cord makes a 37 angle with the vertical. What is the bob’s speed? How
long does it take it to make one revolution? Ans: 3.54 m/s; 1.81 s
A Cadillac of mass 1610 kg is moving at a constant speed of 20 m/s on an unbanked curve with
190 m radius. What must be the minimum coefficient of friction between the tires and the road?
Ans: 0.21
A car of mass m is moving at a constant speed of 20 m/s around a banked curve with 190 m
radius. What angle is needed to make the reliance on friction unnecessary? Ans: 12.11
A Rotor is a ride found in some amusement parks. It consists of a hollow, cylindrical room that
can be made to rotate about a central vertical axis. A person enters, closes the door, and stands up
against the wall. The Rotor starts rotating and gradually increases its speed until the floor is
dropped away. The person does not fall, but remains pinned to the wall, supported by an upwarddirected frictional force. The coefficient of friction between the wall and the person is 0.40. The
radius of the Rotor is 2.1m. At what minimum rotational speed is it safe to drop the floor? At this
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speed, what centripetal acceleration does the person experience? Ans: 7.18 m/s; 24.53 m/s 2 or 2.5
g
A 150 g ball is at the end of a 1.10 m long string. It is swung in a vertical circle. What minimum
speed must the ball have to clear the top of the swing? What tension in the string is required for
the ball to move at twice the minimum speed at the bottom of its swing? Ans: 3.28 m/s; 7.35 N
A spring stretches 0.150 m when a 0.300 kg mass is hung from it. Determine the spring constant.
What would be its period if a 0.400 kg mass is hung from it and set into vibration? Ans; 19.6 N/m;
0.90 sec
A 2M mass is twirled in a vertical circle with radius r and at constant speed v. What is the tension
at the top of the vertical circle? At the bottom of the vertical circle. Ans: 2M[(v 2/r) - g]; 2M[(v2/r)
+ g]