F. O.

Chukwusa

PHY 111: General Physics I (Mechanics) WEEK SEVEN (7)

Lecture Note on: COORDINATE SYSTEMS IN GENERAL PHYSICS I

Introduction
Coordinate systems are used to describe the position of an object in space. A coordinate system is an
artificial mathematical tool that we construct in order to describe the position of a real object. The
purpose of a coordinate system is to uniquely determine the position of an object or data point in
space. By ‘space’ we may literally mean in physical space, but in general it simply refers to what
we might call ‘variable-space’, where each dimension corresponds to one variable. In physics, we
often need to describe the position, motion, and orientation of objects in space. To do this, we use
coordinate systems, which provide a mathematical framework for specifying locations and
directions. In this lecture, we will explore the basics of coordinate systems and their applications in
general physics.

Types of Coordinate Systems

There are several types of coordinate systems, including:
(1.) Cartesian Coordinate System
The Cartesian coordinate system is the most widely or commonly used coordinate system in physics.
It consists of three mutually perpendicular axes (x, y, z) that intersect at a point called the origin.
Cartesian coordinate system is also known as the rectangular coordinate system.

The following is the rectangular coordinate system:

It is made up of two number lines:

1. The horizontal number line is the x- axis.
2. The vertical number line is the y- axis.
The origin is where the two intersect. This is where both number lines are 0. It is split into four
quadrants which are marked on this graph with Roman numerals.

Each point on the graph is associated with an ordered pair. When dealing with an x, y graph, the x
coordinate is always first and the y coordinate is always second in the ordered pair (x, y). It is a
solution to an equation in two variables. Even though there are two values in the ordered pair, be
careful that it associates to ONLY ONE point on the graph, the point lines up with both the x value
of the ordered pair (x-axis) and the y value of the ordered pair (y-axis).

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Example (1): Plot the ordered pairs and name the quadrant or axis in which the point lies. A(2, 3),
B(-1, 2), C(-3, -4), D(2, 0), and E(0, 5).

Solution:

A(2, 3) lies in quadrant I.

B(-1, 2) lies in quadrant II.
C(-3, -4) lies in quadrant III.
D(2, 0) lies on the x-axis.
E(0, 5) lies on the y-axis.

Example (2): Find the x- and y- coordinates of the following labelled Points;

Solution:
Since point A corresponds to 2 on the x-axis and -3 on the y-axis, then A’s ordered pair is (2, -3).
Since point B corresponds to 3 on the x-axis and 2 on the y-axis, then B’s ordered pair is (3, 2).
Since point C corresponds to -2 on the x-axis and 3 on the y-axis, then C’s ordered pair is (-2, 3).
Since point D corresponds to -3 on the x-axis and - 4 on the y-axis, then D’s ordered pair is (-3, -4).
Since point E corresponds to -3 on the x-axis and 0 on the y-axis, then E’s ordered pair is (-3, 0).
Since point F corresponds to 0 on the x-axis and 2 on the y-axis, then F’s ordered pair is (0, 2).

The position of a point on a Cartesian plane is represented by referring to it in terms of a horizontal

line and a vertical line, which are called the x-axis and y- axis respectively. The point of
intersection of the x-axis and the y-axis is called the origin.

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Often, we draw a set of axes on graph paper as shown below.

The position of any point on the Cartesian Plane is described by using two numbers, (x, y), that are
called coordinates. The first number, x, is the horizontal position of the point from the origin. It is
called the x-coordinate. The second number, y, is the vertical position of the point from the origin.
It is called the y-coordinate. Since a specific order is used to represent the coordinates, they are
called ordered pairs.

For example, an ordered pair (4, 5) represents a point 4 units to the right of the origin in the
direction of the x-axis, and 5 units above the origin in the direction of the y-axis as shown in the
diagram below.

We say that: The x-coordinate of point P is 4; and the y-coordinate of point P is 5.

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Or simply, we can say that: The coordinates of point P are (4, 5).

Example (3): State the coordinates of each of the points shown on the Cartesian plane below:

Solution:
A is 1 unit to the right of and 2 units above the origin. So, point A is (1, 2).
B is 2 units to the right of and 3 units above the origin. So, point B is (2,3).
C is 2 units to the right of and 4 units above the origin. So, point C is (2,4).
D is 3 units to the right of and 4 units above the origin. So, point D is (3,4).
E is 5 units to the right of and 2 units above the origin. So, point E is (5,2).
F is 7 units to the right of and 3 units above the origin. So, point F is (7, 3).
G is 4 units to the right of and 5 units above the origin. So, point G is (4,5).
H is 5 units to the right of and 6 units above the origin. So, point H is (5,6).
I is 1 unit to the right of and 7 units above the origin. So, point I is (1, 7).
J is 7 units to the right of and 7 units above the origin. So, point J is (7, 7).

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Example (4): Finding the Distance between Two Points;

Find the distance between the points A (2, 3, 4) and B (6, 8, 10) in a Cartesian coordinate system.

Solution
Using the distance formula:
d = √((X2 - X1)2 + (Y2 - Y1)2 + (Z2 - Z1)2)
d = √((6 - 2)2 + (8 - 3)2 + (10 - 4)2)
d = √(16 + 25 + 36)
d = √77
d = 8.77 units

Example (5): Finding the Angle between Two Vectors;

Find the angle between the vectors A (3, 4, 5) and B (2, 6, 8) in a Cartesian coordinate system.

Solution
Using the dot product formula:
A · B = |A| |B| cos(θ)
First, find the magnitudes of the vectors:
|A| = √(32 + 42 + 52) = √(9 + 16 + 25) = √50
|B| = √(22 + 62 + 82) = √(4 + 36 + 64) = √104

Next, find the dot product:

A · B = (3)(2) + (4)(6) + (5)(8) = 6 + 24 + 40 = 70

Now, solve for θ:

70 = √50 √104 cos(θ)
cos(θ) = 70 / (√50 √104)
θ = arccos(70 / (√50 √104)) ≈ 45.6°

Example (6): Finding the Distance between Two Points

Find the distance between the points A (3, 4, 5) and B (6, 8, 10) in a Cartesian coordinate system.
Solution
d = √((X2 - X1)2 + (Y2 - Y1)2 + (Z2 - Z1)2)
d = √((6 - 3)2 + (8 - 4)2 + (10 - 5)2)
d = √(9 + 16 + 25)
d = √50
d = 7.07 units

Example (7): Finding the Midpoint of a Line Segment

Find the midpoint of the line segment connecting the points A (2, 3, 4) and B (6, 8, 10) in a Cartesian
coordinate system.

Solution
Midpoint = ((X1 + X2)/2, (Y1 + Y2)/2, (Z1 + Z2)/2)
Midpoint = ((2 + 6)/2, (3 + 8)/2, (4 + 10)/2)
Midpoint = (4, 5.5, 7)

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(2.) Polar Coordinate System

The polar coordinate system uses a radial distance (r) and an angular coordinate (θ) to specify a point
in a two-dimensional plane.
Cartesian coordinates are very versatile, but for some applications including many curves,
rotations, and complex numbers, it is simpler to use a coordinate system based on the circle. These
are polar coordinates, and our two parameters are r, the radial distance between the point and the
origin, and θ, the angle between the point and the positive x-axis. The simplest, most fundamental
plots for a Cartesian coordinate system are vertical lines — lines of equal x — and horizontal lines
— lines of equal y. For the polar system, the fundamental plots are circles — lines of equal r —
and straight lines through the origin — lines of equal θ.

The polar coordinate system.

We can use geometry to relate Cartesian coordinates to polar coordinates, just as we can relate
different Cartesian coordinate systems. You’ve seen this with complex numbers, but it naturally
also relates the real Cartesian plane to real polar coordinates and can be used to transform
equations in one system to the other.

The relationship between polar and Cartesian coordinates

Example (8): Converting from Polar to Cartesian Coordinates

Convert the polar coordinates (r, θ) = (5, 30°) to Cartesian coordinates (x, y).

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Solution
x = r cos(θ)
x = 5 cos(30°)
x = 5(√3/2)
x = 4.33

y = r sin(θ)
y = 5 sin(30°)
y = 5(1/2)
y = 2.5

So, the Cartesian coordinates are (x, y) = (4.33, 2.5).

Example (9): Finding the Area of a Circle

Find the area of a circle with a radius of 4 units in a polar coordinate system.
Solution
Area = πr2
Area = π(4)2
Area = 16π
Area = 50.27 square units

(3.) Spherical Coordinate System

The spherical coordinate system uses three coordinates (r, θ, φ) to specify a point in three-
dimensional space.

Example (10): Converting from Spherical to Cartesian Coordinates

Convert the spherical coordinates (r, θ, φ) = (10, 45°, 60°) to Cartesian coordinates (x, y, z).

Solution
Using the conversion formulas:

x = r sin(φ) cos(θ)
y = r sin(φ) sin(θ)
z = r cos(φ)

x = 10 sin(60°) cos(45°) = 10(√3/2)(√2/2) = 4.33

y = 10 sin(60°) sin(45°) = 10(√3/2)(√2/2) = 4.33
z = 10 cos(60°) = 10(1/2) = 5

So, the Cartesian coordinates are (x, y, z) = (4.33, 4.33, 5).

Example (11): Finding the Volume of a Sphere
Find the volume of a sphere with a radius of 6 units in a spherical coordinate system.
Solution
Volume = (4/3) πr3
Volume = (4/3)π(6)3

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Volume = 288π
Volume = 904.78 cubic units

Example (12): Finding the Volume of a Sphere

Find the volume of a sphere with a radius of 3 units in a spherical coordinate system.
Solution
Volume = (4/3)πr3
Volume = (4/3)π(3)3
Volume = 36π
Volume = 113.10 cubic units

Lecture Note on: CONSERVATION OF ANGULAR MOMENTUM IN GENERAL PHYSICS I

Introduction
Angular momentum is a fundamental concept in physics that describes the tendency of an object to
keep rotating or revolving around a central axis. In this lecture, we will explore the concept of
conservation of angular momentum, which is a fundamental principle in physics.

What is Angular Momentum?

Angular momentum (L) is a vector quantity that characterizes the rotational motion of an object. It is
defined as the cross product of the position vector (r) and the linear momentum (p) of the object:

L=rxp

Conservation of Angular Momentum

The law of conservation of angular momentum states that the total angular momentum of a closed
system remains constant over time, provided that there are no external torques acting on the system.
The conservation of angular momentum principle can be applied to solve problems involving
rotating systems, collisions, and more.

Mathematically, this can be expressed as:

ΔL = 0

where ΔL is the change in angular momentum.

Examples and Solved Calculations

Example (1): Conservation of Angular Momentum in a Rotating System

A figure skater is spinning around a vertical axis with an initial angular velocity of 2 rad/s. As she
brings her arms closer to her body, her moment of inertia decreases from 0.5 kg m2 to 0.2 kg m2.
What is her final angular velocity?

Solution

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Initial angular momentum (L1) = I1 ω1

L1 = 0.5 kg m2 x 2 rad/s
L1 = 1 kg m2/s

Final angular momentum (L2) = I2 ω2

L2 = 0.2 kg m2 x ω2
Since the system is closed, the total angular momentum remains constant:
L1 = L2
1 kg m2/s = 0.2 kg m2 x ω2
ω2 = 5 rad/s
Therefore, the figure skater's final angular velocity is 5 rad/s.

Example (2): Conservation of Angular Momentum in a Colliding System

A 2 kg block is moving with an initial velocity of 3 m/s when it collides with a stationary 1 kg block.
After the collision, the blocks stick together and rotate around a fixed axis. If the moment of inertia
of the combined blocks is 0.3 kg m2, what is their final angular velocity?

Solution
Initial angular momentum (L1) = m1 v1 r1
= 2 kg x 3 m/s x 0.5 m
= 3 kg m2/s

Final angular momentum (L2) = I2 ω2

= 0.3 kg m2 x ω2

Since the system is closed, the total angular momentum remains constant:
L1 = L2
3 kg m2/s = 0.3 kg m2 x ω2
ω2 = 10 rad/s

Therefore, the final angular velocity of the combined blocks is 10 rad/s.

Example (3): Conservation of Angular Momentum in a Rotating System
A figure skater is spinning around a vertical axis with an initial angular velocity of 2 rad/s. If the
skater brings their arms closer to their body, reducing their moment of inertia by 30%, what is their
new angular velocity?

Solution:
Given that;
Initial angular velocity (ω1) = 2 rad/s
Initial moment of inertia (I1) = 0.5 kg m2
Final moment of inertia (I2) = 0.35 kg m2 (30% reduction)

Using the conservation of angular momentum principle:

L1 = L2

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I1ω1 = I2ω2
Rearranging the equation to solve for ω2:

ω2 = (I1/I2)ω1
= (0.5/0.35)(2)
= 2.86 rad/s
Therefore, the skater's new angular velocity is 2.86 rad/s.
Example (4): Conservation of Angular Momentum in a Colliding System
A 2 kg disk is rotating around a horizontal axis with an initial angular velocity of 3 rad/s. A 1 kg
block is dropped onto the disk, causing it to rotate with a new angular velocity. If the moment of
inertia of the disk-block system is 0.8 kg m2, what is the final angular velocity of the system?

Solution:
Given that;
Initial angular velocity (ω1) = 3 rad/s
Initial moment of inertia (I1) = 0.5 kg m2
Final moment of inertia (I2) = 0.8 kg m2
Mass of the block (m) = 1 kg

Using the conservation of angular momentum principle:

L1 = L2
I1ω1 = I2ω2

Rearranging the equation to solve for ω2:

ω2 = (I1/I2)ω1
ω2 = (0.5/0.8)(3)
ω2 = 1.88 rad/s
Therefore, the final angular velocity of the disk-block system is 1.88 rad/s.

TAKE HOME:
1. A 5 kg wheel is rotating around a horizontal axis with an initial angular velocity of 2 rad/s. If
a 2 kg block is dropped onto the wheel, causing it to rotate with a new angular velocity, what
is the final angular velocity of the system?
2. A figure skater is spinning around a vertical axis with an initial angular velocity of 3 rad/s. If
the skater extends their arms, increasing their moment of inertia by 20%, what is their new
angular velocity?
3. Find the distance between the points A (3, 4, 5) and B (6, 8, 10) in a Cartesian coordinate
system.
4. Find the volume of a sphere with a radius of 7 units in a spherical coordinate system.
5. State the coordinates of each of the points shown on the Cartesian plane below:

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