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Chapter 6 Projectile and Periodic Motion

In this chapter, we will study the motion in two dimensions further. First, we will analyze the path that any object takes when it is thrown in the air. We will see that all falling objects follows the same path. Then, we will analyze the motion that repeats itself, such as the motion of the propeller.


1. Projectile Motion
2. Circular Motion
3. Chapter 6 Quiz

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Section 1. Projectile Motion

Objects launched are called projectiles. The flight of baseballs and basketballs are some examples of projectile motions. Let's analyze the projectile motion by breaking down the forces acting on the object.

Suppose a ball was thrown horizontally at the velocity of 5 m/s.

Let's think about forces acting on x-direction, or horizontal direction, only. We know that the ball was thrown at the velocity of 5 m/s. Because no other force acts on the ball in the air, we know the velocity will stay constant (because the net force is zero). Therefore, after 5 seconds, the ball is

d = vt = 5 m/s * 5s = 25 m

25 m away. After 10 seconds,

5 m/s * 10 s = 50 m

The ball is 50 m away.

QUESTION: How far will the ball travel in 3 seconds in horizontal direction?
    Let's now think about the forces in y-direction, or vertical direction. Because gravity is the only force acting on it, the ball will accelerate at the rate of -9.8 m/s2, i.e. 9.8 m/s2 downward. After 2 seconds, the ball has traveled 19.6 m because


= 0 + 4.9 * 22 = 19.6 m
QUESTION: How far has the ball traveled after 4 seconds in vertical direction?
    Now, let's combine two forces at x and y directions and see its displacement. When t = 0, i.e. right after the ball was thrown, the displacement on x and y direction are both 0. When t = 1, after 1 second, the displacement on x direction is 5 m and on y direction is -4.9 m. When t = 2, the ball traveled 10 m on x direction and -19.6 m on y direction. To summarize,
Time vs. Displacement
Time Displacement on X Displacement on Y
0 s 0 m 0 m
1 s 5 m -9.8 m
2 s 10 m -19.6 m
QUESTION: What is the displacement of the ball in x direction after 3 seconds?

QUESTION: What is the displacement of the ball in y direction after 3 seconds?

    Now, let's look at velocity rather than displacement. When t = 0, the velocity on x direction is 5 m/s, while 0 m/s on y direction. Therefore, the total velocity of the ball is 5 m/s using Pythagorean Theorem. When t = 1, while the velocity on x direction stays the same, the velocity on y direction becomes -4.9 m/s. Therefore, the total velocity of the ball when t = 1 would be



To summarize,

Time vs. Velocity
Time Velocity on X Velocity on Y Total Velocity
0 5 m/s 0 m/s 5 m/s
1 5 m/s -4.9 m/s 7.0 m/s
QUESTION: What is the velocity of the ball in x direction when t = 2?

QUESTION: What is the velocity of the ball in y direction when t = 2?

QUESTION: What is the total velocity of the ball when t = 2?

How a ball drops The left drawing summarizes what we've learned so far. The red dot represents the positions of a falling ball with 0.5 seconds of interval.

You can see that it moves at constant velocity on x direction, while accelerates on y direction. As a result, the path of the ball will follow the path of parabola.



Section 2. Circular Motion

Motion that repeat itself is called periodic motion. A projectile motion, since it doesn't repeat, is not periodic. The motion of a swinging ball is an example of circular motion. Let's analyze the circular motion.

Consider a planet around the Sun.

Animation of a planet orbiting around Sun

When there is no external force, an object will travel in a straight line (Newton's First Law of Motion). In order for an object to travel in a circle, there has to be a force that makes it travel in a circle.



An object in circular motion always tries to move in a straight line (Law of Inertia). However, there is a force that acts toward the center of the motion. This force is called the centripetal force. It is the centripetal force that makes an object travel in a circular path. This force could be friction or a gravitational force, but we call it a centripetal force.

Centripetal force always acts toward the center

When the centripetal force is too strong, the ball will accelerate toward the center of the circle. When the centripetal force is too weak, the ball will get out of the orbit. An object will maintain a circular motion only when the centripetal force is well balanced.

These formulas can be used to calculate the centripetal force:




Fc: is the centripetal force (N)
m: is the mass (kg)
ac: is the centripetal acceleration (m/s2)
v: is the speed (m/s)
r: is the radius of the circle (m)
t: is the period (the length of time required for one complete rotation) (s)

Too complicated? Let's see a concrete example.

You are rotating with Earth, so you are experiencing a circular motion.

Say you have a mass of 50 kg. We know the radius of the circle (6,378,000 m = radius of Earth) and its period (24 hours = 86400 seconds). From this information, you can calculate velocity, acceleration, and centripetal force acting on you.



To calculate velocity, use the formula .

m/s = 1670 km/h.

To calculate acceleration, use the formula .


You can see that we don't accelerate much (which is obvious).

QUESTION: A ball with a mass of 250 g has a centripetal acceleration of 5 m/s2. What is the centripetal force acting on this ball?

QUESTION: A ball with a speed of 10 m/s is in a circular motion. If the circle has a radius of 20 m, what would the period of the ball be?

QUESTION: The Moon's orbit around earth is nearly circular. The orbit has a radius of about 385,000 km and a period of 27.3 days. Determine the acceleration of the moon toward Earth.



Section 3. Chapter 6 Quiz

Try Chapter 6 Quiz and see how well you can do!

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