Module 2.1 – Projectile Motion

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# Module 2.1 – Projectile Motion - PowerPoint PPT Presentation

Module 2.1 – Projectile Motion. Objects launched in the air follow parabolic trajectories. Both kinematics and dynamics principles can be used to study the motion of these objects. Suitable methods will be learned to predict their behaviour. Horizontal Projectiles. No horizontal forces!.

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Module 2.1 – Projectile Motion

Objects launched in the air follow parabolic trajectories. Both kinematics and dynamics principles can be used to study the motion of these objects. Suitable methods will be learned to predict their behaviour.

Horizontal Projectiles

No horizontal forces!

Falling Objects
• Consider two objects, one dropped and one launched horizontally

Equal times!

Horizontal Motion

No horizontal forces

No horizontal acceleration

Vertical Motion

Only force of gravity

For horizontally launched projectiles

Example

A movie stunt driver drives a car off of a cliff that is 70.0 m high. If the car has a speed of 90.0 km/h, how far away does the car land from the base of the cliff?

A ball thrown horizontally at 12.0 m/s from the roof of a building lands 36.0 m from the base of the building. How high is the building?

Parabolic Projectile Path

if it returns to the same height from which it was launched

Example 1

A football is kicked with a speed of 21.0 m/s at an angle of 37.0o to the ground.

• How much later does it hit the ground?
• How far away does it land?
Example 2

A cricket ball is hit by a bat at a height of 0.85 m. The bat gives the ball an initial velocity of 112 km/h, at an angle of 15.0o above the horizontal. How far away does the ball land?

Solution

A soccer ball is kicked with an initial velocity of 22.4 m/s at an angle of 34.0o above the horizontal. It hits the top of a soccer net that is 2.70 m high.

• How many seconds have passed between the time the ball was kicked and when it hit the net?

Since both times are positive, both are possible answers.

• Why are there two possible answers to a)?

Remember that a projectile follows a parabolic trajectory. There are two possible answers to a) because the ball is at a height of 2.70 m twice – once on the way up and again on the way back down. The ball may have hit the goal after 0.238 s when it was on its way up or it may have hit it after 2.31 s when it was on its way back down.

Module Summary

In this module you learned that

• An object launched horizontally will hit the ground at the same time as an object that is dropped; in other words, the vertical motion of a projectile is not affected by its horizontal motion.
• The kinematics equations that were learned in Unit 2 can be applied separately to the horizontal and vertical motion of a projectile, if the initial velocity of the projectile is first broken up into components.
Module 2.2 – Uniform Circular Motion
• An object undergoing circular motion is another example of two-dimensional motion that can be analyzed using Newton’s Laws and kinematics. This module will identify the conditions necessary for an object to maintain uniform circular motion.
Period and Frequency
• Period – the amount of time to complete one revolution. Since it is a measure of time, the SI unit is seconds (s).
• Frequency –the number of revolutions per second. A revolution per second is referred to as a hertz (Hz).
Example

A child on a merry-go-round is moving with a constant speed of 2.65 m/s when 1.60 m away from the centre of the merry-go-round.

• In what direction is the child accelerating?
• Calculate the acceleration of the child.
• What is the frequency of the spinning merry-go-round?
Solution
• Since the merry-go-round is moving with a constant speed, there is no linear acceleration. The only acceleration is the centripetal acceleration due to moving in a circle. The child’s acceleration is therefore inward, toward the centre of the merry-go-round.

A compact disc spins with a frequency of 8.00 Hz when data is being read at a distance of 2.50 cm from the centre. What is the centripetal acceleration of a point at this location?

• As always, the centripetal acceleration is toward the centre of the disc.
Centripetal Force

Acceleration requires a net force

Centripetal acceleration requires a centripetal force

Centripetal force is not an actual force and should not be included in any free body diagram – it is a force requirement.

Directions

Consider a ball being swung on a string in a circle as shown below:

• Centripetal force is always directed toward the centre of the circle
• Centripetal acceleration is also always directed toward the centre of the circle
• Velocity is perpendicular to the radius of the circle (tangential)
Example 1

What is the maximum speed (in km/h) at which a 1300 kg car can safely travel around a circular track of radius 80.0 m if the coefficient of friction between the tire and the road is 0.30?

Example 2

In the previous example, consider a person sitting in the car who appears to feel a force pushing them toward the outside of the turn. As can be seen in the free body diagram used to solve the problem, however, there is no force toward the outside of the turn. Explain this discrepancy.

Solution

There is no force pushing the person out from the centre of the turn. Since the person is accelerating, they are not in an inertial reference frame. So Newton’s Laws cannot be applied. If we look at the person from an inertial reference frame (the ground), we see that their body is simply trying to continue going forward in a straight line. Since the car is turning, it applies a force on the person to make them turn as well. Because the person’s body is trying to continue going in a straight line (which is toward the outside door of the car), it feels as if there is a force pushing them in this direction. This fictitious force that the person “feels” is referred to as the centrifugal force.

How large must the coefficient of friction be between the tires and the road if a car is to round a level curve of radius 62 m at a speed of 55 km/h?

Vertical Circles

More than one force contributes to the centripetal force

Example 3

A 0.110 kg ball is being swung in a vertical circle on a 58.5 cm string. Calculate the tension in the string at the bottom of the path if the speed is 6.25 m/s.

A 0.140 kg ball is being swung in a vertical circle on a 39.2 cm string. Calculate the tension in the string at the top of the path if the speed is 2.41 m/s.

Module Summary

In this module you learned that

• The centripetal acceleration of an object is always toward the centre of the circle and is given by
• Newton’s Second Law can be applied to circular motion where the net force is inward and is referred to as the centripetal force
Module 2.3 – Universal Gravitation

Newton’s Law of Universal Gravitation can be applied to interactions between all objects and can be used in conjunction with circular motion principles to predict the behavior of planets and other celestial bodies.

Force of Gravity
• What determines the force of gravity?
• Henry Cavendish found the constant G that was needed to form an equation

where

Example 1

Two people are standing 1.3 m apart (centre to centre). One person has a mass of 65.0 kg and the other person has a mass of 78.0 kg.

• What is the force that the heavier person exerts on the lighter person?
• What is the force that the lighter person exerts on the heavier person?
• Why do the two people not move toward one another?
Solution
• Because of Newton’s 3rd Law, the lighter person exerts the same force on the heavier person as the heavier person does on the lighter person. The force is therefore also
Solution
• This force is extremely small compared to other forces acting on the two people, such as friction and the forces of gravity exerted by many other objects.
Example 2

The radius of the Earth was actually calculated over 2000 years ago. Using today’s value of for the radius of the earth, calculate the mass of the earth.

A 710 kg spacecraft is 12800 km above the earth\'s surface.

• Calculate the force of gravity on the spacecraft.
• Calculate the acceleration due to gravity at this location.
Launching a Satellite

Shooting a bullet faster and faster…

Eventually…

Satellite Motion
• Since the satellite is orbiting in a circle, there must be a centripetal force
Example

Suppose that you were trying to put a bullet into orbit near the surface of the earth (ignoring any obstacles and the effects of the atmosphere). What speed would this bullet need to maintain an orbit near the surface of the earth?

The space shuttle orbits the earth at a height of approximately 400. km. Calculate the speed of the space shuttle while it is orbiting the earth.

Kepler’s First Law
• The path of each planet around the sun is an ellipse with the sun at one focus. An ellipse is a set of points such that the sum of the distances to each focal point (F1 and F2 in diagram below) is a constant. A circle is a special case of an ellipse in which the two focal points coincide at the centre of the circle.
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 times. In the diagram below, areas a, b, and c are all equal and were swept out in equal times. As can be seen by the portion of the orbit covered, the planet was travelling faster to complete area c (which was closer to the sun) than it was for area a or b. Planets move fastest in the part of the orbit where they are closest to the sun.
Kepler’s Third Law
• The ratio of the cube of the average orbital radius of a planet (r) to the square of the period of the planet (T) for any planet orbiting the sun is a constant.

When comparing two planets

or

Example

Mars’ period is 1.88 earth years. How many times further away is Mars as compared to earth?

The asteroid Icarus, though only a few hundred meters across, orbits the sun like other planets. Its period is 410 days. What is its average distance from the sun?

Module Summary

In this module you learned that

• The gravitation force of attraction between any two objects can be calculated using the equation
• The acceleration due to gravity due to any planet can be found using the equation
Module Summary
• The speed that a satellite needs in order to maintain a particular orbit is given by the equation
• Planets orbit the sun in elliptical paths, travelling fastest when they are closest to the sun. The ratio of the cube of the orbital radius of a planet to the square of its period is a constant and is the same for any two planets orbiting the sun:
Module 2.4 – Simple Harmonic Motion
• Simple harmonic motion (SHM) is an example of periodic motion in which the restoring force is linear. It is a simple approximation that can be applied to many real world scenarios and allows for a straight forward analysis of these situations. It can be applied to buildings, bridges, and many other objects in the real world.
Hooke’s Law
• Stationary mass on a spring
• Restoring force up
• Force of Gravity down
Hooke’s Law
• The restoring force exerted by a spring is linear with the displacement
• k is the spring constant, measured in N/m
• Spring constant is specific to a particular spring
Example

A piece of rubber is 45 cm long when a weight of 8.0 N hangs from it and is 58 cm long when a weight of 12.5 N hangs from it. What is the spring constant of this piece of rubber?

The spring in a typical Hooke’s Law apparatus has a spring constant of 15.0 N/m and a maximum extension of 10.0 cm. What is the largest mass that can be placed on the spring without damaging it?

Module 7.1 Objective 2

Upon completion of this module, the participant will be able to:

• Calculate the period of a spring exhibiting simple harmonic motion.
Simple Harmonic Motion
• Restoring force acts toward equilibrium, resulting in oscillation
Simple Harmonic Motion
• Oscillation considered SHM if the restoring force is linear
• Position oscillates following sinusoidal pattern
Vertical Motion
• Net force depends only on displacement from the centre of oscillation
• Independent of gravity
• Can be treated same as horizontal SHM since net restoring force is linear
SHM
• Since restoring force is not constant, cannot use kinematics equations!
• Period in SHM only depends on mass and spring constant
• Not amplitude!!
Example 2

A spring has a spring constant of 43.2 N/m. If a 135 g mass is placed on it and made to vibrate, what will the frequency of the vibration be?

When an 80.0 kg person climbs into an 1100 kg car, the car\'s springs compress vertically by 1.2 cm. What will be the frequency of vibration when the car hits a bump?

Module 7.1 Objective 3

Upon completion of this module, the participant will be able to:

• Calculate the elastic potential energy of a system and apply conservation of energy to a system exhibiting simple harmonic motion.
Elastic Potential Energy
• Work is done to stretch or compress a spring
Conservation of Energy

Kinetic Energy

Potential Energy

Example

A spring with a spring constant of 125 N/m is attached to a 245 g mass. The mass is pulled a distance of 2.5 cm from equilibrium and released.

• What is the total energy of this system?
• What is the maximum speed of the mass?

A block of mass 0.50 kg is placed on a level, frictionless surface, in contact with a spring bumper, with a spring constant of 100. N/m that has been compressed by an amount 0.30 m. The spring, whose other end is fixed, is then released. What is the speed of the block at the instant when the spring is still compressed by 0.10 m?