Module 1

PlanningGuide

Lesson 1.4
. TryThis
. Notes
. Concepts
. Examples
. Exercises
. Equations
. Definitions
. Answers

Lesson 1.1
Lesson 1.2
Lesson 1.3
Lesson 1.4
Lab 1.1
Lab 1.2
Lab 1.3
Lab 1.4
Project 1

Module 2

Module 3

Module 4


UNIT 1 - ENERGY & MOTION

Lesson 1.4 ACCELERATION

Overview
This lesson deals with acceleration and the effects of gravity on falling objects. On completion of the lesson, you should be able to calculate the maximum height of a vertically launched projectile and estimate its position at a particular time. You should also be able to calculate the centripetal acceleration of an object moving in a circular path.

ToDo
Watch the video presentation.
Carry out the activities.
Read through the lesson notes and do the exercises.
Refer to the solutions and check your answers.
At home: Prepare for Lab 1.4 by reading the instructions and collecting the necessary materials and equipment.
Prepare for the two activities in Lesson 1.5.


ACTIVITIES

Centripetal Acceleration: Show how a force is needed to keep an object moving in a circular path.

Laboratory Cart & Falling Weight: Use a falling weight to accelerate a cart or skateboard.

Activity 1.4.1: Centripetal Acceleration

Purpose: To show that a force is needed to keep an object moving in a circular path.

Equipment:

Glass tube or rigid plastic tube about 15 cm long with diameter of about 0.5 cm.

Piece of string about 1 m long.

Small lead weight

Paper clip

5 - 10 metal washers

Procedure:

Assembling the device

  1. Tie the lead weight to the string, thread the string through the tube and tie the paper clip to the other end.
  2. Hold the tube vertically. Hang a few washers on the paper clip

Using the device

  1. Holding the tube vertically, swing the weight in a circular path by moving the string in the tube.
  2. While swinging the weight at a constant rate, determine the number of washers that must be attached to the paper clip to keep the weight from moving outwards.
  3. Increase the speed of the weight slightly and show that more washers are needed to keep the weight from moving outwards.

Activity 1.4.2: Use a falling weight to accelerate a laboratory cart

Purpose: To show how a falling weight can be used to accelerate a laboratory cart.

Equipment:

Laboratory cart, skateboard or toy car.

Pulley to be mounted on the edge of a table

About 4 meters of strong twine or nylon line

Weight (about 250g to 500g)

Procedure:

  1. Tie the twine to the cart, thread it through the pulley and tie the other end to the weight.
  2. With the cart at the opposite end of the table, the weight should be near the top of the table.
  3. Holding the cart and the weight at the same time – with the string taught – let both the cart and the weight go at the same time.
  4. The falling weight will pull the cart along the table and cause it to accelerate as the weight accelerates under the influence of gravity.

Questions

  1. Does the cart accelerate at 9.81 m/s2 ?
  2. Why or why not?

Answers

  1. No.
  2. The force of gravity on the weight is 9.81 N/kg. This force is used to accelerate the weight plus the cart. Because the combined mass is greater than that of the weight alone, the cart plus weight will accelerate at less than 9.81 m/s2.

Instantaneous Speed
The speed measured over a very short period of time -an instant. When the speed of an object changes, its instantaneous speed varies or changes from instant to instant.

Average Speed
Average speed refers to the speed while covering a particular path. The path may be curved or change direction a number of times. Distance is measured along the path taken.

Velocity
Velocity is the ratio of the change in position to the time interval over which the change takes place. Speed will always be equal to or greater than the velocity because the path taken to achieve the change in position may be further than the displacement.
The velocity of an object also changes if its direction changes. Velocity is the displacement in a particular direction divided by the time taken - even if it continues to move at a constant speed.

Acceleration
Acceleration is a change in velocity over a particular period of time.
Acceleration is change in speed and/or change in direction.
If something is accelerating, its instantaneous speed = acceleration x elapsed time.

Newton’s First Law
An object that is at rest will tend to stay at rest and an object that is in motion will tend to continue its motion in the same direction with the same speed unless a force or set of unbalanced forces acts upon it.

Force and Acceleration
A force or set of unbalanced forces is needed to change the velocity of an object.

Centripetal Acceleration
When an object follows a curved path, it accelerates towards the center of the portion of a circle that makes up the curve. The amount of acceleration depends on the tangential velocity of the object and the radius of the circle.

Acceleration and deceleration due to gravity
If an object near the earth falls under the influence of gravity, in the absence of any other forces (such as friction with the air), it will accelerate at 9.81 m/s2 until it his the ground.
If an object is launched vertically, it will slow down due to gravity at a rate of 9.81 m/s2.
If it is launched from the ground, it will take the same time to reach its maximum height as it takes to fall from its maximum height to the ground. It will also be traveling at the same speed when it hits the ground as it was launched at.

Instantaneous Speed
When an object moves at a constant speed, its instantaneous speed remains constant. The speed measured over a very short period of time is the same as the speed measured over a longer period. When the speed of an object changes, its instantaneous speed varies.

Average Speed
For an object moving at constant speed or varying speed, the average speed refers to a particular period of time and is the total path distance divided by the time. Average speed refers to the speed while covering a particular path. The path may be curved or change direction a number of times. Distance is measured along the path taken.

Velocity
Velocity is the ratio of the change in position to the time interval over which the change takes place. Speed will always be equal to or greater than the velocity because the path taken to achieve the change in position may be further than the displacement.

The velocity of an object also changes if its direction changes. Velocity is the displacement in a particular direction divided by the time taken - even if it continues to move at a constant speed.

Acceleration
Acceleration is a change in velocity over a particular period of time.
Acceleration is change in speed and/or change in direction.
If something is accelerating, its instantaneous speed = acceleration x elapsed time.

Newton’s First Law
An object that is at rest will tend to stay at rest and an object that is in motion will tend to continue its motion in the same direction with the same speed unless a force or set of unbalanced forces acts upon it. This is sometimes referred to as Newton’s law of inertia. Essentially, a force is needed to get a stationary object to move. Once moving, a force is needed to change the speed or direction of the object.

The following equation is a simple statement of Newton’s First Law:

F = ma . . . . .Force = mass x acceleration

Where: F = force (N)

m = mass (kg)

a = acceleration

Force and Acceleration
Acceleration is a change in velocity. This is a change in speed or direction. A force or set of unbalanced forces is needed to change the velocity of an object. The inertia of an object is a measure of its mass and it is its tendency to resist changes in velocity. The greater the object’s inertia, the greater the force need to achieve a particular change in velocity.

It is possible for two forces to act on an object and not cause acceleration if the forces are balanced. Forces are balanced if their resultant is zero. The two forces must be equal and act in opposite directions.

Centripetal Acceleration
A force is needed to cause an object to accelerate. When an object follows a curved path, it accelerates towards the center of the portion of a circle that makes up the curve. The amount of acceleration depends on the tangential velocity of the object and the radius of the circle.

Acceleration and deceleration due to gravity
If an object near the earth falls under the influence of gravity, in the absence of any other forces (such as friction with the air), it will accelerate at 9.81 m/s2 until it his the ground.

If an object is launched vertically, it will slow down due to gravity at a rate of 9.81 m/s2.

If it is launched from the ground, it will take the same time to reach its maximum height as it takes to fall from its maximum height to the ground. It will also be traveling at the same speed when it hits the ground as the speed it was launched at.

Equations used in calculations involving acceleration:

d = vt ...................Distance = velocity x time

Where: d = distance (m)

v = velocity* (m/s)

t = time (s)

(*If v changes with time, use average velocity)

d = ½at2 ...............Distance = ½ x acceleration x time2
(for accelerating object)

Where: d = distance (m)

a = acceleration (m/s2)

t = time (s)

For a projectile launched vertically from the ground
at a velocity = v0 :

v = v0 - gt ......Velocity = initial upward velocity – (acceleration due to gravity x time)

Where: v = velocity (m/s)

v0 = initial velocity (m/s)

g = acceleration due to gravity (9.81 m/s2)

t = time (s)

tmax = v0 /g .......................Time to reach maximum height = initial velocity ÷ acceleration due to gravity

Where: v0 = initial velocity (m/s)

g = acceleration due to gravity (9.81 m/s2)

tmax = time to reach maximum height (s)

dmax = ½g tmax 2 .................
Maximum height = ½ x acceleration due to gravity x time2

Where: dmax = maximum height (m)

g = acceleration due to gravity (9.81 m/s2)

tmax = time to reach maximum height (s)

dmax = [(v0 + vt)/2] tmax ...............
Maximum height = average velocity x time

Where: dmax = maximum height (m)

v0 = initial velocity (m/s)

vt = velocity at max height = 0 (m/s)

tmax = time to reach maximum height (s)

Example 1.4.1 Acceleration
During a speed test, a car accelerated from rest in a fixed direction at an average rate of 1 meter per secon per second.
a) What was it's speed at the end of the first second?
b) What was it's speed at the end of the second second?
c) What was it's average speed during the first second?
d) What was it's average speed during the first 2 seconds?

Solution

  1. It accelerated at 1 meter per second during the first second. At the end of the first second, it's speed was 1 meter per second.
  2. At the end of the second second, it's speed was increased by another 1 meter per second. It's speed was 2 m/s.
  3. At the beginning of the first second, it's speed was zero. At the end of the first second, it's speed was 1 m/s. The average speed during this period was thus = (0 + 1)/2 m/s
    = 0.5 m/s.
  4. At the beginning of the first second, it's speed was zero. At the end of the second second, it's speed was 2 m/s. The average speed during this period was thus = (0 + 2)/2 m/s
    = 1 m/s.

Example 1.4.2 Acceleration due to gravity
A steel ball is dropped from a point 20 meters above ground.
a) How fast will it be traveling 0.5 seconds after being dropped?
b) How fast will it be dropping 1 second after being dropped?
c) How far will it have dropped 1 second after being dropped?

Solution

  1. Velocity = acceleration x time. The acceleration due to gravity is 9.81 m/s2
    v = 9.81 m/s2 x 0.5 s = 4.9 m/s
  2. v = 9.81 m/s2 x 1 s = 9.81 m/s
  3. Distance = average velocity x time

The velocity at time = 0 is 0 m/s. The velocity at time = 1 s is 9.81 m/s

The average velocity = ½ x 9.81 = 4.905 m/s

The distance covered is thus 4.905 m/s x 1 s = 4.905 m

Example 1.4.3 Conversion of energy
A baseball is thrown vertically and lands 2 seconds later.

  1. What height does it reach?
  2. At what velocity does it strike the ground.

(Assume that it is launched from ground level and neglect air resistance)

Solution

  1. The time taken to reach the maximum height is the same as the time taken to reach the ground from the maximum height. tmax = 1 second.
  • Maximum height = ½ x acceleration due to gravity x time2

    Maximum height = 0.5 x 9.81 m/s2 x 12 = 4.905 m.

    •

    b) tmax = v0 /g therefore v0 = tmax g

    The velocity at which the ball was launched was:

    v0 = 1 s x 9.81 m/s2 = 9.81 m/s

    This is the same velocity at which it strikes the ground – but in the opposite direction.

    Review Questions

    1. What is the difference between average speed and instantaneous speed?
    2. Does a car always accelerate as long as the accelerator is depressed?
    3. Why do the brakes of a car cause acceleration?
    4. A 5-kilogram steel ball falls under the influence of gravity. At what rate will the ball accelerate?
    5. If a steel ball with a mass of 200 grams is dropped from a high building, how far will it fall in 2 seconds? (Neglect air resistance.)
    6. If a steel ball that has been dropped from a high building has an instantaneous speed of 15 m/s, how far has the ball fallen from the point at which it was dropped?
    7. If a soccer ball weighing 0.5 kilograms is dropped from an airplane at an altitude of 1,000 feet, will it accelerate until it reaches the ground? Why?
    8. A steel ball is dropped from a point that is 20 meters above ground level. How fast will it be traveling 0.5 seconds after being dropped?
      How fast will it be dropping 1 second after being dropped?
      How far will it have dropped 1 second after being dropped?
    9. If an object travels at 3 meters per second in a straight line, is it accelerating?
    10. If an object travels at 3 meters per second in a circular path, is it accelerating?
    11. If a water bomb weighing 200 grams (0.2 kg) is dropped onto a person from a height of 10 meters, what will its velocity be at impact? (Ignore the effect of air resistance)
    12. If an object with a mass of 0.5 kilograms travels at 3 meters per second in a circular path with a radius of 2 meter, how fast is it accelerating?
    13. Calculate the instantaneous speed and distance traveled by an apple 0.8 seconds after it started falling out of a tree.
    14. If an arrow is shot vertically upward and it lands 3 seconds later, at what speed did it leave the bow?

    HOMEWORK
    Select one or more of the     recommended activities for Lesson 1.5, collect the items needed and test the procedure before demonstrating the activity during the next theory lesson.

    Path distance: The distance covered while following a particular path.

    Displacement: The smallest distance and direction between the beginning and end of a path.

    Speed: The rate at which something moves. The path distance covered in a particular period of time.

    Average Speed: The path distance covered in a specific period of time. If speed changes with time, the average speed is the total path distance covered divided by time taken.

    Instantaneous Speed: The speed measured over a very short period of time -an instant.

    Velocity: Displacement (in a particular direction) in a specified period of time.

    Acceleration: The rate of change in velocity. Acceleration is a change in velocity in a specific period of time.

    Accelerometer: An instrument used to measure or indicate the rate of increase in velocity.

    Tangential velocity: Instantaneous velocity of an object moving in a curved path. Literally: The velocity at a tangent to the curve of the path.

    Centripetal acceleration: An object moving in a curved path accelerates towards the center of the curve that it is following. This is called centripetal acceleration.

    1. The average speed is a combination of a number of instantaneous speeds. The instantaneous speed is the speed in an instant of time. This can be equal to the average speed if it does not change during the period that the average speed is measured.
    2. No. The accelerator controls the fuel supply to the engine. Fuel is needed even if the car is traveling at constant speed.
    3. The brakes cause deceleration which is negative acceleration.
    4. Neglecting air resistance, the ball will accelerate at 9.81 m/s2
    5. ½ x 9.81 m/s2 x 22 s2 = 19.62 m
    6. v = gt therefore t = v / g = 15 m/ 9.81 m/s2= 1.53 s
    7. No. As it accelerates, the air resistance on it increases until the forces acting on the ball become balanced. Gravitational force will be balanced by air resistance on the ball. The ball will then fall at a constant velocity known as its terminal velocity.
    8. Neglecting air resistance:
      a) v = at = 9.81 x 0.5 = 4.905 m/s
      b) 9.81 m/s
      c) d = ½at2 = 0.5 x 9.81 x 1 = 4.905 m
    9. No.
    10. Yes
    11. d = ½at2 and v = gt
      therefore t = square root of (20/9.81) = 1.428 s
      and v = 9.81 x 1.428 = 14.01 m/s
    12. ac = v2 / r = 32 / 2 = 4.5 m/s2
    13. v = at v = 9.81 x 0.8 = 7.848 m/s
    14. Time to reach max height = 1.5 seconds.
      (dmax = ½at2 therefore max height = 0.5 x 9.81 x (1.5)2
      = 11.036 m)
      Velocity at launch = velocity at impact with ground
      v = gt = 9.81 x 1.5 = 14.715 m/s