• Explain why an object traveling in a circular path at a constant speed has acceleration.

Circular motionObjects traveling in a circular path at a constant speed must be continually changing direction. The object must accelerate towards the centre of the path in order to travel in a circle.

Velocity - tangent to the circular path.

Acceleration – towards the centre of the circle.

Circular path

radius

r

ac

v

fc

Force – towards the centre of the circle.

FrequencyFrequency f, is a measure of how many times an object completes the motion around a circle in one second. The SI unit for frequency is Hertz, Hz.

Time Period The time taken for an object to complete one revolution is called the period and has the symbol, T.

Frequency and time period are related to each other by the following equation.

Tf

1

An object has a frequency of 0.20 Hz. How long does it take this object to complete on revolution?

sHzf

T 0.520.0

11

rd 2

t

dv

VelocityThe velocity of an object undergoing circular motion an be calculated by dividing the distance of the circular path (circumference, 2πr) by the time it takes the object to complete one revolution.

Tt T

rv

2

ExampleWhat is the instantaneous velocity of an object that takes 3.2 s a circular path with a radius of 4.0 m?

sT 2.3mr 0.4

11 9.785.72.3

1.25

2.3

0.42

msmsm

v

Centripetal forceWhen an object follows a circular path at constant speed it must accelerate towards the centre of the path. There must be a force that is directed towards the centre of the circle to produce this acceleration. This force is called centripetal force, Fc. The centripetal force can be produced by friction, tension or gravity.

ExampleWhen a car turns a corner a centripetal force towards the centre of the curve is needed. This force is provided by the friction between the tyres and the road. If this friction does not occur the car will not accelerate towards the centre and will continue to travel at constant speed in the direction of the tangent.

v

fc

• Indicate the direction of the velocity, centripetal force and centripetal acceleration of an object in circular motion.

r

mvfc

2

r

vac

2

• For an object in circular motion calculate period, frequency, radius, speed, centripetal acceleration, or centripetal force using appropriate formulae.

Centripetal accelerationThe centripetal acceleration of an object traveling in a circular path can be calculated using the equation on the right.

ExamplesHow would the acceleration of an object under circular motion change if the velocity was doubled?From the equation on the right we can see that acceleration is proportional to the square of the velocity.If the velocity was doubled then the acceleration would be four times larger

2va

How would the acceleration of an object under circular motion change if the radius was doubled?

From the equation on the right we can see that acceleration proportional to the inverse of the radius.If the radius was doubled then the acceleration would be halved.

ra

1

Momentum

•Define momentum (units and direction).•Solve problems using the relationship between momentum, mass and velocity.•Explain ‘conservation of momentum’ and state when this applies.•Solve problems involving conservation of momentum during collisions.•Calculate change in momentum.•Relate impulse to change in momentum.•Explain the difference between elastic and inelastic collisions.

Momentum has the symbol p, it is a vector quantity that is equal to an objects mass multiplied by its velocity. The unit for momentum is kgms-1.

3.0 kg5.0 ms-1

This object has a mass of 3.0 kg and a velocity of 5.0 ms-1 to the right so it momentum is: 1 13.0 5.0 15p m v kg ms kgms

p=m v

What is the momentum of the following objects?

1.5 kg15.0 ms-1

0.035 kg

600 ms-1

2.0 kg3.0 ms-1

Momentum

In an closed system not effected by external forces momentum does not change.

total momentum before = total momentum after

2.0 kg5.0 ms-1

1.0 kg0.0 ms-1

2.0 kg 1.0 kg

Before After

A 2.0 kg trolley with a velocity of 5.0 ms-1 collides with a stationary trolley. After the collision the two trolley travel together, what is their final velocity?

1

1

2.0 5.0

10

p m v

kg ms

kgms

1

1

1.0 0.0

0.0

p m v

kg ms

kgms

Total momentum = 10 kgms-1

Total momentum = 10 kgms-1

1

1

10

3.0

3.3

pvm

kgms

kg

ms

5.0 kg5.0 ms-1

3.0 kg0.0 ms-1

5.0 kg 3.0 kg

Before After

Trolley A has a mass of 5.0 kg and a velocity of 5.0 ms-1 to the right, it collides with a stationary trolley, B (3.0 kg). After the collision trolley A has a final velocity of 2 ms-1 to the right. What is the final velocity of trolley B?

1

1

5.0 5.0

25

Ap m v

kg ms

kgms

1

1

3.0 0.0

0.0

Bp m v

kg ms

kgms

2.0 ms-1 ? ms-1

1

1

5.0 2.0

10

Ap m v

kg ms

kgms

A A BB

1 1

1

25 0

25

T A B

T

p p p kgms kgms

p kgms

1 1

1

25 10

15

B T A

B

p p p kgms kgms

p kgms

1

1

15

3.0

5

B

B

p kgmsv

m kg

v ms

Power

EP

t

Power is how much energy(work) is produced per second. When using this equation the energy could come from work, kinetic energy or Ep.

ExampleWhat is the power of a motor that does 100 J of work in 4 seconds?Energy = work = 100JP = E ⁄ t = 100J / 4 s = 25 W

Power measured in Watts(W)

Energy measured in Joules(J)

Time measured in seconds(s)

ExampleHow much energy is used when a 100 W light bulb is on for 30 s?E = P x t = 100 W x 30 sE = 3000 J

tPE

P

Et

ExampleHow long does it take a 500 W motor to do 10 000 J work?t = E / P = 500 W / 10 000 jt = 20 s

Rearranging the EquationTo find the energy

To find the time

Kinetic energyKinetic energy is the energy that moving objects have. We use the following equation to work out the kinetic energy an object has.

2k mv2

1E

ExerciseWhat is the kinetic energy of a 8.0 kg ball moving at 5.0 ms-1?

Ek = ½ mv2 = ½ 8.0 x 52

= 4.0 x 25 = 100 J

mass(Kg)

speed(ms-1)

kinetic energy (J)

2k mv2

1E

In this equation speed is squared. This has the effect of increasing the energy by four times when the speed is doubled.

ExampleBall traveling at 4 ms-1

Ek = ½ mv2 = ½ 8 x 42

= 4 x 16 = 64 J

Ball traveling at 8 ms-1 (Double the speed)Ek = ½ mv2 = ½ 8 x 82

= 4 x 64 = 256 J (The kinetic energy is quadrupled)

Gravitational potential energyGravitational potential energy (GPE) is the energy an object above the ground has. We use the following equation to work out the gravitational potential energy an object has.

mghEp

ExerciseWhat is the GPE of a 2.0 kg ball 8.0 m above the ground?

Ep = 2.0 kg x 9.8 ms-2 x 8 m = 160 J

mass(Kg)

Force of gravity(9.8 NKg-1)

Gravitational potential energy (J)

Height of object(m)

Converting energyAny time that an object is lifted the work done to lift the object is equal to the gravitational potential energy (Ep) gained by the object.

2 m

Ep = mgh = 0.5 x 10 x 2 = 10 J

Mass of toy car 0.5 Kg

W = f x d = 5 x 2 = 10 J

What is the gravitational potential energy gained when a box is lifted with a constant force of 50 N for 1.5 m?W = f x d = 50 x 1.5 = 75 JEp = work done = 75J

•When work is done accelerating an object by a force over a distance, this energy is transformed into kinetic energy(Ek). However, because of friction some of this work is turned into heat energy and only some of the work is transformed into kinetic energy.•Any time that energy is transformed some energy is always lost as heat energy.•The amount of heat energy lost can be reduced by decreasing the amount of friction acting on an object.•We can work out the efficiency of an energy transformation using the following equation.

1

%100

_

_

inenergy

outenergyEfficiency

When on object falls, the gravitational potential energy it had at the top of the fall is equal to the kinetic energy it has just before it hits the ground (if we ignore air friction).

4 m

At the topEp = mgh = 2 x 10 x 4 = 80 J

Just before hitting the groundEk = Ep

Ek = 80 J

•Describe Hooke’s law•Interconvert between the force, spring constant or extension of a spring using

kxF -=Hooke’s law states that “the distance a spring extends is proportional to the tension force acting on the spring.”

10 N

0.12 m

The diagram on the right shows what happens when a 10 N force is applied to an un-stretched spring. The tension force (F) is always in the opposite direction to the force extending the spring. For this spring the tension force, F = -10 N and the extension, x = 0.12 m. Hooke’s law can be used to work out the spring constant for this spring.

Hooke’s Law

21

2pE kxThe potential energy gained when a spring is compressed or extended can be calculated by using the equation on the right. The potential energy of a spring is dependent only on the extension of the spring and the spring constant.

ExercisesComplete the following exercises from page 150 of NCEA Level 1 Science book.Questions 1, 2, 3 and 4

Answers