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

Explain why an object traveling in a circular path at a constant speed has acceleration . Circular motion Objects 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.

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

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  1. Explain why an object traveling in a circular path at a constant speed has acceleration. Circular motion Objects 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. Force – towards the centre of the circle. radius Acceleration – towards the centre of the circle. r fc ac Velocity - tangent to the circular path. Circular path v

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

  3. Velocity The 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. Example What is the instantaneous velocity of an object that takes 3.2 s a circular path with a radius of 4.0 m?

  4. Indicate the direction of the velocity, centripetal force and centripetal acceleration of an object in circular motion. Centripetal force When 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. Example When 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

  5. For an object in circular motion calculate period, frequency, radius, speed, centripetal acceleration, or centripetal force using appropriate formulae. Centripetal acceleration The centripetal acceleration of an object traveling in a circular path can be calculated using the equation on the right. Examples How 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 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.

  6. 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.

  7. Momentum 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. This object has a mass of 3.0 kg and a velocity of 5.0 ms-1 to the right so it momentum is: 5.0 ms-1 3.0 kg What is the momentum of the following objects? 15.0 ms-1 1.5 kg 0.035 kg 600 ms-1 3.0 ms-1 2.0 kg

  8. In an closed system not effected by external forces momentum does not change. total momentum before = total momentum 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? 5.0 ms-1 0.0 ms-1 2.0 kg 1.0 kg 2.0 kg 1.0 kg Before After Total momentum = 10 kgms-1 Total momentum = 10 kgms-1

  9. Trolley A has a mass of 5.0 kg and a velocity of 5.0 ms-1to 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? 2.0 ms-1 ? ms-1 A B A B Before After 5.0 ms-1 0.0 ms-1 5.0 kg 3.0 kg 5.0 kg 3.0 kg

  10. Power Energy measured in Joules(J) Power measured in Watts(W) Time measured in seconds(s) Power is how much energy(work) is produced per second. When using this equation the energy could come from work, kinetic energy or Ep. Example What is the power of a motor that does 100 J of work in 4 seconds? Energy = work = 100J P = E ⁄ t = 100J / 4 s = 25 W

  11. Rearranging the Equation To find the energy Example How much energy is used when a 100 W light bulb is on for 30 s? E = P x t = 100 W x 30 s E = 3000 J To find the time Example How long does it take a 500 W motor to do 10 000 J work? t = E / P = 500 W / 10 000 j t = 20 s

  12. Kinetic energy Kinetic energy is the energy that moving objects have. We use the following equation to work out the kinetic energy an object has. mass(Kg) kinetic energy (J) speed(ms-1) Exercise What 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

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

  14. Gravitational potential energy Gravitational 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. mass(Kg) Gravitational potential energy (J) Height of object(m) Force of gravity(9.8 NKg-1) Exercise What 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

  15. Ep = mgh = 0.5 x 10 x 2 = 10 J Converting energy Any time that an object is lifted the work done to lift the object is equal to the gravitational potentialenergy (Ep) gained by the object. 2 m 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 J Ep= work done = 75J Mass of toy car 0.5 Kg

  16. 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.

  17. At the top Ep = mgh = 2 x 10 x 4 = 80 J 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 Just before hitting the ground Ek = Ep Ek = 80 J

  18. Hooke’s Law • Describe Hooke’s law • Interconvert between the force, spring constant or extension of a spring using Hooke’s law states that “the distance a spring extends is proportional to the tension force acting on the spring.” 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. 0.12 m 10 N

  19. The 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.

  20. Exercises Complete the following exercises from page 150 of NCEA Level 1 Science book. Questions 1, 2, 3 and 4 Answers

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