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Circular Motion Master Rev2
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Circular Motion Master Rev2

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  • 1. Circular Motion. Introduction.
  • 2. Observing circular motion. Where do you observe things moving with circular motion? • The hammer swung by a hammer thrower • Clothes being dried in a spin drier • Chemicals being separated in a centrifuge • Cornering in a car or on a bike • A stone being whirled round on a string • A plane looping the loop • A DVD, CD or record spinning on its turntable • Satellites moving in orbits around the Earth • A planet orbiting the Sun (almost circular orbit for many) • Many fairground rides • An electron in orbit about a nucleus Any motion in a curve is an example of circular motion.
  • 3. Linear velocity. The instantaneous linear velocity at a point in the circle is usually given the letter v and measured in metres per second (m s-1). To calculate the speed of a body moving in a circular path, you need to know the distance moved and the time taken i.e. Speed = circumference of circle time to complete one rotation Since the circumference of a circle = 2πr ν = 2πr t
  • 4. Questions. 1. A Child’s roundabout has a diameter of 2.5 m, it revolves at 45 rpm. Calculate the linear velocity of its outside edge. 2. A satellite orbits the earth every 90 min, its speed is approximately 7.8 Km s-1 Calculate its height above the earth (radius of the earth is 6400 km) 3. A bucket is being swung on the end of a piece of rope, it is travelling at 6m/s, the rope is 1.5m long, what is the time period of rotation?
  • 5. Degrees and radians The radian is a more r ‘natural’ unit for measuring angles. r θ=1 One radian (or rad for r rad short) is defined as the angle subtended at the centre of a circle of radius r by an arc of length r.
  • 6. Thus the complete circumference 2πr subtends an angle of 2πr/r radians. Thus in a complete circle of 360 degrees there are 2π radians. Therefore 1 radian = 360° /2π = 57.3°
  • 7. Use a calculator to complete the table of q in degrees and radians, sin θ, cos θ, and tan θ when θ has values in degrees shown in the table: Sin θ cos θ tan θ θ degree θ in radians 0.01 1.75E-4 1.75E-4 0.99 1.75E-4 0.1 1.75E-3 1.75E-3 0.99 1.75E-3 0.5 8.7E-3 8.7E-3 0.99 8.7E-3 1 0.017 0.017 0.99 0.017 5 0.087 0.087 0.99 0.087 10 0.175 0.175 0.99 0.175 20 0.35 0.34 0.94 0.36 70 1.22 0.94 0.34 2.75 When θ (in radians) is small what are suitable approximations for: sin θ, tan θ, cos θ?
  • 8. Angular velocity. The angular velocity is the angle through which the radius to this point on the circle turns in one second. This is usually given the letter ω (Greek omega) and is measured in radians per second (rad s-1) So if a body completes 4 revolutions every second it has an angular velocity of?
  • 9. Time period for one rotation (T) = distance/velocity = 2πr/v = 2π/ω Therefore linear and angular velocity are related by the formula: Linear velocity = radius of circle × angular velocity v = rω The angular frequency is the number of revolutions per second ω = θ/t
  • 10. A complete revolution of a circle = 2π rad ω = 2π/t So Since ƒ = 1/T Angular velocity ω = 2πƒ
  • 11. Questions 1. Calculate the angular velocity of an engine operating at 2500 rpm. 2. A old fashioned record player, playing an album. (33 rpm) 3. A satellite in geostationary orbit. 4. A satellite in low polar orbit (orbit time 90min). 5. An ultra sonic motor operating a 500 kHz 6. A motor has an angular velocity of 23 rad/s what is its time period of rotation?
  • 12. Centripetal force When we secure an object to the end of a piece of string and swing it the tension induced on the string is the centripetal force. This is the force that keeps the object orbiting around the same point.
  • 13. When an object is in a regular orbit the apparent forces due to the momentum of the object are equal to the centripetal force. Some people refer to this apparent outwards force centrifugal force, because this is due to the momentum and is not really a force we do not use the term.
  • 14. Acceleration towards the centre of a circle is the centripetal acceleration. An object can only accelerate if a resultant force is acting on it. – the centripetal force – otherwise the object would fly off at a tangent to the circle (Newton’s first law of motion).
  • 15. In the previous example, the moving object is acted on by the force towards the centre of the circle, but it does not get any closer. The centripetal force needed to make an object follow a circular path depends on. (i) the mass, m of the object (ii) the speed, ν (iii) the radius, r of the circle Centripetal force, F= mν2 r
  • 16. Circular acceleration
  • 17. When an object moves in a circle, its velocity is at a tangent to the circle. Its velocity is changing since the direction is changing ∴ the body is accelerating. ∆ν shows how the velocity vector changes. The arrow shows the direction of the change in velocity ∴ the direction of the acceleration.
  • 18. Since F = ma; a = F/m a = ν2 r ν = rω Now so centripetal acceleration can be calculated using angular velocity: a = rω 2 And centripetal force = ma = mrω2
  • 19. Circular motion. A satellite is orbiting the earth in a low earth orbit of 60 Km and takes 120 mins to complete one complete orbit. Calculate the following: Linear velocity. Angular velocity. Centripetal acceleration. Radius of the earth is 6400 Km. The satellite has a mass of 900 kg calculate the centripetal force.
  • 20. Gravitation and Gravity. When we see pictures or film of astronauts floating we often think that they are experiencing no gravity at all. In fact they are being affected by the gravity of all the things around them, to varying degrees. Newton understood that gravity was a universal force of attraction that affected everything, the gravity that causes us to fall towards the ground is the same force that keeps the moon in orbit.
  • 21. Newton’s law of universal gravitation states: “Every particle in the universe attracts every other with a force which is proportional to the product of their masses and inversely proportional to the square of their separation.” From this statement we can work out that: F ∝ m1m2 F ∝ 1/r2 and
  • 22. When we introduce a constant, the universal gravitational constant, G, we get the equation: Gm1m2 F =− 2 r Where F is the force of gravitational attraction between the two bodies of mass m1 and m2. And G is the universal gravitational constant, which is equal to 6.67 x 10-11 N m2 kg-2.
  • 23. Force on m2 m2 m1 Displacement of m2 Force and displacement are vector quantities. They are in opposite directions to each other; this introduces the –ve sign into the equation.
  • 24. Earth and moon. The Earth has a mass (ME) of 5.98 x 1024 kg The moon has a mass (MM) of 7.35 X 1022 kg. The radius (R) of the earth is 6.38 x 106 m The centres of gravity of moon and earth are separated by 3.84 x 108 m. Calculate the Force of attraction an 80 kg man experiences at the earths surface. Calculate the force of attraction between the moon and the earth.
  • 25. Calculate the force a 100kg satellite will experience at, 100, 200, 300, 400, 500, 600, 700, 800, 900 and 1000 Km above the earths surface. Plot a graph of your results. What does the shape of the graph show?
  • 26. Questions Calculate the force of attraction between two members of the class when they are standing 2 meters apart. – Would you feel the force? A canoeist with a mass of 60 kg paddles past a super tanker weighing 650,000 tonnes at a distance of 4m. – Calculate the force expereinced. – Would the force be felt? A satellite orbits the earth at 700 km, its orbit is changed so that it now orbits at a height of 500 km. – Calculate the factor by which the force will increase. – What will happen to the satellites speed?
  • 27. The universal constant of gravitation. If we rearrange Newton’s formula shows the significance of G. 2 Fr G=− m1m2 Suppose the separation of the masses is 1 metre. The masses m1 and m2 are both 1 kg If this is the case then G = -F.
  • 28. The universal gravitational constant is the attractive force between two 1 kg masses placed 1m apart, this force is extremely small, hence the value of the gravitational constant is small.
  • 29. Gravitational field strength All masses produce a gravitational field. The strength of a gravitational field at a point is defined as the force that would be exerted on a unit mass placed at that point. Gravitational field strength = force per unit mass F g= (Linear field) m The units for gravitational field strength are N kg-1.
  • 30. We can use Newton’s law of gravitation to find an expression for g at the surface of the earth as the force of gravity on an mass on the surface of the Earth is given by: GM E m F =− 2 r F GM E g= =− 2 Radial so field m r The second equation gives the field strength for a radial gravitational field, such as that produced by a spherical mass such as a planet or moon.
  • 31. Calculate g for the Sun, Earth and the Moon. The Sun’s mass is 1.989 x 1030 kg r = 695 500 km The Earth’s mass is 5.978 x 1024 kg r = 6400 km The Moon’s is 7.35 X 1022 kg. r = 1700 km
  • 32. The neutral point While travelling between two celestial bodies you will pass through a point where the forces due to each body are the same, this is called the neutral point. GM M GM E gm = g E ∴ = 2 2 r R 2 R ME = 2 r MM
  • 33. Earth-moon neutral point. When the Apollo missions went to the moon they passes through the neutral point when they where 90% of the way to the moon. The mean separation between the earth and the sun is 1.496 x 1011m. Work out how far from the earth the Earth – sun neutral point is.
  • 34. Gravitational potential • If an object is falling back towards Earth from a neutral point it will gain a velocity as it loses gravitational potential energy 1 mg∆h = mv 2 2 v = 2 g∆h • This works well for a small value of change in height but for large values we need to take account of the change in gravitational field
  • 35. Work done in moving a mass to infinity • The work to move a mass from distance R to infinity is:- GM E m W= R
  • 36. Gravitational potential Gravitational potential is always negative. The potential at a point is the amount of energy needed to move 1 kg from infinity to that point. A distant object has zero PE W GM E V =− =− m r V = -GM/r (In some texts U = -GM/r)
  • 37. 0 Potential J/kg planet Max Distance from Centre / m
  • 38. As the potential is the energy needed to move 1Kg from infinity to the point in question. We can calculate the potential energy of any object at that point by simply multiplying the potential by the mass of the object. Ep = mV
  • 39. Gravitational potential (additional notes) Gravitational potential is a field, this is because it is defined at every point in space, unlike gravitational field strength it is scalar rather than vector. If a mass is positioned at one point where the potential is V1 and is moved to an other point with a potential of V2 then the work done is equal to: W = m(V1-V2) W = m∆V
  • 40. Escape velocity. If a mass at the earths surface can be given enough Kinetic Energy equal to its gravitational potential energy (= mV), it will escape completely from the Earth’s gravitational field (and end up at infinity). This velocity is called the escape velocity. The escape velocity is not dependant on the mass.
  • 41. We can derive the formula for determining the escape velocity using the formulas for Gravitational potential energy and kinetic energy. 1 2 GM E m mv = 2 R 2GM E v= R
  • 42. Calculate the escape velocities for the Moon, Earth and Jupiter. Body Mass (kg) diameter (km) Moon 7.35×1022 3476 Earth 5.97×1024 12,104 Jupiter 1.89×1027 142,800
  • 43. Geostaionary Satellites Geostationary satellites travel with a speed of 3068 m/s Calculate the radius of their orbit. The radius of the earth is 6400 km, what is the altitude of the satellite. Calculate the gravitational force acting on a 150Kg satellite.
  • 44. International space station The ISS is a low earth orbit satellite, it orbits the planet 16 times a day. Its average altitude above the earth is 333.3 km and Its mass is 233,000 km. Calculate the velocity of the ISS Calculate the gravitational force acting on the ISS.
  • 45. Mercury and Venus. Planet Mass Mean R T of orbit Mercury 3.3022×1023 kg 57.9 x 106 m 88 days Venus 4.8685×1024 kg. 108.1 x 106 m 224.7 days • Calculate the gravitational forces between the planets and the Sun. • The mass of the sun is 2 x 1030 kg. • Calculate the velocities of the planets orbits. • Using the formulas F=ma and F = -GMm/r2 • We have already calculated that the time period of the orbit squared is proportional to the radius of the orbit cubed. (T2 ∝ R3) • Use this relationship to calculate the radius of Jupiter’s orbit; Jupiter orbits the sun every 11.86 years.

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