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- 1. CIRCULAR MOTION SPECIFIC LEARNING OUTCOMES 1. Recognize that even though a body in circular motion may have a constant speed, it velocity is changing and hence it is accelerating. 2. Demonstrate that the acceleration is towards the centre, i.e. centripetal. 3. Deduce that all things in circular motion must have a centripetal force. 4. Show graphically using velocity vectors that the acceleration is towards the centre. 5. Remember the equations for uniform circular motion questions, and use them to solve problems: Read Chapter 12 (p139 to 145)
- 2. CENTRIPETAL ACCELERATION Demo: Centripetal acceleration (accelerometer on turntable) The speed of an object in circular motion: • is constant in size but ..... • changing in direction Drawing activity
- 3. CENTRIPETAL ACCELERATION Demo: Centripetal acceleration (accelerometer on turntable) The speed of an object in circular motion: • is constant in size but ..... • changing in direction THEREFORE ..... the object’s velocity is changing continuously Drawing activity
- 4. CENTRIPETAL ACCELERATION Demo: Centripetal acceleration (accelerometer on turntable) The speed of an object in circular motion: • is constant in size but ..... • changing in direction THEREFORE ..... the object’s velocity is changing continuously THEREFORE An object in circular motion is always accelerating Drawing activity
- 5. CENTRIPETAL ACCELERATION Demo: Centripetal acceleration (accelerometer on turntable) The speed of an object in circular motion: • is constant in size but ..... • changing in direction THEREFORE ..... the object’s velocity is changing continuously THEREFORE An object in circular motion is always accelerating The size of the acceleration is given by a = ∆v/∆t where ∆v is the size of the change in velocity vector. Our prior observations indicated that ∆v is constant in size ~ therefore we can expect a to also be constant in size. ~ The direction of the acceleration will be the same as the direction of the change in velocity (a = ∆v/∆t). Since, from observation, the direction of ∆v is always towards ~ ~ ~ the centre we can expect the direction of a to also be towards the centre. ~ Drawing activity
- 6. CENTRIPETAL ACCELERATION Demo: Centripetal acceleration (accelerometer on turntable) The speed of an object in circular motion: • is constant in size but ..... • changing in direction THEREFORE ..... the object’s velocity is changing continuously THEREFORE An object in circular motion is always accelerating The size of the acceleration is given by a = ∆v/∆t where ∆v is the size of the change in velocity vector. Our prior observations indicated that ∆v is constant in size ~ therefore we can expect a to also be constant in size. ~ The direction of the acceleration will be the same as the direction of the change in velocity (a = ∆v/∆t). Since, from observation, the direction of ∆v is always towards ~ ~ ~ the centre we can expect the direction of a to also be towards the centre. ~ Features of this acceleration: Size is constant Direction is towards the centre of the circular path Drawing activity
- 7. “Feel the force” CENTRIPETAL FORCE Since an accelerating object will always do so in response to an unbalanced force, every object that travels in uniform circular motion must have an unbalanced force acting on it which is in the same direction as the acceleration. The unbalanced force acing on an object in uniform circular motion is called Centripetal force, Fcent Features of this force: Size is constant Direction is towards the centre of the circular path If the speed of the circular motion is increased then the force required to change the direction of the velocity will need to increase : Example - the carousel __________ Label the forces force __________ __________ force force Time for a complete rotation = 3s Time for a complete rotation = 2s
- 8. Flash cards EQUATIONS FOR UNIFORM CIRCULAR Consider a mass in uniform circular motion with speed v and a radius of circular path, r: v F Acceleration and force . m At any point: velocity is at a Experiment shows that: acent = v2 tangent to the r circular path Fcent = mv2 r since F = ma Period and frequency • The Period, T is the time the object takes to move through one complete revolution. (Unit: second, s) • The frequency, f is the number of revolutions performed per second. (Unit: Hertz, Hz or s-1)) T=1 f=1 and f T
- 9. Speed (in terms of period) In one complete revolution the distance travelled = the circumference of the circular path, C = 2πr The time taken for a complete revolution is the period, T => v = 2πr T Acceleration (in terms of period) Substituting v = 2πr into acent = v2 gives acent = (2πr/T)2 T r r acent = 4π2r => T2 Force (in terms of period) Since F = ma => Fcent = macent Fcent = m4π2r => T2
- 10. Examples 1. The Apollo 11 space capsule was placed in a parking orbit around the Earth before moving onwards to the Moon. The radius of the orbit was 6.56 x 104 m and the mass of the capsule was 4.4 x 104 kg. to the moon moon parking orbit Earth (a) If the centripetal force on the capsule was 407 kN while it was in the parking orbit, what was its acceleration? (b) What was the speed of the capsule in the parking orbit? (c) How long did it take the capsule to complete one orbit?
- 11. 2. A game of swing-ball is played with a 100 g ball. The effective radius of the circular path of the ball is 1.4 m. Find the tension in the string (centripetal force) when the ball has a velocity of: (a) 7.5 ms-1 (b) 15 ms-1 3. A string has a breaking strain of 320 N. Find the maximum speed that a mass can be whirled around in uniform circular motion with a radius of 0.45 m if the mass is 0.2 kg. 4. An object is in uniform circular motion, tracing an angle of 30o every 0.010 s. Find: (a) the period of this motion. (b) the frequency of this motion If the radius of the object’s path is doubled but the period remains the same, what happens to: (c) its speed? (d) its acceleration?
- 12. 5. In a circular motion experiment, a mass is whirled around a horizontal circle of radius 0.5 m. A student times four revolutions to take 1.5 s. sinker tube (a) Calculate the speed of the mass around the circle (b) What is the direction of the velocity of the mass? (c) Calculate the centripetal acceleration of the mass. (d) What is the direction of this acceleration? (e) How does the value of the centripetal acceleration compare to the acceleration of gravity? Ex.12A All Q’s
- 13. CIRCULAR MOTION SUMMARY An object in circular motion is traveling along a circular path with constant speed. The object’s motion can be described in more detail using vectors: Velocity, v ~ • constant in size but constantly changing in its direction. • At any point P, along the path of motion, the direction is at a tangent to the circular path. Force, F ~ • A force at right angles to the velocity causes the direction of the velocity to change. It cannot cause the size of the velocity to change. This force is constant in size. • The direction of the force is towards the centre of the circular path. Acceleration, ~ a • Acceleration is towards the centre of the circular path because the object is responding to an unbalanced force that acts towards the centre. • Acceleration is constant in size because the object is responding to a force that is constant in size.
- 14. v = velocity (ms-1) v ~ ~ m F = Force (N) F ~ ~ P a = acceleration (ms-2) a ~ ~ r r = radius of the circular path (m) (measured from the centre of the path to the centre of mass of the object. m = mass of the object (kg) Period and frequency Motion that takes place in regular cycles is called periodic motion. • The period, T is the time it takes for the object to complete a full cycle. (units: s) v=2πr => • The circumference, C is the distance travelled in T one cycle. C = 2πr (units, m) • The frequency, f is the number of cycles completed in one s. T=1 f=1 and (units: s-1 or Hz) f T
- 15. Fundamental equations acent = v2 r Use these equations when the speed is given Fcent = mv2 since F = ma r Derived equations acent = 4π2r T2 Use these equations when the period is given Fcent = m4π2r since F = ma T2
- 16. Term Definition Velocity speed in a given direction a constant acceleration directed towards the centre of the circular path Centripetal acceleration for an object which is in uniform circular motion a single force would have the same effect as all the actual forces that Unbalanced force act on an object. The unbalanced force is responsible for the object’s acceleration a constant force directed towards the centre of the circular path for an Centripetal force object which is in uniform circular motion. a line at right angles to the radius of a circle that touches the circle at Tangent one point only. Period the time taken for an object to complete a single revolution. Frequency the number of rotations per second. Circumference the total length of the circular path Revolution the single completion of a rotation Constant unchanging Fundamental unable to expressed in a simpler form Derived obtained from another concept
- 17. 12 PHYSICS CIRCULAR MOTION ASSIGNMENT Name 1. A car is travelling around a bend in the road and for a few seconds is in uniform circular motion. (a) The centripetal force is being provided by the road. Name this force. ________________________________________________________________ The car passes over a patch of oil while it is rounding the bend. (b) Describe the path the car will take after it hits the oil patch and explain why this happens in terms of the forces acting. ________________________________________________________________ ________________________________________________________________ ________________________________________________________________ 2. An object in uniform circular motion completes 10 revolutions in 0.4 seconds (a) Find the frequency of this motion. ________________________________________________________________ ________________________________________________________________ (b) Find the period of this motion. ________________________________________________________________ _______________________________________________________________
- 18. 3. A big wheel at a fair spins in a circular path of radius 20 m. Once the wheel has reached a steady speed, a student times each revolution at 13 seconds. (a) Calculate the circumference of the big wheel. ______________________________________ (b) Hence calculate the speed of the big wheel. ______________________________________ ______________________________________ (c) Calculate the centripetal acceleration of each passenger. ________________________________________________________________ ________________________________________________________________ 4. In a circular motion experiment, a mass is whirled around a horizontal circle which has a 0.50 m radius. A student time 4 revolutions to take 2.0 s. (a) Calculate the speed of the mass around the circle. ________________________________________ ________________________________________ (b) What is the direction of the velocity of the mass? ________________________________________ ________________________________________ (c) Calculate the centripetal acceleration of the mass. ________________________________________ ________________________________________
- 19. (d) What is the direction of this acceleration? _______________________________________________________________ (e) How does the value of the centripetal acceleration compare to the acceleration of gravity? ______________________________________________________ 5. Two Aquinas students go to a fun park for a day where they pay to drive carts around a circular track. The track has a radius of 31.8 m and once the carts are at a maximum speed they complete a lap in 16 s. (a) What is the frequency of the cart’s motion when travelling at maximum speed? __________________________________________ (b) When travelling at maximum speed, calculate the speed of the cart. __________________________________________ __________________________________________ (c) Calculate the acceleration of the cart when travelling at maximum speed. _____________________________________________________________ _____________________________________________________________ _____________________________________________________________ _____________________________________________________________
- 20. The cart has a mass of 150 kg and one of the students, Chris has a mass of 75 kg. (d) Calculate the size of the force acting on Chris and his cart at maximum speed. ________________________________________________________________ ________________________________________________________________ ________________________________________________________________ (e) Chris drives over a patch of oil and loses control of his cart whilst travelling at this maximum speed. On the diagram, draw his path after driving through the oil. 6. Jon and Ana are two ice-skaters. In a practiced skating move, Jon spins Ana around in a horizontal circle. Ana moves in a circle as shown: Jon Ana (a) Draw an arrow on the diagram to show the direction of the tension force that Jon’s arm exerts on Ana at the instant shown. (b) If the radius of the circle is 0.95 m and the tension force in Jon’s arm is 5.00 x 102 N, calculate the speed with which Ana (55 kg) is travelling around the circle. Give your answer to the correct number of significant figures. ________________________________________________________________ ________________________________________________________________ ________________________________________________________________
- 21. (d) While Ana is still moving in a circle on the ice, Jon lets her go. (i) Describe her velocity (speed and direction) after he releases her. ______________________________________________________________ (ii) Explain why Ana travels with this velocity. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

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a very good slide for revision!!!