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Circular Motion

Circular Motion. Centripetal Force. video. Angles can be measured in RADIANS. The angle in radians is s/r which is the length of the arc divided by the radius One full circle = 2

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Circular Motion

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  1. Circular Motion Centripetal Force

  2. video

  3. Angles can be measured in RADIANS The angle in radians is s/r which is the length of the arc divided by the radius One full circle = 2𝜋 radians i.e. 360o = 2𝜋 radians

  4. Angular velocity (𝝎) is the rate of change of angle during circular motion in radians /sec (𝝷/t) For a full circle 𝝎 = 2 ( T is the time period of the rotation 1/T = the frequency of rotation. f = 1/T 𝝎 = 2𝜋f Linear speed (v) and (𝝎) are linked by the equation (v = r𝝎)

  5. The speed is always constant in the direction to the tangent at any point on the circumference but he direction is always changing towards the centre of the circle. The acceleration is always towards the centre of the circle

  6. Centripetal acceleration towards the centre of the circle = v2/r or 𝝎2r According to Newton’s second law F = ma therefore centripetal force = mv2/r or m𝝎2r

  7. centripetal force = mv2/r m = mass of turning object v = velocity r = radius of the turning circle

  8. Example situation A vehicle of mass 1200kg is cornering at a velocity of 30 m/s around a level corner of radius 110m. Calculate the centripetal force exerted on the car by friction between the tyres and the road.

  9. centripetal force = mv2/r Mass = 1200kg v = 30m/s r = 110m Centripetal force on vehicle = (1200 x 302) /110 =9818N

  10. The vehicle is on the limit of adhesion at this velocity. Calculate the coefficient of friction between the tyres and the road (take g as 9.8m/s2).

  11. Force(centripetal) = μ(coefficient of friction) x weight of vehicle (mass x 9.81) Weight = 1200 x 9.81 =11772N μ = Force/weight = 9818/11772 0.83

  12. The corner now becomes banked at an angle of 12o to the horizontal. Calculate the maximum cornering velocity of the vehicle on this part of the curve.

  13. F = mv2/r F(centripetal) = μ(coefficient of friction) x weight of vehicle (mass x 9.81) μ x m x 9.81 = mv2/r (this calculation gives us the centripetal force for cornering on a flat road) The extra centripetal force due to the 12obanking Is found by the equation m x 9.81 x tan 12o (= v2/r) v2 = (r x μ x m x 9.81)/m + (r x m x 9.81 x tan12o)/m = (r x μ x 9.81) + ( r x 9,81 x tan 12o) m’s cancel out

  14. v2 = (r x μ x 9.81) + ( r x 9.81 x tan 12o) = (110 x 0.83 x 9.81) + (110 x 9.81x tan 12o) = 1143.8 V = √1143.8 33.8m/s

  15. The crankshaft of a 125cc two-stroke engine carries a single piston of bore 51.5 mm and stroke 60mm. The effective rotating mass of the connecting rod and piston assembly is 200g and is out of balance with the line of the crankshaft by an eccentricity of half the stroke. The engine is running at 900 rpm. Show that this is equivalent to an angular velocity of 94.2 rad/s and calculate the centripetal force on the crankshaft bearings due to this out of balance load.

  16. ω = angular velocity in radians/second 360o (full circle)= 2π radians ω = 2π/t (full circle) 1/t = frequency (f) ω = 2πf If the engine is running at 900rpm ω = 900 x 2π/60 = 94.2 rad/sec

  17. v = ωr F = mv2/r = mω2r In the question m = 200g =0.2kg r = half of stroke =30 mm F = 0.2 x 94.22 x .03 = 53.24 N

  18. The out-of-balance force on the crank is to be balanced using two identical counterweights either side of the connecting rod at an eccentricity of 0.02m. Calculate the mass of one counterweight.

  19. m1ω2r1= 2m2ω2r2 m2 = m1r1/2r2 m1 = mass of connecting rod and piston assembly, 0.2kg r1 = half stroke, 0.03m r2 = eccentricity of counter weights, 0.02m m2 =(0.2 x 0.03)/(2 x 0.02) = 0.15 kg

  20. Centrifugal clutch

  21. Centrifugal clutch Rotating bobs springs cylinder

  22. Rotating bobs When the angular velocity of the shaft increases the centripetal force Increases causing the springs to extend and the clearance between the bobs and the cylinder to decrease. When the centripetal force is high enough the bobs will engage with the clutch cylinder. springs cylinder

  23. 0.1m 0.01m Rotating bobs A centrifugal clutch similar to the one described has 6 rotating bobs each of mass 150g and a radius of 100mm. The spring strengths are 5kN/m Calculate the angular velocity (ω) as the bobs engage with the clutch cylinder. (The bob clearance is 0.01m and the spring length is 0.1m) springs cylinder

  24. 0.01m 0.1m Rotating bobs r = 0.1 + 0.01 = 0.11 The springs needs to extend by 0.01 m Force required = 5kN/m x 0.01 = 50N mrω2 = 50N ω2 50/(m x r) ω2 = 50/(0.15 x 0.11) springs cylinder

  25. ω2 = 50/(0.15 x 0.11) = 3030.3 ω = 55rad/sec ω = 2πf f = ω/2π =55/2π = 8.75 rps 8.75 x 60 rpm 525 rpm

  26. Calculate the force on the cylinder at a velocity of 1200rpm. • ω = 2πf • =7539.8 rad/min • =125.7 rad/sec • F = mrω2 • 0.15 x 0.11 x 125.72 • 261N • Net force = Centripetal force - spring force • = 261 – 50 = 211N

  27. Explain how the coefficient of friction between the bob and the clutch face will govern the power transmitted

  28. As ω increases the net force between the bobs and the cylinder but will always be 50N less than the centripetal force after engagement. Power transmitted will be affected by the force between the bobs and the cylinder ( F = μ x R(reaction force)) And the velocity of the clutch. As the velocity doubles the power will multiply by 4 until slip occurs

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