3 One full circle = 2𝜋 radians i.e. 360o = 2𝜋 radians Angles can be measured in RADIANSThe angle in radians is s/r which is the length of the arc divided by the radiusOne 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 𝝅 𝑻 ( T is the time period of the rotation1/T = the frequency of rotation.f = 1/T𝝎 = 2𝜋fLinear 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/ror m𝝎2r
7 m = mass of turning object r = radius of the turning circle centripetal force= mv2/rm = mass of turning objectv = velocityr = radius of the turning circle
8 Example situationA 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 on vehicle = mv2/rMass = 1200kgv = 30m/sr = 110mCentripetal force on vehicle= (1200 x 302) /110=9818N
10 The vehicle is on the limit of adhesion at this velocity 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=11772Nμ = Force/weight= 9818/117720.83
12 The corner now becomes banked at an angle of 12o to the horizontal 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/rF (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 12o bankingIs 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)=V = √1143.833.8m/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 of94.2 rad/s and calculate the centripetal force on the crankshaft bearings due to this out ofbalance load.
16 ω = angular velocity in radians/second 360o (full circle)= 2π radians ω = 2π/t (full circle)1/t = frequency (f)ω = 2πfIf the engine is running at 900rpmω = 900 x 2π/60= 94.2 rad/sec
17 v = ωrF = mv2/r = mω2rIn the question m = 200g =0.2kgr = half of stroke =30 mmF = 0.2 x x .03= 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 m2 =(0.2 x 0.03)/(2 x 0.02) = 0.15 kg m1 = mass of connecting rod and piston assembly, 0.2kgr1 = half stroke, 0.03mr2 = eccentricity of counter weights, 0.02mm2 =(0.2 x 0.03)/(2 x 0.02)= 0.15 kg
22 When the angular velocity of the shaft increases the centripetal force springsRotating bobscylinderWhen the angular velocity of the shaft increases the centripetal forceIncreases 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.
23 Rotating bobs springs cylinder 0.01m0.1mA 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/mCalculate 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)
24 The springs needs to extend by 0.01 m Rotating bobscylinder0.1m0.01mr = = 0.11The springs needs to extend by 0.01 mForce required = 5kN/m x 0.01= 50Nmrω2 = 50Nω2 50/(m x r)ω2 = 50/(0.15 x 0.11)
25 ω2 = 50/(0.15 x 0.11)=ω = 55rad/secω = 2πff = ω/2π=55/2π= 8.75 rps8.75 x 60 rpm525 rpm
26 Net force = Centripetal force - spring force Calculate the force on the cylinder at a velocity of 1200rpm.ω = 2πf= rad/min=125.7 rad/secF = mrω20.15 x 0.11 x261NNet 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 the velocity doubles the power will multiply by 4 until slip occurs 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