centripetal force

SOLID MECHANICS DYNAMICS TUTORIAL – CENT RIPETAL FORCE This work covers elements of the syllabus for the Engineering Council Exam D225 – Dynamics of Mechan ical Systems C103 Engineering Science. This tutorial examines the relations hip between inertia and acceleration. On completion of this tutori al you should be able to • Explain and define centripetal and centrifugal acceleration. • Explain and define centripetal and centrifugal force. • Solve problems involving centripet al and centrifugal force. • Derive formulae for th e stress and strain induced in rotating bodies. • Solve problems involving stress a nd strain in rotating bodies. • Analyse problems involvi ng vehicles skidding and overturning on bends. It is assumed that the student is already familiar with the following concepts. • Newton’s laws of Motion. • Coulomb’s laws of friction. • Stiffness of a spring. • The laws relating angular displace ment, velocity and acceleration. • The laws relating angular and linear motion. • Basic vector theory. • Basic stress and strain relationships. All the above may be found in the pre-requisite tutorials. 1. CENTRIPETAL AFFECTS 1.1 ACCELERATION AND FORCE Centripetal acceleration occurs with all rotating bodies. Consider a point P rotating about a centre O with constant angular velocity ω (fig. 1). Figure 1 The radius is the length of the line O-P. The tangential velocity of P is v = ωR. This velocity is constant in magnitude but is continually changing direction. Let’s remind ourselves of the definition of acceleration. t∆va takentimey in velocit changea == Velocity is a vector quantity and a change in direction alone is sufficient to produce a change. It follows that a point travelling in a circle is continuously changing its direction, velocity and hence has acceleration. Next let’s remind ourselves of Newton’s second Law of Motion which in its simplest form states Force = Mass x acceleration. It follows that anything with mass travelling in a circle must require a force to produce the acceleration just described. The force required to make a body travel in a circular path is called CENTRIPETAL FORCE and it always pulls towards the centre of rotation. You can easily demonstrate this for yourself by whirli ng a small mass around on a piece of string. Let’s remind ourselves of Newton’s Third Law. Every force has an equal and opposite reaction. The opposite and equal force of cent ripetal is the CENTRIFUGAL FORCE. The string is in tension and this means it pulls in both directions. The force pulling the ball towards the middle is the centripetal force and the force pulling on your finger is the centrifugal force. The derivation of the formula for centripetal force and acceleration is done by considering the velocity as a vector. Figure 2 Consider the velocity vector before an d after point P has revolved a small angle δθ. The magnitude of v 1 and v 2 are equal so let’s denote it simply as v. The direction changes over a small period of time δt by δθ radians. We may deduce the change by using the vector addition rule. The first vector + the change = Final vector. Figure 3 The rule is v 1 + δv = v 2. This is illustrated below. Figure 4 δv is almost the length of an arc of radius v. If the angle is small, this becomes truer. The length of an arc is radius x angle so it follows that δv = v δθ This change takes place in a corresponding small time δt so the rate of change of velocity is δtθ v δtδv δ= In the limit as δt → dt, δv/δt → dtdv δtδv dt,δt → → dtdv is the acceleration. vωdtdθ v dtdva = = = angle of change of rate dtdθω = = Since v = ωR then substitute for v and a = ω2R and this is the centripetal acceleration. Centripetal acceleration = ω2R Since ω = v/R then substitute for ω and a = v2/R Centripetal acceleration = v 2/R If we examine the vector diagram, we see that as δθ becomes smaller and smaller, so the direction of δv becomes radial and inwards. The acceleration is in the direction of the change in velocity and so centripetal acceleration is radial and inwards. If point P has a mass M, then the force required to accelerate this mass radial inwards is found from Newton's 2nd Law. Centripetal force = M ω2 R or in terms of velocity v Centripetal force = M v 2/R Centrifugal force is the reaction force and acts radial outwards. WORKED EXAMPLE No.1 Calculate the centripetal acceleration and force acting on an aeroplane of mass 1500 kg turning on a circle 400 m radius at a velocity of 300 m/s. SOLUTION Centripetal acceleration = v 2/R = 300 2/400 = 225 m/s 2. Centripetal force = mass x acceleration = 1500 x 225 = 337.5 kN WORKED EXAMPLE No.2 Calculate the centripetal force acting on a small mass of 0.5 kg rotating at 1500 rev/minute on a radius of 300 mm. SOLUTION ω= 2πN/60 = 2 x π x 1500/60 = 157 rad/s Cent. acc. = ω2R = (157) 2 x 0.3 = 7395 m/s 2. Cent. force = Mass x acc. = 0.5 x 7395 = 3697 N WORKED EXAMPLE No.3 A centrifugal clutch is shown in the diagram. The clutch must transmit a torque of 18 Nm at a speed of 142 rev/min. The coefficient of friction ‘ µ’ between the drum and the friction lining is 0.3. The radius to the centre of gravity of each sliding head is 0.21 m and the inside radius of the drum is 0.25 m. Calculate the required mass of the sliding heads. Figure No.5 SOLUTION Torque = 18 Nm, radius = 0.25 m T = Friction force x radius Friction force = 18/0.25 = 72 N This is divided between two friction pads so each must produce 72/2 = 36 N each. From the law of friction, Friction force = µ x normal force Normal force = 36/0.3 = 120 N The normal force acts in a radial direction and must be equal to the centripetal force. Centripetal force = M ω2r Equating forces we have M ω2r = 120 Speed = 142 rev/min or 142/60 rev/s Radius to centre of gravity = 0.21 m Angular velocity ω = 2π x speed = 2 π x 142/60 = 14.87 rad/s M x 14.872x 0.21 =120 M = 120/( 14.872x 0.21) = 2.58 kg for each sliding head. SELF ASSESSMENT EXERCISE No.1 1. A centrifugal clutch is similar to that shown in figure 5. The clutch must transmit a torque of 25 Nm. The coefficient of friction ‘ µ’ between the drum and the friction lining is 0.4. Each sliding head ha s a mass of 0.4 kg acting at a radius of 0.15 m. The inside radius of the drum is 0.18 m. Calculate the minimum speed required. (Answer 53.78 rad/s or 513.7 rev/min) 2. A rotating arm has a sliding mass of 2 kg that normally rests in the position shown with the spring uncompressed. The mass is flung outwards as it revolves. Calculate the stiffness of the spring such that the mass compresses it by 20 mm when the arm revolves at 500 rev/min. (Answer 38.4 N/mm) Figure 6 2 STRESS and STRAIN in ROTATING BODIES 2.1 STRESS Wheels experience stresses in them because the centripetal reaction (centrifugal force) tends to stretch the material along a radius. This can cause the wheel to disintegrate if it runs to fast. The effect might be cat astrophic in a grinding wheel or flywheel. Even if the wheel does not disintegrate, the stress will cause it to expand. The blades on a turbine or compressor rotor will stretch slightly under this strain and might touch the casing. Consider a bar of uniform cross section A rotating about its centre as shown. Figure 7 The radius to the tip is R. Consider a small length δr at radius r. The mass of this element is δm = ρAδr. The centrifugal force produced by this mass is δF. δF = δm ω2r = ρAδrω2r In the limit as δr → dr this becomes dF = ρAdrω2r The centrifugal force acting on this section due to the mass between r and R is then found by integration. [] 2r RρAω dr r ρAω F2 2 2R r2 −= = ∫ The tensile stress induced at this secti on (at radius r) is the force per unit area. [ ] 2Ar RρAωAF2 2 2 −= =σ [ ] 2r Rρω2 2 2 −=σ 2.2 STRAIN Consider the same element (figure 7). Suppose the element δr increases in length by δx. In the limit as δr → dr, the strain in this short length is then []2 22 r R2Eρω Eσ drdxε − == = []2 22 r R2Eρω− =ε The extension of the short length is [] []        − −   − = − = − =− == ∫ 3rrR3RR2Eρωx3rrR2Eρωdrr R2Eρωxagain. g integratinby found is extension theR, r to from length For thedrr R2Eρωσ dx 3 23 32R r3 22 R r2 222 22    + − =3rrR32R 2Eρωx3 23 2 WORKED EXAMPLE No. 4 A bar 0.5 m long with a uniform section is revolved about its centre. The density of the material is 7 830 kg/m 3. The tensile stress in the material must not exceed 600 MPa. Calculate the speed of rotation that produces this stress. Go on to calculate the extension of the bar. The elastic modulus is 200 GPa. SOLUTION [ ] 2r Rρω 600x10σ2 2 2 6 −= = The maximum stress will be at the middle with r = 0 and R = 0.25 m [ ]rad/s 1566ω 10 x 2.452ω 20 0.257830ω 10 x 600σ6 22 2 2 6= =−= = Converting to rev/min N = ω x 60/2 π = 14 953 rev/min The extension of one half of the bar from r = 0 to R = 0.25 m is 0.5m m or 0.005x0030.25 x 2 10 x 200 x 21566 x 7830 3rrR32R 2Eρωx3 92 3 23 2 =   +− =   + − = SELF ASSESSMENT EXERCISE No.2 1a. The blade of a turbine rotor has a unifor m cross section. The root of the blade is attached to a hub of radius r. The tip of the blade has a radius R. Show that the stress at the root of the blade due to centrifugal force is given by [ ] 2r Rρω2 2 2 −=σ 1b. The hub radius is 200 mm and the tip radius is 320 mm. The density of the blade material is 8 200 kg/m 3. Calculate the stress at the root when the rotor turns at 7 500 rev/minute. (157.8 MPa) 1c. Derive the formula for the extension of the blade tip x given below.    + − =3rrR32R 2Eρωx3 23 2 where E is the elastic modulus. 1d. Calculate the extension of the tip given that E = 206 GPa. (0.0495 mm) 3. APPLICATION TO VEHICLES 3.1 HORIZONTAL SURFACE When a vehicle travels around a bend, it is subject to centrifugal force. This force always acts in a radial direction. If the bend is in a horizontal plane then the force always acts horizontally through the centre of gravity. The weight of the vehicle is a force that always acts vertically down through the centre of gravity. The diagram shows these two forces. Figure 8 The centrifugal force tends to make the vehicle slide outwards. This is opposed by friction on the wheels. If the wheels were about to slide, the friction force would be µW where µ is the coefficient of friction between the wheel and the road. Figure 9 The tendency would be for the vehicle to overturn. If we consider the turning moments involved, we may solve the velocity which makes it overturn. If the vehicle mass is M its weight is Mg. The radius of the bend is R. Figure 10 Consider the turning moments about point O The moment of force due to the weight is W x d or M g d The moment due to the centrifugal force is C.F. x h Remember the formula for centrifugal force is C.F. = (Mv 2/R) The moment due to the centrifugal force is C.F. x h = (Mv 2/R) x h When the vehicle is about to overturn the moments are equal and opposite. Equating the moments about point O we get M = (Mv 2/R) x h = Mg d Rearrange to make v the subject. v = (g d R/h) ½ If the vehicle was about to slide sideways without overturning then the centrifugal force would be equal and opposite to the friction force. In this case µW = Mv 2/R µMg = Mv 2/R v = ( µgR)½ Comparing the two equations it is apparent that it skids if µ<d/h and overturn if µ>d/h WORKED EXAMPLE No.5 A wheeled vehicle travels around a circular track of radius 50 m. The wheels are 2 m apart (sideways) and the centre of gravity is 0.8 m above the ground. The coefficient of friction is 0.4. Determine whether it overturns or slides sideways and determine the velocity at which it occurs SOLUTION Overturning v = (g d R/h) ½ = (9.81 x 1 x 50/0.8) ½ = 24.76 m/s Sliding sideways v = ( µgR)½ = (0.4 x 9.81 x 50) ½ = 14 m/s It follows that it will slide sideways when the velocity reaches 14 m/s. 3.2 BANKED SURFACE The road surface is flat and inclined at θ degrees to the horizontal but the bend is still in the horizontal plane. The analysis of the problem is helped if we consider the vehicle simply as block on an inclined plane as shown. The C.F. still acts horizontally and the weight vertically. The essential distances required for moments about point O are the vertical and horizontal distances from O and these are h' and d' as shown. Figure 11 When the vehicle is just on the point of overturning the moments about the corner are equal and opposite as before. By applying trigonometry to the problem you should be able to show that h' = (h – d tan θ) cos θ d' = (h tan θ + d) cos θ Figure 12 Equating moments about the corner we have Mv 2 h' /R = Mg d' v 2 h' = (h – d tan θ) cos θ /R = g (h tan θ + d) cos θ Make v the subject.       −+ = θθ tan 1tan 2 hdhd gR v This is the velocity at which the vehicle overturns. If the vehicle slides without overturning, then the total force parallel to the road surface must be equal to the friction force. Remember the friction force is µN where N is the total force acting normal to the road surface. Resolve all forces parallel and perpendicular to the road. The parallel forces are F cos θ and W sin θ as shown. The normal forces are F sin θ and W cos θ as shown. The friction force opposing sliding is µN Figure 13 The forces acting parallel to the surface must be equal and opposite when sliding is about to occur. Balancing all three forces we have µN + Wsin θ = F cos θ The total normal force N is the sum of the two normal forces. N = W c o s θ + F sin θ Substituting {} {} {} {} {}µtanθ 1tanθ µRg vµ tanθ µtanθ 1Rgvµ tanθ 1 µtanθRgvRgvtanθ tanθRgvµµcosθRvgsinθ sinθRvµ µgcosθcosθRvM Mgsinθ sinθRvµM µMgcosθFcosθ Wsinθ µFsinθ µWcosθFcosθ Wsinθ Fsinθ Wcosθ µ 2222 22 22 2 −+=+ = −− −=−= + += + += + += + += + + This gives the velocity at which the vehicle slides. WORKED EXAMPLE No.6 A motor vehicle travels around a banked circular track of radius 80 m. The track is banked at 15 o to the horizontal. The coefficient of friction between the wheels and the road is 0.5. The wheel base is 2.4 m wide and the centre of gravity is 0.5 m from the surface measured normal to it. Determine the speed at which it overturns or skids. SOLUTION 1. OVERTURNING () ()      −+ =       −+ = 15tan5.02.1115tan5.02.1 8081.9 tan 1tan 2X hdhd gR v θθ Hence v = 76.6 m/s 2. SKIDDING {} {}{ } {} 15tan5.0115tan5.081.980tan 1tan2 −+=−+= x x Rg vθ µθ µ Hence v = 26.3 m/s The vehicle will skid before it overturns. SELF ASSESSMENT EXERCISE No.3 1. A motor vehicle travels around a banked circular track of radius 110 m. The track is banked at 8 o to the horizontal. The coefficient of friction between the wheels and the road is 0.4. The wheel base is 3 m wide and the centre of gravity is 0.6 m from the surface measured normal to it. Determine the speed at which it overturns or skids. ( 66.3 m and 24.9 m ) 2. A vehicle has a wheel base of 2.1 m and the centre of gravity is 1.1 m from the bottom. Calculate the radius of the smallest bend it can negotiate at 120 km/h. The coefficient of friction between the wheels and the track is 0.3. (For skidding R = 377.5 m and for overturning R = 118.7 m hence the answer is 377.5 m) 2. Repeat question 3 given that the bend is banked at 10 o. (For skidding R = 225.2 m and for overturning R = 83.3 m hence the answer is 225.2 m)