Dynamics of Rotation 1
Dynamic of Rotation Angular velocity and acceleration are denoted ω and α respectively and have units of rad/s and rad/s. Relationship between Linear and Angular Motions We can show that a = αr and v = ωr. For proof of this you are referred to the recommended text. Equations of Motion Using the known equations of motion for linear motion we can write those for angular as follows: v av = u+v v = u + at Linear s = ut + 1 at v = u + as Angular ω av = ω 1+ω ω = ω 1 + αt θ = ω 1 t + 1 αt ω = ω 1 + αθ Dynamics of Rotating Particles In the same way that a change in linear motion requires a force, a change in angular motion requires a torque. F ω r α A a Diagram shows a concentrated mass m attached to the end of a light arm OA. Provided there is no friction it will rotate freely when force F applied. O F = ma = mαr However, the moment of F about O is a torque T, where T = Fr. Therefore, T = mr α Where, mr is known as the second moment of mass or more commonly, moment of inertia. This is denoted I and has units of kg m.
Dynamics of a Rotating Mass Consider the body of mass m, accelerated about an axis O. δp A r a A O O To determine the torque required to produce this acceleration consider a small particle of mass δm, distance r from O. Force required at A = δp = δma A perpendicular to OA. Torque δt, required to accelerate A = δpr = δma Ar But a A = αr Therefore, δt = δmαr Total torque T = α δmr Or T = Iα As already stated I is the second moment of mass or moment of inertia. It is useful to imagine the whole mass of the body to be concentrated at a particular radius (k), such that the I of the concentrated mass is the same as that of the actual body. This is called the radius of gyration and denoted the afore mentioned k. It is defined as follows: I = mk For a solid disc k = r where r is the radius of the disc. For a hollow disc or cylinder k = r 1 +r where r 1 = inside radius and r = outside radius. Proof that k = r for a solid disc Consider an elemental disc of thickness δr at radius r of a solid disc thickness b and radius R δr r Mass of elemental disc is given by: δm = ρδv 3
Where elemental volume δv = πrbδr Therefore δm = ρπrbδr Now I = mr Hence δi = ρπr 3 bδr Integrating between the limits of 0 and R will give I Therefore I = πρb r 3 dr = πρb [ r4 0 ] 4 0 Or I = πρb R4 4 R ----------------------------------------------------- (1) Now the mass of the whole disc is m = ρπr b --------------() Substituting () into (1) and cancelling the gives I = m R Since I = mk we see that for a solid disc k = R R Note: k = R = 0.707R - The effective radius at which the mass rotates is clearly not the mid point but is closer to the edge. Tutorial Problem Show that the radius of gyration for a uniform rod of cross section A and length L rotating about one end is given by 0.577L. δx x Inertia Torque Problems that involve angular acceleration may be reduced to a statics type problem by the introduction of an imaginary inertia torque (this is analogous to what we did with problems involving linear motion). α Iα 4
Consider a Hoist ω T α Iα Tf F ma v a mg There are four possible cases: 1. Load rising. Load falling 3. Load being accelerated 4. Load being retarded In each case three equations may be written down: (i) (ii) (iii) Equation for balance of torque on drum Equation for balance of force on load Relationship between linear and angular acceleration of drum In each case, friction torque at bearing opposes motion, inertia torque of drum opposes angular acceleration. 5
Consider the case when an accelerated hoist raises a load. α r Iα T T f P P T = T f + Iα + Pr T f = Friction torque Iα = Inertia torque Pr = Torque due to tension P in rope P = mg + ma m a a = αr mg ma Tutorial Problems From Applied Mechanics by Hannah & Hillier Inertia Torque 1. A flywheel has a moment of inertia of 10 kgm. Calculate the angular acceleration of the wheel due to a torque of 8 Nm if the bearing friction is equivalent to a couple of 3 Nm. (0.5 rad/s ). A light shaft carries a disc 400 mm in diameter, 50 mm thick, of steel (density 7800 kg/m 3 ). Calculate its moment of inertia about an axis through the centre of the disc and perpendicular to the plane of the disc. What torque would be required to accelerate the disc from 60 to 10 rev/min in 1 second, neglecting friction? If a friction torque of 1.5 Nm acts, what braking torque would be required to bring the disc to rest from 60 rev/min in 1 second? (0.98 km/m ; 6.16N; 4.66 N) 3. The rotor of an electric motor of mass 00 kg has a radius of gyration of 150 mm. Calculate the torque required to accelerate it from rest to 1500 rev/min in 6 seconds. Frictional resistance may be ignored. (118 Nm) 4. A light shaft carries a turbine rotor of mass t and a radius of gyration of 600 mm. The rotor requires a uniform torque of 1. knm to accelerate it from rest to 6000 rev/min in 10 min. Find (a) the frictional couple, (b) the time taken to come to rest when steam is shut off. (446 Nm; 16.9 min) 6
Coupled Systems 5. A load of 8 t is to be raised with a uniform acceleration of 1.1 m/s by means of a light cable passing over a hoist drum of m diameter. The drum has a mass of 1 t and a radius of gyration of 750 mm. Find the torque required at the drum if friction is neglected. (87.8 kn) 6. A mine cage of mass 4 t is to be raised with an acceleration of 1.5 m/s using a hoist drum of 1.5 m diameter. The drum s mass is 750 kg and its radius of gyration is 600 mm. The effect of bearing friction is equivalent to a couple of 3 kn at the hoist drum. What is the torque required at the drum? If the driving torque ceases when the load is moving upwards at 6 m/s, find the deceleration of the load and how far it travels before coming to rest. (37.44 knm; 9.64 m/s ; 1.87 m) 7. A hoist drum has a mass of 360 kg and a radius of gyration of 600 mm. The drum diameter is 750 mm. A mass of 1 t hangs from a light cable wrapped round the drum and is allowed to fall freely. If the friction couple at the bearings is.7 knm, calculate the runaway speed of the load after falling for s from rest. (.69 m/s) 8. A flywheel of mass 100 kg and radius of gyration 0.5 m is keyed to a horizontal shaft 0.1 m in diameter carried in bearings. Around the shaft is wound a cord and from its free end is suspended a mass of 10 kg. Calculate the resultant angular acceleration of the flywheel. m = 100 kg k = 0.5 m 0.05 m m = 10 kg (078 rad/s ) 7
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