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Transcript
Advanced Higher Physics Unit 1 Rotational Dynamics Using moments The spanner exert a moment or turning effect on the nut. Turning point distance from force to turning point force If the moment is big enough, it will unscrew the nut. If not, there are two ways of increasing the moment. Using moments-increasing the moment 1. Increase the distance from the Force to the pivot-apply a force at the end of the spanner or use a longer spanner. Turning point distance from force to turning point force If the same force is applied over a greater distance, a larger moment is produced. Using moments-increasing the moment 2. Increasing the Force applied-push/pull harder or get someone stronger to do it! Turning point distance from force to turning point force If a greater force is applied over the same distance, a larger moment is produced. Moment of a force The moment of a force is given by: moment Fd F is the force applied, measured in Newton (N) d is the distance from turning point, measured in metres (m) The moment of a force is therefore measured in Newton metres (Nm). Torque F r A force is applied to the rim of a disc which can rotate around its centre axis. In this case the moment of a force is called the Torque. T Fr In data booklet F is the force applied, measured in Newton (N) r is the radius of the circle, measured in metres (m) T is the Torque associated with force F, measured in Newton metres (Nm). Inertia Inertia can be defined as resistance to change in motion. In linear motion, MASS is a measure of an object’s inertia (since a large mass needs a large force to produce an acceleration). In angular motion, we use MOMENT OF INERTIA. Moment of inertia The moment of inertia of an object is its resistance to change in angular motion. The moment of inertia depends on: •The mass of an object •How the mass is distributed about the axis of rotation. r Consider a mass m at a distance r from the axis of rotation. The moment of inertia can be calculated using: m I mr 2 In data booklet (additional relationships) m is the mass of the object, measured in kg r is the distance from the axis of rotation, measured in m I is the moment of inertia measured in kgm² Moment of inertia : simple situation All the mass can be considered to be at the same distance from the axis. Masses on a very light rod r r Wheel with heavy rim and very light spokes In these cases I=mr² where m is the total mass. Moment of inertia: rods rod about centre 1 I ml 2 12 With: l total length of the rod m total mass of the rod rod about end 1 2 I ml 3 The moment of inertia for a rod rotating about end is 4 times bigger than the moment of inertia for a rod rotating about centre as it is harder to do so. This is because there are now more particles at a greater distance from the axis of rotation. Moment of inertia: solid disc about centre r Where 1 2 I mr 2 In data booklet (additional relationships) m is the total mass of the disc r is the radius of the circle Moment of inertia: Sphere about centre 2 2 I mr 5 Where m is the total mass of the sphere r is the radius of the sphere (all the moment of inertia formulas can be found in the data booklet in Additional Relationships) Newton 2nd Law An unbalanced Torque will produce an angular acceleration. T I With In data booklet I, the moment of inertia in kgm² α, the angular acceleration in radsˉ² T, the Torque in Nm Angular momentum The angular momentum is defined as the moment of the linear momentum. r m w v For this particle of mass m: The linear momentum p = mv The angular momentum = the moment of p = mvr = mr²w, since v=rw. L mvr mr 2 w With In data booklet L angular momentum measured in kgm²sˉ¹. Angular momentum of a rigid body A rigid body is an object in which all the individual parts have the same angular velocity w. The angular momentum of this body is the total of the angular momenta of its particles: L mr 2 w w is constant as all particles must be rotating at the same rate. L w mr 2 L Iw In data booklet Conservation of angular momentum In the absence of external Torque, the total angular momentum before impact equal the total angular momentum after impact. I A w0 A I B w0 B I A wA I B wB Not in data booklet Example: lump of mud stuck to a bike wheel Before: After: Total angular momentum before = total angular momentum after Iwheelw0 = w (Iwheel + Imud) The moment of inertia of the wheel and the mud after impact is larger than the moment of inertia of the wheel before impact. Therefore the angular velocity of the wheel is smaller after impact. Example: pupil spinning on a chair Before: Pupil pushes arms out After: Pupil draws arms in Total angular momentum before = total angular momentum after Ioutw1 = Iinw2 Iin is smaller than Iout because the particles are closer to the axis of rotation. Therefore w2 is larger than w1. Example: mass dropped on a turntable Before: After: Axis of rotation Total angular momentum before = total angular momentum after Idiscw1 = w2(Idisc+Imass) The moment of inertia of the disc and the mass after impact is larger than the moment of inertia of the disc before impact. Therefore the angular velocity of the disc is smaller after impact. Rotational Kinetic energy Erot With 1 2 Iw 2 In data booklet I moment of inertia, measured in kgm² w angular velocity, measured in radsˉ¹ Erot rotational kinetic energy, measured in J Work done Ew T With Not in data booklet T, Torque measured in Nm θ, angular displacement in rad Ew, work done in J Conservation of energy In the absence of frictional torque: Ew Ek 1 2 1 2 T Iw Iw0 2 2 Not in data booklet Objects rolling down an inclined plane Ep = mgh h Erot = ½Iw² w Ek = ½mv² v Potential energy at top = total of linear and angular kinetic energy mgh = ½mv² + ½Iw² Comparison of linear and angular motion The equations of angular motion are similar to those of linear motion. linear motion angular motion dv F m ma dt dw T I I dt mAu A mBuB mAvA mBvB I A w0 A I B w0 B I A wA I B wB 1 2 1 Fs mv mu 2 2 2 1 2 1 2 T Iw Iw0 2 2 Example 1: 10 kg 6m 4rads 2 Frictional torque = 1000 Nm time = 5s Find: a) Total I b) Unbalanced T c) Driving T d) Driving F e) Final w f) Deceleration when driving T removed g) Time taken to move to rest. Example 2: r F Find: a) Driving Torque Solid disc Axle mass = 20 kg b) Moment of inertia of solid disc Radius = 3m c) Angular acceleration radius = 1cm Force applied to axle = 4000 N Frictional Torque = 30 Nm Cord length = 50 cm d) Angular displacement e) Angular velocity Example 3: A turntable of mass 5kg and radius 25 cm is rotating at 10 radsˉ¹. A metal ring of mass 2 kg and radius 10 cm is dropped over the centre of the turntable. a) Find the new angular velocity of the system. b) Using rotational energy determine whether this is an elastic or an inelastic situation.