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Rigid Body Particle Object without extent Point in space Solid body with small dimensions Rigid Body An object which does not change its shape Considered as an aggregation of particles Distance between two points is a constant Suffer negligible deformation when subjected to external forces Motion made up of translation and rotation Motion of a Rigid Body Translational – Every particle has the same instantaneous velocity Rotational – Every particle has a common axis of rotation Centre of Mass Centre of mass of a system of discrete particles: n r m i ri i 1 n mi i 1 , Centre of mass for a body of continuous distribution: r M 0 rdm M , It is the point as if all its mass is concentrated there Located at the point of symmetry Conditions of Equilibrium For particle – Resultant force = 0 For rigid body – – Resultant force = 0 and Total moments = 0 Toppling An object will not topple over if its centre of mass lies vertically over some point within the area of the base Figure Stability Stable Equilibrium – – – The body tends to return to its original equilibrium position after being slightly displaced Disturbance gives greater gravitational potential energy Figure Unstable Equilibrium The body does not tend to return to its original position after a small displacement Disturbance reduces the gravitational potential energy Neutral Equilibrium The body remains in its new position after being displaced No change in gravitational potential energy Rotational Motion about an Axis The farther is the point from the axis, the greater is the speed of rotation (v r) Angular speed, , is the same for all particles Rotational K.E. ERot 1 2 mv 2 1 m(r ) 2 2 1 2 mr 2 2 E Rot 1 2 I 2 The term I mr 2 is known as the moment of inertia Moment of Inertia (1) Unit: kg m2 A measure of the reluctance of the body to its rotational motion Depends on the mass, shape and size of the body. Depends on the choice of axis For a continuous distribution of matter: I r 2 dm Experimental Demonstration of the Energy Stored in a Rotating Object Moment of Inertia (2) A body composed of discrete point masses I mi ri , 2 i A body composed of a continuous distribution of masses M I r 2dm , 0 Moment of Inertia (3) A body composed of several components: – Algebraic sum of the moment of inertia of all its components A scalar quantity Depends on – – – mass the way the mass is distributed the axis of rotation Radius of Gyration If the moment of inertia I = Mk2, where M is the total mass of the body, then k is called the radius of gyration about the axis Moment of Inertia of Common Bodies (1) Thin uniform rod of mass m and length l – M.I. about an axis through its centre perpendicular to its length 1 I ml 2 12 – M.I. about an axis through one end perpendicular to its length 1 2 I ml 3 Moment of Inertia of Common Bodies (2) Uniform rectangular laminar of mass m, breadth a and length b – About an axis through its centre parallel to its breadth 1 I mb 2 12 – About an axis through its centre parallel to its length 1 I ma 2 12 Moment of Inertia of Common Bodies (3) – About an axis through its centre perpendicular to its plane I 1 m (a 2 b 2 ) 12 Moment of Inertia of Common Bodies (4) Uniform circular ring of mass m and radius R – About an axis through its centre perpendicular to its plane I mR2 Moment of Inertia of Common Bodies (5) Uniform circular disc of mass m and radius R – About an axis through its centre perpendicular to its plane 1 I mR 2 2 – The same expression can be applied to a cylinder of mass m and radius R Moment of Inertia of Common Bodies (6) Uniform solid sphere of mass m and radius R 2 I mR 2 5 Theorems on Moment of Inertia (1) Parallel Axes Theorem I I G Md 2 Theorems on Moment of Inertia (2) Perpendicular Axes Theorem I Ix Iy Torque (1) A measure of the moment of a force acting on a rigid body – T = F·r Also known as a couple A vector quantity: direction given by the right hand corkscrew rule Depends on – – Magnitude of force Axis of rotation Torque (2) Work done by a torque – – Constant torque: W = T Variable torque: W Td 0 Kinetic Energies of a rigid body (1) Translational K.E. KE tran 1 Mv 2 2 Rotational K.E. – – It is the sum of the k.e. of all particles comprising the body For a particle of mass m rotating with angular velocity : KE rot 1 m ( r )2 2 Kinetic Energies of a rigid body (2) = 1 mr 2 2 1 2 I 2 If a body of mass M and moment of inertia IG about the centre of mass possesses both translational and rotational k.e., then KE 1 1 Mv 2 I G 2 2 2 Moment of inertia of a flywheel (1) Determination of I of a flywheel – – – – – – Mount a flywheel Make a chalk mark Measure the axle diameter by using slide calipers Hang some weights to the axle through a cord Wind up the weights to a height h above the ground Release the weights and start a stop watch at the same time Moment of inertia of a flywheel (2) – Measure: – the number of revolutions n of the flywheel before the weights reach the floor – the number of revolutions N of the flywheel after the weights have reached the floor and before the flywheel comes to rest Moment of inertia of a flywheel (3) Theory potential energy lost by the falling weight kinetic energy kinetic energy work done gained by the gained by the against friction falling weight flywheel at the axle 1 2 1 2 mgh mv I nf 2 2 1 2 2 1 2 mr I nf 2 2 ….. (1) where f = work done against friction per revolution Moment of inertia of a flywheel (4) – When the flywheel comes to rest: Loss in k.e. = work done against friction 1 2 I Nf ….. (2) 2 1 2 1 2 n (2) In (1) mgh mv I [1 ] 2 2 N 1 2 I n v [m 2 (1 )] …. (3) 2 r N Moment of inertia of a flywheel (5) – The hanging weights take time t to fall from rest through a vertical height h Total vertical displacement = average vertical velocity time v0 h t 2 2h v t Knowing v, I can be calculated from (3) Applications of flywheels In motor vehicle engines In toy cars Angular momentum The angular momentum of a particle rotating about an axis is the moment of its linear momentum about that axis. A ( mr 2 ) mr 2 I Conservation of angular momentum (1) The angular momentum about an axis of a given rotating body or system of bodies is constant, if the net torque on the object is zero d d As T I ( I ) dt dt – If T = 0, I = constant Conservation of angular momentum (2) Examples – High diver jumping from a jumping board Conservation of angular momentum (3) – Dancer on skates – Mass dropped on to a rotating turntable Experimental verification using a bicycle wheel Conservation of angular momentum (4) Application – Determination of the moment of inertia of a turntable Set the turntable rotating with an angular velocity Drop a small mass to the platform, changes to a lower value ’ If there is no frictional couple, the angular momentum is conserved, I = I’ ’ = (I + mr2) ’ , ’ can be determined by measuring the time taken for the table to make a given number of revolutions and I can then be solved Rotational motion about a fixed axis (1) T=I d’Alembert’s Principle – The rate of change of angular momentum of a rigid body rotating about a fixed axis equals the moment about that axis of the external forces acting on the body d ( I ) ( Fp) dt Rotational motion about a fixed axis (2) d I T dt i.e. I = T Compound pendulum (1) Applying the d’Alembert’s Principle to the rigid body d 2 I s 2 Mgh sin . dt But I s M (k 2 h2 ) where k is the radius of gyration about its centre of mass G Compound pendulum (2) 2 d 2 2 M ( k h ) 2 Mgh sin , dt For small oscillations d 2 gh 2 2 2 dt k h SHM with period k 2 h2 T 2 hg Compound pendulum (3) It has the same period of oscillation as the simple pendulum of length k 2 h2 l h l is called the length of the equivalent simple pendulum Compound pendulum (4) The point O, where OS passes through G and has the length of the equivalent simple pendulum, is called the centre of oscillation S and O are conjugate to each other The period T is a minimum when h = k (see expt. results) Torsional pendulum c I where c = torsional constant I = moment of inertia SHM with period I T 2 c Rolling objects (1) d 2r v r T T Rolling objects (2) P has two components: – – v parallel to the ground r(=v) perpendicular to the radius OP If P coincides with Q, the two velocity components are oppositely directed. Thus Q is instantaneously at rest Rolling objects (3) Hence, for pure rolling, there is no work done against friction at the point of contact Kinetic energy of a rolling object Total kinetic energy = translational K.E. + rotational K.E. 1 2 1 2 = mv I 2 2 Stable Equilibrium No toppling Compound pendulum