MMB222 D Dynamics i Kinetics of rigid body: work & energy gy Work & energy equation for rigid body Kinetic energy of rigid body Impulse & momentum of rigid body Prof. Jacek Uziak Rigid Body It is an idealized model of an object that does not deform or change shape. h h Distance between any two points on the body is constant. ng w n w n Angle between any two lines on or in the body is constant. Because of the latter fact, whatever angular velocity a given line has at an instant, that angular velocity is also that of all the other lines on/in the body body. In other words, a body only has one velocity (which would have one parameter in planar motion, three in 3D motion). 2 Laws of motion for a rigid body are known as Euler’s laws. Euler gave two laws for the motion of a rigid body. Euler’s laws (generalized 2nd Newton’s Law) Equations of motion for system of particles i =a ∑ Fi = Ma G & = HG n Describes how the forces control the “translational” motion of the rigid body Angular momentum of a system of particles: ∑ MG (G - centre of mass) Describes how the change of angular momentum of the rigid body is controlled by the moment of forces and couples applied on the body. 3 WORK & ENERGY The principle of work and energy for a rigid body is expressed in the form: T1 + U1 2= T2 where T1 and T2 represent the initial and final values of the p kinetic energy of the rigid body and U1 2 the work of the external forces acting on the rigid body. The work of a force F applied at a point A is s2 U1 2 = ∫ ( (F cos α) ds ) s1 where F is the magnitude of the force, α - angle it forms with g g the direction of motion of A, and s the variable of integration measuring the distance traveled by A along its path. 4 The work of a couple of moment M applied to a rigid body during r t ti n durin a rotation in q of the ri id b d : f rigid body: θ2 U1 2 = ∫ M ds θ1 1 2 The kinetic energy of a rigid body in plane motion: T= G ω v 1 2 mv + Iω2 2 where v is the velocity of the mass center G of the body, w the angular velocity of veloc ty the body, and I its moment of inertia about an axis through G perpendicular to the plane of reference reference. 5 T= G ω v 1 2 mv 2 + 1 Iω2 2 The ki Th kinetic energy of a rigid b d i i f i id body in plane motion may be separated into two parts: (1) the kinetic energy ½mv 2 associated with the gy motion of the mass center G of the body, and (2) the kinetic energy ½Iω2 associated with the rotation of the body about G. t ti f th b d b t 6 Kinetic energy of a rigid body Note that in the general plane motion of each body, it is necessary to use those y y parameters which refer to the translation of the mass centre G and the rotation of the body b t the i b d about th axis passing th i through G. h Kinetic energy of a body in pure translation f d l or pure rotation can now be considered as a special case of general plane motion 7 G v The translating rigid body has a mass m and g g y all of its particles have a common velocity v. Body does not have any angular velocity and posses only kinetic energy of translation 1 2 mv 2 For a rigid body rotating about a fixed axis through O with an angular velocity ω: O 1 2 2 I Oω ω where IO is the moment of inertia of the body about the fixed axis. 8 Kinetic Energy: General Plane Motion Kinetic energy of a body in general plane motion may also b expressed i t ti l be d in terms of f rotational velocity about the instantaneous centre C of zero velocity velocity. In this approach, the general plane motion is treated as pure rotation about the instantaneous centre C and the kinetic energy can be expressed as gy p 1 ICω 2 2 9 When a rigid body, or a system of rigid bodies, moves under the action of conservative forces, the principle of forces work and energy may be expressed in the form which is referred to as the principle of conservation of energy. This principle may be used to solve problems involving conservative forces such as the force of gravity or the force exerted by a spring. The concept of power is extended to a rotating body subjected to a couple T1 + V1 = T2 + V2 dU M dθ Power = = = Mω dt dt where M is the magnitude of the couple and w is the angular velocity of the body. 10 Example A uniform disk of mass m and radius R rolls without slip on a horizontal surface. Determine the kinetic energy of the disk when its centre O has a velocity vo as shown opposite 11 vo O The disk is in general plane motion and its kinetic energy can be calculated as as, K.E. = Linear K.E. (at centre of mass) + Rotational K.E. (about R t ti l K E ( b t centre of mass) t f ) or 1 1 2 K .E. = mv o + I o ω 2 2 2 vo where ω = R or 1 I o = mR 2 2 12 Hence, Hence the kinetic energy: 2 vo 1 1 1 1 1 3 2 2 2 2 2 K .E. = mv o + ⋅ mR = mv o + mv o = mv o 2 2 2 4 4 R2 2 13 The alternative method is to consider the motion as pure rotation about the instantaneous center of zero velocity i t t t f l it which, in this case, is the point of contact between the disk and the surface In this surface. approach the kinetic energy is given by the expression, 1 K .E . = I C ω 2 2 where IC is the moment of inertia about point C, which can be calculated using the parallel axis theorem as 14 IC 1 3 2 2 = I o + mR = mR + mR = mR 2 2 2 2 Angular velocity is still Substituting, Substituting vo ω= R 1 1 3 3 2 2 K .E. = I C ω = ⋅ mR = mR 2 2 2 2 R2 4 15 2 vo Example The velocity of the 8-kg y f g cylinder is 0.3 m/s at a certain instant. What is its speed v after dropping an additional 1.5 m? The mass of the grooved drum is 12 kg, its centroidal radius of gyration is k = 210 mm, and the radius of its g groove is ri = 200 mm. The frictional moment at O is constant 3 Nm. 16 Answer: 3.01 m/s Example p The 15-kg wheel is released from rest and rolls on its hubs without slipping. Calculate the velocity v of the centre O of the wheel after it has moved a distance x = 3 m down the incline. th in lin The radius of gyration of the wheel about O is 125 mm. Answer: 0.85 m/s 17 Impulse & Momentum: Rigid Body p g y The principle of impulse and momentum derived for f a system of particles can be applied t th t f ti l b li d to the motion of a rigid body. Syst Momenta1 + Syst Ext Imp1 (∆m)v mv P Iω 18 2= Syst Momenta2 Syst Momenta1 + Syst Ext Imp1 2= Syst Momenta2 For a rigid body the system of the momenta of the particles forming the body is equivalent to a vector mv attached to the mass c nt r m ss center G of th b d and a c upl Iω. f the body nd couple The vector mv is associated with translation of the body with G and represents the linear momentum of the body, while the couple Iω corresponds to the rotation of the body about G and represents the angular momentum of the body about an axis through G. (∆m)v mvG P Iω 19 The principle of impulse and momentum can be expressed graphically b d h ll by drawing three diagrams representing h d respectively the system of initial momenta of the body, y f f y, the impulses of the external forces acting on it, and the system of the final momenta of the body. Summing and equating respectively the x components, the y components, and the moments about any given point of the vectors shown in the figure, we obtain three equations of motion which may be solved for the desired i f i hi h b l df h d i d unknowns. ∫ Fdt mv2 y y y mv1 G Iω1 O x O x O G Iω2 x 20 1.The 1 The law of conservation of momentum, which is similar to that for a particle and which could be formulated in the following way: Provided no forces external to the system of bodies are acting on the body, the total momentum of the system remains constant. In th I other words, m G is constant where, m ds mv st t h is the total mass of the system, and vG is the l it f th m ss nt G. th velocity of the mass centre G 21 2.The 2 The law of conservation of angular momentum, which could be expressed as follows: Provided no external moments act on a system, the sum of the angular y momenta of the bodies in the system remains constant 22 The Th second l d law i useful i mechanical is f l in h i l analysis of situations such as o E Engagement of rotating parts when a t f t ti t h clutch is released. o It also describes how the angular velocity of a body changes if its moment of inertia is changed by some internal means (e.g. a gyrating dancer spins faster by pulling p y p g her/his arms inwards). 23 Example p 500 m/s Answer: 0.703 rad/s The 28 g bullet has a horizontal velocity of 500 m/s as it strikes the 25 kg / i ik h k compound pendulum, which has a radius of gyration k = 925 mm. If the distance h = 1075 mm, calculate the angular velocity of the pendulum y p with its embedded bullet immediately after the impact. 24 Example The 30 g bullet has a horizontal velocity of 500 m/s as it strikes g the 10 kg slender bar OA, which is suspended from point O and is y initially at rest. It takes 0.001 s for the bullet to embed itself in the bar. Calculate the time average of the horizontal force Ox exerted by the pin on the bar at O during the interaction between the bullet and the bar. 25 Answer: 3750 N Example The constant 40 N force is applied to f ppli d t the 36 kg stepped cylinder as shown shown. The centroidal radius of gyration of the cylinder is k = 200 mm, and it rolls on the incline without pp g slipping. If the cylinder is at rest when the force is first applied, determine its angular velocity 8 s later. Answer: 24.18 rad/s 26
Comments
Report "Kinetics Body Work&Energy Impulse&Momentum"