Rigid Body Kinetics :: Force/Mass/Acc

Transcription

1 Rigid Body Kinetics :: Force/Mass/Acc General Equations of Motion G is the mass center of the body Action Dynamic Response 1

2 Rigid Body Kinetics :: Force/Mass/Acc Fixed Axis Rotation All points in body move in a circular rotation axis All lines of the body have the same ω and α Accln comp of mass center: a n = r ω 2 and a t = r α Two scalar comp of force eqns: ΣF n = m r ω 2 and ΣF t = m r α Moment of the rotn axis O: Using parallel axis theorem: 2

3 Rigid Body Kinetics :: Force/Mass/Acc General Plane Motion Combines translation and rotation Choice of moment eqn. :: mass centre G :: point P with known acc 3

4 Example (1) on general motion A metal hoop with radius r = 150 mm is released from rest on the 20 o incline. Coefficients of static and kinetic friction are given. Determine the angular accln α of the hoop and the time t for the hoop to move a distance of 3 m down the incline. Solution: Draw the FBD and the KD 4

5 Example (1) on general motion If the wheel slips when it rolls, a rα Check whether the hoop slips or not when it rolls Assume that the hoop rolls without slipping a = r α (Pure rolling) Eliminating F and substituting a = rα a = m/s 2 a is independent of both m and r Alternatively we may use: same relation Check for the assumption of no slipping: Calculate F and N, and compare F with its limiting value 5

6 Example (1) on general motion Check for the assumption of no slipping Maximum possible friction force is: Since the limiting value F max < F The assumption of pure rolling was incorrect. The hoop slips as it rolls a rα The friction force becomes the kinetic value Repeat the calculations with this F a = 2.25 m/s 2 α = 7.37 rad/s 2 (α dependent on r but not on m) at x = 3 m t = s 6

7 Example (2) on general motion 7

8 Example (2) on general motion Solution :: FBD and KD Horizontal component of acceleration of point D Assuming rolling without slipping, Acceleration of mass centre G, 8

9 Example (2) on general motion 9

10 Advantages of Work Energy Method These principles are especially useful in describing motion resulting from the cumulative effect of forces acting through distances. If the forces are conservative: velocity changes can be determined by analyzing the energy conditions only at the beginning and at the end of the motion interval. For finite displacements: no need to compute acceleration; leads directly to velocity changes as functions of forces, which do work. Involves only those forces, which do work, and thus, produces change in magnitudes of velocities. Simplifies calculations 10

11 Work done by force F: or ds is the magnitude of the vector displacement dr Work done by couple M: Work done by both forces (F) during translation part of motion cancel each other (opposite dirn) Net work done will be the work done by the couple: (due to rotational part of motion) During finite rotation, work done: 11

12 Kinetic Energy: Three classes of rigid body plane motion Translation All particles will have same velocity For entire body: T = ½ m i v 2 (both rectilinear and curvilinear) Fixed Axis Rotation For a particle m i : T i = ½ m i (r i ω) 2 For entire body: T = ½ ω 2 m i r i 2 = ½ ω 2 I o General Plane Motion Third summation: since T = Also, KE in terms of rotational the instantaneous center C of zero vel 12

13 Potential Energy and Work-Energy Equation Work-Energy relation for the motion of a general system of particles: (U 1-2 is work done by all external forces) (U 1-2 is work done by all external forces other than weight and spring forces, which are taken care of by means of potential energy rather than work) In case of interconnected system of rigid bodies, the work-energy equations include the effect of stored elastic energy in the connections. U 1-2 includes the negative work of internal friction forces. Work-energy method is most useful for analyzing conservative systems of interconnected bodies, where energy loss due to -ve work of friction is negligible. 13

14 Power Time rate at which work is performed Work done at a given instant by a force F acting on a rigid body in plane motion: (v is the velocity at the point of application of the force) Work done at a given instant by a couple M acting on a rigid body in plane motion: (ω is the angular velocity of the body) M and ω have same senses +ve Power. Energy is supplied to the body M and ω have opposite senses -ve Power. Energy is removed from the body Total instantaneous power if both F and M acting simultaneously: P = F v + M ω 14

15 Power Power can also be expressed as rate of change of total mechanical energy of the system of rigid bodies: Work-energy relation for an infinitesimal displacement: du = dt + dv du is the work done by the active forces and couples applied to the bodies Dividing by dt: Total power developed by the active forces and couples equals the rate of change of the total mechanical energy of the bodies or system of bodies Since R is resultant of all forces acting on body, & M is the resultant G of all forces Dot product accounts for the case of curvilinear motion of the mass center, where v and a are not in the same direction. 15

16 Example Solution Wheel has a general plane motion Draw the FBD of the wheel 16

17 Example Solution Draw the FBD of the wheel Only 100 N and 40x9.81 = 392 N forces do work C is the instantaneous center of zero velocity Vel of point A: v A = [( )/100]v = 3v Point A on the cord moves a dist of: ( )/100 = 3 times that of O = 3x3 = 9m Including the effect of weight in U term: The work done by the wheel: U 1-2 = 100(9) 392sin15(3) = 595 J For general plane motion, KE: T 1 = 0; T 2 = ½ (40)(0.1ω) 2 + ½ (40)(0.15) 2 ω 2 = 0.65 ω 2 since vel of the center of the wheel v = r ω = 0.1 ω 17

18 Example Solution Work-energy eqn: = 0.65 ω 2 ω = 30.3 rad/s We may also calculate the KE of the wheel using KE in terms of rotational the instantaneous center C of zero vel I c = I + m OC 2 and I = I o = mk o 2 Same relation. Power input from 100 N force when ω = 30.3 rad/s: P = F v P = 100(0.3)(30.3) = 908 W 18

19 Example Solution Plane Motion Conservative since friction forces can be neglected Choosing the datum for zero gravitational potential energy v g through O. 19

20 Example Solution Let us define three states: at θ = 45 o, θ = 0 o, & max spring deflection At θ = 45 o the wheel starts from rest At θ = 0 o the wheel momentarily comes to rest For the interval from θ = 45 o to θ = 0 o, the initial and final KE of the wheels = 0 At State 2: Each link is O Total KE: During this interval, the collar B drops a distance 0.375/ 2 = m PE: V 2 is zero since the collar and the links reaches the datum. There are no active forces that are doing work (other than the weights, which are included in PE) U 1-2 = 0 Work-Energy equation 20

21 Example Solution At the third state (max deflection of spring), all parts of the system are momentarily at rest. KE, T 3 = 0 Using the Work-Energy equation between states 1 and 3 Maximum deformation of the spring: x = 60.1 mm (positive root) Try solving this problem without using the work-energy equation. 21