5. Plane Kinetics of Rigid Bodies

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5. Plane Kinetics of Rigid Bodies 5.1 Mass moments of inertia 5.2 General equations of motion 5.3 Translation 5.4 Fixed axis rotation 5.5 General plane motion 5.

1 5. Plane Kinetics of Rigid Bodies 5.1 Mass moments of inertia 5.2 General equations of motion 5.3 Translation 5.4 Fixed axis rotation 5.5 General plane motion 5.6 Work and energy relations 5.7 Impulse and momentum Mass moments of inertia The moment-of-inertia may be expressed as I = r i 2 m i Or about the axis O-O: I = න r 2 dm 2 1

2 5.1 Mass moments of inertia (contd.) If the density is constant throughout the body, the moment of inertia becomes I = ρ න r 2 dv Radius of Gyration If all the mass m of a body could be concentrated at a distance k from the axis, the moment of inertia would be unchanged. The radius of gyration k of a mass m about an axis for which the moment of inertia is I is defined as k = I m or I = k2 m Mass moments of inertia (contd.) Transfer of Axes If moment of inertia of a body is known about an axis passing through mass center, it may be determined easily about any parallel axis. I = ҧ I + md 2 4 2

5.2 General Equations of Motion The force equation, F = mതa tells us that the resultant ΣF of the external forces acting on the body equals the mass m of the body times the acceleration തa of its

3 5.2 General Equations of Motion The force equation, F = mതa tells us that the resultant ΣF of the external forces acting on the body equals the mass m of the body times the acceleration തa of its mass center G. The moment equation taken about the mass center, M G = ሶ H G shows that the resultant moment about the mass center of the external forces on the body equals the time rate of change of the angular momentum of the body about the mass center General Equations of Motion (contd.) 6 3

ሶ 09.05.2016 5.2 General Equations of Motion (contd.

) Plane Motion Equations The mass center G has an acceleration തa and the body has an angular velocity ω = ωk and an angular

4 ሶ General Equations of Motion (contd.) Plane Motion Equations General Equations of Motion (contd.) Plane Motion Equations The mass center G has an acceleration തa and the body has an angular velocity ω = ωk and an angular acceleration α = αk, both taken positive in the z-direction. Because the z-direction of both ω and α remains perpendicular to the plane of motion, we may use scalar notation ω and α = ω to represent the angular velocity and angular acceleration. F = mതa and M G = ҧ Iα 8 4

5 5.2 General Equations of Motion (contd.) Alternative Moment Equations Alternatively M P = ҧ Iα + mതad Translation For a translating body, then, our general equations for plane motion equation may be written F = mതa M G = ҧ Iα =

5.3 Translation Sample Problem (1) The pickup truck weighs 1500kg and reaches a speed of 50 km/h from rest in a distance of 60m up the 10-percent incline with constant acceleration.

6 5.3 Translation Sample Problem (1) The pickup truck weighs 1500kg and reaches a speed of 50 km/h from rest in a distance of 60m up the 10-percent incline with constant acceleration. Calculate the normal force under each pair of wheels and the friction force under the rear driving wheels. The effective coefficient of friction between the tires and the road is known to be at least

7 5.3 Translation Sample Problem (2) The uniform 30-kg bar OB is secured to the accelerating frame in the 30 position from the horizontal by the hinge at O and roller at A. If the horizontal acceleration of the frame is a=20 m/s 2 compute the force F A on the roller and the x- and y-components of the force supported by the pin at O

5.4 Fixed Axis Rotation Our general equations for plane motion are directly applicable and are repeated here.

4 Fixed Axis Rotation Sample Problem (3) The 300 kg concrete block is elevated by the hoisting mechanism shown,

The drums, which are fastened together and turn as a single unit about their mass center at O, have a combined

8 5.4 Fixed Axis Rotation Our general equations for plane motion are directly applicable and are repeated here. F = mതa, M G = ҧ Iα, M O = I O α Fixed Axis Rotation Sample Problem (3) The 300 kg concrete block is elevated by the hoisting mechanism shown, where the cables are securely wrapped around the respective drums. The drums, which are fastened together and turn as a single unit about their mass center at O, have a combined mass of 150 kg and a radius of gyration about O of 450 mm. If a constant tension P of 1.8 kn is maintained by the power unit at A, determine the vertical acceleration of the block and the resultant force on the bearing at O. 16 8

9 Fixed Axis Rotation Sample Problem (4) The uniform slender bar is released from rest in the horizontal position shown. Determine the value of x for which the angular acceleration is a maximum, and determine the corresponding angular acceleration. 18 9

10 General Plane Motion The dynamics of a rigid body in general plane motion combines translation and rotation. F = mതa M G = ҧ Iα 20 10

5.5 General Plane Motion (contd.) Choice of Moment Equation. The application of the alternative relation for moments about any point P, M P = ҧ Iα + mതad 21 5.

11 5.5 General Plane Motion (contd.) Choice of Moment Equation. The application of the alternative relation for moments about any point P, M P = ҧ Iα + mതad General Plane Motion Sample Problem (5) A metal hoop with a radius r = 150 mm is released from rest on the 20 incline. If the coefficients of static and kinetic friction are μ s = 0.15 and μ k = 0.12, determine the angular acceleration of the hoop and the time t for the hoop to move a distance of 3 m down the incline

12 23 (Ctnd.) 24 12

brakes are applied to give the car a constant rearward acceleration a.

13 ҧ General Plane Motion Sample Problem (6) A car door is inadvertently left slightly open when the brakes are applied to give the car a constant rearward acceleration a. Derive expressions for the angular velocity of the door as it swings past the 90 position. The mass of the door is m, its mass center is a distance r from the hinge axis O, and the radius of gyration about O is k O

B to roll on the horizontal surface by means of the connecting cable, which wraps around the inner hub of the spool.

14 (Ctnd.) General Plane Motion Sample Problem (7) The drum A is given a constant angular acceleration α 0 of 3 rad/s 2 and causes the 70-kg spool B to roll on the horizontal surface by means of the connecting cable, which wraps around the inner hub of the spool. The radius of gyration തk of the spool about its mass center G is 250 mm, and the coefficient of static friction between the spool and the horizontal surface is 0.1. Determine the tension T in the cable and the friction force F exerted by the horizontal surface on the spool

16 5.5 General Plane Motion Sample Problem (8) The uniform steel beam of mass m and length l is suspended by the two cables at A and B. If the cable at B suddenly breaks, determine the tension T in the cable at A immediately after the break occurs. Treat the beam as a slender rod and show that the result is independent of the length of the beam

6 Work and Energy Kinetic Energy (a) Translation.

17 5.6 Work and Energy Work of Forces and Couples The work done by a force F is given by U = න F. dr or U = න(F cos α) ds U = න M dθ Work and Energy Kinetic Energy (a) Translation. The translating rigid body of Figure (a) has a mass m and all of its particles have a common velocity v. T = 1 2 mv

18 5.6 Work and Energy Kinetic Energy(contd.) (b) Fixed-axis rotation. T = 1 2 I Oω 2 35 (c) General plane motion. 5.6 Work and Energy Kinetic Energy(contd.) T = 1 2 mv2 ҧ + 1 Iω 2 ҧ

19 5.6 Work and Energy Kinetic Energy(contd.) The kinetic energy of plane motion may also be expressed in terms of the rotational velocity about the instantaneous center C of zero velocity. Because C momentarily has zero velocity, T = 1 2 I Oω 2 for the fixed point O holds equally well for point C, so that, alternatively, we may write the kinetic energy of a rigid body in plane motion as T = 1 2 I Cω Work and Energy Potential Energy and the Work-Energy Equation Gravitational potential energy V g and elastic potential energy V e were covered in detail before. Recall that the symbol U (rather than U) is used to denote the work done by all forces except the weight and elastic forces, which are accounted for in the potentialenergy terms. The work-energy relation was introduced before for particle motion and was generalized to include the motion of a general system of particles. This equation U 1 2 = T + V g + V e Power If the force F and the couple M act simultaneously, the total instantaneous power is P = F. v + Mω 38 19

If the wheel starts from rest, compute its angular velocity ω after its center has moved a distance of 3 m up the incline.

20 5.6 Work and Energy Sample Problem (10) The wheel rolls up the incline on its hubs without slipping and is pulled by the 100-N force applied to the cord wrapped around its outer rim. If the wheel starts from rest, compute its angular velocity ω after its center has moved a distance of 3 m up the incline. The wheel has a mass of 40 kg with center of mass at O and has a centroidal radius of gyration of 150 mm. Determine the power input from the 100-N force at the end of the 3-m motion interval

a mass of 30 kg and a centroidal radius of gyration of 100 mm.

The 7-kg collar at B slides on the fixed vertical shaft with negligible friction.

the links reach the horizontal position.

21 (Cntd.) Work and Energy Sample Problem (9) In the mechanism shown, each of the two wheels has a mass of 30 kg and a centroidal radius of gyration of 100 mm. Each link OB has a mass of 10 kg and may be treated as a slender bar. The 7-kg collar at B slides on the fixed vertical shaft with negligible friction. The spring has a stiffness k=30 kn/m and is contacted by the bottom of the collar when the links reach the horizontal position. If the collar is released from rest at the position θ = 45 and if friction is sufficient to prevent the wheels from slipping, determine (a) the velocity v B of the collar as it first strikes the spring and (b) the maximum deformation x of the spring

22 43 (Cntd.) 44 22

23 ሶ Impulse and Momentum Linear Momentum The linear momentum of any mass system, G = mതv F = G and t 2 G 1 + න F dt = G2 t 1 In planar applications: F x = Gሶ x F y = Gሶ y and t 2 (G x ) 1 + න Fx dt = (G x ) 2 t 1 45 t 2 (G y ) 1 + න Fy dt = (G y ) 2 t Impulse and Momentum Angular Momentum Angular momentum about the mass center: H G = ҧ Iω t 2 M G = Hሶ G and H G 1 + න MG dt = H G 2 t

at any particular instant of time about O, which may be a fixed or moving point on or off the body.

24 5.7 Impulse and Momentum Angular Momentum (contd.) The angular momentum H O about any point O is easily written as H O = Iω ҧ + mvd ҧ Impulse and Momentum Angular Momentum (contd.) This expression holds at any particular instant of time about O, which may be a fixed or moving point on or off the body. When a body rotates about a fixed point O on the body or body extended, H O = I O ω t 2 M O = Hሶ O and H O 1 + න MO dt = H O 2 t

25 5.7 Impulse and Momentum Interconnected Rigid Bodies F = Gሶ a + Gሶ b + M O = ( ሶ H O ) a +( ሶ H O ) b + න t 1 t 2 න Fdt = G system t 1 t 2 MO dt = H O system Impulse and Momentum InterconnectedRigid Bodies (contd.) We note that the equal and opposite actions and reactions in the connections are internal to the system and cancel one another so they are not involved in the force and moment summations. Also, point O is one fixed reference point for the entire system

5.7 Impulse and Momentum Conservation of Momentum Principles for the particles are applicable to either a single rigid body or a system of interconnected rigid bodies.

26 5.7 Impulse and Momentum Conservation of Momentum Principles for the particles are applicable to either a single rigid body or a system of interconnected rigid bodies. Thus, if ΣF=0 for a given interval of time, then G 1 = G 2 Similarly, if the resultant moment about a given fixed point O or about the mass center is zero during a particular interval of time for a single rigid body or for a system of interconnected rigid bodies, then H O 1 = H O 2 or H G 1 = H G Impulse and Momentum Sample Problem (11) The grooved drums in the two systems shown are identical. In both cases, (a) and (b), the system is at rest at time t=0 Determine the angular velocity of each grooved drum at time t=4s. Neglect friction at the pivot O

28 5.7 Impulse and Momentum Sample Problem (12)

57 5.7 Impulse and Momentum Sample Problem (13) The uniform rectangular block of dimensions shown is sliding to the left on the horizontal surface with a velocity v 1 when it strikes the small step

29 Impulse and Momentum Sample Problem (13) The uniform rectangular block of dimensions shown is sliding to the left on the horizontal surface with a velocity v 1 when it strikes the small step at O. Assume negligible rebound at the step and compute the minimum value of v 1 which will permit the block to pivot freely about O and just reach the standing position A with no velocity. Compute the percentage energy loss n for b=c

30 59 (Cntd.) 60 30

5.7 Impulse and Momentum Sample Problem (14) The sheave E of the hoisting rig shown has a mass

The 40-kg load D which is carried by the sheave has an initial downward velocity v 1 = 1.

essentially a constant force F =380 N in the cable at B.

31 5.7 Impulse and Momentum Sample Problem (14) The sheave E of the hoisting rig shown has a mass of 30 kg and a centroidal radius of gyration of 250 mm. The 40-kg load D which is carried by the sheave has an initial downward velocity v 1 = 1.2 m/s at the instant when a clockwise torque is applied to the hoisting drum A to maintain essentially a constant force F =380 N in the cable at B. Compute the angular velocity ω 2 of the sheave 5 seconds after the torque is applied to the drum and find the tension T in the cable at O during the interval. Neglect all friction

32 (Cntd.) 63 32

The University of Melbourne Engineering Mechanics

The University of Melbourne Engineering Mechanics The University of Melbourne 436-291 Engineering Mechanics Tutorial Eleven Instantaneous Centre and General Motion Part A (Introductory) 1. (Problem 5/93 from Meriam and Kraige - Dynamics) For the instant