Chapter 8 Kinematics Kinetics of Rigid Body Chapter

Chapter 8: Kinematics & Kinetics of Rigid Body

Chapter Outline • Rigid Body Motion • Translation • Rotation about a Fixed Axis • Relative-Motion Analysis: Velocity • Force & Acceleration • Kinetic Energy • The Work of a Force • The Work of a Couple • Principle of Work & Energy • Conservation of Energy © 2007 Pearson Education South Asia Pte Ltd

8. 1 Rigid Body Motion • When all the particles of a rigid body move along paths which are equidistant from a fixed plane, the body is said to undergo planar motion. • There are three types of rigid body planar motion: 1) Translation – This type of motion occurs if every line segment on the body remains parallel to it original direction during the motion. © 2007 Pearson Education South Asia Pte Ltd

8. 1 Rigid Body Motion Rectilinear translation occurs when the paths of motion for any two particles of the body are along equidistant straight lines. Curvilinear translation occurs when the paths of motion are along curves lines which are equidistant. © 2007 Pearson Education South Asia Pte Ltd

8. 1 Rigid Body Motion 2) Rotation about a fixed axis – When a rigid body rotates about a fixed axis, all the particles of the body, except those which lie on the axis of rotation, move along circular paths. © 2007 Pearson Education South Asia Pte Ltd

8. 1 Rigid Body Motion 4) General plane motion – When a body is subjected to general plane motion, it undergoes a combination of translation and rotation. The translation occurs within a reference plane, and the rotation occurs about an axis perpendicular to the reference plane. © 2007 Pearson Education South Asia Pte Ltd

8. 2 Translation • Consider a rigid body which is subjected to either rectilinear

8. 2 Translation Position. The locations of points A and B in the body are defined from the fixed x, y reference frame by using position vectors r. A and r. B. The translating x’, y’ coordinate system is fixed in the body and has its origin located at A, hereafter referred to as the base point. The position of B with respect to A is denoted by the relative-position vector r. B/A. By vector addition, © 2007 Pearson Education South Asia Pte Ltd

8. 2 Translation Velocity. A relationship between the instantaneous velocities A and B is obtained by taking the time derivative of the position equation, which yields v. B = v. A + dr. B/A/dt The term dr. B/A/dt = 0, since the magnitude of r. B/A is constant by definition of a rigid body. Therefore, © 2007 Pearson Education South Asia Pte Ltd

8. 2 Translation Acceleration. Taking time derivative of the velocity equation yields a similar relationship between the instantaneous accelerations of A and B: It indicate that all points in a rigid body subjected to either rectilinear or curvilinear translation move with the same velocity and acceleration. © 2007 Pearson Education South Asia Pte Ltd

8. 3 Rotation About a Fixed Axis When a body is rotating about a fixed axis, any point P located in the body travels along a circular path. Angular Motion. A point is without dimension, and so it has no angular motion. Only lines or bodies undergo angular motion. Consider the body shown and the angular motion of a radial line r located with the shaded plane and directed from point O on the axis of rotation to point P. © 2007 Pearson Education South Asia Pte Ltd

8. 3 Rotation About a Fixed Axis Angular Position. At the instant shown in the figure, the angular position of r is defined by the angle θ, measured between a fixed reference line and r. Angular Displacement. The change in the angular position, which can be measured as a differential dθ, is called the angular displacement. © 2007 Pearson Education South Asia Pte Ltd

8. 3 Rotation About a Fixed Axis Angular Velocity. The time rate of change in the angular position is called the angular velocity ω. Since dθ occurs during an instant of time dt, then, ( +) © 2007 Pearson Education South Asia Pte Ltd

8. 3 Rotation About a Fixed Axis Angular Acceleration. The angular acceleration α measure the time rate of change of the angular velocity. The magnitude of this vector may be written as ( +) The line of action of α is the same as that for ω. However, it sense of direction depends on whether ω is increasing or decreasing. © 2007 Pearson Education South Asia Pte Ltd

8. 3 Rotation About a Fixed Axis Constant Angular Acceleration. If the angular acceleration of the body is constant, α = αc, ( +) Constant Angular Acceleration © 2007 Pearson Education South Asia Pte Ltd

8. 3 Rotation About a Fixed Axis Motion of Point P. As the rigid body rotates, point P travels along a circular path of radius r and center at point O. This path is contained within the shaded plane © 2007 Pearson Education South Asia Pte Ltd

8. 3 Rotation About a Fixed Axis Position. The position of P is defined by the position vector r, which extends from O to P. Velocity. The velocity of P has a magnitude which can be found from its polar coordinate components and. Since r is constant, the radial components and so © 2007 Pearson Education South Asia Pte Ltd

8. 3 Rotation About a Fixed Axis The direction of v is tangent to the circular path. Both the magnitude can direction of v can also be accounted for by using the cross product of ω and rp. Here rp is directed from any point on the axis of rotation to point P © 2007 Pearson Education South Asia Pte Ltd

8. 3 Rotation About a Fixed Axis Acceleration. The acceleration of P can be expressed in terms of its normal and tangential components The tangential component of acceleration represents the time rate of change in the velocity’s magnitude. © 2007 Pearson Education South Asia Pte Ltd

8. 3 Rotation About a Fixed Axis Since at and an are perpendicular to one another, if needed the magnitude of acceleration can be determined from the Pythagorean theorem, © 2007 Pearson Education South Asia Pte Ltd

Example 16. 1 A cord is wrapped around a wheel which is initially at rest. If a force is applied to the cord and gives it an acceleration a = (4 t) m/s 2, where t is in seconds, determine as a function of time (a) the angular velocity of the wheel, and (b) the angular position of the line OP in radians. © 2007 Pearson Education South Asia Pte Ltd

Example 16. 1 Solution Part (a). The wheel is subjected to rotation about a fixed axis passing through point O. Thus, point P on the wheel has a motion about a circular path, and the acceleration of this point has both tangential and normal components. The tangential component is (a. P)t = (4 t) m/s 2, since the cord is wrapped around the wheel and moves tangent to it. © 2007 Pearson Education South Asia Pte Ltd

Example 16. 1 Hence the angular acceleration of the wheel is + Using this

Example 16. 1 Integrating, with the initial condition that ω = 0, t =

Example 16. 1 Using this result, the angular position θ of OP can be found from ω = dθ/dt, since this equation relates θ, ω, and t. Integrating, with the initial condition θ = 0 at t = 0, + © 2007 Pearson Education South Asia Pte Ltd

8. 4 Relative–Motion Analysis: Velocity • The general plane motion of a rigid body can be described as a combination of translation and rotation. • To view these “component” motions separately, we use a relative-motion analysis involving two sets of coordinate axes. • The x, y coordinate system is fixed and measures the absolute position of two points A and B on the body. © 2007 Pearson Education South Asia Pte Ltd

8. 4 Relative–Motion Analysis: Velocity • The origin of the x’, y’ coordinate system will be attached to the selected “base point” A, which generally has a known motion. • The axes of this coordinate system do not rotate with the body; rather they will only be allowed to translate with respect to the fixed frame. © 2007 Pearson Education South Asia Pte Ltd

8. 4 Relative–Motion Analysis: Velocity Position. • The position vector r. A specifies the location of the “base point” A, and the relative-position vector r. B/A locates point B with respect to point A. • By vector addition, the position of B is © 2007 Pearson Education South Asia Pte Ltd

8. 4 Relative–Motion Analysis: Velocity Displacement. • During an instant of time dt, point A and B undergo displacements dr. A and dr. B. • If we consider the general plane motion by its component parts then the entire body first translates by an amount dr. A so that A, the base point, moves to its final position and point B to B’. © 2007 Pearson Education South Asia Pte Ltd

8. 4 Relative–Motion Analysis: Velocity • The body is then rotated about A by an amount dθ so that B’ undergoes a relative displacement dr. B/A and thus moves to its final position B. © 2007 Pearson Education South Asia Pte Ltd

8. 4 Relative–Motion Analysis: Velocity • Due to the rotation about A, dr. B/A = r. B/A dθ, and the displacement of B is due to rotation about A due to translation and rotation © 2007 Pearson Education South Asia Pte Ltd

8. 4 Relative–Motion Analysis: Velocity. • To determine the relationship between the velocities of points A and B, it is necessary to take the time derivative of the position equation, or simply divide the displacement equation by dt. This yields, © 2007 Pearson Education South Asia Pte Ltd

8. 4 Relative–Motion Analysis: Velocity • The terms dr. B/dt = v. B and dr. A/dt = v. A are measured from the fixed x, y axes and represent the absolute velocities of points A and B, respectively. • The body appears to move as if it were rotating with an angular velocity ω about the z’ axis passing through A © 2007 Pearson Education South Asia Pte Ltd

8. 4 Relative–Motion Analysis: Velocity • v. B/A has a magnitude of v. B/A

Example 16. 6 The link is guided by two block A and B, which move in the fixed slots. If the velocity of A is 2 m/s downward, determine the velocity of B at the instant θ = 45°. © 2007 Pearson Education South Asia Pte Ltd

Example 16. 6 Solution (Vector Analysis) Kinematic Diagram. • Since points A and B are restricted to move along the fixed slots and v. A is directed downward, the velocity v. B must be directed horizontally to the right. © 2007 Pearson Education South Asia Pte Ltd

Example 16. 6 • This motion causes the link to rotate CCW; by right-hand-rule the angular velocity ω is directed outward, perpendicular to the plane of plane. • Knowing the magnitude and direction of v. A and the lines of action of v. B and ω, it is possible to apply the velocity equation to points A and B in order to solve for the two unknown magnitudes v. B and ω. © 2007 Pearson Education South Asia Pte Ltd

Example 16. 6 Velocity Equation. Equating the i and j components gives © 2007

8. 5 Planar Kinetics of a Rigid Body: Force and Acceleration • Since a body has a definite size and shape, an applied non-concurrent force system may cause the body to both translate and rotate • The translational aspects of the motion are governed by the equation F = ma • The rotational aspects, caused by moment M, are governed by the equation M = Iα • I, the moment of inertia, is a measure of the resistance of a body to angular acceleration © 2007 Pearson Education South Asia Pte Ltd

8. 5 Planar Kinetics of a Rigid Body: Force and Acceleration • Define moment of inertia as the integral of the second moment about an axis of all the elements of mass dm, which compose the body Example • For body’s moment of inertia about the z axis, I = ∫m r 2 dm © 2007 Pearson Education South Asia Pte Ltd

8. 5 Planar Kinetics of a Rigid Body: Force and Acceleration • Moment arm, r, is the perpendicular distance from the z axis to the arbitrary element dm • Since the formulation involves r, the value of I is different for each axis about which it is computed • The axis chosen fort analysis passes through the body’s center of gravity G and is always perpendicular to the plane of motion • Moment of inertia computed about this axis is IG • Mass moment of inertia always a positive quantity © 2007 Pearson Education South Asia Pte Ltd

8. 5 Planar Kinetics of a Rigid Body: Force and Acceleration • If the body consists of a material having a variable density ρ = ρ(x, y, z), for elemental mass dm of the body, dm = ρ d. V • Using volume elements for integration, I = ∫V r 2 ρ dm • When ρ is a constant, I = ρ ∫V r 2 dm • When the elemental volume chosen for integration has infinitesimal dimensions in all 3 directions, d. V = dx dy dz © 2007 Pearson Education South Asia Pte Ltd

8. 5 Planar Kinetics of a Rigid Body: Force and Acceleration • Determine the moment of inertia for elemental with infinitesimal dimensions in all 3 directions with triple integration • Simplify the integration process to a single integration provided that the chosen elemental volume has a differential size or thickness in only one direction • Shell and disk elements are often chosen for this purpose © 2007 Pearson Education South Asia Pte Ltd

8. 5 Planar Kinetics of a Rigid Body: Force and Acceleration Procedure for Analysis • Consider only symmetric bodies having surfaces which are generated by revolving a curve about an axis • An example of such a body which is generated about the z axis • Two types of differential elements can be chosen © 2007 Pearson Education South Asia Pte Ltd

8. 5 Planar Kinetics of a Rigid Body: Force and Acceleration Procedure for Analysis Shell Element • A shell element with height z, radius r = y, and thickness dy is chosen for integration, volume, d. V = (2πy)(z) dy • Used to determine moment of inertia Iz of the body about the z axis, since the entire element, due to its thinness, lies at the same perpendicular distance r = y from the z axis © 2007 Pearson Education South Asia Pte Ltd

8. 5 Planar Kinetics of a Rigid Body: Force and Acceleration Procedure for Analysis Disk Element • A disk element with radius y, and thickness dz is chosen for integration, volume, d. V = πy 2 dz • Finite in the radial direction and its parts do not lie at the same radial distance r from the z axis • Determine moment of inertia of the element about the z axis and then integrate the result © 2007 Pearson Education South Asia Pte Ltd

8. 5 Planar Kinetics of a Rigid Body: Force and Acceleration Example 17. 1 Determine the moment of inertia of the cylinder about the z axis. The density of the material, ρ, is constant. © 2007 Pearson Education South Asia Pte Ltd

8. 5 Planar Kinetics of a Rigid Body: Force and Acceleration Solution Shell Element • Volume, d. V = (2πr)(h) dr • Mass, dm = ρ d. V = ρ(2πhr dr) • Since the entire element lies at the same distance r from the z axis, for moment of inertia of the element, d. Iz = r 2 dm = ρ(2πhr 3 dr) © 2007 Pearson Education South Asia Pte Ltd

8. 5 Planar Kinetics of a Rigid Body: Force and Acceleration Solution • Integrating over entire region of the cylinder, Iz = ∫r 2 dm = ρ2πh∫R 0 r 3 dr = (ρπ/2)R 4 h • For mass of the cylinder, m = ∫m dm = ρ2πh∫R 0 r dr = ρπh R 2 so that Iz = ½ m. R 2 © 2007 Pearson Education South Asia Pte Ltd

8. 6 Kinetic Energy Translation • When a rigid body of mass m is subjected to either rectilinear or curvilinear translation, the kinetic energy due to rotation is zero. • The kinetic energy of the body is therefore Where v. G is the magnitude of the translational velocity v at the instant considered © 2007 Pearson Education South Asia Pte Ltd

8. 6 Kinetic Energy Rotation About a Fixed Axis • When the rigid body is rotating about a fixed axis passing through point O, the body has both translational and rotational kinetic energy as defined by © 2007 Pearson Education South Asia Pte Ltd

8. 6 Kinetic Energy • The body’s kinetic energy may also be formulated by noting that v. G = r. Gω, in which case T = ½ (IG + mr. G 2)ω. • By parallel-axis theorem, the terms inside the parentheses represent the moment of inertia Io of the body about an axis perpendicular to the plane of motion and passing through point O. © 2007 Pearson Education South Asia Pte Ltd

8. 6 Kinetic Energy General Plane Motion • When the rigid body is subjected to general plane motion, it has an angular velocity ω and its mass center has a velocity v. G • Hence the kinetic energy is defined by © 2007 Pearson Education South Asia Pte Ltd

8. 6 Kinetic Energy • The total kinetic energy of the body consists of the scalar sum of the body’s translational kinetic energy, ½ mv. G 2, and rotational kinetic energy about its mass center, ½ IGω2 © 2007 Pearson Education South Asia Pte Ltd

EXAMPLE 18. 1 The system of three elements shown consists of a 6 -kg block B, a 10 -kg disk D and a 12 -kg cylinder C. If no slipping occurs, determine the total kinetic energy of the system at the instant shown. © 2007 Pearson Education South Asia Pte Ltd

EXAMPLE 18. 1 View Free Body Diagram Solution First determine ωD, ωC and v. G. From the kinematics of the disk, Since the cylinder rolls without slipping, the instantaneous center of zero velocity is at the point of contact with the ground © 2007 Pearson Education South Asia Pte Ltd

EXAMPLE 18. 1 Block Disk © 2007 Pearson Education South Asia Pte Ltd

EXAMPLE 18. 1 Cylinder Therefore the total kinetic energy of the system is ©

8. 7 The Work of a Force • Several types of forces encountered in planar kinetic problem are summarized as below. Work of a Variable Force • If an external force F acts on a rigid body, the work done by the force when it moves along the path s, © 2007 Pearson Education South Asia Pte Ltd

8. 7 The Work of a Force • Here θ is the angle between

8. 7 The Work of a Force Work of a Constant Force • If an external force Fc acts on a rigid body, and maintain a constant magnitude Fc and constant direction θ, while the body undergoes a translation s, the work becomes • Fc cos θ represents the magnitude of the component of force in the direction of displacement © 2007 Pearson Education South Asia Pte Ltd

8. 7 The Work of a Force Work of a Weight • The weight of a body does work only when the body’s center of mass G undergoes a vertical displacement Δy. • If this displacement is upward, the work is negative, since the weight and displacement are in opposite directions. © 2007 Pearson Education South Asia Pte Ltd

8. 7 The Work of a Force Work of a Spring Force • If a linear elastic spring is attached to a body, the spring force Fs = ks acting on the body does work when the spring either stretches or compresses from s 1 to a further position s 2. © 2007 Pearson Education South Asia Pte Ltd

8. 7 The Work of a Force • In both cases the work will

8. 7 The Work of a Forces That Do Not Work • These forces can either act at fixed points on the body, or they can have a direction perpendicular to their displacement. • Example includes the weight of a body when the center of gravity of the body moves in a horizontal plane. © 2007 Pearson Education South Asia Pte Ltd

8. 8 The Work of a Couple • When a body subjected to a couple undergoes general planar motion, the two couple forces do work only when the body undergoes a rotation. • To show this, consider the body as shown, which is subjected to a couple moment M = Fr. © 2007 Pearson Education South Asia Pte Ltd

8. 8 The Work of a Couple • When the body translates, such that the component of displacement along the line of action of the forces is dst • Clearly the positive work of one force cancels the negative work of the other © 2007 Pearson Education South Asia Pte Ltd

8. 8 The Work of a Couple • If the body undergoes a differential rotation dθ about an axis which is perpendicular to the plane of the couple and intersects the plane at point O, then each force undergoes a displacement dsθ = (r/2) dθ in the direction of the force. © 2007 Pearson Education South Asia Pte Ltd

8. 8 The Work of a Couple • Furthermore, the resultant work is positive when M and dθ have the same sense of direction and negative if these vectors have an opposite sense of direction. • When the body rotates in the plane through a finite angle θ measured in radians, from θ 1 to θ 2, the work of a couple is © 2007 Pearson Education South Asia Pte Ltd

8. 8 The Work of a Couple • If the couple moment is M

8. 9 Principle of Work and Energy • By applying the principle of work and energy to each of the particles of a rigid body and adding the results algebraically, since energy is a scalar, the principle of work and energy for a rigid body becomes © 2007 Pearson Education South Asia Pte Ltd

8. 9 Principle of Work and Energy • This equation states that the body’s initial translational and rotational kinetic energy, plus the work done by all the external forces and couple moments acting on the body as the body moves from its initial to final position, is equal to the body’s final translational and rotational kinetic energy © 2007 Pearson Education South Asia Pte Ltd

8. 9 Principle of Work and Energy • Note that the work of the body’s internal forces does not have to be considered since the body is rigid. • These forces occur in equal but opposite collinear pairs, so that when the body moves, the work of one force cancels that of its counterpart. • Furthermore, since the body is rigid, no relative movement between these forces occurs, so that no internal work is done. © 2007 Pearson Education South Asia Pte Ltd

8. 10 Conservation of Energy • When a force system acting on a rigid consists only of conservative forces, the conservation of energy theorem may be used to solve a problem which otherwise would be solved using the principle of work and energy. • This theorem is easier to apply since the work of a conservative force is independent of the path and depends only on the initial and final positions of the body. © 2007 Pearson Education South Asia Pte Ltd

8. 10 Conservation of Energy Gravitational Potential Energy • Since the total weight of a body can be considered concentrated at it center of gravity, the gravitational potential energy of the body is determined by knowing the height of the body’s CG above or below a horizontal datum. © 2007 Pearson Education South Asia Pte Ltd

8. 10 Conservation of Energy • Measuring y. G as positive upward, the gravitational potential energy of the body is thus, • Here the potential energy is positive when is positive, since the weight has the ability to do positive work when the body is moved back to the datum. © 2007 Pearson Education South Asia Pte Ltd

8. 10 Conservation of Energy • Likewise, if the body is located below the datum (-y. G), the gravitational potential energy is negative, since the weight does negative work when the body returned to the datum Elastic Potential Energy • The force developed by an elastic spring is conservative force. © 2007 Pearson Education South Asia Pte Ltd

8. 10 Conservation of Energy • The elastic potential energy which a spring imparts to an attached body when the spring is elongated or compressed from an initial undeformed position (s = 0) to a final position s, is • In the deformed position, the spring force acting on the body always has the capacity for doing work when the spring is returned back to its original undeformed position. © 2007 Pearson Education South Asia Pte Ltd

8. 10 Conservation of Energy • In general, if a body is subjected to both gravitational and elastic forces, the total potential energy is expressed as a potential function V represented as the algebraic sum • Here measurement of V depends on the location of the body with respect to selected datum. © 2007 Pearson Education South Asia Pte Ltd

8. 10 Conservation of Energy • Realizing that the work of conservative forces can be written as a difference in their potential energies, (∑U 1 -2)cons = V 1 – V 2, we can write the principle of work and energy for a rigid body as • Here (∑U 1 -2)noncons represents the work of the nonconservative forces such as friction. If this term is zero, © 2007 Pearson Education South Asia Pte Ltd

8. 10 Conservation of Energy • This equation is referred to as the conservation of mechanical energy. • Its states that the sum of the potential and kinetic energies of the body remains constant when the body moves from one position to another. © 2007 Pearson Education South Asia Pte Ltd

PROCEUDRE FOR ANALYSIS Potential Energy • Draw two diagrams showing the body located at its initial and final positions along the path. • If the center of gravity, G, is subjected to a vertical displacement, establish a fixed horizontal datum from which to measure the body’s gravitational potential energy Vg. © 2007 Pearson Education South Asia Pte Ltd

PROCEUDRE FOR ANALYSIS • Data pertaining to the elevation y. G of the body’s center of gravity from the datum and the extension or compression of any connecting springs can be determined from the problem geometry and listed on the two diagrams. • Recall that the potential energy V = Vg + Ve. Here Vg = W y. G, which can be positive or negative, and Ve = ½ ks 2, which is always positive. © 2007 Pearson Education South Asia Pte Ltd

PROCEUDRE FOR ANALYSIS Kinetic Energy • The kinetic energy of the body consists of two parts, namely translational kinetic energy, T = ½ mv. G 2, and rotational kinetic energy, T = ½ IGω2 • Kinematic diagrams for velocity may be useful for determining v. G and ω for establishing a relationship between these quantities. Conservation of Energy • Apply the conservation of energy equation T 1 + V 1 = T 2 + V 2 © 2007 Pearson Education South Asia Pte Ltd

EXAMPLE 18. 7 The 10 -kg rod AB is confined so that its ends move in the horizontal and vertical slots. The spring has a stiffness of k = 800 N/m and is unstretched when θ = 0°. Determine the angular velocity of AB when θ = 0°, if the rod is released from rest when θ = 30°. © 2007 Pearson Education South Asia Pte Ltd

EXAMPLE 18. 7 Potential Energy • The two diagrams of the rod, when it is located at its initial and final positions as shown • The datum, used to measure the gravitational potential energy, is placed in line with the rod when θ = 0°. © 2007 Pearson Education South Asia Pte Ltd

EXAMPLE 18. 7 • When the rod is in position 1, the center of gravity G is located below the datum so that the gravitational potential energy is negative. • (positive) elastic potential energy is stored in the spring, since it is stretched a distance of s 1 = (0. 4 sin 30°) m, thus © 2007 Pearson Education South Asia Pte Ltd

EXAMPLE 18. 7 • When the rod is in position 2, the potential energy of the rod is zero, since the spring is unstretched, s 2 = 0, and the center of gravity G is located at the datum. Thus, Kinetic Energy • The rod is released from rest from position 1, thus (v. G)1 = 0 and ω1 = 0, and © 2007 Pearson Education South Asia Pte Ltd

EXAMPLE 18. 7 • In position 2, the angular velocity is ω2 and the rod’s mass center has a velocity of (v. G)2. Thus, • Using kinematics, (v. G)2 can be related to ω2 as shown © 2007 Pearson Education South Asia Pte Ltd

EXAMPLE 18. 7 • At the instant considered, the instantaneous center of zero velocity (IC) for the rod is at point A; hence (v. G)2 =(r. G/IC)ω2 = (0. 2)ω2 • Substituting into the previous expression and simplfying, we get © 2007 Pearson Education South Asia Pte Ltd