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Answer: Exerdses 76 82 are concemed ll'itlr tire linear

Elementary Linear Algebra: A Matrix Approach | 2nd Edition | ISBN: 9780131871410 | Authors: Lawrence E. Spence ISBN: 9780131871410 187

Solution for problem 78 Chapter 2.8

Elementary Linear Algebra: A Matrix Approach | 2nd Edition

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Elementary Linear Algebra: A Matrix Approach | 2nd Edition | ISBN: 9780131871410 | Authors: Lawrence E. Spence

Elementary Linear Algebra: A Matrix Approach | 2nd Edition

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Problem 78

Exerdses 76 82 are concemed ll'itlr tire linear transformations T : n2 -+ n2 and U: n2 ~ n2 defined as T ([:~ ]) = [ :~: = ~~] and u ([ :~]) = [2x~~ Xl] De1em1ine 1he standard matrice' A and 8 of T and U. especlively

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ME012Study Guide –Midterm2 C HAPTER 15– K INEMATICS OF A PARTICLE : MPULSE AND M OMENTUM Section 15.1 – Principle of Linear Impulse and Momentum Integrating the equation of motion with respect to time obtains the principle of impulse and momentum 2 2 ∑∫ = ∫ ⃑ = ⃑2− 1 1 1 This equation provides a direct means of obtaining the particle’s final velocit⃑ after a 2 specified time period when the particle’s initial velocity is known and the forces acting on the particle are either constant or can be expressed as functions of time Linear Momentum = ⃑ Since m is a positive scalar, the linear momentum vector has the same direction as ⃑ Units of kg*(m/s) or slug*(ft/s) Linear Impulse = ∫ A vector quantity which measures the effect of a force during the time the force acts. Impulse acts in the same direction as the force Units of N*s or lb*s If force is expressed as a function of time: 2 = ∫ = (− 2 ) 1 1 Principle of Linear Impulse and Momentum – can be written in x, y, or z directions 2 ⃑ ⃑1+ ∑∫ = ⃑2⃑⃑⃑ 1 Section 15.2 – Principle of Linear Impulse and Momentum for a System of Particles 2 ∑ (⃑ ⃑ 1 + ∑∫ =∑ (⃑ ⃑ 2 1 taking time derivative with mass center G and then combined with first equation yields: 2 (⃑⃑ 1) + ∑∫ ( ⃑ 2⃑) 1 Section 15.3 – Conservation of Linear Momentum for a System of Particles When the sum of the external impulses acting on a system of particles is zero, Linear Momentum is conserved Conservation of Linear Momentum ∑ () 1 ( ) 2 Which therefore means ( 1 ( ) 2 Section 15.4 – Impact Impact occurs when two bodies collide with each other during a very short period of time, causing relatively large (impulsive) forces to be exerted between the bodies. Central impact occurs when the direction of motion off the mass centers of the two colliding particles is along a line passing through the mass centers of the particles. This line passing through the mass centers is called the line of impact, which is perpendicular to the plane of contact. Oblique impact is when the motion of one or both of the particles make an angle with the line of impact. Central Impact Momentum for the system of particles is conserved since during collision, the internal impulses (deformation/restitution impulses) cancel. ( ) + ( ) = ( ) + ( ) 1 1 2 2 Coefficient of Restitution = ( 2− ( ) 2 ( 1− ( 1 Provided a value for e exists, these two equations may be used to solve for ( ) and ( ) . 2 2 It has been found that e varies appreciably with impact velocity as well as with the size and shape of colliding bodies. Because of this, the coefficient of restitution is reliable only when used with data which closely approximates the conditions which were known to exist which measurements of it were made. In general, e has a value between 0 and 1. Elastic Impact (e=1) – If the collision between the two particles is perfectly elastic, the deformation impulse ( ) is equal and opposite to the restitution impulse ( ). Although ∫ ∫ in reality, this can never be achieved, e=1 for an elastic collision. Plastic Impact (e=0) – The impact is said to be inelastic or plastic when e=0. In this case there is no restitution impulse∫( =0), so that after collision both particles stick together and move with a common velocity. NOTE: The principle of work and energy cannot be used for the analysis of impact problems since it is not possible to know how the internal forces of deformation and restitution vary or displace during the collision. The energy loss during the collision can be calculated by knowing the particle’s velocities before and after collision: ∑ − = ∑ − If the impact is perfectly elastic, no energy is lost in the collision, whereas if the collision is plastic, the energy lost is a maximum. Oblique Impact – When oblique impact occurs between two smooth particles, the particles move away from each other at velocities with unknown directions and magnitudes. If initial velocities are known, then four unknown are present (( 2,( 2, 2∅ 2. Procedure for Analysis  Central Impact o Provided the coefficient of restitution, mass of each particle, and initial velocity of each particle are known, the solution (final velocities) may be obtained by the conservation of momentum and coefficient of restitution. o If the solution yields a negative magnitude, the velocity acts in the opposite sense.  Oblique Impact o If y axis established within plane of contact, x axis along line of impact o Then impulsive forces of deformation and restitution act only in the x direction. o By resolving the velocity or momentum vectors into components along the x and y axes, it is possible to write four independent scalar equations in order to determine the initial and final velocities of each object in the x and y directions.  Momentum is conserved along line of impact, x axis, so velocity in the x direction initially is equal to velocity in the x direction finally.  The coefficient of restitution relates the relative velocity components of the particles along the line of impact (x axis)  If these two equations are solved simultaneously, we obtain the final velocities.  Momentum of particle A is conserved along the y axis, perpendicular to the line of impact, since no impulse acts on A in this direction. Therefore, the velocity of particle A initially is equal to its velocity finally. The same is true for Particle B. Section 15.5 – Angular Momentum Angular Momentum is denoted . Scalar: = ()() Angular Momentum about point O about the z-axis = the moment arm (the perpendicular distance from O to line of action) times mass times velocity Vector: ⃑ ⃑ × ⃑ Where ⃑ is the position vector drawn from O to the particle. Section 15.6 – Relation between Moment of a Force and Angular Momentum The moments about point O of all forces acting on the particle can be related to the particle’s angular momentum by applying the equation of motion if mass is constant. The moments of the forces about point O can be obtained by performing a cross-product multiplication of each side of this equation by position vector r. ∑ = ̇ ∑ = ̇ where is the time rate of change of linear momentum. Section 15.7 – Principle of Angular Impulse and Momentum 2 ∑∫ =( ) − ( 2⃑ ⃑ 1⃑⃑ 1 where ∑ ∫ 2 is the angular impulse which can be rewritten as 1 2 2 ∫ = ∫ ( ⃑ × ) 1 1 Vector Formulation is written as 2 ⃑ ⃑1+ ∫ = ⃑2 1 2 ( 1 ∑∫ = ()⃑⃑ ⃑⃑⃑ 2⃑⃑⃑⃑ 1 C HAPTER 16– P LANAR KINEMATICS OF A R IGIDB ODY Section 16.1 – Planar Rigid-Body Motion The planar motion of a body occurs when all the particles of a rigid body move along paths which are equidistant from a fixed plane. There are three types of rigid-body planar motion:  Translation: this type of motion occurs when a line in the body remains parallel to its original orientation throughout the motion. o When the paths of motion for any two points on the body are parallel lines, the motion is called rectilinear translation. If the paths of motion are curved lines; curvilinear translation.  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.  General plane motion: Combination of translation and rotation. o The translation occurs within a reference plane, and the rotation occurs about an axis perpendicular to the reference plane. Section 16.2 – Translation Position: = + ⁄ Velocity: = Acceleration: = Section 16.3 – Rotation about a Fixed Axis When a body rotates about a fixed axis, any point P located in the body travels along a circular path. Angular motion – Only lines or bodies undergo angular motion. Angular position – The angular position of r is defined by the angle θ, measured from a fixed reference line to r. Angular Displacement – Change in angular position, dθ. The vector has magnitude dθ. Since motion is about a fixed axis, the direction of dθ is always along this axis. Direction is determined by the right-hand rule. Angular Velocity – time rate of change in the angular position, ω (omega). Since dθ occurs during an instant of time dt, then θ = Angular Acceleration - time rate of change in the angular velocity, α (alpha) = Similarly to our linear equations: = Equations of Constant Angular Acceleration = + 1 2 = 2 = + 2 ( − ) Motion of Point P – As the rigid body rotates, point P travels along a circular path of radius r with center at point O. Position and Displacement – The position of P is defined by the position vec⃑, which extends from O to P. If the body rotates dθ then P will displace ds=rdθ. Velocity – the velocity of P has a magnitude which can be found by dividing ds=rdθ by dt so that = Both magnitude and direction of ⃑ ⃑=⃑⃑⃑ ⃑⃑⃑ Where ⃑⃑⃑ is directed from any point on the axis of rotation to pt. P. NOTE: ⃑⃑⃑ ⃑⃑ ≠ ⃑⃑⃑ ⃑⃑⃑ Acceleration – The acceleration of P can be expressed in terms of its normal and tangential components where recall = 2 = = The tangential component of acceleration represents the time rate of change in the velocity’s magnitude. If speed of P is increasing, then the tangential acceleration acts in the same direction as velocity, if speed is decreasing, it acts in the opposite direction. 2 = The normal component of acceleration represents the time rate of change in the velocity’s direction. The direction of the normal acceleration is always towards O, the center of the circular path. Vector form: 2 ⃑ = ⃑⃑ ⃑⃑⃑ =⃑ + ⃑ − ⃑ If two rotating bodies contact one another then the points in contact move along different circular paths, and the velocity and tangential components of acceleration of the points will be the same; however, the normal components of acceleration will not be the same. Procedure for Analysis – Velocity and Acceleration of a point located on a rigid body, rotating about a fixed axis. Angular Motion  Establish the positive sense of rotation about the axis of rotation and show it alongside each kinematic equation as it is applied.  If a relation is known between any two of the four variables; ω,α,θ, and t, then a third variable can be obtained by using one of the kinematic equations which relates all three variables.  If the body’s angular acceleration is constant, then you can use the related equations to solve for t, ω, θ, etc.  Once the solution is obtained, the sense of θ, ω, and α is determined from the algebraic signs of their numeric quantities. Motion of Point P  In most cases the velocity of P and its two components of acceleration can be determined from the scalar equations relating v to ω, and the components of a to α and ω with both equations including r.  If the geometry of the problem is difficult to visualize, use the vector versions of the previous equations. Section 16.4 – Absolute Motion Analysis A body subjected to general plane motion undergoes a simultaneous translation and rotation. The motion can be completely specified by knowing both the angular rotation of a line fixed in the body, and the motion of a point on the body. Procedure for Analysis The velocity and acceleration of a point P undergoing rectilinear motion can be related to the angular velocity and angular acceleration of a line contained with a body: Position Coordinate Equation  Locate point P on the body using a position coordinated S, which is measured rom a fixed origin and is directed along the straight-line path of motion of point P.  Measure from a fixed reference line the angular position θ of a line lying in the body.  From the dimensions of the body, relate s to θ, s = f(θ), using geometry and/or trigonometry. Time Derivatives  Take the first derivative of s = f(θ) with respect to time to get a relation between v and ω.  Take the second time derivative to get a relation between a and α  In each case, the chain rule must be used. Section 16.5 – Relative-Motion Analysis: Velocity The general plane motion of a rigid body can be described as combination of translation and rotation. = + ⁄ = + × ⁄ The relative velocity equation can be applied either by using Cartesian vector analysis, or by writing the x and y scalar component equations directly. Section 16.7 – Relative –Motion Analysis Acceleration = + ( × ) − ⁄ ⁄ where the cross product accounts for the tangential acceleration and the omega-squared term accounts for the normal/centripetal acceleration Section 16.8 – Relative Motion Analysis using Rotating Axes We are familiar with the equation: = ⁄ taking the time derivative; we gt: = This equation with much manipulation turns into: + × ⁄ + (⁄ ) Where capital omega is the angular velocity of the x,y,z reference, measured from the X,Y,Z, reference. CHAPTER 17– PLANAR KINEMATICS OF A RIGIDB ODY : ORCE AND A CCELERATION Section 17.1 – Mass Moment of Inertia 2 2 = ∫ = ∫ Parallel Axis Theorem = + 2 Radius of Gyration √ = ∑ = ∑ = Section 17.4 – Equations of Motion: Rotation about a Fixed Axis Center of Mass: ∑ = ̃ ∑ CHAPTER 18– PLANAR KINEMATICS OF R IGIB ODY: WORK AND ENERGY Section 18.1 – Kinetic Energy Kinetic Energy for a rotational system: = + 2 2 2 Work of a weight: = ∆ −1( ) Work of a spring force: = −( − ) 2 2 2 2 1 Work Energy Principle: 1+ ∑ 1−2= 2 Section 18.5 – Conservation of Energy 1+ 1= 2 2

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Chapter 2.8, Problem 78 is Solved
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Textbook: Elementary Linear Algebra: A Matrix Approach
Edition: 2
Author: Lawrence E. Spence
ISBN: 9780131871410

The full step-by-step solution to problem: 78 from chapter: 2.8 was answered by , our top Math solution expert on 12/27/17, 07:57PM. This textbook survival guide was created for the textbook: Elementary Linear Algebra: A Matrix Approach, edition: 2. The answer to “Exerdses 76 82 are concemed ll'itlr tire linear transformations T : n2 -+ n2 and U: n2 ~ n2 defined as T ([:~ ]) = [ :~: = ~~] and u ([ :~]) = [2x~~ Xl] De1em1ine 1he standard matrice' A and 8 of T and U. especlively” is broken down into a number of easy to follow steps, and 48 words. Since the solution to 78 from 2.8 chapter was answered, more than 236 students have viewed the full step-by-step answer. This full solution covers the following key subjects: . This expansive textbook survival guide covers 34 chapters, and 2741 solutions. Elementary Linear Algebra: A Matrix Approach was written by and is associated to the ISBN: 9780131871410.

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Answer: Exerdses 76 82 are concemed ll'itlr tire linear