# Dynamics Lecture2 General Curvilinear Motion - Rectangular Components and Projectile Motion

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### Transcript of Dynamics Lecture2 General Curvilinear Motion - Rectangular Components and Projectile Motion

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General Curvilinear MotionCurvilinear motion occurs when the particle moves along a curved pathPosition.The position of the particle, measured from a fixed point O, is designated by the position vector r = r(t).

Displacement.Suppose during a small time interval t the particle moves a distance s along the curve to a new position P`, defined by r` = r + r. The displacement r represents the change in the particles position.

General Curvilinear MotionVelocity.During the time t, the average velocity of the particle is defined as

The instantaneous velocity is determined from this equation by letting t 0, and consequently the direction of r approaches the tangent to the curve at point P. Hence,

General Curvilinear Motion Direction of vins is tangent to the curve Magnitude of vins is the speed, which may be obtained by noting the magnitude of the displacement r is the length of the straight line segment from P to P`.

General Curvilinear MotionAcceleration.If the particle has a velocity v at time t and a velocity v` = v + v at time t` = t + t. The average acceleration during the time interval t is

General Curvilinear Motiona acts tangent to the hodograph, therefore it is not tangent to the path

General Curvilinear MotionCurvilinear Motion: Rectangular ComponentsPosition.Position vector is defined by r = xi + yj + zkThe magnitude of r is always positive and defined as

The direction of r is specified by the components of the unit vector ur = r/r

Velocity.

whereThe velocity has a magnitude defined as the positive value of

and a direction that is specified by the components of the unit vector uv=v/v and is ALWAYS tangent to the path.Curvilinear Motion: Rectangular ComponentsAcceleration.

The acceleration has a magnitude defined as the positive value of

whereCurvilinear Motion: Rectangular Components The acceleration has a direction specified by the components of the unit vector ua = a/a. Since a represents the time rate of change in velocity, a will NOT be tangent to the path.Curvilinear Motion: Rectangular ComponentsPROCEDURE FOR ANALYSISCoordinate System A rectangular coordinate system can be used to solve problems for which the motion can conveniently be expressed in terms of its x, y and z components. Curvilinear Motion: Rectangular ComponentsKinematic Quantities Since the rectilinear motion occurs along each coordinate axis, the motion of each component is found using v = ds/dt and a = dv/dt, or a ds = v ds Once the x, y, z components of v and a have been determined. The magnitudes of these vectors are found from the Pythagorean theorem and their directions from the components of their unit vectors.Curvilinear Motion: Rectangular ComponentsEXAMPLE 12.9At any instant the horizontal position of the weather balloon is defined by x = (9t) m, where t is in seconds. If the equation of the path is y = x2/30, determine the distance of the balloon from the station at A, the magnitude and direction of both the velocity and acceleration when t = 2 s.

Solution:Position.When t = 2 s, x = 9(2) m = 18 m and y = (18)2/30 = 10.8 m

The straight-line distance from A to B is

mVelocity.

EXAMPLE 12.9When t = 2 s, the magnitude of velocity is

The direction is tangent to the path, where

Acceleration.

EXAMPLE 12.9

The direction of a is

EXAMPLE 12.9The motion of box B is defined by the position vector r = {0.5sin(2t)i + 0.5cos(2t)j 0.2tk} m, where t is in seconds and the arguments for sine and cosine are in radians ( rad = 180). Determine the location of the box when t = 0.75 s and the magnitude of its velocity and acceleration at this instant.

EXAMPLE 12.10Solution:Position.Evaluating r when t = 0.75 s yields

The distance of the box from the origin is

EXAMPLE 12.10The direction of r is obtained from the components of the unit vector,

EXAMPLE 12.10Velocity.

Hence at t = 0.75 s, the magnitude of velocity is

Acceleration.The acceleration is not tangent to the path.

At t = 0.75 s, a = 2 m/s2EXAMPLE 12.10 Free-flight motion studied in terms of rectangular components since projectiles acceleration always acts vertically Consider projectile launched at (x0, y0) Path defined in the x-y plane Air resistance is neglected The only force acting on the projectile would be its weight, resulting in constant downwards acceleration ac = g = 9.81 m/s2Motion of a Projectile

Motion of a ProjectileHorizontal MotionSince ax = 0,

Horizontal component of velocity remain constant during the motionMotion of a ProjectileVertical.Positive y axis is directed upward, then ay = - g

Motion of a Projectile Problems involving the motion of a projectile have at most three unknowns since only three independent equations can be written: - one in the horizontal direction - two in the vertical direction Velocity in the horizontal and vertical direction are used to obtain the resultant velocity (via Pythagorean theorem) Resultant velocity is always tangent to the path

Motion of a ProjectilePROCEDURE FOR ANALYSISCoordinate System Establish the fixed x, y, z axes and sketch the trajectory of the particle Specify the three unknowns and data between any two points on the path Acceleration of gravity always acts downwards Express the particle initial and final velocities in the x, y componentsMotion of a Projectile Positive and negative position, velocity and acceleration components always act in accordance with their associated coordinate directionsKinematics Equations Decide on the equations to be applied between the two points on the path for the most direct solutionMotion of a ProjectileHorizontal Motion Velocity in the horizontal or x directions is constant (vx) = (vo)xx = xo + (vo)x t

Vertical Motion Only two of the following three equations should be usedMotion of a Projectile

Eg: if final velocity is not needed, first and third of the equations would not be neededMotion of a ProjectileEXAMPLE 12.11A sack slides off the ramp with a horizontal velocity of 12 m/s. If the height of the ramp is 6 m from the floor, determine the time needed for the sack to strike the floor and the range R where the sacks begin to pile up.

Coordinate System. Origin of the coordinates is established at the beginning of the path, point A.Initial velocity of a sack has components (vA)x = 12 m/s and (vA)y = 0Acceleration between point A and B ay = -9.81 m/s2Since (vB)x = (vA)x = 12 m/s, the three unknown are (vB)y, R and the time of flight tAB

EXAMPLE 12.11Vertical Motion.Vertical distance from A to B is known

The above calculations also indicate that if a sack is released from rest at A, it would take the same amount of time to strike the floor at CEXAMPLE 12.11Horizontal Motion.

EXAMPLE 12.11The chipping machine is designed to eject wood at chips vO = 7.5 m/s. If the tube is oriented at 30 from the horizontal, determine how high, h, the chips strike the pile if they land on the pile 6 m from the tube.

EXAMPLE 12.12Coordinate System.Three unknown h, time of flight, tOA and the vertical component of velocity (vB)y. Taking origin at O, for initial velocity of a chip,

(vA)x = (vO)x = 6.5 m/s and ay = -9.81 m/s2

EXAMPLE 12.12Horizontal Motion.

Vertical Motion.Relating tOA to initial and final elevation of the chips,

EXAMPLE 12.12The track for this racing event was designed so that the riders jump off the slope at 30, from a height of 1m. During the race, it was observed that the rider remained in mid air for 1.5 s. Determine the speed at which he was traveling off the slope, the horizontal distance he travels before striking the ground, and the maximum height he attains. Neglect the size of the bike and rider.

EXAMPLE 12.13Coordinate System.Origin is established at point A. Three unknown are initial speed vA, range R and the vertical component of velocity vB.Vertical Motion.Since time of flight and the vertical distance between the ends of the paths are known,

EXAMPLE 12.13Horizontal Motion

For maximum height h, we consider path ACThree unknown are time of flight, tAC, horizontal distance from A to C and the height hAt maximum height (vC)y = 0

EXAMPLE 12.13Since vA known, determine h using the following equations

Show that the bike will strike the ground at B with velocity having components of

EXAMPLE 12.13

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