BDA 2013 Dynamics En. Saifulnizan Bin Jamian Faculty of Mechanical Engineering and Manufacturing Kole Kolejj Univ Univer ersi siti ti Tekn Teknol olog ogii Tun Tun Huss Hussei ein n Onn 86400 86400 Parit Parit Raja, Raja, Johor Johor [C16-101-11
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Contents Introduction Semester Overview Mechanics Vectors Kinematics
of Particles
Rectilinear
Motion
Introduction
Sinopsis: Pengenalan dinamik, kinematik zarah, kinematik jasad tegar, kinetik zarah dan kinetik jasad tegar.
Objektif: Kurs Kursus us ini ini dira dirang ngka ka untu untuk k memba embant ntu u pela pelaja jar r mema memaha hami mi perk perkar ara a berik berikut: ut:
Mema Memaham hamii konsep konsep kinem kinemati atik k zarah zarah.. Memaha hami mi kons konsep ep kine kineti tik k zara zarah. h. Mema Memaham hamii konsep konsep kinem kinemati atik k jasad jasad tegar. tegar. Mema Memaham hamii konsep konsep kinet kinetik ik jasad jasad teg tegar. ar. Mema
Introduction Rujukan Utama : Bear F.P. and Johnson E. R., 2004. Vector Mechanics for Engineers – Dynamics, 7th S.I. Edition, Edition, Mc Graw Hill Hibbeler Hibbeler R.C., R.C., 2004. Engineering Mechanics –Dynamics, –Dynamics, 3rd S.I. Edition, Prentice Hall. Rujuka Rujukan n lain lain : David H. Myszka, 2005. Machines and Mechanisms, Mechanisms , 3rd Edition, Prentice Hall. Meriam Meriam J.L. and Kraige Kraige L. G., G., 1997. 1997. Engineering mechanics mechanics Vol. 2 – Statics and Dynamics, Dynamics, 4th Edition, John Wiley & Sons, Inc. Bedford A. & Wallace T. Fowler, 2005. Engineering Mechanics-Dynamics, Mechanics-Dynamics, S.I. Edition, Prentice Hall. Abdul Ghani Mohamad, 1996. Mekani Mekanik k Badan Badan Tegar Tegar Dinami Dinamik k , Penerbi Penerbitt UTM. UTM. PENILAIAN (PERATUS TABURAN MARKAH) (a) Tugasan & Tutorial : 5% (b) Projek Mini : 5% (c) Ujian 1, 2 & 3 : 40% (d) Peperiksaan Akhir : 50%
Introduction
Sinopsis: Pengenalan dinamik, kinematik zarah, kinematik jasad tegar, kinetik zarah dan kinetik jasad tegar.
Objektif: Kurs Kursus us ini ini dira dirang ngka ka untu untuk k memba embant ntu u pela pelaja jar r mema memaha hami mi perk perkar ara a berik berikut: ut:
Mema Memaham hamii konsep konsep kinem kinemati atik k zarah zarah.. Memaha hami mi kons konsep ep kine kineti tik k zara zarah. h. Mema Memaham hamii konsep konsep kinem kinemati atik k jasad jasad tegar. tegar. Mema Memaham hamii konsep konsep kinet kinetik ik jasad jasad teg tegar. ar. Mema
Introduction Rujukan Utama : Bear F.P. and Johnson E. R., 2004. Vector Mechanics for Engineers – Dynamics, 7th S.I. Edition, Edition, Mc Graw Hill Hibbeler Hibbeler R.C., R.C., 2004. Engineering Mechanics –Dynamics, –Dynamics, 3rd S.I. Edition, Prentice Hall. Rujuka Rujukan n lain lain : David H. Myszka, 2005. Machines and Mechanisms, Mechanisms , 3rd Edition, Prentice Hall. Meriam Meriam J.L. and Kraige Kraige L. G., G., 1997. 1997. Engineering mechanics mechanics Vol. 2 – Statics and Dynamics, Dynamics, 4th Edition, John Wiley & Sons, Inc. Bedford A. & Wallace T. Fowler, 2005. Engineering Mechanics-Dynamics, Mechanics-Dynamics, S.I. Edition, Prentice Hall. Abdul Ghani Mohamad, 1996. Mekani Mekanik k Badan Badan Tegar Tegar Dinami Dinamik k , Penerbi Penerbitt UTM. UTM. PENILAIAN (PERATUS TABURAN MARKAH) (a) Tugasan & Tutorial : 5% (b) Projek Mini : 5% (c) Ujian 1, 2 & 3 : 40% (d) Peperiksaan Akhir : 50%
Introduction (cont.) KEHADIRAN/PERATURAN SEMASA KULIAH
Pelajar mesti hadir tidak kurang dari 80% masa pertemuan yang ditentukan bagi sesuatu mata pelajaran termasuk mata pelajaran Hadir Wajib (HW) dan matapelajaran Hadir Sahaja (HS). Pelajar yang tidak memenuhi perkara (1) diatas tidak dibenarkan menghadiri kuliah dan menduduki sebarang bentuk penilaian selanjutnya. Markah sifar (0) akan diberikan bagi mata pelajaran yang berkenaan atau Hadir Gagal (HG) bagi mata pelajaran Hadir Wajib (HW) Pelajar perlu mengikut dan patuh kepada peraturan berpakaian yang berkuatkuasa dan menjaga displin diri masing-masing untuk mengelakkan dari tindakan tatatertib diambil oleh pelajar. Pelaja Pelajarr perlu perlu mematu mematuhi hi peratu peraturan ran kesela keselamat matan an semasa semasa pengajaran dan pembelajaran
Overview of Semester Topics Kinematics - the study of how objects move without regard to the forces associated with that motion
Particles (Test 1) rectilinear kinematics curvilinear kinematics Rectangular normal-tangential Polar
dependent motion relative motion
Rigid bodies (Test 2) fixed-axis rotation absolute motion relative motion velocity acceleration
instantaneous centers of instantaneous
zero velocity
Overview of Semester Topics (cont.) Kinetics - the study of the t he relationship between the motion and the forces that cause the motion
ParticlesU ParticlesUsing sing Newton's Newton's laws Using Newton’s laws Using
work and energy methods Using impulse and moments methods (Test 3)
Rigid bodies Using Newton’s laws
translation fixed-axis rotation general plane motion Using
work and energy methods Using impulse and moments methods
Mechanics Terms mechanics - the study of forces and their effects matter matter - any substance substance that occupies occupies space space body - matter matter bounded bounded by a closed surface surface mass - quantitative measure measure of a body's body's resistance resistance to a change in its motion; used to measure inertia weight - the force of gravity acting on a body; depends depends on location (W=mg) force - the action action of one body body on another another concentrated force - represen represents ts a loading loading which which is assumed to act at a point on a body particle - an object with with mass but whose size can be ignored
Rigid Body A
rigid body is a collection of particles that remain at fixed distances from each other at all times and under all conditions of loading. The rigid-body concept represents an idealization of the true situation, since all real bodies will change shape to a certain extent when they are subjected to a system of forces. When it is assumed that the body is rigid (free of deformation), the material properties of the body are not required for the analysis of forces and their effects on the body.
Three Branches of "Mechanics" rigid
body mechanics (Classical mechanics)
Statics-
the study of bodies in equilibrium (zero acceleration -- at rest or moving moving with constant constant velocity) velocity)
Dynamics Dynamics-
the study of the accelerated motion of
bodies Kinematics Kinetics
deformable
body mechanics (Mechanics of
materials) fluid
mechanics
Newton's Laws "The genesis of his principal ideas, both mathematical and physical, took place in a two-year period when he was in his twenties." -- David Berlinski
paraphrased: A particle at rest or in motion will remain unless acted upon by an external force. An unbalanced force causes an acceleration in the same direction as the applied force. (F=ma) Every action has an equal but opposite reaction.
Vectors vector - a quantity that has both magnitude and direction (e.g. position vector, r AB) scalar - physical quantity completely described by a real number, r AB magnitude - non-negative real number, |r AB| unit vector - a vector whose magnitude is one, e or u = r AB / |r AB|
Vectors (cont.) A vector is represented graphically by an arrow, which is used to define its magnitude, direction, and sense. The magnitude of the vector is the length of the arrow, the direction is defined by the angle between a reference axis and the arrow's line of action, and the sense is indicated by the arrowhead. The end with the arrowhead is call the tip or head of the vector, and the other end is called the tail. A = Au A = A cos(a) i + A cos(b) j + A cos(g) k = Ax i + Ay j + Az k I am picky about notation and units.
Vector Algebra See
Vector Summary or your calculus book.
Dot
products, cross products, and mixed triple products will be used to determine moment vectors (torque).
Vector Components See
Coordinate Systems. In statics, it is easier to work with the rectangular components (Cartesian coordinates) of a vector. Right-handed, 3D Cartesian coordinate systems: Lay fingers in direction of +x-axis, bend fingers in direction of +y-axis, and thumb will point in direction of +z-axis.
Vector Components (cont.)
Right-handed, 3D Cartesian coordinate systems:
Lay fingers in direction of +x-axis, bend fingers in direction of +yaxis, and thumb will point in direction of +z-axis.
Kinematics of Particles Terms particle - an object whose size can be ignored when studying its motion position vector of point P with respect to origin O is written as r P/O(t). displacement - the difference in position of a particle at two instants in time velocity - the time rate change of a particle's position acceleration - the time rate of change of a particle's velocity Displacement, velocity, and acceleration are independent of the origin location. Velocity is in the same direction as displacement.
Kinematics of Particles (cont.)
In terms of a fixed, rectangular, Cartesian coordinate system, Position
Displacement
Velocity
Acceleration
Types of Motion rectilinear
motion - motion in a coordinate system such that y- and z-components of r , v, and a are all zero for all time i.e. motion along a straight line with varying speed and acceleration plane curvilinear motion - motion in a coordinate system such that z-components of r , v, and a are all zero for all time general curvilinear or space curvilinear motion - motion where no coordinate system can be found that makes at least on component of r , v, and a zero for all time
Rectilinear Motion
If the coordinate system is oriented such that the x-axis coincides with the line of motion, the general equ ations for position, displacement, velocity, and acceleration simplify to
Position
Displacement
Velocity
Acceleration
Rectilinear Motion (cont.)
Velocity acts in the same direction as displacement. Accelerations acts colinearly with velocity. If v and a have the same sign, then v is increasing (particle is accelerating). If v and a have opposite signs, then v is decreasing (particle is decelerating). Distance traveled (a scalar) is a measure of the total length of a particle's path. Displacement (a vector) is a measure of the difference between the beginning and the end of a particle's path.
Examples 1 ( Hibbeler 12-1) A
bicyclist starts from rest and after traveling along a straight path a distance of 20 m reaches a speed of 30 km/h. Determine his acceleration if it is constant. Also, how long does it take to reach the speed of 30 km/h? Answer.
Examples 2 (Hibbeler 12-3, 10th Ed.) A
baseball is thrown downward from a 50 ft tower with an initial speed of 18 ft/s. Determine the speed at which it hits the ground and the time of travel.
Answer
Example 3 (Hibbeler 12-13, 10th Ed.) The
position of a particle along a straight line is given by s=(1.5t 3 – 13.5t2 + 22.5t)ft, where t is in seconds. Determine the position of the particle when t = 6s and the total distance it travels during the 6 s time interval. Hint : Plot the path to determine the total distance traveled. Answer
Example 4 (Hibbeler 12-19, 10th Ed.) A stone A is dropped from rest down a well, and in 1 s another stone B is dropped from rest. Determine the distance between the stones another second later. Answer
Example 5 (Hibbeler 12-5, 10th Ed.) Traveling
with an initial speed of 70 km/hr, a car accelerates at 6000 km/hr 2 along a straight road. How long will it take to reach a speed of 120 km/h? Also, through what distance does the car travel during this time?
Answer
Example 5 (Hibbeler 12-31, 10th Ed.) Determine
the time required for a car to travel 1 km along a road if the car starts from rest, reaches a maximum speed at some intermediate point, and then stops at the end of the road. The car can accelerate at 1.5 m/s 2 and decelerate at 2 m/s2.
Answer a, b
Rectilinear Motion: Position, Velocity & Acceleration • Particle moving along a straight line is said to be in rectilinear motion. • Position coordinate of a particle is defined by positive or negative distance of particle from a fixed origin on the line. • The motion of a particle is known if the position coordinate for particle is known for every value of time t . Motion of the particle may be expressed in the form of a function, e.g., = 6t 2 − t 3 or in the form of a graph x vs. t .
Rectilinear Motion: Position, Velocity & Acceleration • Consider particle which occupies position P at time t and P’ at t +∆t , Average velocity =
∆ ∆t
∆ x ∆t →0 ∆t • Instantaneous velocity may be positive or negative. Magnitude of velocity is referred to as particle speed . • From the definition of a derivative, ∆ dx v = lim = ∆t →0 ∆t dt Instantaneous velocity = v = lim
e.g., x = 6t 2 − t 3 v=
dx dt
= 12t − 3t 2
Rectilinear Motion: Position, Velocity & Acceleration • Consider particle with velocity v at time t and v’ at t +∆t , ∆v Instantaneous acceleration = a = lim ∆t →0 ∆t • Instantaneous acceleration may be: - positive: increasing positive velocity or decreasing negative velocity - negative: decreasing positive velocity or increasing negative velocity. • From the definition of a derivative, ∆v dv d 2 x a = lim = = ∆t →0 ∆t dt dt 2 2 e.g. v = 12t − 3t
a=
dv dt
= 12 − 6t
Rectilinear Motion: Position, Velocity & Acceleration • Consider particle with motion given by
= 6t 2 − t 3 v=
a=
dx dt dv dt
= 12t − 3t 2
=
d 2 x dt 2
= 12 − 6t
• at t = 0, x = 0, v = 0, a = 12 m/s2 • at t = 2 s, x = 16 m, v = vmax = 12 m/s, a = 0 • at t = 4 s, x = xmax = 32 m, v = 0, a = -12 m/s2 • at t = 6 s, x = 0, v = -36 m/s, a = 24 m/s2
Determination of the Motion of a Particle • Recall, motion of a particle is known if position is known for all time t . • Typically, conditions of motion are specified by the type of acceleration experienced by the particle. Determination of velocity and position requires two successive integrations. • Three classes of motion may be defined for: - acceleration given as a function of time, a = f (t ) - acceleration given as a function of position, a = f( x) - acceleration given as a function of velocity, a = f(v)
Determination of the Motion of a Particle • Acceleration given as a function of time, a = f (t ): dv dt dx dt
v (t )
= a = f (t )
t
∫ dv = ∫ f (t ) dt
dv = f (t ) dt
v0 x (t )
= v(t )
0 t
∫ dx = ∫ v(t ) dt
dx = v (t ) dt
x0
t
v (t ) − v0 = ∫ f (t ) dt 0 t
x(t ) − x0 = ∫ v (t ) dt
0
0
• Acceleration given as a function of position, a = f ( x): v=
dx dt
or dt =
dx
a=
v v ( x )
v dv = f ( x )dx
dv dt x
or a = v
∫ v dv = ∫ f ( x )dx v0
x0
dv dx
= f ( x )
1 v ( x )2 2
− 12 v02
x
= ∫ f ( x )dx x0
Determination of the Motion of a Particle • Acceleration given as a function of velocity, a = f (v): dv dt
= a = f (v )
v (t )
∫ v0
v
dv
dv f (v )
dv dx
f (v )
v (t )
= dt
∫ v0
dv f (v )
t
= ∫ dt 0
= t
= a = f (v ) v (t )
x(t ) − x0 =
∫ v0
dx =
v dv
x (t )
f (v )
v (t )
∫ dx = ∫ x0
v0
v dv f (v )
v dv f (v )
Example 6: Sample Problem 11.2 (Beer) SOLUTION: • Integrate twice to find v(t ) and y(t ). • Solve for t at which velocity equals zero (time for maximum elevation) and evaluate corresponding altitude.
Ball tossed with 10 m/s vertical velocity from window 20 m above ground. Determine: • velocity and elevation above ground at time t , • highest elevation reached by ball and corresponding time, and • time when ball will hit the ground and corresponding velocity.
• Solve for t at which altitude equals zero (time for ground impact) and evaluate corresponding velocity.
Example 6: Sample Problem 11.2 (Beer) SOLUTION: • Integrate twice to find v(t ) and y(t ). dv dt
= a = −9.81 m s 2
v (t )
t
∫ dv = − ∫ 9.81 dt v0
v(t ) − v0 = −9.81t
0
v(t ) = 10 dy dt
m s
− 9.81
m
t
s 2
= v = 10 − 9.81t
y (t )
t
∫ dy = ∫ (10 − 9.81t )dt y0
0
y (t ) − y0 = 10t − 1 9.81t 2 2
m t 4.905 m t 2 − s s 2
y (t ) = 20 m + 10
Example 6: Sample Problem 11.2 (Beer) • Solve for t at which velocity equals zero and evaluate corresponding altitude. v (t ) = 10
m m − 9.81 t = 0 2 s s t = 1.019 s
• Solve for t at which altitude equals zero and evaluate corresponding velocity. m m t − 4.905 2 t 2 s s
y (t ) = 20 m + 10
m m 2 (1.019 s) − 4.905 2 (1.019 s) s s
y = 20 m + 10
y = 25.1 m
Example 6: Sample Problem 11.2 (Beer) • Solve for t at which altitude equals zero and evaluate corresponding velocity. m m t − 4.905 2 t 2 = 0 s s
y (t ) = 20 m + 10
t = −1.243 s (meaningless ) t = 3.28 s v(t ) = 10
m m − 9.81 t 2 s s
v(3.28 s ) = 10
m m − 9.81 (3.28 s ) 2 s s v = −22.2
m s
Example 7 : Sample Problem 11.3 (Beer) SOLUTION: a = − kv
• Integrate a = dv/dt = -kv to find v(t ). • Integrate v(t ) = dx/dt to find x(t ).
Brake mechanism used to reduce gun recoil consists of piston attached to barrel moving in fixed cylinder filled with oil. As barrel recoils with initial velocity v0, piston moves and oil is forced through orifices in piston, causing piston and cylinder to decelerate at rate proportional to their velocity. Determine v(t ), x(t ), and v( x).
• Integrate a = v dv/dx = -kv to find v( x).
Example 7 : Sample Problem 11.3 (Beer) SOLUTION: • Integrate a = dv/dt = -kv to find v(t ). a=
v (t )
dv dt
∫
= −kv
v0
dv v
t
= −k ∫ dt
ln
0
v(t )
= − kt
v0
v(t ) = v0 e
− kt
• Integrate v(t ) = dx/dt to find x(t ). v(t ) = x (t )
dx dt
= v0 e − kt t
∫ dx = v0 ∫ e 0
t
1 −kt x(t ) = v0 − e k 0
− kt
dt
0
x(t ) =
v0 k
(1 − e − ) kt
Example 7 : Sample Problem 11.3 (Beer) • Integrate a = v dv/dx = -kv to find v( x). dv = − kv a=v dx
v
x
v0
0
∫ dv = −k ∫ dx
dv = −k dx
v − v0 = −kx v = v0 − kx
• Alternatively, with x(t ) =
v0
− kt ( 1− e ) k − kt
and
v(t ) = v0 e
then
v v(t ) x(t ) = 0 1 − k v0
or e
− kt
=
v(t ) v0
v = v0 − kx
Uniform Rectilinear Motion For particle in uniform rectilinear motion, the acceleration is zero and the velocity is constant. dx dt
= v = constant
x
t
x0
0
∫ dx = v ∫ dt
x − x0 = vt x = x0 + vt
Uniformly Accelerated Rectilinear Motion
For particle in uniformly accelerated rectilinear motion, the acceleration of the particle is constant. dv dt
v
= a = constant
t
∫ dv = a ∫ dt
v0
v − v0 = at
0
v = v0 + at dx dt
= v0 + at
x
t
x0
0
∫ dx = ∫ (v0 + at )dt
x − x0 = v0 t + 1 at 2 2
x = x0 + v0 t + 1 at 2 2 v
dv dx
= a = constant
v 2 = v02 + 2a ( x − x0 )
v
x
v0
x0
∫ v dv = a ∫ dx
1 2
(v 2 − v02 ) = a( x − x0 )
Motion of Several Particles: Relative Motion
• For particles moving along the same line, time should be recorded from the same starting instant and displacements should be measured from the same origin in the same direction. x B A = x B − x A = relative position of B with respect to A x B = x A + x B A
v B A = v B − v A = relative velocity of B with respect to A v B = v A + v B A
a B A = a B − a A = relative acceleration of B with respect to A a B = a A + a B A
Example 8 :Sample Problem 11.4(Beer) SOLUTION: • Substitute initial position and velocity and constant acceleration of ball into general equations for uniformly accelerated rectilinear motion. • Substitute initial position and constant velocity of elevator into equation for uniform rectilinear motion. Ball thrown vertically from 12 m level in elevator shaft with initial velocity of 18 m/s. At same instant, open-platform elevator passes 5 m level moving upward at 2 m/s. Determine (a) when and where ball hits elevator and (b) relative velocity of ball and elevator at contact.
• Write equation for relative position of ball with respect to elevator and solve for zero relative position, i.e., impact. • Substitute impact time into equation for position of elevator and relative velocity of ball with respect to elevator.
Example 8 :Sample Problem 11.4(Beer) SOLUTION: • Substitute initial position and velocity and constant acceleration of ball into general equations for uniformly accelerated rectilinear motion. v B = v0 + at = 18
m m − 9.81 2 t s s m m t − 4.905 2 t 2 s s
2 y B = y 0 + v0 t + 12 at = 12 m + 18
• Substitute initial position and constant velocity of elevator into equation for uniform rectilinear motion. v E = 2
m s
m t s
y E = y 0 + v E t = 5 m + 2
Example 8 :Sample Problem 11.4(Beer) • Write equation for relative position of ball with respect to elevator and solve for zero relative position, i.e., impact. 2 y B E = 12 + 18t − 4.905t − (5 + 2t ) = 0
t = −0.39 s (meaningless ) t = 3.65 s • Substitute impact time into equations for position of elevator and relative velocity of ball with respect to elevator. y E = 5 + 2(3.65) y E = 12.3 m v B E = (18 − 9.81t ) − 2
= 16 − 9.81(3.65) v B E = −19.81
m s
Example 9 (Hibbeler 12-7, 10th Ed.) The
position of particle along a straight line is given by s=(t 3-9t2+15t)ft where t is in seconds. Determine its maximum acceleration and maximum velocity during the time interval 0 ≤t ≤10 s.
Answer
Example 10 (Hibbeler 12-11, 10th Ed.) The
acceleration of particle as it moves along a straight line given by a=(2t-1)m/s 2, where t is in seconds. If s =1 m and v = 2m/s when t = 0, determine the particle’s velocity and position when t = 6s. Also, determine the total distance the particle travels during this time period.
Answer
Example 11 (Hibbeler 12-2, 10th Ed.) A
car starts from rest and reaches a speed of 20 fps after traveling 500 ft along a straight road. Determine its constant acceleration and the time of travel.
Answer .
Example 12 (Hibbeler 12-9, 10th Ed.) A
car is to be hoisted by elevator to the fourth floor of a parking garage, which is 48 ft above the ground. If the elevator can accelerate at 0.6 fps 2, decelerate at 0.3 fps2, and reach a maximum speed of 8 ft/s, determine the shortest time to make the lift, starting from rest and ending at rest.
Answer
Example 13 (Hibbeler 12-17, 10th Ed.) Two
particles A and B start from rest at the origin s=0 and move along a straight line such that a A=(6t-3) m/s2, and aB=(12t2-8) m/s2, where t is in seconds. Determine the distance between them when t = 4s and the total distance each has been traveled in t=4s.
Answer
Motion of Several Particles: Dependent Motion • Position of a particle may depend on position of one or more other particles. • Position of block B depends on position of block A. Since rope is of constant length, it follows that sum of lengths of segments must be constant. x + 2 x B = constant (one degree of freedom)
• Positions of three blocks are dependent. 2 x A + 2 x B + xC = constant (two degrees of freedom) • For linearly related positions, similar relations hold between velocities and accelerations. 2 2
dx A dt dv A dt
+2 +2
dx B dt dv B dt
+ +
dxC dt dvC dt
= 0 or 2v A + 2v B + vC = 0 = 0 or 2a A + 2a B + aC = 0
Example 14 : Sample Problem 11.5 (Beer) SOLUTION: • Define origin at upper horizontal surface with positive displacement downward. • Collar A has uniformly accelerated rectilinear motion. Solve for acceleration and time t to reach L. • Pulley D has uniform rectilinear motion. Pulley D is attached to a collar which Calculate change of position at time t . is pulled down at 75 mm/s. At t = 0, collar A starts moving down from K • Block B motion is dependent on motions of collar A and pulley D. Write motion with constant acceleration and zero relationship and solve for change of block initial velocity. Knowing that B position at time t . velocity of collar A is 300 mm/s as it passes L, determine the change in • Differentiate motion relation twice to elevation, velocity, and acceleration develop equations for velocity and of block B when block A is at L. acceleration of block B.
Example 14 : Sample Problem 11.5 (Beer) SOLUTION: • Define origin at upper horizontal surface with positive displacement downward. • Collar A has uniformly accelerated rectilinear motion. Solve for acceleration and time t to reach L. v A2 = (v A )0 + 2a A [ x A − ( x A )0 ] 2
(300 mm/s)2 = 2a A (200 mm)
a A = 225mm / s 2
v A = (v A )0 + a At 300 mm = 225t mm / s 2
t = 1.333 s
Example 14 : Sample Problem 11.5 (Beer) • Pulley D has uniform rectilinear motion.
Calculate
change of position at time t . x D = ( x D
0
+ v D t
x D − ( x D )0 = (75 mm/s)(1.333s) = 100 mm • Block B motion is dependent on motions of collar A and pulley D. Write motion relationship and solve for change of block B position at time t . Total length of cable remains constant,
) + 2( D )0 + ( x B )0 [ x A − ( x A )0 ] + 2[ x D − ( x D )0 ] + [ x B − ( x B )0 ] = 0 (200 mm) + 2(100 mm) + [ x B − ( x B )0 ] = 0 A
+2
D
+
B
=(
A 0
x B − ( x B )0 = −400 mm
Example 14 : Sample Problem 11.5 (Beer)
• Differentiate motion relation twice to develop equations for velocity and acceleration of block B. x A + 2 x D + x B = constant v A + 2v D + v B = 0
(300 mm/s) + 2(75 mm/s) + v B = 0 v B = −450 mm / s
a A + 2a D + a B = 0
(225 mm / s^2) + a B = 0
a B = −225 mm/s 2
Graphical Solution of RectilinearMotion Problems
• Given the x-t curve, the v-t curve is equal to the x-t curve slope. • Given the v-t curve, the a-t curve is equal to the v-t curve slope.
Graphical Solution of RectilinearMotion Problems
• Given the a-t curve, the change in velocity between t 1 and t 2 is equal to the area under the a-t curve between t 1 and t 2. • Given the v-t curve, the change in position between t 1 and t 2 is equal to the area under the v-t curve between t 1 and t 2.
Other Graphical Methods • Moment-area method to determine particle position at time t directly from the a-t curve: x1 − x0 = area under v − t curve v1
= v0t 1 + ∫ (t 1 − t )dv v0
using dv = a dt , v1
x1 − x0 = v0t 1 + ∫ (t 1 − t ) a dt v0 v1
∫ (t 1 − t ) a dt =
v0
first moment of area under a-t curve with respect to t = t 1 line.
x1 = x0 + v0 t 1 + (area under a-t curve)(t 1 − t ) t = abscissa of centroid C
Other Graphical Methods • Method to determine particle acceleration from v-x curve: dv dx = AB tan θ
a=v
= BC = subnormal to v-x curve
Curvilinear Motion: Position, Velocity & Acceleration • Particle moving along a curve other than a straight line is in curvilinear motion. • Position vector of a particle at time t is defined by a vector between origin O of a fixed reference frame and the position occupied by particle. • Consider particle which occupies position P defined r r by r at time t and P’ defined by r ′ at t + ∆t , r
r
∆r d r v = lim = ∆t →0 ∆t dt r
= instantaneous velocity (vector) ∆ s ds = ∆t →0 ∆t dt
v = lim
= instantaneous speed (scalar)
Curvilinear Motion: Position, Velocity & Acceleration r
• Consider velocity v of particle at time t and velocity r v′ at t + ∆t , r
r
∆v d v a = lim = ∆t →0 ∆t dt r
= instantaneous acceleration (vector) • In general, acceleration vector is not tangent to particle path and velocity vector.
Derivatives of Vector Functions r
• Let P (u ) be a vector function of scalar variable u, r
r
r
r
P (u + ∆u ) − P (u ) ∆ P = lim = lim du ∆u →0 ∆u ∆u →0 ∆u
d P
• Derivative of vector sum, r
r
d ( P + Q ) du
r
r
=
d P d Q
+
du
du
• Derivative of product of scalar and vector functions, r
d ( f P ) du
r
df r d P = P + f du du
• Derivative of scalar product and vector product, r
r
d ( P • Q ) du
r
=
d P du
r
r
d ( P × Q ) du
d P du
r
r
d Q
• Q + P •
r
=
r
r
du r
r
× Q + P ×
d Q du
Rectangular Components of Velocity & Acceleration • When position vector of particle P is given by its rectangular components, r
r
r
r
r = xi + y j + z k
• Velocity vector, r r r dx r dy r dz r r v = i + j + k = x&i + y& j + z &k dt dt dt r
r
r
= v x i + v y j + v z k • Acceleration vector, r r r d 2 x r d 2 y r d 2 z r a= i + j + k = x&&i + y && j + z &&k dt 2 dt 2 dt 2 r
r
r
r
= a x i + a y j + a z k
Rectangular Components of Velocity & Acceleration • Rectangular components particularly effective when component accelerations can be integrated independently, e.g., motion of a projectile, a x = x a y = y a z = &z && = 0 && = − g &=0 with initial conditions, (v x )0 , v y x0 = y0 = z 0 = 0 Integrating twice yields v x = (v x )0 v y = (v y )0 − gt
0
, (v z )0 = 0 v z = 0
x = (v x )0 t y = (v y )0 t − 12 gt 2 z = 0
• Motion in horizontal direction is uniform. • Motion in vertical direction is uniformly accelerated. • Motion of projectile could be replaced by two independent rectilinear motions.
Example 15(Hibbeler 12-67, 10th Ed.) The
velocity of a particle is given by v ={16t2i+4t3 j+(5t+2)k} m/s, where t is in seconds. If the particle is at the origin when t =0, determine the magnitude of the particle’s acceleration when t=2 s. Also, what is the x, y, z coordinate position of the particle at this instant?
Answer
Example 16(Hibbeler 12-74, 10th Ed.) particle moves along the curve y=e2x such that its velocity has constant magnitude of v=4 ft/s. Determine the x and y components of velocity when the particle is at y=5 ft.
A
Answer
Example 17(Hibbeler 12-78, 10th Ed.) The
particle travels along the path defined by the parabola y=0.5x 2. If the component of velocity along the x axis is v x=(5t)ft/s, where t is in seconds, determine the particle’s distance from the origin O and the magnitude of its acceleration when t=1s. When t=0, x=0, y=0.
Answer
Example 18(Hibbeler 12-79, 10th Ed.) When
a rocket reaches an altitude of 40 m it begins to travel along the parabolic path (y-40) 2=160x, where the coordinates are measured in meters. If the component of velocity in the vertical direction is constant at vy=180m/s, determine the magnitudes of the rocket’s velocity and acceleration when it reaches an altitude of 80 m. Answer
Example 19(Hibbeler 12-95, 10th Ed.) Determine
the horizontal velocity v A of a tennis ball at A so that it just clears the net at B. Also, find the distances s where the ball strikes the ground.
Answer
Example 20(Hibbeler 12-98, 10th Ed.) The
ball is thrown from the tower with a velocity of 20 ft/s as shown. Determine the x and y coordinates to where the ball strikes the slope. Also, determine the speed at which the ball hits the ground.
Answer
Example 21(Hibbeler 12-83, 10th Ed.) Determine
the maximum height on the wall to which the firefighter can project water from the hose, if the speed of the water at the nozzle is v c=48 ft/s.
Answer
Example 22(Hibbeler 12-84, 10th Ed.) Determine
the smallest angle θ, measured above the horizontal, that the hose should be directed so that the water stream strikes the bottom of the wall at B. The speed of the water at the nozzle is v C=48 ft/s.
Answer
Motion Relative to a Frame in Translation • Designate one frame as the fixed frame of reference. All other frames not rigidly attached to the fixed reference frame are moving frames of reference. • Position vectors for particles A and B with respect to r r the fixed frame of reference Oxyz are r A and r B . r
• Vector r B A joining A and B defines the position of B with respect to the moving frame Ax’y’z’ and r r r r B = r A + r B A • Differentiating twice, r
r
r
r
r
r
v B = v A + v B A a B = a A + a B A
r
v B A = velocity of B relative to A. r
a B A = acceleration of B relative to A.
• Absolute motion of B can be obtained by combining motion of A with relative motion of B with respect to moving reference frame attached to A.
Tangential and Normal Components • Velocity vector of particle is tangent to path of particle. In general, acceleration vector is not. Wish to express acceleration vector in terms of tangential and normal components. r
r
• et and et ′ are tangential unit vectors for the particle path at P and P’ . When drawn with r r r respect to the same origin, ∆et = et ′ − et and ∆θ is the angle between them.
∆et = 2 sin (∆θ 2 ) r
∆et sin (∆θ 2 ) r r lim en = en = lim ∆θ →0 ∆θ ∆θ →0 ∆θ 2 r
en =
r
d et d θ
Tangential and Normal Components r
r
• With the velocity vector expressed as v = vet the particle acceleration may be written as r r r d v dv r d e dv r d e d θ ds r a= = et + v = et + v dt dt dt dt d θ ds dt but r d et r ds = en =v ρ d θ = ds d θ dt After substituting, dv r v 2 r r a= et + en dt ρ
at =
dv dt
an =
v2
ρ
• Tangential component of acceleration reflects change of speed and normal component reflects change of direction. • Tangential component may be positive or negative. Normal component always points toward center of path curvature.
Tangential and Normal Components • Relations for tangential and normal acceleration also apply for particle moving along space curve. dv r v 2 r a= et + en dt ρ r
at =
dv dt
an =
v2
ρ
• Plane containing tangential and normal unit vectors is called the osculating plane. • Normal to the osculating plane is found from r
r
r
eb = et × en r
en = principal normal r
eb = binormal
• Acceleration has no component along binormal.
Example 23(Hibbeler 12-102, 10th Ed.) At
a given instant the jet plane has a speed of 400 ft/s and an acceleration of 70 ft/s2 acting in the direction shown. Determine the rate of increase in the plane’s speed and the radius of curvature ρ of the path.
Answer
Example 24(Hibbeler 12-100, 10th Ed.) A
car is traveling along a circular curve that has a radius of 50 m. If its speed is 16 m/s and is increasing uniformly at 8 m/s 2, determine the magnitude of its acceleration at this instant.
Answer
Example 25(Hibbeler 12-107, 10th Ed.) Starting
from rest, motorboat travels around the circular path, ρ, at a speed v=(0.8t) m/s, where t is in seconds. Determine the magnitudes of the boats velocity and acceleration when it has traveled 20 m.
Answer
Example 26(Hibbeler 12-115, 10th Ed.) The
truck travels in a circular path having a radius of 50 m at a speed is increased by a = (0.05s) m/s 2, where s is in meters. Determine its speed and the magnitude of its acceleration when it has moved s=10m. Answer
Example 27(Hibbeler 12-109, 10th Ed.) A
car moves along a circular track of radius 250 ft, and its speed for a short period of time 0 ≤t ≤2 s is v =3(t+t2) ft/s, where t is in seconds. Determine the magnitude of its acceleration when t=2s. How far has it traveled in t=2s?
Answer
Example 28(Hibbeler 12-121, 10th Ed.) The
box of negligible size is sliding down along a curved path defined by the parabola y=0.4x 2. When it is at A(x A=2m, y A=1.6m), the speed is v B=8m/s and the increase in speed is dv B/dt = 4m/s2. Determine the magnitude of the acceleration of the box at this instant.
Answer
Example 29(Hibbeler 12-128, 10th Ed.) A
boy sits on a merry-go-round so that he is always located at r=8 ft from the center of rotation. The merry-go-round is originally at rest, and then due to rotation the boy’s speed is increased at 2 ft/s 2. Determine the time needed for his acceleration to become 4 ft/s 2.
Answer
Radial and Transverse Components • When particle position is given in polar coordinates, it is convenient to express velocity and acceleration with components parallel and perpendicular to OP .
r
• The particle velocity vector is r d er dr r d r dr r d θ r r = er + r eθ v = (r er ) = er + r dt dt dt dt dt r r &e = r & er + r θ θ
r
r = r er r
d er d θ
= eθ
r
d er dt dt
d eθ
r
=
r
d eθ
r
r
d er d θ d θ dt r
=
d eθ d θ d θ dt
d θ r
r
= −er
= eθ
d θ
r
dt
= −er
d θ dt
• Similarly, the particle acceleration vector is r
a=
d dr r d θ r er + r eθ dt dt dt r
2
r
2
d r r dr d er dr d θ r d θ r d θ d eθ = + e + eθ + r eθ + r 2 r dt dt 2 dt dt dt dt dt dt
(
)r
r
&& + 2 r & )e = &r & − r θ & 2 er + (r θ &θ θ
Radial and Transverse Components • When particle position is given in cylindrical coordinates, it is convenient to express the velocity and acceleration vectors using the unit r r r vectors e R , eθ , and k . • Position vector, r
r
r
r = R e R + z k • Velocity vector, r r d r & r r r = R e R + Rθ & eθ + z v= & k dt • Acceleration vector, r
a=
r
d v
dt
(
)r
r
r
& 2 e + ( Rθ && + 2 R & )e + z && − Rθ & θ = R R θ && k
Example 30 (Sample Problem 11.10) SOLUTION: • Calculate tangential and normal components of acceleration. • Determine acceleration magnitude and direction with respect to tangent to curve. A motorist is traveling on curved section of highway at 100 km/h. The motorist applies brakes causing a constant deceleration rate. Knowing that after 8 s the speed has been reduced to 75 km/h, determine the acceleration of the automobile immediately after the brakes are applied.
Example 30 (Sample Problem 11.10) SOLUTION: • Calculate tangential and normal components of acceleration. at = an = 100 km / h = 27.8 m/s 75 km/ h = 20.8 m/s
∆v (20.8 ms − 27.8 m/s) = = −0.875 m/s 2 8s ∆t v
2
ρ
=
(27.8 m/s)2 750 m
= 1.03 m/s 2
• Determine acceleration magnitude and direction with respect to tangent to curve. a=
2
2 2 2 at + an = (− 0.875) + 1.03
α = tan −1
an at
= tan −1
1.03 0.875
a = 1.35 m/s
α = 49.7°
2
Example 31(Sample Problem 11.12) SOLUTION: • Evaluate time t for θ = 30o. • Evaluate radial and angular positions, and first and second derivatives at time t . Rotation of the arm about O is defined by θ = 0.15t 2 where θ is in radians and t in seconds. Collar B slides along the arm such that r = 0.9 - 0.12t 2 where r is in meters.
• Calculate velocity and acceleration in cylindrical coordinates. • Evaluate acceleration with respect to arm.
After the arm has rotated through 30o, determine (a) the total velocity of the collar, (b) the total acceleration of the collar, and (c) the relative acceleration of the collar with respect to the arm.
Example 31(Sample Problem 11.12) SOLUTION: • Evaluate time t for θ = 30o.
θ = 0.15 t 2 = 30° = 0.524 rad
t = 1.869 s
• Evaluate radial and angular positions, and first and second derivatives at time t . r = 0.9 − 0.12 t 2 = 0.481 m r & = −0.24 t = −0.449 m s &r &=
−0.24 m s 2
θ = 0.15 t 2 = 0.524 rad θ & = 0.30 t = 0.561rad s && = 0.30 rad s 2 θ
Example 31(Sample Problem 11.12) • Calculate velocity and acceleration. & = −0.449 m s vr = r
vθ = r θ & = (0.481m )(0.561rad s ) = 0.270 m s v β = tan −1 θ
2 + vθ 2 v = vr
vr v = 0.524 m s
β = 31.0°
ar = &r & − r θ & 2
= −0.240 m s 2 − (0.481m )(0.561rad s )2 = −0.391m s 2 && + 2r & aθ = r θ &θ
(
)
= (0.481m ) 0.3 rad s 2 + 2(− 0.449 m s )(0.561rad s ) = −0.359 m s 2 2 + aθ 2 a = ar
a γ = tan −1 θ
ar a = 0.531m s
γ = 42.6°
Example 31(Sample Problem 11.12) • Evaluate acceleration with respect to arm. Motion of collar with respect to arm is rectilinear and defined by coordinate r . a B OA = r && = −0.240 m s
2
Example 32(Hibbeler 12-140, 10th Ed.) If
a particle moves along a path such that r=(2cos t) ft and θ=(t/2) rad, where t is in seconds, plot the path r =f( θ) and determine the particle’s radial and transverse components of velocity and acceleration.
Answer
Example 33(Hibbeler 12-150, 10th Ed.) A
block moves outward along the slot in the platform with speed of (dr/dt)=(4t) m/s, where t is in seconds. The platform rotates at a constant rate of 6 rad/s. If the block starts from rest at the center, determine when t=1 s.
Answer
Example 34(Hibbeler 12-154, 10th Ed.) Because
of telescopic action, the end of the industrial robotic arm extends along the path of the limacon r=(1+0.5 cosθ)m. At the instant θ=π/4, the arm has an angular rotation (dθ/dt)=0.6 rad/s, which is increasing at (d2 θ/dt2)=0.25 rad/s2. Determine the radial and transverse components of the velocity and acceleration of the object A held in its grip at this instant.
Answer
Example 35(Hibbeler 12-163, 10th Ed.) A
particle P moves along the spiral path r=(10/ θ) ft, where θ is in radians. If it maintains a constant speed of v=20 ft/s, determine the magnitudes v r and vθ as functions of θ and evaluate each at θ=1 rad.
Answer.
Rectangular Coordinates (x,y) In a rectangular coordinate system, the position of a particle is described by the distance from two orthogonal lines (x- and y-axes). This coordinate system is convenient to use when the x- and ycomponents of motion are specified separately from each other and/or do not depend on each other.
Rectangular Coordinates (x,y) (cont.) Velocity acts tangent to the path. Acceleration generally does not act tangent to the path.
Normal/Tangential Coordinates (n,t)
