Kinematics of Particles
Short Description
Lecture...
Description
Ch. 7: Kinematics of Particles
316
7.0 Outline
316
Introduction Rectilinear Motion Plane Curvilinear Motion Rectangular Coordinates (x-y) Normal and Tangential Coordinates (n-t) Polar Coordinates (r-θ) Relative Motion (Translating Axes)
317 319 341 349 365 381 404
7.0 Outline
Ch. 7: Kinematics of Particles
317
7.1 Introduction Kinematics is the study of the motion of bodies with no consideration to the forces that accompany the motion. It is an absolute prerequisite to kinetics, which is the study of the relationships between the motion and the corresponding forces that cause the motion or are generated as a result of the motion. A particle is a body whose physical dimensions are so small compared with the radius of curvature of its path. This makes the body rotation effect insignificant and the motion of the body can be treated as that of the particle.
7.1 Introduction
Ch. 7: Kinematics of Particles
318
Position of P rectangular coordinates x, y, z cylindrical coordinates r, θ, z spherical coordinates R, θ, Φ Motion of P absolute motion analysis relative motion analysis Absolute motion analysis: coordinates measured from fixed reference axes, e.g. motion of the piston described by the frame fixed to the ground Relative motion analysis: coordinates measured from moving reference axes, e.g. motion of the piston described by the frame attached to the car 7.1 Introduction
Ch. 7: Kinematics of Particles
319
7.2 Rectilinear Motion: motion along a straight line If change in the position coordinate during ∆t is the displacement ∆s ( ± ) , v av =∆s/∆t ds = s ∆t → 0 dt velocity = time rate of change of the position coord., s If change in the velocity during ∆t is ∆v, a av =∆v/∆t
(1) __ instantaneous velocity, v= lim ∆s/∆t=
dv d 2s ∆v/∆t = = v = = s ( 2 ) __ instantaneous acceleration, a = ∆lim 2 t →0 dt dt ( 3) __ vdv = ads
7.2 Rectilinear Motion
Ch. 7: Kinematics of Particles
320
Displacement vs. Distance displacement: vector quantity involving initial and ending position distance: positive scalar quantity Both velocity and acceleration are vector quantities generally their changes include 1) change in magnitude and 2) change in direction For rectilinear motion, direction is the constant straight line path algebraic problem Integration of basic differential relations
7.2 Rectilinear Motion
Ch. 7: Kinematics of Particles s2
= ∫ ds s1
v2
= ∫ dv v1
t2
−s ∫ vdt, s = 2
1
area under v-t curve
t1
t2
−v ∫ adt, v = 2
1
area under a-t curve
t1
v2
s2
v1
s1
= ∫ vdv
321
2 2 ads, v − v ( ) /=2 area under a-s curve 2 1 ∫
graphic/numerical vs. algebraic approach
Relationships among several motion quantities 7.2 Rectilinear Motion
Ch. 7: Kinematics of Particles d dt
d dt
s v a ( v, s, t ) ∫ dt ∫ dt
kinetic relation
←
322
F ( v, s, t )
common problems: know a, find s by integration
7.2 Rectilinear Motion
Ch. 7: Kinematics of Particles
323
a) a = constant, e.g. G-force, dry friction force at the beginning, t 0,= s so = , v vo = v
at time t,
t
∫ dv = a ∫ dt
vo v
s
vo
so
→ v = v o + at
0
2 2 vdv a ds v v = → = o + 2a ( s − s o ) ∫ ∫ s
t
t
so
0
0
2 ds vdt v at dt s s v t at /2 = = + → = + + ( ) o o ∫ ∫ ∫ o
7.2 Rectilinear Motion
Ch. 7: Kinematics of Particles
324
b) a = f(t), e.g. synthetic force, piston force v
t
vo
0
s
t
so
0
∫ dv = ∫ f ( t ) dt
∫ ds = ∫ vdt
t
→ v = v o + ∫ f ( t ) dt 0
t
t t
0
0 0
→ s = s o + ∫ vdt = s o + v o t + ∫ ∫ f ( t ) dtdt
or s f= ( t ) with i.c. t o 0, so , vo
7.2 Rectilinear Motion
Ch. 7: Kinematics of Particles
325
c) a = f(v), e.g. viscous drag force, damping force
[a =
dv/dt ]
t
v
dv t = ∫ dt = ∫ → v = g(t) f v 0 vo ( ) inv
∫ dt → s = h (t)
or
[ vdv = ads]
v
s
v
v v ∫v f ( v ) dv = s∫ ds → s =so + v∫ f ( v ) dv = g ( v ) o o o
7.2 Rectilinear Motion
Ch. 7: Kinematics of Particles
326
d) a = f(s), e.g. spring force, attraction force v
s
s
vo
so
so
2 2 vdv = f s ds → v = v ( ) o + 2 ∫ f ( s ) ds → v = g ( s ) ∫ ∫ s
ds inv = →= s h (t) [ v ds/dt ]=t ∫ g (s ) so
7.2 Rectilinear Motion
Ch. 7: Kinematics of Particles P. 7/1
327
Small steel balls fall from rest through the opening at A at the steady rate of 2 per second. Find the vertical separation h of two consecutive balls when the lower one has dropped 3 meters. Neglect air resistance.
7.2 Rectilinear Motion
Ch. 7: Kinematics of Particles
328
P. 7/1 = [a s= ] a g from gravitational force, downward v =v o + gt and s =s o + v o t + gt 2 / 2 (3-h) m = v o 0= and define s o 0
∴ s= gt 2 / 2 3m @ tu
2 lower= ball: 3 gt= 0.782 s l / 2, t l
upper ball: t u =t l − 0.5 =0.282 s 2.61 m s u = 3 − h = gt 2u / 2 → h =
@ tl
7.2 Rectilinear Motion
Ch. 7: Kinematics of Particles
329
P. 7/2 In traveling a distance of 3 km between points A and D, a car is driven at 100 km/h from A to B for t seconds and at 60 km/h from C to D also for t seconds. If the brakes are applied for 4 s between B and C to give the car a uniform deceleration, calculate t and the distance s between A and B.
7.2 Rectilinear Motion
Ch. 7: Kinematics of Particles
330
a
P. 7/2
t t sec
t sec 4 sec
v A
B
C
D t
sB
t/3600
sA
0
= = [ v ds/dt ] ∫ ds sD
t/3600
sC
0
= ∫ ds
∫
sC
4/3600
sB
0
= ∫ ds
= ∫ vdt, sB t/36
vdt, = 3 − s C t/60
= ∫ vdt area under v-t curve,
1 s C − s B = × 4 / 3600 × (100 + 60 ) =4 / 45 2 t 65.5 sec, = s s= 1.819 km = B
7.2 Rectilinear Motion
Ch. 7: Kinematics of Particles P. 7/3
331
The 350-mm spring is compressed to a 200-mm length, where it is released from rest and accelerates the sliding block A. The acceleration has an initial value of 130 m/s2 and then decreases linearly with the x-movement of the block, reaching zero when the spring regains its original 350-mm length. Calculate the time t for the block to go a) 75 mm and b) 150mm.
7.2 Rectilinear Motion
Ch. 7: Kinematics of Particles
332
P. 7/3 v
x
= = [ vdv ads ] ∫ vdv
∫
ads
−0.15
0
v2 = −866.7x 2 + 19.5, v = 29.44 0.0225 − x 2 t
= = [ v ds/dt ] ∫ dt 0
( assume block move → )
x
ds ∫ v −0.15
a
t 0.034 {sin −1 ( x/0.15 ) + π / 2} =
130
@x = 0.0356 s −0.075 m, t = @ x 0= m, t 0.0534 s =
x -0.15
0
130 a= − x= −866.7x 0.15
[a =s ]
x + 866.7x =0
solution of the unforced harmonic equation x= Asinω t + Bcosω t, ω = 866.7 = 29.44 rad/s −0.15 m, x = −0.15, A = i.c.: t = 0, x = 0 m/s → B = 0 ∴ x =−0.15cos 29.44t
7.2 Rectilinear Motion
Ch. 7: Kinematics of Particles P. 7/4
333
A train that is traveling at 130 km/h applies its brakes as it reaches point A and slows down with a constant deceleration. Its decreased velocity is observed to be 96 km/h as it passes a point 0.8 km beyond A. A car moving at 80 km/h passes point B at the same instant that the train reaches point A. In an unwise effort to beat the train to the crossing, the driver ‘steps on the gas’. Calculate the constant acceleration a that the car must have in order to beat the train to the crossing by 4 s and find the velocity v of the car as it reaches the crossing.
7.2 Rectilinear Motion
Ch. 7: Kinematics of Particles
334
P. 7/4 const acceleration: v =v o + at, v 2 =v 02 + 2a ( s − s o ) , s =s o + v o t + at 2 / 2 Train: 962 = 1302 + 2a × 0.8, a = −4802.5 km/h 2 1.6 = 130t − 4802.5t 2 / 2, t = 0.0189 h or 68.11 s ( check with v = v o + at ) Car: to beat the train by 4 sec → t = 64.11 s or 0.0178 h 2 = 80 × 0.0178 + a × 0.01782 / 2, a = 3628.3 km/h 2 = 0.28 m/s 2 v= 80 + 3628.3 × 0.0178 = 144.6 km/h = 40.2 m/s
7.2 Rectilinear Motion
Ch. 7: Kinematics of Particles P. 7/5
335
The horizontal motion of the plunger and shaft is arrested by the resistance of the attached disk that moves through the oil bath. If the velocity of the plunger is vo in the position A where x = 0 and t = 0, and if the deceleration is proportional to v so that a = -kv, derive expressions for the velocity v and position coordinate x in terms of the time t. Also express v in terms of x.
7.2 Rectilinear Motion
Ch. 7: Kinematics of Particles
336
P. 7/5 = [a dv/dt ] = [ v dx/dt ]
v
x
dx ∫= 0
[ vdv=
ads ]
t
dv = ∫ −kv vo
v
− kt dt, v v e = o ∫ 0
t
− kt v e dt,= x o ∫ 0
vdv ∫v −kv= o
vo 1 − e − kt ) ( k
x
∫ ds,
= v v o − kx
0
7.2 Rectilinear Motion
Ch. 7: Kinematics of Particles P. 7/6
337
The electronic throttle control of a model train is programmed so that the train speed varies with position as shown in the plot. Determine the time t required for the train to complete one lap.
7.2 Rectilinear Motion
Ch. 7: Kinematics of Particles
338
P. 7/6 ∆s/v = 0 − 2 km: constant velocity, t = 2 / 0.25 = 8 sec −0.125 / (π / 2 ) = −0.25 / π 2 − ( 2 + π / 2 ) km: dv/ds = = = a dv/dt ] [ vdv ads, ∆t
0.125
/ π ) dt ∫ ∫ ( −0.25= 0
= v ( dv/ds ) dv/dt
= ∆t 8.71 sec dv/v,
0.25
lap time = 8 × 2 + 8.71× 4 = 50.84 sec
7.2 Rectilinear Motion
Ch. 7: Kinematics of Particles P. 7/7
339
A bumper, consisting of a nest of three springs, is used to arrest the horizontal motion of a large mass that is traveling at 40 m/s as it contacts the bumper. The two outer springs cause a deceleration proportional to the spring deformation. The center spring increases the deceleration rate when the compression exceeds 0.5 m as shown on the graph. Determine the maximum compression x of the outer spring.
7.2 Rectilinear Motion
Ch. 7: Kinematics of Particles
340
P. 7/7 0
= = [ vdv ads ] ∫ vdv area under a-s curve 40
0 − 402 1 1 =− × 0.5 × 1000 − × ( x − 0.5 ) × (1000 + 1000 + 4000 ( x − 0.5 ) ) 2 2 2 x = 0.831 m
7.2 Rectilinear Motion
Ch. 7: Kinematics of Particles
341
2.3 Plane Curvilinear Motion: motion along a curved path that lies in a single plane Vector quantity is independent of any particular coordinate system
7.3 Plane Curvilinear Motion
Ch. 7: Kinematics of Particles
342
Time derivative of a vector (described in fixed coord.) change in both magnitude and direction
at time t, the particle is at A located by r at time t + ∆t, the particle moves to B located by = r' r + ∆r displacement (vector) during time ∆t is ∆r (independent of coordinate system) distance traveled (scalar) during time ∆t is ∆s (measured along the path)
7.3 Plane Curvilinear Motion
Ch. 7: Kinematics of Particles
343
average velocity, v av = ∆r / ∆t average speed = ∆s / ∆t ∆r dr v lim = = r instantaneous velocity,= ∆t → 0 ∆t dt v includes the effect of change both in magnitude and direction of r as ∆t → 0, direction of ∆r approaches that of the tangent to the path ∴ average velocity → velocity, v av → v ∴ v is always a vector tangent to the path Consider only the magnitude of the velocity ds speed, = v v= = s dt as ∆t → 0, A → A' ∴ average speed → speed, ∆r → ∆s, v av → v , and v av → v
7.3 Plane Curvilinear Motion
Ch. 7: Kinematics of Particles
344
Magnitude of the derivative magnitude of the velocity=
dr = v = r = s= v= speed dt
Derivative of the magnitude dr dt
=
dr = r= rate at which the length of position vector r is changing dt
Derivative of the direction Derivative of the magnitude and Derivative of the direction of the vector contribute to Derivative of that vector
7.3 Plane Curvilinear Motion
Ch. 7: Kinematics of Particles
345
average acceleration, a av = ∆v / ∆t ∆v dv a lim = = v instantaneous acceleration,= ∆t → 0 ∆t dt a includes the effect of change both in magnitude and direction of v Because the magnitude v at any point can be arbitrary, generally the direction of the acceleration is niether tangent nor normal to the path but its normal component always points toward the center of curvature of the path
7.3 Plane Curvilinear Motion
Ch. 7: Kinematics of Particles
346
The acceleration bears the same relation to the velocity as the velocity bears to the position vector.
7.3 Plane Curvilinear Motion
Ch. 7: Kinematics of Particles
347
Derivatives and Integrations of vectors dP = P = Px i + P y j + P z k dt d ( Pu ) = Pu + P u dt d ( PQ ) + P Q = PQ dt d (P × Q) + P × Q = P×Q dt if V = V ( x, y, z ) and an element of volume, dτ = dxdydz
∫ Vdτ = i ∫ V ( x, y, z ) dτ + j∫ V ( x, y, z ) dτ + k ∫ V ( x, y, z ) dτ x
y
z
7.3 Plane Curvilinear Motion
Ch. 7: Kinematics of Particles
348
Three common coordinate systems for plane curvilinear motion – rectangular, normal & tangential, and polar Choose the appropriate coordinate system according to the manner in which the motion is generated or by the form in which the data are specified
7.3 Plane Curvilinear Motion
Ch. 7: Kinematics of Particles
349
2.4 Rectangular Coordinates (x-y) Suit for motions where the x- and y- components of the acceleration are independently generated or determined, i.e. x- and y- coordinates at a specific point are related by the same instant of time only. It looks like the superposition of two perpendicular rectilinear motions, in x- and y- directions, simultaneously. Their combination generate the curvilinear motion.
7.4 Rectangular Coordinates (x-y)
Ch. 7: Kinematics of Particles
350
r = xi + yj = x ( t ) i + y ( t ) j v = r = x i + y j = v x i + v y j (v x and v y are independent) xi + yj = a x i + a y j (a x and a y are independent) a = v = r = direction of velocity,= tanθ v= dy / dx y / vx speed, v = a =
v 2x + v 2y
a 2x + a 2y
= If x f= f 2 ( t ) , elimination of time t between these two 1 ( t ) and y parametric equations gives the equation of the curved path y = f ( x )
7.4 Rectangular Coordinates (x-y)
Ch. 7: Kinematics of Particles
351
Projectile motion: motion of the thrown object Neglect aerodynamic drag and the earth’s curvature and rotation, and assume the altitude range is small enough
a x = 0 and a y = −g
7.4 Rectangular Coordinates (x-y)
Ch. 7: Kinematics of Particles a= 0 x
v= x
( v x )o
x x o + ( v x )o t =
ay = −g v y = ( v y ) − gt o
v 2y =
352
y= y o + ( v y ) t − gt 2 / 2 o
( v y ) − 2g ( y − yo ) 2
o
x- and y- motions are independent elimination of time t of x ( t ) and y ( t ) gives the parabolic path
7.4 Rectangular Coordinates (x-y)
Ch. 7: Kinematics of Particles P. 7/8
353
A particle is ejected from the tube at A with a velocity v at an angle θwith the vertical y-axis. A strong horizontal wind gives the particle a constant horizontal acceleration a in the x-direction. If the particle strikes the ground at a point directly under its released position, determine the height h of point A. The downward y-acceleration may be taken as the constant g.
7.4 Rectangular Coordinates (x-y)
Ch. 7: Kinematics of Particles P. 7/8
354
ax = a
ay = g
since a is constant, the instant formulas can be used 1 ax = a, v x = − vsinθ + at, x = − vtsinθ + at 2 2 1 a y = g, v y = vcosθ + gt, y = vtcosθ + gt 2 2 when x 0= & y h, = 0= t ( at/2 − vsinθ ) , t = 2vsinθ /a 2v 2 g equation, h sin θ cos θ + sin θ = substitute t into y-coord a a
7.4 Rectangular Coordinates (x-y)
Ch. 7: Kinematics of Particles P. 7/9
355
Electrons are emitted at A with a velocity v at the angle θinto the space between two charged plates. The electric field between the plates is in the direction E and repels the electrons approaching the upper plate. The field produces an acceleration of the electrons in the E-direction of eE/m, where e is the electron charge and m is its mass. Determine the field strength E that will permit the electrons to cross one-half of the gap between the plates. Also find the distance s.
7.4 Rectangular Coordinates (x-y)
Ch. 7: Kinematics of Particles P. 7/9
u
y
eE m v 2y =+ v 02 2a y ( y − y o )
ax = 0
356
ay = −
x
at the peak, y = b/2 0 =
( usinθ )
2
mu 2 eE E sin 2 θ − 2 × b/2, = m eb
ay = -eE/m
1 eE x utcosθ = y utsinθ − × t 2 = 2 m s at (= s, 0 ) : s utcos and substitue into y-equation θ, t = ucosθ s 2b 0= stanθ 1 − tan θ , s = 0, tanθ 2b
7.4 Rectangular Coordinates (x-y)
Ch. 7: Kinematics of Particles P. 7/10
357
Water is ejected from the nozzle with a speed vo = 14 m/s. For what value of the angle θwill the water land closest to the wall after clearing the top? Neglect the effects of wall thickness and air resistance. Where does the water land?
(2)
(1)
7.4 Rectangular Coordinates (x-y)
Ch. 7: Kinematics of Particles
358
P. 7/10 Treat water as stream of particles. At one particular particle, @ just above the wall (19, 1) x (= = v x )o t 19 14t cos θ 1 2 1 2 1 2 = + + = + − + y y v t a t 1 0.3 14t sin gt , 0.7 gt = 14t sin θ θ ( ) o y y o 2 2 2 2
1 142 t 2 =192 + 0.7 + gt 2 , t =2.14, 1.81 sec and θ =50.64°, 41.43° 2 = θ 50.64° makes the water land closest to the wall water will land at ( x, 0 ) 1 0= 0.3 + 14t sin 50.64 − gt 2 , t = 2.234 sec 2 = = x 14t cos 50.64 19.835 m ∴ water lands 0.835 m to the right of B
7.4 Rectangular Coordinates (x-y)
Ch. 7: Kinematics of Particles P. 7/11
359
A projectile is ejected into an experimental fluid at time t = 0. The initial speed is vo and the angle to the horizontal is θ. The drag on the projectile results in an acceleration term a D = −kv , where k is a constant and v is the velocity of the projectile. Determine the x- and y-components of both the velocity and displacement as functions of time. What is the terminal velocity? Include the effects of gravitational acceleration.
7.4 Rectangular Coordinates (x-y)
Ch. 7: Kinematics of Particles P. 7/11
360
a = −kv = −k ( v x i + v y j) dv − kv x =x ax = dt vx
∫
vo cosθ t
1 = dv x − kv x
t
dt, v x ∫=
dx dt
θ ) e− kt ( vo cos=
0
x
dx, x ) e dt ∫= ∫ ( vo cos θ = − kt
0
0
v o cos θ 1 − e − kt ) ( k
dv y a y =− kv y − g = dt vy
∫
vo sin θ
1 = dv y − kv y − g
g − kt g dy = + − dt, v v sin θ y o e = ∫0 k k dt t
g − kt g + v sin θ e − dt= ∫0 o k k t
terminal velocity:
( v x )t
∞
= 0,
y
∫ dy, =y 0
(v )
y t ∞
1 g g − kt e v sin + 1 − − t θ ) o ( k k k
= −
g k
7.4 Rectangular Coordinates (x-y)
Ch. 7: Kinematics of Particles
361
P. 7/12 A projectile is launched with speed vo from point A. Determine the launch angle θthat results in the maximum range R up the incline of angle α (where 0 ≤ α≤ 90°). Evaluate your results for α = 0, 30, and 45°.
7.4 Rectangular Coordinates (x-y)
Ch. 7: Kinematics of Particles P. 7/12
362
= a x 0,= v x v o cos θ= , x v o t cos θ ay = −g, v y = v o sin θ − gt, y = v o t sin θ − gt 2 / 2 at B, Rcos α v o t cos θ= = →t
Rcosα v o cos θ
Rcosα Rsinα v o = v o cos θ
g Rcosα sin θ − 2 v o cos θ gRcos 2α sin α cos α tan θ − 2 = 2v o cos 2 θ
2
2v o2 cos 2 θ tan α = 2v o2 sin θ cos θ − gRcosα dR 0 = dθ v o2 ( 2 cos 2θ dθ ) − gcosα dR −2v o2 tan α ( 2 cos θ sin θ dθ ) =
change in θ causes change in R, ∴ max R when
2v o2 dR = ( cos 2θ + sin 2θ tan α )= 0 dθ gcosα tan 2θ = −1/ tan α 1 1 2θ = tan −1 − 180 − tan −1 = tan α tan α = 180 − ( 90 − α= ) 90 + α
θ=
90 + α 2
7.4 Rectangular Coordinates (x-y)
Ch. 7: Kinematics of Particles P. 7/13
363
Determine the equation for the envelope a of the parabolic trajectories of a projectile fired at any angle but with a fixed muzzle velocity u. (Hint: Substitute m = tanθ, where θis the firing angle with the horizontal, into the equation of the trajectory. The two roots m1 and m2 of the equation written as a quadratic in m give the two firing angles for the two trajectories shown such that the shells pass through the same point A. Point A will approach the envelope a as the two roots approach equality.) Neglect air resistance and assume g is constant.
7.4 Rectangular Coordinates (x-y)
Ch. 7: Kinematics of Particles a P. 7/13 = x
364
0,= v x ucos= θ , x utcosθ
ay = usinθ − gt, y = utsinθ − gt 2 / 2 −g, v y = eliminate t to get the trajectory equation
g x2 y xtanθ − 2 = 2 u cos 2 θ given point ( x, y ) , find θ by letting m = tan θ sec 2 θ = 1 + m2 1 + tan 2 θ =
gx 2 m 2 − 2xu 2 m + ( 2yu 2 + gx 2 ) = 0
two roots ∴ point ( x, y ) can be reached from two distinct paths point ( x, y ) will approach the envelope a as those two paths approach each other → two distinct roots become two repeated roots ∴ discriminant must be zero
( 2xu )
2 2
− 4gx 2 ( 2yu 2 + gx 2 ) = 0
u 2 gx 2 = y − 2g 2u 2 ∴ point ( x, y ) on the envelope must obey this relation → envelope equation
7.4 Rectangular Coordinates (x-y)
Ch. 7: Kinematics of Particles
365
2.5 Normal and Tangential Coordinates (n-t) It is a moving coordinate system, along the path with the particle. The +n direction points toward the center of curvature. At point A, e n = unit vector in the n-direction e t = unit vector in the t-direction
ρ = radius of curvature of the path
7.5 Normal and Tangential Coordinates (n-t)
Ch. 7: Kinematics of Particles
366
position of the particle = ( 0, 0 ) ds ρ dβ = dt dt v = ve t = ρβ e t along the t-axis β > 0 when particle moves along + t direction
= v
dv a = = ve t + v e t dt
7.5 Normal and Tangential Coordinates (n-t)
Ch. 7: Kinematics of Particles
367
Determine e t
β e n de= dβ e n → e= e t dβ e= t n t de t detour : e n = and e n = − β e t dβ ∴ a = vβ e + v e n
t
β ρβ= a= v= n 2
v2
ρ
a t = v = s= ρβ + ρβ vdv = a t ds v = v x i + v y j = ve t a = a x i + a y j = a n en + a t et
7.5 Normal and Tangential Coordinates (n-t)
Ch. 7: Kinematics of Particles
368
a n = change in direction of v dv n vdβ = = v= β ρβ 2 dt dt component a n always directed toward the center of curvature a= n
a t = change in magnitude of v dv t dv = = v= s dt dt component a t will be in + t direction if speed v is increasing a t=
component a t will be in − t direction if speed v is decreasing v has the direction along the motion of the object a may not have the direction along the motion of the object
7.5 Normal and Tangential Coordinates (n-t)
Ch. 7: Kinematics of Particles P. 7/14
369
The camshaft drive system of a four-cylinder automobile engine is shown. As the engine is revved up, the belt speed v changes uniformly from 3 m/s to 6 m/s over a 2 second interval. Calculate the magnitudes of the accelerations of points P1 and P2 half way through this time interval.
7.5 Normal and Tangential Coordinates (n-t)
Ch. 7: Kinematics of Particles
370
P. 7/14 dv ∆v = = ( 6 − 3) / 2 = 1.5 m/s 2 dt ∆t belt speed v = 4.5 m/s at the time half of the interval
uniform speed change → a t =
a P1 =
a n2 + a 2t =
( 4.52 / 0.06 ) + 1.52 = 337.5 m/s2 2
a P2= a= 1.5 m/s 2 t ∴ the sprocket and the belt must be very strong!
7.5 Normal and Tangential Coordinates (n-t)
Ch. 7: Kinematics of Particles P. 7/15
371
A baseball player releases a ball with the initial conditions shown in the figure. Determine the radius of curvature of the trajectory (a) just after release and (b) at the apex. For each case, compute the time rate of change of the speed.
7.5 Normal and Tangential Coordinates (n-t)
Ch. 7: Kinematics of Particles
372
P. 7/15 t v = 30 m/s
Just after release: v 2 302 a= gcos30 , = = = ρ 105.9 m n
ρ
a=g
v = at = −gsin30 = −4.905 m/s 2
n t v = 30cos30 m/s a=g n
ρ
At the apex: a n= g=
v
2
ρ
=
( 30 cos 30 ) ρ
2
, ρ= 68.8 m
2 a= v= 0 m/s t
7.5 Normal and Tangential Coordinates (n-t)
Ch. 7: Kinematics of Particles P. 7/16
373
Pin P in the crank PO engages the horizontal slot in the guide C and controls its motion on the fixed vertical rod. Determine the velocity y and the acceleration y of guide C for a given value of the angle θ if (a) θ = ω and θ = 0 and (b) if θ = 0 and θ = α .
7.5 Normal and Tangential Coordinates (n-t)
Ch. 7: Kinematics of Particles
374
P. 7/16 n
v = rω
t
θ n
transmitted by those of pin P along y direction
sin θ , y rω 2 cos θ = ∴ y rω =
n t
v=0 θ n
a = at = rα θ
2 for pin= P: v r= , at 0 ω , a n rω=
guide C must have its velocity and acceleration, y and y,
a = an = rω2 t θ
(a) , θ 0 θ ω= =
t
(b) θ 0,= θ α = for pin P: = v 0,= a n 0,= a t rα guide C must have its velocity and acceleration, y and y, transmitted by those of pin P along y direction ∴ y= 0, y= rα sin θ
7.5 Normal and Tangential Coordinates (n-t)
Ch. 7: Kinematics of Particles P. 7/17
375
The pin P is constrained to move in the slotted guides that move at right angles to one another. At the instant represented, A has a velocity to the right of 0.2 m/s which is decreasing at the rate of 0.75 m/s each second. At the same time, B is moving down with a velocity of 0.15 m/s which is decreasing at the rate of 0.5 m/s each second. For this instant determine the radius of curvature ρ of the path followed by P. Is it possible to determine also the time rate of change of ρ.
7.5 Normal and Tangential Coordinates (n-t)
Ch. 7: Kinematics of Particles
376
P. 7/17 Guides A and B move perpendicular to each other,
y
at
x an n
v t
∴ motion of guides A and B are independent and their motions are imparted to pin P. = v 0.2i − 0.15 j a = −0.75i + 0.5j in n-t description n t =− 0.8i 0.6 j, n n ⊥ n t and along a n −0.9 m/s 2 , a t = −0.72i + 0.54 j at = an t = −0.03i − 0.04 j a n =− a at = v2 0.252 = = an , ρ = 1.25 m ρ 0.05 = ρβ + ρ ( v/ρ ) ρ appears only in eq.: a t = v = ρβ + ρβ which has 2 unknowns: ρ and β ∴ ρ cannot be determined until β is known
7.5 Normal and Tangential Coordinates (n-t)
Ch. 7: Kinematics of Particles P. 7/18
377
As a handling test, a car is driven through the slalom course shown. It is assumed that the car path is sinusoidal and that the maximum lateral acceleration is 0.7g. If the testers wish to design a slalom through which the maximum speed is 80 km/h, what cone spacing L should be used?
7.5 Normal and Tangential Coordinates (n-t)
Ch. 7: Kinematics of Particles
378
P. 7/18 max a n 0.7g and max speed 80 km/h = = π ω x 3sin x sinusoidal path = → y 3sin= L y' = 3ω cos ω x and y'' = −3ω 2 sin ω x max a n at peak of the wave a n =
v2
ρ
and ρ min at peak
by sense: fast car needs big L → use max speed to find lower bound of L 3/ 2 1 + y'2 ρ = y''
ρ min = At peak, ρ = 71.9 m (80 ×10 / 36 ) / ( 0.7g ) = 2
±1, ± 3, → y' = 3ω 2 ωx = nπ /2, n = 0, y'' = π 1 71.9 = = = → L = 46.14 m ω , 0.0681 3ω 2 L
7.5 Normal and Tangential Coordinates (n-t)
Ch. 7: Kinematics of Particles P. 7/19
379
A particle starts from rest at the origin and moves along the positive branch of the curve y = 2x3/2 so that the distance s measured from the origin along the curve varies with the time t according to s = 2t3, where x, y, and s are in millimeters and t is in seconds. Find the magnitude of the total acceleration of the particle when t = 1 s.
7.5 Normal and Tangential Coordinates (n-t)
Ch. 7: Kinematics of Particles
380
P. 7/19 s= 2t 3 , v= s= 6t 2 , a t = v = 12t 3/ 2
1 + ( y')2 2 = a n v= /ρ & ρ y'' 3 3/2 = y 2x = , y' 3= x , y'' 2 x ∴ x must be known to determine ρ
∫ ds =∫
( dx ) + ( dy ) 2
2
=∫ 1 + ( y') dx 2
t =→ 0 1, s =→ 0 2, x =→ 0 x x
2 =+ 0.913 mm y = 1.746 mm ∫ 1 9xdx, x = 0
2 = At t 1= s, v 6 mm/s, = a t 12 mm/s = , y' 2.8665, = y'' 1.57, = ρ 17.8 mm
= a n 62 /17.8 = 2.02 mm/s 2 = → a 12.17 mm/s 2
7.5 Normal and Tangential Coordinates (n-t)
Ch. 7: Kinematics of Particles
381
2.6 Polar Coordinates (r-θ) The particle is located by the radial distance r from a fixed pole and by an angle θto fixed radial line.
7.6 Polar Coordinates (r-θ)
Ch. 7: Kinematics of Particles
382
r = re r from the figure, de r = eθ dθ and deθ = −e r dθ θe and e = ∴ e = −θe r
θ
θ
r
v = r = re r + re r = re r + rθeθ v r = r due to the stretch of r v = rθ due to the rotation of r θ
v =
v 2r + vθ2
7.6 Polar Coordinates (r-θ)
Ch. 7: Kinematics of Particles
(
) (
)
383
(
)
a = v = (re r + re r ) + rθeθ + rθeθ + rθeθ =r − rθ 2 e r + rθ + 2rθ eθ a r = r − rθ 2
( )
1 d 2 aθ =rθ + 2rθ = r θ , 2rθ combines two effects r dt a =
a 2r + aθ2
Note: a r ≠ v r and aθ ≠ v θ , i.e. we must also account for the change in direction of both v and v , which are rθ along e r
θ
θ
and − rθ 2 along e r , respectively
7.6 Polar Coordinates (r-θ)
Ch. 7: Kinematics of Particles P. 7/20
384
The rocket is fired vertically and tracked by the radar shown. When θreaches 60°, other corresponding measurements give the values r = 9 km, r = 21 m/s2, and θ = 0.02 rad/s. Calculate the magnitudes of the velocity and acceleration of the rocket at this position.
7.6 Polar Coordinates (r-θ)
Ch. 7: Kinematics of Particles
385
P. 7/20 30
r
θ
v and a of the rocket is in the vertical direction v = rθ = 180 m/s = vsin30 → v = 360 m/s θ
v r = r= vcos30= 311.77 m/s a =r − rθ 2 =17.4 m/s 2 = acos30 → a = 20.09 m/s 2 r
7.6 Polar Coordinates (r-θ)
Ch. 7: Kinematics of Particles P. 7/21
386
Link AB rotates through a limited range of the angle β, and its end A causes the slotted link AC to rotate also. For the instant represented where β= 60°and β = 0.6 rad/s constant, determine the corresponding values of r, θ, and θ. r,
7.6 Polar Coordinates (r-θ)
Ch. 7: Kinematics of Particles
387
P. 7/21
Description of position, velocity, and acceleration of A in r-θ when they are given in n-t coordinate system v = ρβ = 0.15 × 0.6 = 0.09 m/s along + t-direction
r 30
30
θ
t
β =0 → a t =0 2 = a a= 0.054 m/s 2 along + n-direction ρβ= n
n
From geometry, = r 0.15 m and θ= 60° v = v r + vθ = vcos30e r − vsin30eθ = re r + rθeθ r = 0.078 m/s θ = −0.3 rad/s
(
)
(
)
r rθ 2 e r + rθ + 2rθ eθ a = a r + aθ = −asin30e r − acos30eθ =− r = 2.31E-4 rad/s 2 −0.0135 m/s 2 θ =
7.6 Polar Coordinates (r-θ)
Ch. 7: Kinematics of Particles P. 7/22
388
The slotted arm OA forces the small pin to move in the fixed spiral guide defined by r = Kθ. Arm OA starts from rest at θ=π/4 and has a constant counterclockwise angular acceleration θ = α . Determine the magnitude of the acceleration of the pin when θ= 3π/4.
7.6 Polar Coordinates (r-θ)
Ch. 7: Kinematics of Particles
389
P. 7/22 pin is constrained to move in the guide ∴ path of the pin: r = Kθ = θ and r Kθ hold and differential relations r K= θ =→ π / 4 3π / 4, θ = α= 0 → θ, θ = constant θ 2 = θo2 + 2θ(θ − θ o ) = π / 4, θ 2 πα , θ = α At θ 3= r 3K = = π /4, r K= πα , r Kα a = r − rθ 2 e r + rθ + 2rθ eθ
(
)
(
)
a = 10.753Kα
7.6 Polar Coordinates (r-θ)
Ch. 7: Kinematics of Particles P. 7/23
390
The circular disk rotates about its center O with a constant angular velocity ω = θ and carries the two spring-loaded plungers shown. The distance b that each plunger protrudes from the rim of the disk varies according to b = bosin2πnt, where bo is the maximum protrusion, n is the constant frequency of oscillation of the plungers in the radial slots, and t is the time. Determine the maximum magnitudes of the r- and θ-components of the acceleration of the ends A of the plungers during their motion.
7.6 Polar Coordinates (r-θ)
Ch. 7: Kinematics of Particles
391
P. 7/23 location of point A described by ( r, θ ) r = ro + b = ro + bo sin 2π nt r = 2π nb o cos 2π nt, r = − ( 2π n ) b o sin 2π nt = θ ω= , θ 0 r − rθ 2 = −r ω 2 − ( 4π 2 n 2 + ω 2 ) b sin 2π nt a = 2
r
o
o
aθ =rθ + 2rθ =4π nω b o cos 2π nt ar
max
= roω 2 + ( 4π 2 n 2 + ω 2 ) b o
aθ
max
= 4π nω b o
7.6 Polar Coordinates (r-θ)
Ch. 7: Kinematics of Particles P. 7/24
392
The small block P starts from rest at time t = 0 at point A and moves up the incline with constant acceleration a. Determine r and θ as a function of time.
s
7.6 Polar Coordinates (r-θ)
Ch. 7: Kinematics of Particles P. 7/24
393
1 block moves with constant acceleration → s = at 2 2 at 2 by geometry, x = R + scosα = R+ cos α , x = atcosα 2 at 2 y ssin sin α= , y atsinα = = α 2 a2 4 2 2 2 2 2 r = x + y = R + Rat cos α + t 4 at ( 2Rcosα + at 2 ) 2 3 2rr= 2Ratcosα + a t , r= a2 4 2 2 2 R + Rat cos α + t 4 y xy − yx = , θsec 2θ x x2 1 r2 2 sec = θ = cos 2 θ x 2 xy − yx Ratsinα θ = = 2 a2 4 r 2 2 R + Rat cos α + t 4
= tan θ
7.6 Polar Coordinates (r-θ)
Ch. 7: Kinematics of Particles P. 7/25
394
The block P slides on the surface shown with constant speed v = 0.6 m/s and passes point O at time t = 0. If R = 1.2 m, determine the following quantities at time t = 2(1+π/3): θ, r, and θ r, θ , r,
7.6 Polar Coordinates (r-θ)
Ch. 7: Kinematics of Particles P. 7/25 θ
15
t 30
n a ds v = dt
s
2(1+π / 3)
0
0
∫ ds=
∫
395
v r 15
vdt, = s 1.2 + 0.4π
0.4π = π / 3= 60° 1.2 at this position, x = R + Rcos30 = 2.239 m y = R − Rsin30 = 0.6 block P moves up the quarter for
r = x 2 + y 2 =2.318 m θ =tan −1 ( y/x ) =15° v r = vcos45= r= 0.424 m/s , θ 0.183 rad/s v vsin45 = = rθ= θ
v 2 0.62 a t= v= 0 a n= = = 0.3= a ρ 1.2 r − rθ 2 , r = a = −0.3cos 45 = −0.134 m/s 2 r
aθ = 0.3sin 45 = rθ + 2rθ, θ = 0.0245 rad/s 2
7.6 Polar Coordinates (r-θ)
Ch. 7: Kinematics of Particles P. 7/26
396
The slotted arm OA oscillates about O within the limits shown and drives the crank CP through the pin P. For an interval of the motion, θ = K , a constant. Determine the magnitude of the corresponding total acceleration of P for any value of θ within the range for which θ = K . Use polar coordinates r and θ. Show that the magnitudes of the velocity and acceleration of P in its circular path are constant.
7.6 Polar Coordinates (r-θ)
Ch. 7: Kinematics of Particles P. 7/26
397
b β
θ
bθ
r
= θ K= θ 0 = β 2= θ β 2K = β 0 motion of pin P is the circular path with radius b 2bK = v ρβ = = constant a n= v 2 / ρ= 4bK 2 a t= v= 0,
a= 4bK 2= constant
7.6 Polar Coordinates (r-θ)
Ch. 7: Kinematics of Particles P. 7/27
398
The earth satellite has a velocity v = 17,970 km/h as it passes the end of the semiminor axis at A. Gravitational attraction produces an acceleration a = ar = -1.556 m/s2 as calculated from the gravitational law. For this position, calculate at which the speed of the satellite is changing and the rate v
r. the quantity
7.6 Polar Coordinates (r-θ)
Ch. 7: Kinematics of Particles P. 7/27
399
r
v
t
60° θ
a n
at = v = −acos60 = −0.778 m/s 2 r − rθ 2 but θ is unknown too a = a= −1.556 = r
vθ = rθ = vcos30, θ = 2.7E-4 rad/s, r = −0.388 m/s 2
7.6 Polar Coordinates (r-θ)
Ch. 7: Kinematics of Particles P. 7/28
400
Pin A moves in a circle of 90 mm radius as crank AC revolves at the constant rate β = 60 rad/s. The slotted link rotates about point O as the rod attached to A moves in and out of the slot. For the position β= 30°, r, θ, and θ . determine r,
7.6 Polar Coordinates (r-θ)
Ch. 7: Kinematics of Particles θ
P. 7/28
401
t 30
v r
r
θ a n
2 r= 3002 + 902 − 2 × 300 × 90 cos 30, = r 226.57 mm 90 r = = , θ 11.456° sinθ sin30 2 v A 5.4 m/s and = ρβ= = 324 m/s 2 v= a a= A n
ρ
v r = r= v A cos 48.54= 3.575 m/s θ v sin 48.54, = θ 17.86 rad/s v= r= θ
A
r − rθ 2 , r = ar = a n cos 41.456 = 315 m/s 2 a = −a sin 41.456 = rθ + 2rθ, θ = −1510 rad/s 2 θ
n
7.6 Polar Coordinates (r-θ)
Ch. 7: Kinematics of Particles P. 7/29
402
If the slotted arm is revolving CCW at the constant rate of 40 rev/min and the cam is revolving clockwise at the constant rate of 30 rev/min, determine the magnitude of the acceleration of the center of the roller A when the cam and arm are in the relative position for which θ= 30°. The limacon has the dimensions b = 100 mm and c = 75 mm. (Caution: Redefine the coordinates as necessary after noting that the θin the expression r = b – ccosθis not the absolute angle appearing in Eq. 2/14.)
7.6 Polar Coordinates (r-θ)
Ch. 7: Kinematics of Particles
403
P. 7/29 pin A must always be in the slot → θ = 40 rev/min CCW if β is the angle the cam is revolving at the rate β = 30 rev/min CW, then r = b − ccos (θ + β ) π /6
dθ ω = = ∫0 dθ dt β
0.125
0
0
∫ dβ = ∫
t
θdt, t ∫=
0.125 s
0
β dt, = β 0.3927 rad = 22.5°
= and β 0 θ 0= r= 0.1 − 0.075cos ( 30 + 22.5 ) = 54.3 mm
( ) c (θ + β )
r = c θ + β sin (θ + β ) = 0.436 r =
2
cos (θ + β ) = 2.453
2 r − rθ 2 = ar = 2.453 − 0.0543 ( 40 × 2π / 60 ) = 1.5 m/s 2 a = rθ + 2rθ =× 2 0.436 × ( 40 × 2π / 60 ) = 3.653 m/s 2
θ
= a
a 2r + aθ2 = 3.95 m/s 2
7.6 Polar Coordinates (r-θ)
Ch. 7: Kinematics of Particles
404
2.7 Relative Motion (Translating Axes) It is in many cases easy and practical to observe motion of an object w.r.t. a moving reference system. And its absolute motion can be determined by combining the relative motion with the absolute motion of the moving reference frame. Fixed coordinate system ? has no motion in space -- earth-ground -- non-rotating coordinate system with origin on the earth’s axis of rotation -- non-rotating coordinate system fixed to the sun
7.7 Relative Motion (Translating Axes)
Ch. 7: Kinematics of Particles
405
At this moment, we will study the relative plane motion which employs the pure-translation moving reference system. That is the relative motion will be observed on the moving reference frame that has no rotation. Observe motion of A from point B. Measurements are made referenced with the pure translating moving coordinate. Find the absolute motion of A described in fixed coordinate. Two chosen frames involved: 1. fixed frame X-Y, e.g. 2. pure translating moving frame x-y, e.g. Note : Coordinates of X-Y and x-y may not parallel to each other ++Separation of the frame, vector, and its description ++Consistency of the description for the vector quantities in a particular equation ++Transformation between two descriptions
7.7 Relative Motion (Translating Axes)
Ch. 7: Kinematics of Particles
406
rA and rB are measured referenced with the fixed coordinate frame rA/B is measured referenced with the pure translating moving coordinate frame. rA = rB + rA/B
vA = v B + v A/B
aA = a B + a A/B
describe rA/B in a specific pure translating moving frame x-y = rA/B x= i + yj v A/B x= i + y j a A/B xi + yj For the relative motion using the pure translating coordinate system a A/B = −a B/A rA/B = −rB/A v A/B = − v B/A if the pure-translating moving frame has constant velocity, a B = 0 ∴ a A = a A/B → determination of the absolute acceleration can be made on the inertial system, a pure-translating system that has no acceleration → Newton's 2nd law holds in the inertial as well as in the fixed reference frame
7.7 Relative Motion (Translating Axes)
Ch. 7: Kinematics of Particles P. 7/30
407
The car A has a forward speed of 18 km/h and is accelerating at 3 m/s2. Determine the velocity and acceleration of the car relative to observer B, who rides in a nonrotating chair on the ferris wheel. The angular rate ω= 3 rev/min of the ferris wheel is constant.
7.7 Relative Motion (Translating Axes)
Ch. 7: Kinematics of Particles
408
P. 7/30 observer B rides in a nonrotating chair → car A is observed through the pure translating moving coord frame 10 = 18 × 5i m/s v = v + v v i= [ A B A/B ] A 36 2i − 2 j m/s vB = ρβ ( cos 45i − sin 45 j) = v A/B =v A − v B =3i + 2 j m/s
[aA = aB + aA/B ] aA = 3i m/s2 ρβ 2 ( cos 45i − sin 45 j) = a B =− −0.628i − 0.628 j m/s 2 a A/B = a A − a B = 3.628i + 0.628 j m/s 2
7.7 Relative Motion (Translating Axes)
Ch. 7: Kinematics of Particles P. 7/31
409
Hockey player A carries the puck on his stick and moves in the direction shown with a speed vA = 4 m/s. In passing the puck to his stationary teammate B, by what shot angle α should the direction of his shot trail the line of sight if he launches the puck with a speed of 7 m/s relative to himself?
7.7 Relative Motion (Translating Axes)
Ch. 7: Kinematics of Particles
410
P. 7/31
player A saw the puck moves in ( 45 + α ) ° direction but in fact, the puck moved to stationary player B which stood in 45° direction v P = v A + v P/A 7 4 = , = α 23.8° sin 45 sin α
7.7 Relative Motion (Translating Axes)
Ch. 7: Kinematics of Particles P. 7/32
411
The aircraft A with radar detection equipment is flying horizontally at 12 km and is increasing its speed at the rate of 1.2 m/s each second. Its radar locks onto an aircraft flying in the same direction and in the same vertical plane at an altitude of 18 km. If A has a speed of 1000 km/h at the instant that θ= 30°, determine the values of r and θ at this same instant if B has a constant speed of 1500 km/h θ r
7.7 Relative Motion (Translating Axes)
Ch. 7: Kinematics of Particles
412
P. 7/32 = = r 6000 / sin 30 12000 m 10 10 = 416.67i m/s v A = 1000 × = 277.78i m/s 36 36 = v B − v A = 138.89i m/s = rθeθ + re r
v B = 1500 × v B/A
v r = r = v B/A cos 30 → r = 120.28 m/s − v sin 30 → θ = −5.787E-3 rad/s v = rθ = θ
B/A
= a B 0= a A 1.2i m/s 2
(
)
(
)
r − rθ 2 e r + rθ + 2rθ eθ a B/A = a B − a A = −1.2i m/s 2 = a r =r − rθ 2 =− a B/A cos 30 → r =−0.637 m/s 2 aθ = rθ + 2rθ = a B/A cos 60 → θ = 1.66E-4 rad/s 2
7.7 Relative Motion (Translating Axes)
Ch. 7: Kinematics of Particles P. 7/33
413
A batter hits the baseball A with an initial velocity of vo = 30 m/s directly toward fielder B at an angle of 30°to the horizontal; the initial position of the ball is 0.9 m above the ground level. Fielder B requires ¼ s to judge where the ball should be caught and begins moving to that position with constant speed. Because of great experience, fielder B chooses his running speed so that he arrives at the “catch position” simultaneously with the baseball. The catch position is the field location at which the ball altitude is 2.1 m. Determine the velocity of the ball relative to the fielder at the instant the catch is made.
7.7 Relative Motion (Translating Axes)
Ch. 7: Kinematics of Particles
414
P. 7/33 Ball: a = -gj = constant v 2 = v o2 + 2a ( s − s o ) v 2y = ( 30sin 30 ) − 2g ( 2.1 − 0.9 ) vy = −14.19 j m/s v x = 30 cos 30i = 25.98i m/s ∴ v A = 25.98i − 14.19 j m/s 2
1 2 1 2 = + + = + − s s v t at 2.1 0.9 30sin 30 t gt , t =2.976 s ( ) o o 2 2 2.976 × 30 cos 30 = 77.32 m from batter ∴ horizontal displacement of the ball = ∴ catcher B must move 77.32 − 65 = 12.32 m to the right, in 2.976 − 0.25 = 2.726 s → = v B 4.52i m/s v A/B = v A − v B = 21.46i − 14.19 j m/s
7.7 Relative Motion (Translating Axes)
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