1. If a particle’s position is ... -...
18
1. If a particle’s position is described by the polar coordinates =4 1 + sin m and = 2 − rad, where t is in seconds and the argument for the sine is in radians, determine the radial and transverse components of the particle’s velocity and acceleration when t = 2 s.
Transcript of 1. If a particle’s position is ... -...
1. If a particle’s position is described by the polar coordinates 𝑟 = 4 1 + sin 𝑡 m and𝜃 = 2𝑒−𝑡 rad, where t is in seconds and the argument for the sine is in radians,determine the radial and transverse components of the particle’s velocity andacceleration when t = 2 s.
2. The slotted link is pinned at O, and as a result
of the constant angular velocity ሶ𝜃 = 3 𝑟𝑎𝑑/𝑠 itdrives the peg P for a short distance along thespiral guide 𝑟 = (0.4𝜃) 𝑚 where is 𝜃 in radians.Determine the radial and transversecomponents of the velocity and acceleration ofP at the instant q = p/3 rad.
2
222
/2.732.1202
/77.334189.002
smrra
smra
eaeaa
r
rr
q
q
smrsradrr dt
d
/2.134.0/34.04.0 qqq
mrradat 4189.03
4.03
ppq
qq 4.04.0 rr dt
d
00 rcst qq
smrv
smrv
evevv
r
rr
/26.134189.0
/2.1
SOLUTION
Velocity Acceleration
3. The sphere P travels in a straight line with a constant
speed of v=100 m/s. For the instant shown, determine
the corresponding values of as measured
relative to the fixed Oxy coordinate system.
qqq ,,,,, rrr
+ r+ q
Position
Velocity
o
m.r
45
137113280
q
v
rv
qv
s/rad..
.
r
v
s/m.sinvrv
s/m.cosvrv
evevv
r
rr
2290137113
88225
8822515
5939615
q
q
q
q
SOLUTION
The sphere P travels in a straight line with a constant speed of v=100 m/s.
+ r+ q
Acceleration
v
rvqv
2
222
22
39102
02
9250
00
s/rad.r
rrra
s/m.rrrra
aaaaaacstv
r
rr
q
qqq
q
4. The particle P moves along the parabolic surface shown.When x =0.2 m, the particle speed is v=5 m/s. For this
instant, determine the corresponding values of
Both x and y are in meters..,, qq andrr
5. At the bottom of a loop in the vertical (r-q) plane at an altitude of 400 m,
the airplane P has a horizontal velocity of 600 km/h and no horizontal
acceleration. The radius of curvature of the loop is 1200 m. For the radar
tracking at O, determine the recorded values of and for this instant. r q
+ r+ q
qo.tana
m.r
8211000
400
0310774001000 22
q
Position
Velocity
s/rad.r
vrv
s/m..sin.sinvv
s/m..cos.cosvrv
s/m..
v
r
05750
8966182167166
7515482167166
6716663
600
q
q
q
vq
q
qv
rv
+ r+ q
q
222
15231200
67166s/m.
.va
o.
m.r
821
031077
q
s/rad.
s/m.r
05750
75154
q
v
a
Acceleration (no horizontal acceleration)
a
1200 m (radius of curvature – in normal & tangential coordinates)r= 1077.03 m (radial distance measured from a fixed point (pole) to particle – in polar coordinates)
22
2222
2
/036.02
2/49.218.21cos15.23cos
/158.120575.003.1077597.8
/597.88.21sin15.23sin
sradr
rarrasmaa
smrarrra
smaa
rr
r
qqqqq
q
qqq
q
q
6. The hydraulic cylinder gives pin A a constant velocity v=2 m/s along its axis for
an interval of motion and, in turn,causes the slotted arm to rotate about O.
Determine the values of and for the instant when q=30o . q ,, rr q
v = 2 m/s (cst), determine when q = 30°.q ,, rr +r
r
+q
q
=30°
b
vvr
vq
b
Geometry:
B
s/rad..r
vs/msinsinvrv
s/m.coscosvrvr
33330
11302
7321302
qq qbq
b
Acceleration:
2
2222
453830
333732122202
323333300
00
s/rad..
..
r
rrrrra
s/m.r..rrrra
aaa
r
r
qqqqqq
q
q
Velocity:
Pin A: (Piston: rectilinear motionAO: in polar coordinates)
3012030180 b
r
300 mm
30°
r = 300 mm
30°O B
A
isosceles triangle
7. The partial surface of the cam is that of a logarithmic spiral (r = 40e0.05q ) mm, where q is in
radians. If the cam has an angular acceleration ሷ𝜃 = 2 rad/s2 when its angular velocity is ሶ𝜃 = 4 rad/s
at q =𝜋
6rad. Determine,
a) the velocity and acceleration of point C on the cam that contacts the follower rod,
b) the radius of curvature of the path,
c) the time rate of change of the speed at the instant q =𝜋
6rad.
8. When the yoke A is at the position d = 0.27 m, it has a velocity of v = 2 m/s towards
right which is increasing at a rate of 0.6 m/s each second. Pin P is forced to move in the
vertical slot of the yoke and the parabolic surface. For the instant depicted, determine
the velocity and acceleration of pin P in
a) Cartesian Coordinates,
b) Normal and Tangential
Coordinates,
c) Polar Coordinates.
A
x = 2 m x (m)
y (m)9. Particle A is moving along a parabolicpath. At the instant when the abscissa of itsposition is x = 2 m, its velocity is 6.45 m/sand it decreases at a rate of 15 m/s persecond. Determine the velocity andacceleration of the particle for this instant in
a) Cartesian coordinates,
b) Normal and tangential coordinates,
c) Polar coordinates.
2
16
3xy
87.364
3
16
6tan
2
bb xdx
dy
x
ttt eaev
1545.6
Solution
(Given)
A
x (m)
y (m)
+n+t
t
tev
45.6
tt ea
15
na
b
b8
3
16
6
2
2
2
xdx
yd
m
dx
yd
dx
dy
2083.5
8
3
4
311
2/32
2
2
2/32
222
/98.72083.5
45.6sm
van
in normal and tangential coordinates
ntt e.eae.v
98715456
A
x (m)
y (m)
+n+t
t
tev
45.6
tt ea
15
na
b
b
2
16
3xy
222 /99.1698.715/45.6 smasmv
in Cartesian coordinates
jijiv
87.316.5sin45.6cos45.6 bb
2/78.16
87.36cos1587.36sin98.7cossin
sma
aaa
x
tnx
bb
2/616.2sincos smaaa tny bb
jia
616.278.16
in polar coordinates
A
x (m)
y (m)
+r
q
v
ta
na
b
b
my 75.0216
3 2
q
q
q
smvvr /19.6cos qb
x = 2 m
y = 0.75 m
oa 55.202
75.0tan
q
smvv /812.1sin qbq
qeev r
812.119.6
2/638.16cossin smaaa tnr qbqb
2/443.3sincos smaaa tn qbqbq
qeea r
443.3638.16
Magnitudes of velocity and acceleration of particle A
10. The peg moves in the curved slot defined by the equation r2 = 4sin(2q) [m2], and
through the slot in the arm. At q = 30°, the angular velocity and angular acceleration of
the arm are = 2 rad/s and = 1.5 rad/s2, respectively. Determine the magnitudes of the
velocity and acceleration of the peg at this instant,
a) in polar coordinates,
b) in Cartesian coordinates,
c) in normal and tangential
coordinates. Also determine
the radius of curvature
for this instant.
q q
qq ,
at q = 30° = 2 rad/s , = 1.5 rad/s2q q
qq ,
mrr 86.1302sin42 qq
Solution
qqqq 2cos42cos242 rrrrdt
d
smrsrad o /15.230,/2 qq
smrvsmrv r /15.2,/72.3 qq
smveev r /297.472.315.2 q
in polar coordinates
qqqq 2sin22cos4 22
2
2
rrrdt
d
*
**
22 /77.15/5.1,/15.2,/2,86.1,30 smrsradsmrsradmro qqq
2
22
/39.112
/11.23
smrra
smrrar
q
q
2/85.2539.1111.23 smaeea r q
smveev r /297.472.315.2 q
in Cartesian coordinates
2/85.2539.1111.23 smaeea r q
A
+r
q
v
b
q
q30o
qvr
vq v
q30o
oa 97.5715.2
72.3tan
b
b
jiv
jiv
294.4152.0
30sin297.430cos297.4
bb
jia
jia
695.179.25
30sin85.2530cos85.25
aq
q30o
ar
a
a
oa 24.2611.23
39.11tan
in normal and tangential coordinates
+t
b+n
tev
297.4
nt
nt
eea
eea
718.25608.2
76.303.2cos85.2576.303.2sin85.25
o76.3
o03.2
ma
v
n
718.0718.25
297.4 22
o03.2
o76.3
