KINETICS OF PARTICLESdebis.deu.edu.tr/userweb/binnur.goren/Dynamics2016G/11P...5. The light rod is...
Transcript of KINETICS OF PARTICLESdebis.deu.edu.tr/userweb/binnur.goren/Dynamics2016G/11P...5. The light rod is...
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1. The two small 0.2-kg sliders are connected by a light rigid bar and are constrained to
move without friction in the circular slot. The force P=12 N is constant in magnitude and
direction and is applied to the moving slider A. The system starts from rest in the
position shown. Determine the speed of slider A as it passes the initial position of slider
B if (a) the circular track lies in a horizontal plane and if (b) the circular track lies in a
vertical plane. The value of R is 0.8 m.
12121221 eegg VVVVTTU
smv
vm
/1.8
2
128.030sin128.030cos12
2
2
2
2.0
Nothing changes.
a) In horizontal plane
b) In vertical plane
Reference line for part (b)
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2. Determine the constant force P required to cause the 0.5 kg slider to have a speed v2 = 0.8 m/s
at position 2. The slider starts from rest at position 1 and the unstretched length of the spring of
modulus k = 250 N/m is 200 mm. Neglect friction.
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l1 l2
Reference line1
2
m=0.5 kg v2 = 0.8 m/s rest at position 1 and k = 250 N/m lo=200 mm
h2=0.2sin15=0.052 m
Length of cable
1
2
12121221 eegg VVVVTTU
ml 472.025.04.0 22
1
ml 32.025.02.0 22
2
Work by the cable
PPlPU 152.032.0472.021
222
2
1
2
22
2
2
2.025.02502
12.045.0250
2
1052.081.95.08.05.0
2
1152.0
2
1
2
1
2
1152.0
P
xkxkmghmvP
P=52.07 N
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3. The 2 kg collar is released from rest at A and
slides down the inclined fixed rod in the vertical
plane. The coefficient of kinetic friction is 0.4.
calculate (a) the velocity v of the collar as it strikes
the spring and (b) the maximum deflection x of the
spring.
FBD of Collar
Motion
x
y
mg
N
Ff=mkN
N...NFN.N
cosmgNF
kf
y
924381930819
0600
m
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(a) the velocity v of the collar as it strikes the spring
N.Ff 9243
(b) the maximum deflection x of the spring.
1
2
3Reference
12121221 eegg VVVVTTU
s/m.v
sin..v..
5562
6050819222
1509243
2
2
2
23232332 eegg VVVVTTU
0533606613800
16002
160819255622
2
19243
2
22
.x.x
xsinx..x.
x
m.x.
....x max, 09890
080
09890
8002
533680040661306613 2
21
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4. The 1.2 kg slider is released from rest in position A and slides without friction
along the vertical-plane guide shown. Determine (a) the speed vB of the slider as it
passes position B and (b) the maximum deflection d of the spring.
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m=1.2 kg determine vB and the maximum deflection d of the spring.
1
2
3
d
12121221 eegg VVVVTTU
a)
13131331 eegg VVVVTTU
b)
Datum
smv
mgmvVT g
/396.9
05.42
10
2
2
212
mmm
kmgmgVVV egg
2.540542.0
0240002
15.481.92.15.181.92.1
02
15.45.10
2
2
313
d
d
d
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5. The light rod is pivoted at O and carries the 2- and 4-kg particles. If the rod is
released from rest at q =60o and swings in the vertical plane, calculate (a) the
velocity v of the 2 kg particle just before it hits the spring in the dashed position and
(b) the maximum compression x of the spring. Assume that x is small so that the
position of the rod when the spring is compressed is essentially horizontal.
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released from rest at q =60o (a) the velocity v of the 2 kg particle just before it hits the spring in the dashed position and
(b) the maximum compression x of the spring.
12
Reference
3 (maximum compression)
22211121 0 egeg VVTVVTU
22 22
14
2
160sin45.081.9260sin3.081.94 BA vv
a) 222111 egeg VVTVVT
A
B
45.03.0 BBAA rvrv smvsrad B /16.1/58.2
2
max
350002
160sin45.081.9260sin3.081.94 xk
b) 333111 egeg VVTVVT
mmmx 07.1201207.0max
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6. Two springs, each of stiffness k=1.2 kN/m, are of equal length and undeformed
when q =0. If the mechanism is released from rest in the position q =20o, determine
its angular velocity ሶ𝜃 when q =0. The mass m of each sphere is 3 kg. Treat the spheres
as particles and neglect the masses of the light rods and springs.
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k=1.2 kN/m, are of equal length and undeformed when q =0. mechanism is
released from rest when q =20o, determine ሶ𝜽 when q =0. m =3 kg.
Reference line
Ref.
22211121 0 egeg VVTVVTU
1
2
l1
l2
m
ml
m
ml
056.0225.041.0
41.055sin25.02
0668.0287.0225.0
287.035sin25.02
1
2
1
1
d
d
We ignore the equal and opposite potential energy changes for masses (a) and (b).
a
b
c
22
1
2
1
2
11
056.012002
10668.01200
2
120cos25.081.93
2
1
2
120cos25.0
E
kkmgE dd
25.025.032
13
2
2
2 mgE
v
q
sradEE /22.421 q
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7. The two right-angle rods with attached spheres are released from rest in the
position q = 0. If the system is observed momentarily come to rest when q = 45°,
determine the spring constant k. The spring is unstretched when q =0. Treat the
spheres as particles and neglect friction.
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released from rest when q = 0. System is momentarily stationary
when q = 45°, determine the spring constant k. Spring is unstretched
when q =0.
1
2
Reference
180 mm
60 mm
a
oa 56.7160
180tan
a
22211121 0 egeg VVTVVTU
45oa
71.5645=26.56o From conservation of energy:
mx 219.006.056.26cos1897.022
2219.02
156.26sin1897.081.92218.081.922 k
Deformation of spring in Case 2:
k=155.7 N/m
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8. The light quarter-circular rod is pivoted at O and carries the 3 kg particle. When the
system is released from rest at the position (1), it moves to position (2) under the action
of the constant force F=250 N applied to the cable. The spring of stiffness k=1500 N/m has
an unstretched length of 200 mm. Calculate the speed of the particle and the angular
velocity of the circular rod as the particle passes the position (2).
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9. The 0.6-kg slider is released from rest at A and slides down the smooth parabolic
guide (which lies in a vertical plane) under the influence of its own weight and of the
spring of constant 120 N/m. Determine the speed of the slider as it passes point B
and the corresponding normal force exerted on it by the guide. The unstretched
length of the spring is 200 mm.
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Work and Energy Principle
1
Datum2
smv
v
VVTVVT
VVVVTTU
egeg
eegg
/92.5
2.025.01202
16.0
2
12.05.025.0120
2
15.081.96.0
2
22
2
222
222111
12121221
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Newton’s Second Law (Normal&Tangential Coordinates)
NN
.
......N
vmmgFNmaF
m.
dx
yd
dx
dy
dx
yd,x
dx
dyxy
k.k.kxy:guideparabolicofEquation
springnn
/
/
x
84
250
925608196020250120
2504
011
4042
25050
22
23
2
2
232
2
2
0
2
22
FBD of slider (at point B)
+t
+n
Fspring mg
N
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10. Calculate the horizontal velocity v with which the 20 kg carriage must strike the spring
in order to compress it a maximum of 100 mm. The spring is known as a “hardening”
spring, since its stiffness increases with deflection as shown in the accompanying graph.
smvxxv
dxxxvVT
VVVVTTU
e
eegg
/38.203
205100010
010002010202
10
1
1.0
0
322
1
1.0
0
22
121
12121221
12
v1