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Unformatted text preview: *194. The space capsule has a mass of 1200 kg and a
moment of inertia [G = 900 kg  m2 about an axis passing
through G and directed perpendicular to the page. If it is
traveling forward with a speed 125 = 800 m/s and
executes a turn by means of two jets, which provide a constant thrust of 400 N for 0.3 s, determine the capsule‘s  5
 "‘ 15' angular velocity just after the jets are turned off. Ls... won
t+ (HG), + ZIMGd = (HG), ﬁt] “00M " 35' 0 + 2[400¢0515°(0.3)(1.5)] = 900:», m, = 0.386 rad/s Ans 195. Solve Prob. 1755 using the principle of impulse
and momentum. <(+> ("(2)1 +£IMo dz=(Ha)z 0
(“lo 3(lr‘°"’)a!:= (o. 13).» 3(r+52‘°'“)3 = 0.13.» «D: 20.8 rad/s A... _'___________—_————_———————o 6. Solve Prob. 17—54 using the principle of impulse
d momentum;
(5.4)] + zJ’MAﬁ 8' (HA): 0 + I’sm = [10(o.2)’]m
0 mu
5 1 _ — 0.411) a) = 56.2 rad/s An: “’
m(v¢;)l + EIFdr = m(vG), M (3+) 0+A,(3) = o A, =0 Ans vvt~v‘f’2“>'ﬁ<1;gw immanwa'swgn‘vv‘m") (+1) 0 + A,(3) — 98.1(3) = o A, = 98.1 N Am 509 1911. The pilot of a crippled FlS fighter was able to
control his plane by throttling the two engines. If the
plane has a weight of 17 000 lb and a radius of gyration
of k(; = 4.7 ft about the mass center G, determine the
angular velocity of the plane and the velocity of its mass
center Gin I: 5 s if the thrust in each engine is altered
to T, = 5000 lb and T2 = 800 lb as shown. Originally the
plane is ﬂying straight at 1200 ft/s. Neglect the effects of
drag and the loss of fuel. ((+l (HG)1+2JMGdl=(HG)z 0+ 500015)” 25>— 800(5)(l.25) = [(%0%0)(4.7)1]m a): 225 rad/s Ans "10101): +ZJdel=m(Vax)z
17000 17000
(322 )(1200)+5800(5)‘(32,2 )(Vch
(VG). =1.25(10’)n/s Am *1912. The spool has a mass of 30 kg and a radius of
gyration k0 = 0.25 m. Block A has a mass of 25 kg, and
block 8 has a mass of 10 kg. If they are released from
rest, determine the time required for block A to attain a
speed of 2 m/s. Neglect the mass of the ropes. v‘  2m/s 2
m —6.667nd/a M 0.18m v. 1: 6.667(0.18) = 1.20 m/s u (He), + that: = (H0). 0 + 25(9.81)(0.3)r — 10(9.81)(0.18)(!) = 25(2)(o.3) + 30(o.25)‘(6.667) + toumxo. 1913. The man pulls the rope off the reel with a
constant force of 8 lb in the direction shown. If the reel
has a weight of 250 lb and radius of gyration kg = 0.8 ft
about the trunnion (pin) at A, determine the angular
velocity of the reel in 3 5 starting from rest. Neglect
friction and the weight of rope that is removed. H (11.). + tIMAdI = (HA). 0 + 30.7.90) = [32ng(0.8)2]¢» a) = 6.04 rad/a Am 512 1917. The drum has a mass of 70 kg, a radius of
300 mm, and radius of gyration k0 = 125 mm. If the
coefﬁcients of static and kinetic friction at A are [1,, = 0.4
and pk = 0.3, respectively, determine the drum’s angular
y 2 s after it is released from rest. Take 0 = 30° velocit
L+ (HA)l + EIMA dz = (HA), 0 + 70(9.8l)(sin30°)(0.3)(2) = (70(o.125)z + 70(03):)(0 m = 27.863 rad/s = 27.9 rad/s Ans mm). + :15 d: = mm), x +/ 0 + 70(9.81)sin30°(2) — F(2) = 70(27.863)(0.3) F = 50.79 N +'\ 0 + ND(2)  70(9.81)cos30°(2) = 0 ND = 594.7 N If",  0.4[(594.7] = 237.37 N > 50.79 N —___—_______________________ 1918. The double pulley consists of two wheels which
are attached to one another and turn at the same rate.
The pulley has a mass of 15 kg and a radius of gyration
k0 = 110mm. If the block at A has a mass of 40 kg
determine the speed of the block in 3 s after a constant
force F = 2 kN is applied to the rope wrapped around
the Inner hub of the pulley.The block is originally at rest.
Neglect the mass of the rope. «4) (Ho), + zluod: = (Ho), 0 + momma) — 40(9.81)(0.2)(3) = manor.» + 40(02m)(0.2) a, = 120.4 ml: V‘ = 02020.4) = 24.111115 An 1930. The square plate has a mass m and is suspended
at its corner A by a cord. lfit receives a horizontal impulse
I at corner B, determine the location y of the point P
about which the plate appears to rotate during the impact. (to (Hal + EIMad: = (HG): T
a "I z z 5
0+[(__) = —(a +a)a)
ﬁ 12
K —)v
(3*) “(Vax): '9' 2154‘ = “(Va1): 0+I=mva 6! a):— fzan Puke Which the car exerts on the ole whil '
. e AC 15
mntrally vertical. P 4r“ l
G = 2.25 m. Determine the horizontal l
l *)
(HG) +2IMG d; = (ﬂab 0+UFdr](3.5)= i75(2.25)2<60) IFdf=15.2k.Ns Ans 57.1 1943. A thin disk of mass m has an angular velocity w,
while rotating on a smooth surface. Determine its new
angular velocity just after the hook at its edge strikes the
peg P and the disk starts to rotate about P without
rebounding. HI :3], I 1
(Enr’m. = [Enr’ + mrzlmz ml = "'0’; Ann * a:::4.10_1I;ie pendulum consists of a 5lb slender rod AB ﬁred. t h wooden block. A projectile weighing 0.2 lb is
in o t 6 center of the block with a velocity of 1000 ft/ If the pendulum is initially at rest, and the projectile embedss itself into the block determin
I , e the an ul '
pendulum just after the impact. g m velOClty Of the Mavs Moment of Inertia: The mass moment inertia of the pendulum and
the embeded bullet about point A is 5 >m+3:_.<12> = 2.239 slug ['11 fans Prvati’an of Angular Momentum : Since force que to the impact is inlrmai tn the system consisting of the pendulum and the bullet. it will cancel
out Thus. angular momentum is conserved about point/4. Applying Eq l9— 17, we have (HAM = (HA )2
(mi, Uh)(’b) = (IA)! 0’;
0.2
(3:5)(1000H25) =1.239m2
(u2 = 6.94 rad/s Ans 528 1945. A thin ring having a mass of 15 kg strikes the
20mmhigh step. Determine the largest angular velocity
u“ the ring can have so that it will not rebound off the
step at A when it strikes it. The weight is non — impulsive “’ 1’;
(HA)1=(HA)z ) 1,.
z z ‘ /”""m
15a», )(0.18)(0.18—0.02)+[15(0. IS)’](aa)=[15(o.1s) +15(o.rs) ]ah 9’
an =0.9444m [gnomm 3.00""
+\zr_ = m(aa),.: (15)(9.81)cos O—NA = lSaﬂOJB) [gunk]
160
When hoop is abouuorcbound. NA 0. Also. c059: 1—86. and so 3
ya» =6.9602rad/s k 9 FA
4’
=T4=737 rad/s Ann (M3M A)‘ 529 1946. Determine the height h at which a billiard ball of
mass m must be struck so that no frictional force develops
between it and the table at A. Assume that the cue C only exerts a horizontal force P on the ball. For lheball
(L) M. + ZIFd: = mv,
0 + HA!) = luv, (1) £+ "1401*sz 4’: (HA): 2
o + (nun) = [311”) + "Mint; (2) WV; = tulr (3) Solving qu.(l)—(3) {all yield: h=~r Au 1947. The disk has a mass of 15 kg. If it is released from
rest when 0 = 30°, determine the maximum angle 0 of rebound after it collides with the wall.The coefficient of
$1 ‘ restitution between the disk and the wall is e = 0.6. When t9 = 0°, the disk hangs such that it just touches the wall,
‘1 Neglect friction at the pin C. Datum at lower position of G.
n+w=n+w o l 3 2 o
0+(15)(9_81)(0.15)(l —cos30 ) = i[5(15>(0‘15) 10;" a): 3,4l8 rad/s ,' o(0.15m’)
i”) e =0’6Z3.41t3(015)0 a} = 2,0508 rad/s
T: + ‘6 = T3 + ‘5
l[2(‘5)(0>15)l](2.0503)l + 0 = 0+ 15(9.81)(0. lS)(l ~0050)
2 2 9 2 17 9° Ans 530 ...
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 Fall '08
 Hudyma
 Dynamics

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