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Unformatted text preview: Problem 16.11 (2011] Problem 16.“), use the princi—
ple of impulse and momentum to determine the collar’s
velocity as a function 0" time. (b) Use the result of (a) to
determine the time at which the collar reaches the right
hand end of the bar. Solution: ’2
(it) Thu impulse uciing on the collar is j ZFdr : Huh2 — my”:
'l f I a If — a} d. d L L. — ‘7; I1 )l _r r _ )_! Ur __. ‘ n W1 1 I, 7 K
K H .1 ‘ r g l tn 312 4 * 322
[h] Start with m r A : t", Integrate] (M :f r 11!.
n'! ()0 u U (10 32.2 r4
and evaluate 4 : ( ) («n chcc r L 2.34 s_ ()0 4 Problem 16.33 The "ITkg skier is traveling at IO m/s
at position I, from which he goes to position ] to 2 in
0.7 s. (a) Neglecting friction and aerodynamic drag, what is
the time average of the tangential component of
force exerted on the skier as he moves from posi—
tion ] to position 2? (b) If his actual velocity is measured at position 2 and
determined to be 13.] m/s, what is the time average
of the tangential component of force exerted on the
skier as he moves from position 'I to position 2? Solution: (21) The impulse is
!2 f th = Fawm — Ii) = mm 7 mm.
11 If friction and aerodynamic drag are neglected, the velocity at 2
is determined from energy considerations: 1 l
Emu; = Emu? + High. front which u; 2 Jr)? +2g : J10? + 23(4) : 13.35 ms. The time average of the tangential force is (b) [f v; : I3‘I this, the average tangential force is Problem 16.45 Suppose that the railroad track in Prob
lem 16.44 has a constant slope of 0.2 degrees upward
toward the right. If the cars are 6 ft apart at the instant
shown, what is the velocity of their common center of
mass immediately after the impact? Solution: Time to couple (both accelerate at the same rate) is t =
6 ft
6 s. (2ft/s—lftfs) : Impulse — momentum is now 12 l .
( 0,000 2b)(2 flI,S)+(T’00001I2))(1 ms)
32.2 fth 32.2 ftfs . D [90,000 lb
— (190,000 lb) sm 0.2 (6 s) = “centururmuss “centernl'mass = 0957 W5 l mfs Problem 16.61 Two cars with energyabsorbing bum
pers collide headon. Their speeds are 1),; :3 mi/h
and v3 = 2 rni/h in the directions shown. The weights
of the cars are WA 2 2800 lb and W3 =4400 lb. If
the coefﬁcient of restitution of the impact is e = 0.2,
what are the velocities of the cars immediately after
the collision? Solution:
(2800b )(3 h)+( 4400 lb )( 2m h)
m ﬂ _
32.2 his? p 32.2 M;2 p (2800lb)u,+(4400b) ,
= ———“ U
32.2 «152 A 32.2 Ms2 B 0.2(3 — [—2]) mph : vg’ — UA’ Solving we ﬁnd of = —0.667 mph. vg’ : 0.333 mph M Problem 16.80 The one gives the cue ball A a velocity 3;
parallel to the y axis. The cue ball hits the eight ball
B and knocks it straight into the corner pocket. If the
magnitude of the velocity of the cue ball just before the
impact is 2 mls and the coefﬁcient of restitution is e = 1,
what are the velocity vectors of the two balls just after
the impact? (The halls are of equal mass.) Solution: Denote the line from the Sball to the corner pocket by
3?. This is an oblique central impact about HP. Resolve the cue ball
velocity into components parallel and normal to HP. For a 45G angle,
the unit vector parallel to BP is 83}: = ﬁhi +j). and the unit vector normal to HP is egpn : fa +j), Resolve the cue ball velocity before
impact into components: VA = UAPEEP + 'uApnegpn. The magnitudes
0,”: and 1),; p” are determined from Vlvnpenplz + ivnpnespnlz = I‘m = 2 W5 and the condition of equality imposed by the 45° angle, from which
11,“: : UAPn : x/i rn/s. The cue ball velocity after impact is v14 2
v’Apegp + UApnegpn. (since the component of VA that is at right
angles to HP will be unchanged by the impact). The velocity of the
Bball after impact is v’BP 2 Upreap. The unknowns are the magni“
tudes u’BP and vi”, These are determined from the conservation of
linear momentum along BP and the coefﬁcient of restitution. mitt)“; = mAu’AP +msv’BP. and I n‘
1: "BF _ “Ar 1MP For mA 2 ms, these have the solution 1):“, = 0, 1);”, : 1MP, from
which V’A : UAPneBPn = i +.i (W5)
and VIE : 11,4983}: 2 —i+j Problem 16.89 The bar rotates in the horizontal plane
about a smooth pin at the origin. The 2kg sleeve A
slides on the smooth bar, and the mass of the bar is
negligible in comparison to the mass of the sleeve.
The spring constant k :40 N/m, and the spring is
unstretched when r = 0. At t: 0, the radial position
of the sleeve is r 2 0.2 m and the angular velocity of
the bar is we : 6 rad/s. What is the angular velocity of
the bar when r = 0.25 m? Solution: Since the spring force is radial, it does not affect the
angular momentum which is constant. (0.2 m)(2 kg)[(0.2 m){6 rad/5)] = (0.25 m)(2 kg)[(0.25 m)wI (u r 3.84 rad/s Problem 16.93 A lkg disk slides on a smooth
horizontal table and is attached to a string that passes
through a hole in the table. (a) If the mass moves in a circular path of constant
radius r = 1 m with a velocity of 2 m/s, what is
the tension T? (b) Starting from the initial condition described in
part (a), the tension T is increased in such a way
that the string is pulled through the hole at a
constant rate until r z 0.5 m. Determine the value
of T as a function ofr while this is taking place. Solution: (a) Circular motion
T : intvz/rer
T={l)(2)2/1= 4 N (b) By conservation of angular momentum, I'llvll
mmun I mrur. UT : n—
r and from Newton's second law mm: 2
lTl:mv%/r =m( ) /r i" 2
T=(l)(<11(2))/r24lrva ...
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This note was uploaded on 03/20/2008 for the course EM 311 M taught by Professor N/a during the Spring '08 term at University of Texas at Austin.
 Spring '08
 N/A
 Dynamics

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