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Unformatted text preview: Problem 21.8 The mass of the disk is 2 kg. The length
of the bar is L = 60 mm and its mass is negligible. At
I = 0, the pendulum is vertical and it has a counterclock
wise angular velocity of 35 degrees per second. What is
the amplitude of the resulting vibrations in degrees? Solution: See 2LT. r : 0.05 m. L 20.06 m. The equation of motion is (Fa 'lg[L+ r] 1126 l — ——.— 0 =0 — 8.99 ad! (i :0
m2 +(r2+tr+Lil) :’ (1:2 H r S) The solution is H z A cost[3.99 radish) + B sin([8.99 rad/5]!) g : (3.99 md/sH—A sin([8.99 rad/sir) + B emits99 rad/sir)! Using the initial conditions we have
0 : A 35 (E6) 11 radls : B(8.99 rad/3)] :> A : 0. B = 0.0679 mills The amplitude of the motion is :30“ a
3 = (0.0679 rad) (I rad) _ 3.89 'oblem 21.10 The suspended Object A weighs 10 lb.
:1: radius of the pulley is R = 4 in. and its moment
inertia is 0.08 slugfig. The spring constant is k =
. lb/i't. The spring is unstretched when x = 0. At I = 0.
3 system is released from rest with x = 0. What is the
.lue of x at r : 3 s? )Iulion: See the solution to 21.09. Using the numben; we have 12. , _ 7
(T: + (3!“ rad!s)“x = 9.70 ms
t . m sulution is
.l = A sin([3.~ll 13de]!le Beos([3.4l rad/slit + 0.833 it
,ing the initial conditions we have
0 = B + 0.833 l‘l
=> A :0. B : —U.S33 fl
U : Ailll rad/S) 1e motion is thus x : (0.833 l'tHl — C05([3.41 rad/51H] :> .l'U : 3 s) : l.4l l't Problem 21.16 The 2lb bar is pinned to the 5H) disk.
The disk rolls on the circular surface. What is the fre
quency of small vibrations of the system relative to its
vertical equilibrium position? Solution: Use energy methods. II My 15.271 Slb l5 22
r=7 — —— —:  — —n 2 (3i32.2rus3ii12 1] )wb“+2(32.2ms1)(12 ) “’b‘" 1 I Slb 4 3 I52
+ 5 (2 [3—HT] [72 “i )(7) “’b : ““93 'b'sz’mmi“ .5 [S
V : ~t2 lb) (7Ti fl) c056 r (5 lb) (E ft) cost! = 47.5 ftlb)cost+ Differentiating and linearizing we ﬁnd , ([26 {ﬂu 2
(0.396 Ibis7ft) n, + (7.5 ftlb)H : 0 =5 W + (4.35 rad/s) s = 0
{ 
. 5 d!
'_ 13—”: = 0.692 H2 # 211 rad Problem 21.50 The spring constant is k : 800 Mm Solution: Sec 2.49
and the damping coefﬁcient is c = 40 Ns/m. The radius
of the pulley is IZO mm and its moment of inertia is
0.03 N—mz. The spring is unstretched when .t' 2 0. At
1:0,.x = 0 and the 20kg mass is moving downward
at i this. What is the value (if x at I = 2 s? I : “WWWWW + 55min!” + 07345 I" We have (I = 0,334 [3015. {I'd : 3.90 rndls The solutions are t. —_ trim —l.4{r.!,[ + Bdlsinmnnt + [Bard ~ Aiilumwdn
Using the initial conditiuns
U : A + 0.245 m
:> .4 : —0.245 m. B : 0.232 tn
I mix : Brad ~ Ad The motion is given by x = rm“ ”LUW {40.245 tn]cos([3.l)i) rad/sh) + [0.232 m]sint[3.90 radish” + 0.245 m roblem 21.52 In Problem 2.SI. the spring is
nstrelchcd at! : 0 and the disk has a clockwise angular
:locity of 2 rad/s. What is the angular velocity of the
isk when i : 3 s? olution: From the solution to Problem 2 .5 l. the canonical form
" lhe equation of motion is tilt + 71th + 7 0
—— _ — c it : .
dfl l ‘1: I)
t‘ ‘ 2R 1
'th1: u‘ : — : 0.322 radls and w‘ : — : 7.47 (radfsi'.
3m .m he system is subcritically damped. so that the solution is of
is form it = e '“(A sin (ad: + B cos am). when: ab! = x/w2 , (if :
.623 mdfs. Apply the initial conditions: 1;, : U. and from
incnmtics. ii.) = iii/R : —2 radls, from which .1}. : 2 ills. from
rhich B : 0 and A :,i1./w,,i:0.4327. The solution is xi!) =
—m (i) sinv.11. and .i'ui = —t.’.t‘ + .inF’“ cosmdi. Al I = 3 5, mi"
I r
: 0.58611 and .i : 0.I528 ills. Front kinematics. Hir) : —:(—~). R
u r : 3 s. i} : 415:; rad/s clockwise. ...
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 Spring '08
 N/A
 Dynamics

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