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PROBLEM 19.41
A 1.2-kg uniform rod AB is attached to a hinge at A and to two springs, each of constant k = 450 Nlm . ( a ) Determine the mass m of block C for which the period of small oscillations is 0.6 s. (b) If end B is depressed 60 mm and released, determine the maximum velocity of block C.

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SOLUTION
' b

\ , 2 m4

(a)

C+ T MA
1.2mg Where At equilibrium,8 = 0 And

= Z(M,),:

+ 0.825mRg - 2 ( 0 . 7 5 ) ~ ,= jh. + 0.825mR(a,)G+ 1.2m(a,),
Fs = k(0.758 + 6) ,

(1)

EM, = 0: 1.2mg + 0.825mRg - 2(0.75)kS, = 0
Substitute ( 2 ) into ( 1 )

(2)

Ta + 0.825mR( a t ) , + 1.2m(at),

= -2(0.75)k(0.75)8

a = e, (a, )c = 0.8250, (a,, = I .2a )

- = -m12 - (1.2 kg)(1.65 rn)2 = 0.2725 kg - m2 I 12 12 [0.2725 +(0.825)' (I .2) + ( 1 . 2 ) ' m l ~ 2(0.75)' (450)8 = 0 + [1.08925 + 1.44rnle + 506.258 = 0 e+ 506.25 8=0 1 .08925 + 1.44m

PROBLEM 19.41 CONTINUED
Now 506.25 r 274' = = (2nfn) = wn = 1.08925 + 1.44m (rn )
2

4(E)
2

=

109.66

506.25 = 109.66(1.08925 + 1.44m) m = 2.4495 kg

(b)

(VC

=

1.2(8)m a , 0, =

1.65

=-= I

0 06 1.65

0.03636 rad

ern Bmwn= (0.03636rad)(109.66 sK2)' = 0.38075 rads =
( v ~ = )1.2(0.38075 radls) = 0.45691 m/s ~ ~ ~

( v ~ = )457 mm/s 4 ~ ~

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PROBLEM 19.42
A 12-lb uniform cylinder can roll without sliding on a horizontal surface and is attached by a pin at point C to the 8-lb horizontal bar AB. The bar is attached to two springs, each of constant k = 30 Iblin. as shown. Knowing thatthe bar is moved 0.5 in. to the right of the equilibrium position and released, determine (a)the period of vibration of the system, (h) the magnitude of the maximum velocity of the bar.

SOLUTION

byP
Qx

r

Wd45 = a31
mil=-2kx-p

M=P-F
1 -M,-" 2 Eliminate P, F and 6
=

F~

o,2=-=2 t m + 3M

2 0 0 1blin.)(l2 in./f1)(32.2 8Ib+18Ib on= 29.861 radls

WS'

)

(a)

In

=2~
o n

- 0.21041 5 ,

= 0.210 s 4

(b)

Iv, , ,~

= (0.5 in.)% = 14.93 in.1~ 4

PROBLEM 19.43
A 12-lb uniform cylinder is assumed to roll without sliding on a horizontal surface and is attached by a pin at point C to the 8-lb horizontal bar AB. The bar is attached to two springs, each of constant k = 20 lblin. as shown. Knowing that the coefficient of static friction between thecylinder and the surface is 0.5, determine the maximum amplitude of the motion of point C which is compatible with the assumption of rolling.

SOLUTION
From 19.42 2t a:=-= m + y 2(20 lb/ii.)(12 in~f&)(32.2 ft/s2) 8lb+18Ib

= 24.382 radls

max - =

F N

20 Ib

,

x =A

2

;

A = Amplitude

ps = 0.5 =

Ib (24.382 radls)' A 40 lb(32.2 f't/s2)
A = 0.090278 f&

'

A =1.083 in. 4

PROBLEM 19.44
A 270-kg flywheel has a diameter of 1.2 m and a radius of gyration of 0.5 m. A belt is placed over the rim and attached to two springs, each of constant k = 13 kN/m. The initial tension in the belt is sufficient to prevent slipping. If the end C of the belt is pulled 25 mrn to the right and released, determine (a) the period of vibration, (b) the maximum angular velocityof the flywheel.

SOLUTION
(a)

Fc
CM, = C(M,:,, )

F = k[(asT), c

- 0.618],

F = k[0.60 + (asr)B] B
= I

0.6Fc - 0.6FB 0.6k[(6,)~ At equilibrium

3
+ (hT),] F e =
F = C

- 0.61~1 0.4k[0.68 -

(1)

(e = o),

F = k(6,,),, ,

k(&), (2)

CMA = 0 = 0.6k(S,),
Substitute (2) into (1)

- 0.6k(~,,),

1 + 0.72k8 = 0 8
7 = mh2 = (270 kg)(0.5 m)2 = 67.5 kg.m2
0.72k =(0.72 rn2)(13000 ~ / m == 9360 N . m ) Thus

w n = -27r

2

0.72k I

9360 N - m = 138.67 s-z 67.5 kg-m2
= 0.5335 s

r =-=

"

4

~~
27r

zn = 0.534 s 4

PROBLEM 19.44 CONTINUED
(b)

e = em (writ + 4), B = em%cos (%t + 4) sin
Qmax = e m w n

wn =
0m r

Jss-' 1 1.776
= =

= .% =

0.6 m

0.04167 rad

emax (0.04167 rad)(11.776 s-l) = 0.49067 rads =
w = 0.491...