Question

r(m) = 0.36 k(N/m) = 414 m(kg) = 23 x₁(m)= 0.350 x₂(m)=0.350 x₃(m) = 0.700 Q₀(deg)=50 Q₁(deg)=7 ()₀ (rad/s)=2 () (rad/s)=7

4. (25 points) The lever OĂ, which has mass m and radius of gyration r about O, has an angular velocity w = w, when 0 = 0. The attached spring has a stiffness k and it is unstretched when 0 =(when the lever is vertical). Calculate the constant moment M applied to the lever which will give the lever an angular velocity of w as the lever reaches the position 0 = 01. k G, M X3 X2 X 1

Answer 1

Position 1 Position 2 l2 sw. chao Thao Δω. hi at 0 WE, Iw?s my?w3. 13/09 (2) 2 K.E = 5.9616 J P. Es = mgho & Kn хо xo = le-li ho =% sindo lis unstretched tength la length at Oo = 0.7826m de = Tap th} = 10.35² +0.72 & la 2 2 artaz 28,23 Os (180-00) la = 10.35² +0.72 – 200.35)(0.7) Cos (1309) l2 = 0.963 0.963 -0.7826 = 0.1804 m Xo

PEA = (23)9:301 0.35 sin so) + 41460-18047 P.EL= 67. 231 J. at Positiona K. Edga I w? = 022(0.36373? 73.0296 J P. Ez = mgh, + 4 x 2 where l3 = √6.35² l3 = 1.0482 X 1 = l3 - l, + 0.72 260.35) (0.7) COS(180–7) X1 1.0482 0.78263 0.265 6.m & ha=Xzsinde = 0.35 sin 7° = 0.0426 P.E. (23)19.8)Y 0.35 Sinzº) 414 (0.265032 24. 2143 J + so, work done Ez – EL KE2 +P.Eg -KE, -PEL M = (73.0290 +24.2143 - 5.9616 - 67.22 ) Mхо 23 a 1800 M= 32.048 N-m
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