Question

The 17-kg wheel has an eccentric mass which places the center of mass G a distance r¯ = 142 mm away from the geometric center O. A constant couple M = 11.8 N∙m is applied to the initially stationary wheel, which rolls without slipping along the horizontal surface and enters the curve of radius R = 1110 mm. Determine the normal force N_C under the wheel just before it exits the curve at C. The wheel has a rolling radius r = 100 mm and a radius of gyration k_O = 141 mm.

M Answer: Nc = i

Answer 1

- Page noir Answer! 1. The objective is to determine the normal force No. under the wheel just before it exits the cuore at c. given that! mass m =17 kg couple, M = 4.8 N-m. roller rading, Cuove Radius; Rolllomm = litm. rolling sadius, 8=100 mm = 0.1m. Radius of gyration, a ko = 141 mm. :0.141 m. diagram: e = 142 mm 0.142 mi lus By the work energy principle. MXB = { mCV?-12)+55 (62-60,-). 11.8x0 = { (in) (0-12X482]dz () x ovu, kto? 15.8 x 7. - 0.0856?+ 0.169 w 2. 37.070 : 0.254 w? . . V: Sewe. V707 w²= 37.000 - 0.254 wo siusiau = 12.08 Gad/s.

page noi.2. length of the path Be in 9 . 71 BCE RO 0111 : 0.87 length of BC = 90 0.87 90 0.87 :010 = 087 0: 8.7 h=on ģion 1,11-01 luson Moll - (6.142 +011) - loim X L R 50868m. hocool - i INT- lo -0.613 h = 0.397 m. datum litke. . :*0.868 Cosus re: 0.613 m oom Look- energy thoosen (b/w B and c) Mixo-mgh = 1 / 1 m (u?_v2²) + I (1²-16) 11.5487 - 17x981xCo-391) = {(n) (0-12 w?- O.? 12.08)+ 4x17x 0.142%(?12.084)

Page ио'3 33-94 2 = 8:5 (о? «- (-чѕ4)) + •5 o-ozo)(u' - 1454) 33-42 = o-oss u° - 12 мет) + (-то (8° — 1us •1)] : o-ogs w' -2 иы + o-170 ° - 24 Воз ъ3-8ч : 0-255 w* – 37.20ч. o.gѕѕ w2 = 33 34 4 37. 20ч. * = 11 • очи о•2STS w* - 27 8. 6. 9 = 27ge 3 = 16- 64 8а а/s, F : mk us 2 : 11x o: 142, x16:29 2. : 612, 43 0 с : о. 61 км. Ғ. : у хм. V: я 9 - Co-1 + 04 (2) x 14.64 : 4:538 6. м. : 4:032 О-262 x 17. Е, Е 314 34 м. тє, з о-314 км

N = [m xgx cos us] I fz tfi N: 17x9.81 x cos 45 + 319.34 + 672-43 page no. u. N: 235.16. kn. N: 0.235 kn 2. The normal force Ne is 0.235 kn