Question

abc123 _okc 875 4. [15 points] The 1-lb ball is guided along the vertical circular path r= 2r. cos e (ft) using the arm OA. The arm OA rotates such that 0 = t (rad) where t is the time. Neglect friction and the size of the ball. Assume that gravity g = 32.2 ft/s and r. = 0.25 ft. (a) Draw a Free Body Diagram (FBD) of the ball. (b) The magnitude of the acceleration when 0 - 30° is - .36 Hst (f/s®). (c) The magnitude of the reaction force exerted by the arm OA on the ball when e = 30° is (Ib). 30°=6 306 r=arc coso 8:30°

Answer 1

Orawing Fare body diagram af the ball : . Olc= 0.25 { Зо 2 (0.25) cos = o-S Cos0 FOA - o.5 sinoé 30° - o-5 cos e 6 %3D - o-8 siho e Given - = 0-523 sad/s = 16 At e = 30 O.S COA 3 0 o:433 - o.S sin 30° x < (0-s 23) = - 013075 /s ji = -o:5 Cos 30 (0.523)² - 0.8 sin30 (0) = - 0. 1184 H Componenta of acceleration!- ji - r62 = - o- 118 4 (o.433) (0.523) ? agr = - ar =-0 - 2368 tls 2 8.

2 si o 90 = + o-433 (0) + 2(-0.13075)(.523) 90 = -0.13 67 ft/s2 O: 2734 JH From the Shown above, woriting egh af motiond - + 1 E For = masr; (- 0-2368) N Cos 30 I sin 30 32.2 o.5688 lb N = St E Fe = mag : FOA N Sin 30 1 CoA 30 (- 0.1367) 32-2 FOA = 0.5773 lb