Question

T 9p The 1 lb ball rotates in a horizontal plane in a circular path B that lies in a plane h = 2.5 ft below A and has a cord length LB 3.5 ft. (a) Determine the ball's velocity in path B in fps. Later the ball rotates in circular path C when the1 cord is shortened by pulling it through the hole at A until the angle phi = 36 deg. (b) Determine the ball's velocity in path C in fps wt Lb Lc 2.5 3.5 36.0

Answer 1

Given that The 1/6 ball rotates in a Horizontal plane in a Circular Path B that lies in a Plane h = 2.5 ft below and Cord length (B = 3.5 ft When angle Phi = 36 degree So, given that the value of ball (m) = 1 lb h = 2.5 ft L_b = 3.5 ft emptyset = 36 degree Radius (gamma_B) = squareroot (L_B)^2 - (h)^2 = squareroot (3.5)^2 - (2.5)^2 = squareroot (12.25) - (6.25) = squareroot 6 gamma_B = 2.49 ft Draw free Body diagram of ball B. \r\n Consider Equilibrium of forces in Vertical direction Tsin theta = mg T (2.5/3.5) = 1 times 32 lb ft/s^2 T = 1 times 32 times 3.5/2.58 lb F/S^2 = 112/2.5 = 448 lb F/S^2 Consider Equilibrium of Force in Horizontal direction T cos theta = F_c Here F_c = Centrifugal force = m V_B^2 /V_B 44.8 times (2.49/3.5) 1 times V_B^2/2.49 44.8 times 0.711 = V_B^2/2.49 31.872 = V_B^2/2.49 V_B^2/2.49 = 31.872