Question

The cables at A and B are wrapped securely around the rims and the hub of the integral pulley as shown. If the cables at A and B are given upward velocities of 4.8 ft/sec and 5.5 ft/sec, respectively, calculate the velocity vo (positive if upward, negative if downward) of the center O and the angular velocity w (positive if counterclockwise, negative if clockwise) of the pulley. 4.8 ft/sec 5.5 ft/sec B BA B Answers: Vo = ft/sec W = rad/sec

Answer 1

Draw the integral pulley as shown below calculate the angular velocity of the pulley omega = V_Q - V_o/OQ Here, V_Q is the velocity at Q, V_0 is the velocity at 0, OQ is the radius of the outer pulley since the cables are assumed to in extensible, V_Q = V_B omega = V_B - V_o/OQ substitute V_B = 5.5 ft/sec, 6.7'' for OQ omega = (5.5 ft/sec) - V_0/6.7'' (1) \r\n Angular velocity of the pulley can also be written as, omega = V_0 - V_p/OP Here, V_P is the velocity at P, V_0 is the velocity at 0 OP is the radius of the inner pulley since the cables are assumed to inextensible, V_p = V_A omega = V_0 - V_A/OP Substitute V_A = 4.8 ft/sec, 3.9'' for OP omega = V_0 - (9.8 ft/sec)/3.9'' (2) solve equation (1) and (2) V_0 - (4.8 ft/sec)/3.9'' = (5.5 ft/sec) - V_0/6.7'' V_0 = 5.057547 ft/sec \r\n substitute V_0 = 5.05 ft/sec 5.5 ft/sec - 5.05 ft/sec/6.7'' = 5.5 ft/sec - 5.05 ft/sec/0.55 ft omega = 0.7924 rad/sec