Question

In wheel A, the mass of the wheel assembly is concentrated at the center. (Therefore the centroidal radius of gyration is 0.) However, in wheel B, the mass is concentrated at the outer radius, r. (Therefore, the centroidal radius of gyration for wheel B is k-r.) The wheel is released from rest. For without slipping in both cases. Your answer must contain the initial and final momentum diagrams and the FBD. (Partial Answer: ? = _10gsin6 for the first case)

Answer 1

First case: I_A = I_0 + mr^2 I_P = mk^2_0 + mr^2 I_A = 0 + mr^2 I_A = mr^2 centroidal radius of gyration is '0' k_0 = 0 Alembert principle sigma M_a = I_A alpha mg sing theta times r = mr^2 alpha g sin theta = r alpha alpha = g sin theta/r dw/dt = g sin theta/r integral^w_w_0 = 0 dw = g*sin theta/r integral^t = 10 sec_0 (w-0) = g sin theta/r (10-0) w = 10g sin theta/r rad/sec \r\n second case: I_A = I_0 + mR62 = mk^2_0 + mr^2 = mr^2 + mr62 I_A = 2 mr^2 D Alembert principle sigma M_A = I_A alpha mg sin theta times r = 2mr^2 alpha g sin theta = 2r alpha alpha = g sin theta/2r dw/dt = g sin theta/2r integral^w_w_0 = 0 dw = g sin theta/2r integral^t = 10 sec_0 dt w = g sin theta/2 r (10) w = 5g sin theta/r rad/sec