Question

(3) In a real experiment with setup of Figure 2, initially the disk is at rest, and the hanging mass is placed 0.700 m above the floor. Then the hanging mass falls down until finally touches the floor. Measurements show the hanging mass mh-0212 kg, the spindle radius r = 0.0251 m, the disk's angular accleration α 6.612 rad/s, and the disk's final angular velocity wf 18.79 rad/s. Please compute the quantities below: (hint: Use the formulae in the lab manual and the relations derived in the previous question. Take g 9.8 m/s.) 1. Hanging mass's acceleration a 2. String tension T 3, Torque exerted on disk τ 4. Disk's moment of inertia I 5. Hanging mass's initial potential energy U 6. Hanging mass's final velocity Vh 7. Hanging mass's final kinetic energy Kh 8. Disk's final rotational kinetic energy Kd 9. Total energy loss rate

Answer 1

alpha middot R = a a = 6.612 times 0.0251 a = 0.1659612 m/s^2 M_n g - T = M_n a T = M_n (g - a) T = 0.212 (9.8 - 0.6159612) T = 2.0424 N tau = r times tau = 0.051 times 2.0424 tau = 0.05126 N-m tau = I alpha I = tau /alpha I = 0.05126/6.612 I = 0.0077438 Kg m^2 \r\n U = m g h U = 0.212 times 9.8 times 0.7 U = 1.45432 J V^2_n - U^2 = 2 a H V^2_n = 2 times 0.16596 times 0.7 V_n = squareroot 1.4 times 0.16596 V_n = 0.482 m/s K_n = 1/2 M V^2_n = 1/2 times 0.212 times = 0.02462 J K_d = 1/2 I w^2 w = V_n/r = 0.482/0.0251 = 19.203 rad/s K_d = 1/2 times 0.0077438 times (19.203)^2 K_d = 1.42781 J \r\n S = ut + 1/2 at^2 0.7 = 0 + 1/2 times 0.1659612 t^2 t^2 tau^2 = 1.4/0.1659612