A yo-yo-shaped device mounted on a horizontal frictionless axis is used to lift

Question

A yo-yo-shaped device mounted on a horizontal frictionless axis is used to lift a 39 kg box as shown in the figure. The outer radius R of the device is 1.1 m, and the radius r of the hub is 0.39 m. When a constant horizontal force (Fapp) of magnitude 210 N is applied to a rope wrapped around the outside of the device, the box, which is suspended from a rope wrapped around the hub, has an upward acceleration of magnitude 0.87 m/s2. What is the rotational inertia of the device about its axis of rotation? this is the link for the image: http://edugen.wileyplus.com/edugen/courses/crs4957/art/qb/qu/c10/fig10_62.gif

Explanation / Answer

The device rotates in counterclockwise direction in order for the box to move up. Take counterclockwise direction as positive. Let, alpha = angular acceleration of the device, a = linear acceleration of a point on the rim of the hub a = acceleration of the box = 0.80 m/s^2 alpha = a/r = 0.80/0.20 Or alpha = 4 rad/s^2 -------------------------------------(1) Let T = tension in the rope Forces on the box are 1. Tension T upward 2. Weigth mg downward Net upward force = T - mg Upward acceleration of box = a Therefore, T - mg = ma Or T = mg + ma = m(g + a) Or T = 30*(9.8 + 0.80) = 30*10.6 Or T = 318 N---------------------------(2) Calculate torques on the device around the axis of rotation taking counterclockwise torque as +ve and clockwise as -ve. Torque by applied force F = FR = 140 * 0.50 = 70 Nm Torque by T = -Tr = -318 * 0.20 [using value of T from equation(2)] = -63.6 Nm Net torque on the device = 70 - 63.6 Nm = 6.4 Nm-------(3) Let rotational inertial of the device = I I = net torque/alpha Using equations (1) and (3) in the above, I = 6.4/4 kg-m^2 = 1.6 kg-m^2 Ans: 1.6 kg-m^2

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