A yo-yo shaped device mounted on a horizontal frictionless axis is used to lift a 30 kg box as shown. The outer radius R of the device is 0.50 m, and the radius r of the hub is 0.20 m. When a constant horizontal force Fappof magnitude 140 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 0.80 m/s2. What is the rotational inertia of the device about its axis of rotation?
2006-11-27 06:56:03 · 1 answers · asked by Galaxy D 2 in Science & Mathematics ➔ Physics
The applied torque is Fapp*R=70 Nm.
The tension on the rope supporting the 30 kg box is m*(g+0.8) = 318.2 N.
The torque producing this tension is 318.2*r = 63.64 Nm.
The remaining torque, 6.36 Nm, accelerates the wheel at 0.8/r = 4 rad/s^2.
The inertia of the wheel is torque/ang. accel. = 6.36/4 = 1.59 kg-m^2.
2006-11-29 02:29:11 · answer #1 · answered by kirchwey 7 · 4⤊ 0⤋