PHYSICS

A yo-yo-shaped device mounted on a horizontal frictionless axis is used to lift a 19-kg box. 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 !@#$%^&p of magnitude 154 N is applied to a rope wrapped around the outside of the device, the box, which is suspended from a rope wrapped around the hub, has an upward acceleration of magnitude 0.80 m/s2 What is the rotational inertia of the device about its axis of rotation?
>>>>>>>>>>>>>>>>>>>>>>>>>.
Do I first need to calculate Fg where Fg = m*g = (19)(9.8) = 186.2 N??? Then Fsum = !@#$%^&p + Fg = 154 + 186.2 = 340.2 N. I know that I can also find the tilt angle theta where theta = arctan (!@#$%^&p/Fg) = 39.59 degrees but I am not sure if I need that. What should I do next???

I am not certain of the picture. What I see is a rope on the inner hub pulling a weight.

force*innerradius= torque slowing down

Outer force*outer radius - lifting force*inner radius= I*angularacceleration

Now, be careful here: the lifting force is 19g + 19*acceleration, where acceleration is innerradius*angularacceleration

so solve for I, you know all else. I have no idea of a tilt angle entering here.

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asked by Calvin

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