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 Fapp 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?
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Do I first need to calculate Fg where Fg = m*g = (19)(9.8) = 186.2 N??? Then Fsum = Fapp + Fg = 154 + 186.2 = 340.2 N. I know that I can also find the tilt angle theta where theta = arctan (Fapp/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.
To find the rotational inertia of the yo-yo-shaped device, we can use the equations of rotational motion. Here's how you can proceed:
1. Start by calculating the weight of the box, which is equal to the force due to gravity. You correctly calculated it as Fg = m*g = (19 kg) * (9.8 m/s^2) = 186.2 N.
2. Next, calculate the net force on the device. The net force is the sum of the applied force (Fapp) and the force due to gravity (Fg). So Fnet = Fapp + Fg = 154 N + 186.2 N = 340.2 N.
3. Now, let's consider the torques acting on the device. The torque slowing down the rotation is given by the product of the force applied at the outer radius (Fapp) and the inner radius (r): Torque = Fapp * r.
4. The torque accelerating the rotation is the product of the lifting force (19g + 19 * acceleration, where acceleration = r * angular acceleration) and the inner radius (r): Torque = (19 * g + 19 * r * angular acceleration) * r.
5. Since the device is in rotational equilibrium, the sum of the torques is equal to the moment of inertia (I) multiplied by the angular acceleration (α). Therefore, we have the equation: (Fapp * r) - [(19 * g + 19 * r * α) * r] = I * α.
6. Rearrange the equation to isolate the moment of inertia (I): I = [(Fapp * r) - (19 * g + 19 * r * α) * r] / α.
7. Plug in the given values: Fapp = 154 N, r (inner radius) = 0.20 m, g (acceleration due to gravity) = 9.8 m/s^2, and α (angular acceleration) = 0.80 m/s^2.
8. Substitute the values into the equation and calculate the rotational inertia (I).
I = [(154 N * 0.20 m) - (19 * 9.8 m/s^2 + 19 * 0.20 m * 0.80 m/s^2) * 0.20 m] / 0.80 m/s^2.
After performing the calculation, you will determine the rotational inertia (I) of the yo-yo-shaped device about its axis of rotation.