Big Yo-Yo problem (Rotational Dynamics with Kinematics? Maybe?)

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SUMMARY

The discussion centers on solving the Big Yo-Yo problem using rotational dynamics and kinematics. The user calculated the angular acceleration of a yo-yo by applying the formula τ = Iα, resulting in an angular acceleration of 3.065625 rad/s² and a translational acceleration of 2.4525 m/s². However, attempts to derive velocity using kinematic equations were unsuccessful, indicating a misunderstanding of the torque origin and moment of inertia. The conversation emphasizes the importance of correctly identifying the axis for torque calculations and the role of tension in the string versus gravitational forces.

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Homework Statement
A giant yo-yo has a mass of 3 kg and a moment of inertia of 7.68 kg⋅m^2. The central spool has a radius of 0.8 m. As it falls, the string unwinds from the central spool without slipping. If the yo-yo is released from rest, how fast will it be moving when it has fallen a distance of 1.5 meters?
Relevant Equations
## \tau=I\alpha ##
## \tau = r F ##
## x = x_o + vt + 1/2at^2 ##
## (V_ƒ)^2 = (V_o)^2 + 2ax ##
What I attempted to do first was find alpha and turn that into translational acceleration.

Taking mass of yoyo * radius of spool * gravity, (3kg)(0.8m)(9.81m/s^2) yielded 23.544 N*m, and dividing by I = 7.68 kg * m^2 yielded 3.065625 rad/s^2. Finally, multiplying by r = 0.8m gave me 2.4525 m/s^2.

I assumed I could simply use the kinematics equations then, after translating into linear motion, to find velocity at the distance, but every attempt I've made has been wrong. I tried using (Vƒ)^2 = (Vo)^2 + 2ax to give me a velocity, setting vo to 0. But that was incorrect. I also tried some weird thing with [ tex ] x = x_o + v*t + 1/2*a*t^2 [ /tex ] , setting x = 1.5m, x_o = 0, to find t, and using that result again to find v*t. Nothing has worked so far, and I'm not really sure where my thought process is going wrong! Thank you for reading!
 
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Using energy concepts is a good approach to this problem.

If you want to stick with the torque approach, then note the point of space that you are taking to be the origin for your torque calculation. What is the moment of inertia of the yo-yo about this origin?
 
To elaborate on TSny's reply...
What axis are you taking for the torque and angular acceleration? If centre of spool then the force exerting the torque is the tension in the string, not gravity; if a fixed point in the vertical line of the string then the MoI is not what you used; if the point of contact of spool with string, as a dynamic concept, that is not an inertial frame so may mislead.
 

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