Angular acceleration of a yo-yo pulled with constant force

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The discussion revolves around calculating the angular acceleration of a yo-yo being pulled with a constant tension while rolling without slipping. The key equations involved are α = τnet / I and τ = Fsinθr, with tension T creating a negative torque on the yo-yo. There is confusion regarding whether to include the moment of inertia I alone or I + mR² in the calculations. Participants clarify that the yo-yo and spool should be treated as a single rigid object, emphasizing that they cannot rotate in opposite directions. The conversation highlights the importance of analyzing moments about a specific point to resolve the forces acting on the yo-yo.
ShilpaM
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Homework Statement


A yo-yo is pulled with a constant tension T. The string is horizontal and parallel to the table and unwinding from the bottom of the spool, as shown. The yo-yo's outer radius is R and the spool radius is r. The mass of the yo-yo is m and the moment of inertia of the yo-yo around the axis through its centre of mass is I. What is the angular acceleration of the yo-yo, assuming it rolls without slipping?

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2. Homework Equations

α = τnet / I
τ = Fsinθr

The Attempt at a Solution


I think the spool rotates in the same direction as T, but the rest of the yo-yo has to rotate in the opposite direction in order to roll without slipping. T applies a negative torque, since it causes rotation in the CW direction, so τT = -Tr
This is where I'm stuck. Is the inertia of the yo-yo just I or is it I + mR2? And I'm not sure if it's only tension applying a torque to the yo-yo, or if the floor is exerting a backward force on the yo-yo to make it roll, and if so, is the force proportionate to the normal force. Can anyone please help me figure this one out?
 
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ShilpaM said:
I think the spool rotates in the same direction as T, but the rest of the yo-yo has to rotate in the opposite direction
The yo-yo and spool should be taken as one rigid object. They cannot rotate in opposite directions.
The easiest way is to take moments about a point through which the unknown forces pass. What point would that be?
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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