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Angular acceleration of a yo-yo pulled with constant force
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[QUOTE="ShilpaM, post: 4944039, member: 535289"] [h2]Homework Statement [/h2] A yo-yo is pulled with a constant tension [I]T[/I]. 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 [I]R [/I]and the spool radius is [I]r[/I]. The mass of the yo-yo is [I]m[/I] and the moment of inertia of the yo-yo around the axis through its centre of mass is [I]I[/I]. What is the angular acceleration of the yo-yo, assuming it rolls without slipping? [B] [ATTACH=full]174753[/ATTACH] 2. Homework Equations [/B] α = τ[SUB]net[/SUB] / I τ = Fsinθr [h2]The Attempt at a Solution[/h2] I think the spool rotates in the same direction as [I]T, [/I] but the rest of the yo-yo has to rotate in the opposite direction in order to roll without slipping. [I]T [/I]applies a negative torque, since it causes rotation in the CW direction, so τ[SUB]T[/SUB] = -Tr This is where I'm stuck. Is the inertia of the yo-yo just [I]I[/I] or is it [I]I + mR[SUP]2[/SUP][/I]? 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? [B][/B] [/QUOTE]
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Angular acceleration of a yo-yo pulled with constant force
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