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What is the speed when a disk has reached maximum extension?
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[QUOTE="freutel, post: 5134278, member: 554085"] Ok, so the force balance at disk A is the gravitational force downwards and an equal tension force upwards. There will be torque clockwise because the tension force is excerted at distance r from the rotation axis. Just as I solved it earlier i get for acceleration downwards [I]a[SUB]cm[/SUB]=2/3g[/I]. Then will the force applied on disk B be [I]F=ma[SUB]cm[/SUB][/I]? If so then after subbing and using the same equations i get for [I]a[SUB]tan[/SUB]=3/2g[/I]. Total downwards acceleration is now 2g. I don't know if this is correct but if so can I use the equation [I]v[SUP]2[/SUP]=v[SUB]0[/SUB][SUP]2[/SUP]+2aL[/I] to say that the final speed is [I]v=√(4gL)[/I]? I believe you're referring to a yoyo. So disk A will roll back up and then down again? I thought this was too easy because the original questions says yoyo instead of disk so saying it will behave like a yoyo seemed too simple for me. [/QUOTE]
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What is the speed when a disk has reached maximum extension?
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