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UTexas Physics problem loop-the-loop, please help?

  • So I haven't done this before but I honestly don't know where to start. A bead slides without friction around a loop-the-loop. The bead is released from a height 14.1 m from the bottom of the loop-the-loop which has a radius 5 m. g = 9.8 m/s^2 What is its speed at the top of the loop-the-loop? Please help, normally i would know how to do this but i don't know how to do it when mass is not given. Any response will help, thanks.

  • Answer:

    by conservation of energy: KEi + PEi = KEf + PEf KEi = 0 0 + mgh_1 = (1/2)mv^2 + mgh_2 Divide both sides by m gh_1 = (1/2)v^2 + gh_2 isolating v... 2g(h_1 - h_2) = v^2 sqrt(2g(h_1 - h_2)) = v h_1 = 14.1m h_2 = diameter = 5m*2 = 10m g = 9.8 m/s^2 plug in. i think this is right xD

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Always use conservation of energy in problems when you can't use kinematics, first part, as there is no non conservative forces acting, we shall use conservation of energy U1 + K1 =U2 + K2 where U=mgh and K=1/2mv² taking h=0 at bottom, and v1=0 mgh1=1/2mv²+mgh2 so, v=√[2g(h1-h2)] =8.964 m/s for your problem, second part, from FBD, mg+N=mv²/R where V is speed at that moment, answer=37.6 N

Ravi

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