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23 Nov 2019
I don`t agree with the solution on this web, but I don´t surethat my solution is right.
Please, comment my solution:
The Newton equation for the mass is:
-N-mgsinθ=-mRÏ2
The mass push to top-loop with a force N' equal to its weight,then N'=N:
-mg-mgsinθ=-mRÏ2
But, at the top-loop θ=Ï/2:
-mg - mg = -2mg = -mRÏ2
Finally, I make conservation of energy between two points:bottom-loop and top-loop.
mgz= 1/2mR2Ï2 + mg2R = mgR + mg2R
Z = 3R.
And that is all.
I don`t agree with the solution on this web, but I don´t surethat my solution is right.
Please, comment my solution:
The Newton equation for the mass is:
-N-mgsinθ=-mRÏ2
The mass push to top-loop with a force N' equal to its weight,then N'=N:
-mg-mgsinθ=-mRÏ2
But, at the top-loop θ=Ï/2:
-mg - mg = -2mg = -mRÏ2
Finally, I make conservation of energy between two points:bottom-loop and top-loop.
mgz= 1/2mR2Ï2 + mg2R = mgR + mg2R
Z = 3R.
And that is all.
0
answers
0
watching
47
views
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