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I do not consider the time I gave freely to you wasted. If you are willing to pick up at post #26 and reconsider your expressions for the initial and final potential energy, please do so and I will continue guiding you. If not, not.
are you sure conservation will lead me to finding the time of impact?kuruman said:I do not consider the time I gave freely to you wasted. If you are willing to pick up at post #26 and reconsider your expressions for the initial and final potential energy, please do so and I will continue guiding you. If not, not.
some1 else told me to llok at the relaton between the accaleration o A in x axes withs its component to the direction AG and then the relation between Gs accelaration and that componentManolisjam said:are you sure conservation will lead me to finding the time of impact?
Ok I am trying to do what you said i get the same with i did already .what is different? MY potential energy was zero when G was at L distance so if its zero at the same height i sa the final potential energy is the initial i found.kuruman said:I do not consider the time I gave freely to you wasted. If you are willing to pick up at post #26 and reconsider your expressions for the initial and final potential energy, please do so and I will continue guiding you. If not, not.
Laso could you classify the diffculty of the problem 1-10. for a math undergrad .kuruman said:I do not consider the time I gave freely to you wasted. If you are willing to pick up at post #26 and reconsider your expressions for the initial and final potential energy, please do so and I will continue guiding you. If not, not.
Solved it!kuruman said:I do not consider the time I gave freely to you wasted. If you are willing to pick up at post #26 and reconsider your expressions for the initial and final potential energy, please do so and I will continue guiding you. If not, not.
This is a duplicate ofManolisjam said:Solved it!
At the 0 point using conservation energy i get something like u_a^2+u_g^2=2gsinθl now i know dx/dt=u_a=lsinθdθ/dt . find same way u_G now squaring those. andi plugging them in conservation i get Ldθ/dt=sqrt(2sinθ) this is a differential eq separable. find θ(t) and solve for θ=π/2 but i can't solve the integralharuspex said:This is a duplicate of
https://www.physicsforums.com/threads/classical-mechanics-problem-with-balls.940763/
@Manolisjam , please do not duplicate threads to garner a wider audience. If you wish to bring others in you can either use the "@" link to bring in specific people, or ask your current respondent (me, in this case) to do it. Or even "report" the thread to the mentors.
Anyway, you claim to have solved it, and I think that may be true for the collision velocity if you figured out how to write the energy equation correctly, but I do not see how you will have found the time that way.
Have you found time to collision?
if you can show me another way for the time. i still haven't understant what you are trying to help me do.Manolisjam said:At the 0 point using conservation energy i get something like u_a^2+u_g^2=2gsinθl now i know dx/dt=u_a=lsinθdθ/dt . find same way u_G now squaring those. andi plugging them in conservation i get Ldθ/dt=sqrt(2sinθ) this is a differential eq separable. find θ(t) and solve for θ=π/2 but i can't solve the integral
Ok, you have found another route to the same equation. (Your final equation is not quite right: check the powers of L and g in it.)Manolisjam said:if you can show me another way for the time. i still haven't understant what you are trying to help me do.
It seems to involve elliptic integrals. Nasty.haruspex said:I'll get back to you on solving the integral
Yes, it is nasty. It's a bit simpler if all masses start from rest and along the horizontal line through the origin, but still an elliptic integral. Considering that OP is a math undergrad, this is perhaps a math exercise that assumes understanding of physics to get to the math. I would be curious to see what the solution is according to the person who assigned the problem.haruspex said:It seems to involve elliptic integrals. Nasty.