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Physics
Classical Physics
Mechanics
Deriving Kinetic Energy in a Double Pendulum System
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[QUOTE="Andreas C, post: 5481320, member: 593662"] Ok, I found the answer, here it is for anyone who might be interested: Using the chain rule, we write say dx[SUB]1[/SUB]/dt=sinθ as dθ/dt ⋅ d(sinθ)/dθ = dθ/dt ⋅ cosθ. If we plug this (and y[SUB]1[/SUB]) into the equation for kinetic energy in this case (which is T[SUB]1[/SUB]=(x^2+y^2)/2), and do some algebra, we eventually get the anticipated solution of T[SUB]1[/SUB]=(dθ/dt)^2/2. It's the same thing for T[SUB]2[/SUB], only a bit more complicated. The point is that you're meant to apply the chain rule for derivatives. [/QUOTE]
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Physics
Classical Physics
Mechanics
Deriving Kinetic Energy in a Double Pendulum System
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