Recent content by imjess

Thank you so much! And is v= -2lmgcosθ. Just to check if that's right...

KE = 1/2[Mx.^2 + θ.^2 Ml^2 sin^2(θ)] I have seemed to get this...could any1 please check if I have done this right

KE = 1/2[Mx.^2 + θ.^2 Ml^2 sin^2(θ) + 1/3 θ.^2 Ml^2] How did you get the last part of KE...no can only get KE = 1/2[Mx.^2 + θ.^2 Ml^2 sin^2(θ)] I don't know where to go from there? Please help me

Homework Statement Consider particle of mass m on a spring of length k, suspended vertically( subject to gravity). Choose an appropriate coordinate for this 1 degree of freedom system and find the langragian. Derive the equation of motion and find the equilibrium point. Using a new...

Homework Statement A uniform rod AB of mass M and length 2l attatched to a slider at A which is constrained to move along a smooth horizontal wire. The rod, which is subject to gravity, is free to swing in the vertical plane containing the wire. Using Ω (the angle between rod and vertical)...

hi...what was your x and y? was it y=-2lcosθ and x? and isn't your potential energy suppose to be v = -mg2lcosθ