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Wombat11

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I start out by substituting rcos(Θ) and rsin(Θ) for x and y respectively. This gives me z=(b/2)r^2. The Lagrangian of this system is (1/2)m(rdot^2+r^2⋅Θdot^2+zdot^2)-mgz. (rdot and such is the time derivative of said variable). I then find the time derivative of z, giving me zdot=br⋅rdot and plug it into the Lagrangian giving me (1/2)m( rdot^2 + r^2⋅Θdot^2 + b^2r^2⋅rdot^2) - (1/2)mgbr^2. (plugged in regular z too). Finding the Lagrange equation of motion for 'r' gives me (ddot will be the second time derivative of said variable) 0= rddot(1+b^2r^2)-r⋅Θdot^2 +

Sorry about the notation, I have no idea how to put the equations into the computer.

__b^2⋅rdot^2⋅r__+ gbr . This is not correct though, the right answer has a negative term of__b^2⋅rdot^2⋅r__. Any help would be greatly appreciated.Sorry about the notation, I have no idea how to put the equations into the computer.

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