Can you please tell me how the gravitational force affects the lagrangian equation? A particle of mass m is constrained to move on the inside surface of asmooth cone of half-angle a (i.e., z =r cot a) as shown in the figure to agravitational force. The particle is subject to a gravitational force (i.e.,g=-gk ).(a) Determine the Lagrangian for this particle in terms of the generalizedcoordinates r and 0.(b) Determine Lagrange's equations of motion for this particle. so the Langrangian is1L =m (² csc²a + rö*) – mgr cot a%3D(7.28)

Lagrangian The Lagrangian of a system characterizes the state of a system. For a conservative system, that a system with no dissipative forces, the Lagrangian is the difference between the kinetic energy and the potential energy of the system. L=T-V The Lagrangian of a particle is a function of generalized coordinates and generalized velocities and time. L=Lqi, qi˙, t The equation of motion of a particle is given by Lagrange's equation. There is a Lagrange's equation associated with every generalized coordinate. Lagrange's equation of motion for the generalized coordinate qi is ddt∂L∂qi˙-∂L∂qi=0The Lagrangian of a particle moving in the inner surface of a zone of half-angle characterized by the generalized coordinates r and θ is given by L=12mr˙2cosec2α+r2θ˙2-mgrcotα We will get two Lagrange's equations, one for r and one for θ. Let us first find Lagrange's equation for the generalized coordinate r ∂L∂r˙=22mr˙cosec2α=mr˙cosec2α ∂L∂r=22mrθ˙2-mgcotα=mrθ˙2-mgcotα Lagrange's equation for r is thus ddt∂L∂r˙-∂L∂r=0 This gives ddtmr˙cosec2α-mrθ˙2+mgcotα=0⇒mr¨cosec2α-mrθ˙2+mgcotα=0⇒r¨cosec2α-rθ˙2+gcotα=0 For Lagrange's equation for the generalized coordinate θ we have ∂L∂θ˙=22mr2θ˙=mr2θ˙ ∂L∂θ=0 For θ ddt∂L∂θ˙-∂L∂θ=0 This gives ddtmr2θ˙=0⇒ddtr2θ˙=0⇒2rr˙θ˙+r2θ¨=0Answer: Lagrange's equation of motion for the generalized coordinate r is r¨cosec2α-rθ˙2+gcotα=0 Lagrange's equation of motion for the generalized coordinate θ is 2rr˙θ˙+r2θ¨=0

Answered: (b) Determine Lagrange's equations of…