A Cart–Inverted-Pendulum System (Energy Method) Reconsider the mechanical system, as shown in Figure 5.79. Derive the equations of motion of the system by using Lagrange’s equations.

Question

A Cart–Inverted-Pendulum System (Energy Method)

Reconsider the mechanical system, as shown in Figure 5.79. Derive the equations of motion of the system by using Lagrange’s equations.

Accepted Answer

As shown in Figure 5.88a, the motion of the cart is purely translational and

[latex]T_ ext{cart} = \frac{1}{2}m\dot{x}^2[/latex]

The inverted pendulum undergoes both translation and rotation and

[latex]T_ ext{pendulum} = \frac{1}{2}M\upsilon_C^2 + \frac{1}{2}I_C\dot{θ}^2[/latex]

where the velocity at the center of mass can be found by

[latex]\upsilon_C^2= \left\lgroup \dot{x} + \frac{L}{2}\dot{θ} ext{cos}θ ight group^2 + \left\lgroup \frac{L}{2}\dot{θ} ext{sin}θ ight group^2 = \dot{x}^2 + L\dot{x}\dot{θ} ext{cos}θ + \frac{1}{4}L^2 \dot{θ}^2[/latex]

and the mass moment of inertia [latex]I_C = \frac{1}{12} ML^2[/latex]. Thus, the kinetic energy of the system is

[latex]T = T_ ext{cart} + T_ ext{pendulum} = \frac{1}{2}(m + M)\dot{x}^2 + \frac{1}{6}ML^2 \dot{θ}^2 + \frac{1}{2}ML\dot{x}\dot{θ} ext{cos}θ[/latex]

Note that no spring elements are involved in the system. This implies that [latex]V_e = 0[/latex] and [latex]V = V_ ext{g}[/latex]. Using the datum defined in Figure 5.88b, we can obtain the gravitational potential energy

[latex]V_ ext{g} = M ext{g}h = M ext{g}\frac{L}{2} ext{cos}θ [/latex]

As shown in Figure 5.88c, the system is subjected to a horizontal force [latex]f[/latex], by which the work done is not zero. Thus, it is a nonconservative system. Same as Example 5.14, the displacement of the cart, [latex]x[/latex], and the angular displacement of the pendulum, [latex]θ[/latex], are chosen as the generalized coordinates. We then apply Lagrange’s equations

[latex]\frac{d}{dt}\left\lgroup \frac{∂T}{∂\dot{q}_i} ight group - \frac{∂T}{∂q_i} + \frac{∂V}{∂q_i}=Q_i, i = 1,2 [/latex]

For [latex]i = 1, q_1 = x, \dot{q}_1 = \dot{x}[/latex],

[latex]\frac{∂T}{∂\dot{x}} = (m + M)\dot{x} + \frac{1}{2}ML\dot{θ} ext{cos}θ[/latex]

[latex]\frac{d}{dt}\left\lgroup \frac{∂T}{∂\dot{x}} ight group = (m + M)\ddot{x} + \frac{1}{2}ML\ddot{θ} ext{cos}θ - \frac{1}{2}ML\dot{θ} ext{sin}θ\dot{θ}[/latex]

[latex]\frac{∂T}{∂x} = 0[/latex]

[latex]\frac{∂V}{∂x} = 0[/latex]

Applying Equation 5.51 gives the generalized force

[latex]Q_i = \sum\limits_{j=1}^{N}\mathbf{F}_j \frac{∂\mathbf{r}_j}{∂q_i} j = 1,2,...,N[/latex] (5.51)

[latex]Q_1 = \mathbf{F}\frac{∂\mathbf{r}}{∂x} = (f\mathbf{i})\frac{∂}{∂x}(x\mathbf{i}) = f[/latex]

Substituting into Lagrange’s equation results in

[latex](m + M)\ddot{x} + \frac{1}{2}ML\ddot{θ} ext{cos}θ - \frac{1}{2}ML\dot{θ}^2 ext{sin}θ = f[/latex]

Similarly, for [latex]i = 2, q_2 = θ, \dot{q}_2 = \dot{θ}[/latex],

[latex]\frac{∂T}{∂\dot{θ}} = \frac{1}{3}ML^2\dot{θ} + \frac{1}{2}ML\dot{x} ext{cos}θ[/latex]

[latex]\frac{d}{dt}\left\lgroup \frac{∂T}{∂\dot{θ}} ight group = \frac{1}{3}ML^2\ddot{θ} + \frac{1}{2}ML\ddot{x} ext{cos}θ - \frac{1}{2}ML\dot{x} ext{sin}θ\dot{θ}[/latex]

[latex]\frac{∂T}{∂θ} = - \frac{1}{2}ML\dot{x}\dot{θ} ext{sin}θ[/latex]

[latex]\frac{∂V}{∂θ} = - \frac{1}{2}M ext{g}L ext{sin}θ[/latex]

Applying Equation 5.51 gives the generalized force

[latex]Q_2 = \mathbf{F}\frac{∂\mathbf{r}}{∂θ} = (f\mathbf{i})\frac{∂}{∂θ}(x\mathbf{i}) = 0[/latex]

Note that the external force [latex]f[/latex] does not generate a moment on the pendulum. Substituting into Lagrange’s equation gives

[latex]\frac{1}{3}ML^2\ddot{θ} + \frac{1}{2}ML\ddot{x} ext{cos}θ - \frac{1}{2}M ext{g}L ext{sin}θ = 0[/latex]

The result is the same as the one obtained previously in Example 5.14.

Question 5.18: A Cart–Inverted-Pendulum System (Energy Method) Reconsider t...

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