Rotational Motion: Pendulum
1. Write the equations of motion for the simple pendulum shown in Fig. 2.14, where all the mass is concentrated at the end point and there is a torque, $T_c,$ applied at the pivot.
2. Use MATLAB to determine the time history of $θ$ to a step input in $T_c$ of $1 \ N·m.$ Assume $l = 1 \ m, \ m = 1 \ kg,$ and $g = 9.81 \ m/sec^2.$

Question

Rotational Motion: Pendulum
1. Write the equations of motion for the simple pendulum shown in Fig. 2.14, where all the mass is concentrated at the end point and there is a torque, [latex] T_c, [/latex] applied at the pivot.
2. Use MATLAB to determine the time history of [latex] θ [/latex] to a step input in [latex] T_c [/latex] of [latex] 1 \ N·m. [/latex] Assume [latex] l = 1 \ m, \ m = 1 \ kg,[/latex] and [latex] g = 9.81 \ m/sec^2.[/latex]

Accepted Answer

1. Equations of motion: The moment of inertia about the pivot point is [latex] I = ml^2.[/latex] The sum of moments about the pivot point contains a term from gravity as well as the applied torque [latex] T_c. [/latex] The equation of motion, obtained from Eq. (2.14), is

[latex] M = I \alpha [/latex] (2.14)

[latex] T_{c}-m g l \sin heta=I \ddot{ heta}, [/latex] (2.21)

which is usually written in the form

[latex] \ddot{ heta}+\frac{g}{l} \sin heta=\frac{T_{c}}{m l^{2}} [/latex] (2.22)

This equation is nonlinear due to the [latex] \sin θ [/latex] term. A general discussion of nonlinear equations is contained in Chapter 9; however, we can proceed with a linearization of this case by assuming the motion is small enough that [latex] \sin heta \cong heta [/latex] Then Eq. (2.22) becomes the linear equation

[latex] \ddot{ heta}+\frac{g}{l} heta=\frac{T_{c}}{m l^{2}} [/latex] (2.23)

With no applied torque, the natural motion is that of a harmonic oscillator with a natural frequency of [latex] ^3 [/latex]

[latex] \omega_{n}=\sqrt{\frac{g}{l}} [/latex] (2.24)

The transfer function can be obtained as described for Eq. (2.7), yielding

[latex] \frac{V (s)}{U(s)} = \frac{\frac{1}{m}}{s+ \frac{b}{m}}. [/latex] (2.7)

[latex] \frac{\Theta(s)}{T_{c}(s)}=\frac{\frac{1}{m l^{2}}}{s^{2}+\frac{g}{l}} [/latex] (2.25)

2. Time history: The dynamics of a system can be prescribed to MATLAB in terms of row vectors containing the coefficients of the polynomials describing the numerator and denominator of its transfer function. In this case, the numerator (called num) is simply one number, since there are no powers of [latex] s, [/latex] so that

[latex] ext { num }=\frac{1}{m l^{2}}=\frac{1}{(1)(1)^{2}}=[1] ,[/latex]

and the denominator (called den) contains the coefficients of the descending powers of [latex] s [/latex] in [latex] (s^2 + ^g/_l) [/latex] and is a row vector with three elements:

[latex] ext { den }=\left[\begin{array}{lll} 1 & 0 & \frac{g}{l} \end{array} ight]=\left[\begin{array}{lll} 1 & 0 & 9.81 \end{array} ight]. [/latex]

The desired response of the system can be obtained by using the MATLAB step response function, called step. The MATLAB statements

Question 2.EX.5: Rotational Motion: Pendulum 1. Write the equations of motion...

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