0
$\begingroup$

I was thinking about a really simple inverted pendulum of length $L$, mass $M$ and made this free-body diagram:

1

I decided to apply Newton's second law relative to the origin (pivot) and to the centroid (location of the force $F_g$), giving equations (1) and (2) respectively:

$$\sum \tau_{o}=-F_{g}\cdot \frac{L}{2}\sin\left( \theta \right)=I_0 \ddot{\theta} \tag{1}$$ $$\sum \tau_{c}=H\cdot \frac{L}{2}\cos\left (\theta\right)-V\cdot \frac{L}{2}\sin\left(\theta \right)=I_{c}\ddot{\theta} \tag{2}$$

If I consider the pendulum as a rod, $I_c=ML^2/12$ and $I_0=ML^2/3$. Are both approaches valid? Do they give the same response of the system?

$\endgroup$
10
  • $\begingroup$ What is the meaning of the symbols you use here? Most are undefined. $\endgroup$
    – nasu
    Commented Jul 23, 2021 at 22:33
  • $\begingroup$ V and H are the vertical and horizontal reaction forces at the pivot, Fg is the weight of the rod, theta is the angle between the rod and vertical axis, Ic and Io are the inertia moments with respect to the origin and centroid $\endgroup$ Commented Jul 23, 2021 at 22:37
  • $\begingroup$ So, the pivot is at the bottom end of the rod? $\endgroup$
    – nasu
    Commented Jul 23, 2021 at 22:40
  • $\begingroup$ Why don't you simply develop the equations (solve the ODEs)? $\endgroup$
    – Gert
    Commented Jul 23, 2021 at 22:40
  • $\begingroup$ Yes, it's an inverted pendulum. I will try to edit the main post. @Gert I will, I'm almost sure equation (1) will give a sensible response, but I'm wondering about equation (2) from a conceptual standpoint $\endgroup$ Commented Jul 23, 2021 at 22:53

1 Answer 1

0
$\begingroup$

No, the two approaches, as you have presented them, would not be equivalent. The first one is accurate. The second is not. You are missing the Newton's equations for the motion of the center of mass of the rod. With their help you would be able to express the components $H$ and $V$ of the reaction force and plug them in the torque equation to reach an equation, equivalent to the one from the first approach.

\begin{align} & x_c \,\hat{i} \, + \, y_c\,\hat{j} \, = \, \frac{L}{2}\,\sin(\theta) \, \hat{i} \, +\, \frac{L}{2}\,\cos(\theta) \, \hat{j} \end{align}

The full set of equations of motion are

\begin{align} & M\frac{d^2}{dt^2} \big(\, x_c \,\hat{i} \, + \, y_c\,\hat{j} \,\big) \,=\, H \, \hat{i} \, + \,V \, \hat{j} \,- \,Mg\,\hat{j}\\ &\\ &I_c\frac{d^2\theta}{dt^2} \,\hat{k} \, = \, \big(\, - x_c \,\hat{i} \, - \, y_c\,\hat{j} \,\big)\times \big(\,H \, \hat{i} \, + \,V \, \hat{j}\,\big) \end{align}

Plug expressions

\begin{align} & \frac{ML}{2}\,\frac{d^2}{dt^2} \big(\, \sin(\theta) \, \hat{i} \, +\,\cos(\theta) \, \hat{j} \,\big) \,=\, H \, \hat{i} \, + \,\big(\,V \,- \,Mg\,\big)\hat{j}\\ &\\ &I_c\frac{d^2\theta}{dt^2} \,\hat{k} \, = \, -\,\frac{L}{2}\big(\, \sin(\theta) \, \hat{i} \, +\,\cos(\theta) \, \hat{j} \,\big)\times \big(\,H \, \hat{i} \, + \,V \, \hat{j}\,\big) \end{align}

Perform most of the operations \begin{align} \frac{ML}{2}\, &\Big[\,-\sin(\theta)\frac{d^2\theta}{dt^2} - \cos(\theta)\Big(\frac{d\theta}{dt}\Big)^2\,\Big] \hat{i} \, \\ +\, \frac{ML}{2}\,&\Big[\,\,\,\,\,\,\,\cos(\theta)\frac{d^2\theta}{dt^2} - \sin(\theta)\Big(\frac{d\theta}{dt}\Big)^2\,\Big] \, \hat{j} \,=\, H \, \hat{i} \, + \,\big(\,V \,- \,Mg\,\big)\hat{j}\\ &\\ &I_c\frac{d^2\theta}{dt^2} \,\hat{k} \, = \, \frac{L}{2}\big(\, H\cos(\theta) - V\sin(\theta) \,\big) \hat{k} \end{align}

Solve for the horizontal and vertical components of the reaction force \begin{align} &H \, =\, \frac{ML}{2}\, \Big[\,-\sin(\theta)\frac{d^2\theta}{dt^2} - \cos(\theta)\Big(\frac{d\theta}{dt}\Big)^2\,\Big] \\ &V \, = \, \frac{ML}{2}\,\Big[\,\,\,\,\,\,\,\cos(\theta)\frac{d^2\theta}{dt^2} - \sin(\theta)\Big(\frac{d\theta}{dt}\Big)^2\,\Big] \, +\, Mg\\ &\\ &I_c\frac{d^2\theta}{dt^2} \, = \, \frac{L}{2}\, H\cos(\theta) - \frac{L}{2}\, V\sin(\theta) \end{align}

and plug them in the third (torque) equation, simplify and apply the appropriate trigonometric identities. The result is

$$ I_c\frac{d^2\theta}{dt^2} \, = \, -\,\frac{ML^2}{4}\,\frac{d^2\theta}{dt^2} \, -\, \frac{MgL}{2} \sin(\theta) $$

or re-expressed $$ \left(I_c\, + \,\frac{ML^2}{4}\right)\frac{d^2\theta}{dt^2} \, = \, -\, \frac{MgL}{2} \sin(\theta) $$

where $$I_o \, =\, I_c\, + \,\frac{ML^2}{4}$$ which is in line with the parallel axis theorem.

$\endgroup$
1
  • $\begingroup$ Thank you very much for the answer, Futurologist. Indeed, when the H and V are plugged in equation (2), the resultant forces in the right side and the inertia moment in the left side match equation (1). Once again, thank you for the excellent demonstration! $\endgroup$ Commented Jul 31, 2021 at 1:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.