Critical Tip Velocity


By Rob Siegwart
Category: design

The following describes an object moving at constant velocity and impacting a rigid obstacle, leading to pivoting and tip-over. The objective is to find the translational velocity that causes the object to rise to the "just-tipping" condition (maximal rise in height).

The equations are based on conservation of momentum and conservation of energy. The linear momentum is converted to an angular momentum by considering its moment arm about the pivot point. After impact, the angular momentum is converted from its initial value to zero at the "just tipping point". Energy is also conserved as the rotational kinetic energy is converted to potential energy.

Variables

  • \(v_0\) – initial translational velocity
  • \(H_O\) – angular momentum about pivot point O
  • \(m\) – mass
  • \(y\) – vertical distance from pivot point O to center of gravity
  • \(I_O\) – moment of inertia about pivot point O
  • \(KE\) – kinetic energy
  • \(PE\) – potential energy
  • \(g\) – gravitational acceleration constant
  • \(h\) – CG height change
  • \(\theta_0\) – resting angle between the CG vector relative to the pivot point and the horizontal plane
  • \(\omega\) – rotational velocity

Equations

Translational momentum moment,

\begin{equation*} H = mv_0y \end{equation*}

Angular momentum,

\begin{equation*} H_O = I_O\omega \end{equation*}

Rotational kinetic energy,

\begin{equation*} KE = \frac{1}{2}I_O\omega^2 \end{equation*}

Potential energy,

\begin{equation*} PE = mgh \end{equation*}

Critical Tipping Velocity

The initial angular velocity, \(\omega_0\), can be solved by equating the two momentum quantities.

\begin{equation*} H = H_O \end{equation*}
\begin{equation*} \omega_0 = \frac{mv_0y}{I_O} \end{equation*}

After impact, for the object to rise to the just-tipping point (maximum CG height change), the rotational kinetic energy is completely converted to potential energy:

\begin{equation*} \frac{1}{2}I_O\omega^2 = mgh \end{equation*}
\begin{equation*} \frac{1}{2}I_O \left( \frac{mv_0y}{I_O} \right)^2 = mgh \end{equation*}

Rearranging for the initial velocity, \(v_0\):

\begin{equation*} v_0 = \sqrt{\frac{2ghI_O}{my^2}} \end{equation*}

Furthermore, the CG height change \(h\) is known and can be expressed in terms of the rest angle, \(\theta_0\):

\begin{equation*} h = r - r\sin\theta_0 = r(1-\sin\theta_0) \end{equation*}

\(v_0\) then becomes:

\begin{equation*} v_0 = \sqrt{\frac{2gr(1-\sin\theta_0)I_O}{my^2}} \end{equation*}