### Part B : Dynamics of Pumpkin

The pumpkin dynamics has three distinct phases which must be treated separately - while it is still in contact with the shute, while it is in the air but connected by a rope to the slinger at the tip of the trebuchet's beam and finally while following a ballistic trajectory.

#### Pumpking sliding along on the shute

Task is two-fold : establishing a connection between $$\ddot{\gamma}$$ and $$\ddot{\alpha}$$ and secondly to find an equation for the force S the rope is exerting on the pumpkin and on the top of the trebuchet's beam.

In Fig.(B.1) the coordinates of the pumpkin is given by :

\begin{equation*} x = b \sin\alpha - L_p\sin\gamma \end{equation*} \begin{equation*} y = H + b \cos\alpha - L_p\cos\gamma \end{equation*}

Our interest is with the acceleration of the pumpkin :

\begin{equation*} \ddot{x} = -b \sin(\alpha) \dot{\alpha}^2 + b \cos(\alpha) \ddot{\alpha} + L_p \sin(\gamma) \dot{\gamma}^2 - L_p \cos(\gamma) \ddot{\gamma} \tag{B.1a} \end{equation*} \begin{equation*} \ddot{y} = -b \cos(\alpha) \dot{\alpha}^2 - b \sin(\alpha) \ddot{\alpha} + L_p \cos(\gamma) \dot{\gamma}^2 + L_p \sin(\gamma) \ddot{\gamma} \tag{B.1b} \end{equation*}

Because the acceleration in the y-direction is zero for the pumpkin while in touch with the shute Eq.(B.1b) gives us immediately the desired result for $$\ddot{\gamma}$$ :

\begin{equation*} \ddot{\gamma} = \left(- \cos(\gamma) \dot{\gamma}^2 + \frac{b}{L_p} \left( \cos(\alpha) \dot{\alpha}^2 + \sin(\alpha) \ddot{\alpha} \right) \right) / \sin\gamma \tag{B.2} \end{equation*}

This equation is directly being used in the program, see file diffeqs.js for the mark B.2. In Eq.(B.2) the division by $$\sin(\gamma)$$ is worrisome unless we can show that the pumpkin always separates from the shute before γ goes to zero.

F=m a in the x-direction will give us an equation for the force S :

\begin{equation*} S \sin\gamma - \mu ( M_p g - S \cos\gamma ) = M_p \ddot{x} \end{equation*}

Here μ is the coefficient of dynamic friction between the pumpkin and the shute , Mp is the pumpkin's mass and g = 9.81m/sec2. Solving for S :

\begin{equation*} S = \frac{M_p ( \mu g + \ddot{x} )}{\sin\gamma + \mu \cos\gamma}\tag{B.3} \end{equation*}

In Fig (B.1) the forces Np and Pp represent the normal (in line with the beam axis) and perpendicular component of rope force S as they are acting on the beam at the tip of the beam.

\begin{align} N_p &= - S \cos(\alpha{-}\gamma) \\ P_p &= S \sin(\alpha{-}\gamma)\tag{B.4} \end{align}

A positive Np will cause tension in the beam ; a positive Pp will rotate the beam clock-wise causing a positive $$\ddot{\alpha}$$.

We now perform a series of substitution to obtain an equation for Pp which does not contain $$\ddot{\gamma}$$. First we substitute S from Eq.(B.3) , then $$\ddot{x}$$ from Eq.(B.1a) and eliminate from the $$\ddot{\gamma}$$ using Eq.(B.2).

\begin{equation*} \frac{P_p}{a M_p}\;=\;\underbrace{ \frac{\sin(\alpha{-}\gamma)}{\sin\gamma + \mu_p \cos\gamma} \left\{ \mu_p\frac{g}{a} + \frac{\frac{L_P}{a}\dot{\gamma}^2 - \frac{b}{a}\cos(\alpha{-}\gamma)\dot{\alpha}^2} {\sin\gamma}\right\} }_{= p_1} \;-\; \underbrace{ \frac{\frac{b}{a} \sin^2(\alpha{-}\gamma)}{\sin\gamma (\sin\gamma + \mu_p \cos\gamma)} \ddot{\alpha} }_{= p_2} \tag{B.5} \end{equation*} Separation of the pumpkin from the shute takes place when the normal force from the shute onto the pumpkin becomes zero, i.e. : \begin{equation*} S \cos\gamma - m g = 0 \end{equation*} Feel free to follow the column labeled Nshute of the table provided by the simulation.

#### Pumpkin off shute but attached to rope

The pumpkin is now exposed to only the rope force S and gravity. F = m a separately for the x- and the y-component :

\begin{align*} S \sin\gamma &= M_p \ddot{x} \\ S \cos\gamma - M_p g &= M_p \ddot{y} \end{align*}

Substituting to eliminate S and after some algebra including the Eqs.(B.1a) and (B.1b) :

\begin{equation*} \ddot{\gamma} = - \frac{b}{L_p} \left( \sin(\alpha{-}\gamma) \dot{\alpha}^2 - \cos(\alpha{-}\gamma) \ddot{\alpha} \right) - \frac{g}{L_p} \sin\gamma \tag{B.6} \end{equation*}

Eq.(B.6) is coded directly into diffeqs.js.

A problem may arise in the program in calculating force S when the angle γ approaches either 0 or 90 degress, respectively. Hence we use :

\begin{equation*} S ( \cos\gamma + \sin\gamma ) = M_p (g + \ddot{x} + \ddot{y}) \end{equation*}

We now use Eqs.(B.1a) and (B.1b) to substitute for $$\ddot{x}$$ and $$\ddot{y}$$, respectively, and then use Eq.(B.6) to eliminate $$\ddot{\gamma}$$. The result is :

\begin{eqnarray*} S ( \cos\gamma + \sin\gamma ) / M_p &=& g[1+(\cos\gamma-\sin\gamma) \sin\gamma] \\ &-& b[\sin\alpha+\cos\alpha - (\cos\gamma-\sin\gamma) \sin(\alpha{-}\gamma)] \dot{\alpha}^2 \\ &+& L_p [ \sin\gamma + \cos\gamma ] \dot{\gamma}^2 \\ &+& b[\cos\alpha-\sin\alpha - (\cos\gamma - \sin\gamma) \cos(\alpha{-}\gamma)] \ddot{\alpha} \end{eqnarray*}

The following identities come in handy, some a harder to prove than others. \begin{align} 1+(\cos\gamma-\sin\gamma) \sin\gamma &= \cos\gamma ( \cos\gamma + \sin\gamma )\\ \sin\alpha+\cos\alpha - (\cos\gamma-\sin\gamma) \sin(\alpha{-}\gamma) &= \cos(\alpha{-}\gamma) ( \cos\gamma + \sin\gamma ) \\ \cos\alpha-\sin\alpha - (\cos\gamma - \sin\gamma) \cos(\alpha{-}\gamma) &= - \sin(\alpha{-}\gamma) ( \cos\gamma + \sin\gamma ) \end{align}

With that we get :

\begin{equation*} \frac{P_p}{a M_p} =\; \underbrace{ \sin(\alpha{-}\gamma) \left[\frac{g}{a} \cos\gamma - \frac{b}{a} \cos(\alpha{-}\gamma) \dot{\alpha}^2 + \frac{L_p}{a} \dot{\gamma}^2\right] }_{=p_1} \;-\;\underbrace{ \frac{b}{a} \sin^2(\alpha{-}\gamma) }_{=p_2} \;\ddot{\alpha} \tag{B.7} \end{equation*}

Usig now Eqs.(B4) we have now expressions for the action of the pumpkin at the tip of the beam in terms of Np and Sp which enable us to set up the moment equation for the beam.

Zig Herzog; hgnherzog@yahoo.com
Last revised: xx