Problems

Problems 6.7 Problems

1.

A slender rod \(\cG\) is in motion relative to a referential \(\cE (O, \bx, \by , \bz)\) in the following way:

its endpoints \(A\) and \(B\) move in a vertical plane \((O, \bu , \bz)\) which rotates relative to \(\cE\) about vertical axis \((O, \bz)\) with angular velocity \(\dpsi \bz\text{,}\)
point \(B\) slides along axis \((O, \bz)\) while point \(A\) slides along axis \((O, \bu)\text{.}\)

The orientation of rod \(\cG\) relative to referential \(\cF (O, \bu , \bv = \bz\times\bu, \bz)\) is defined by angle \(\theta (t)\) as shown. A small ring \(P\) slides along line \(AB\) in such a way that \(\br_{PB} = \rho (t) \bi\text{.}\) Denote \(\bj = \bv\times \bi\) and \(L = |AB|\text{.}\)

Find velocity \(\vel_{P/\cF}\) from the expression of velocity \(\vel_{P/\cG}\text{.}\) Then find \(\vel_{P/\cE}\) from the expression of \(\vel_{P/\cF}\text{.}\)
Repeat for acceleration of \(P\text{:}\) find \(\ba_{P/\cF}\) from \(\ba_{P/\cG}\text{,}\) then find \(\ba_{P/\cE}\) from \(\ba_{P/\cF}\text{.}\)

Solution.

First we find \(\vel_{P/\cG} = \frac{d}{dt} \br_{BP} \Big|_{\cG}= - \dro \bi\) since both \(A\) and unit vectors are attached to \(\cG\text{.}\)

We then find \(\vel_{P/\cF}\) from \(\vel_{P/\cG}\) by using

\begin{equation*} \vel_{P/\cF} = \vel_{P/\cG} + \vel_{P \in \cG/ \cF} \end{equation*}

To find transport velocity \(\vel_{P \in \cG/ \cF}\text{,}\) we first find the expression of kinematic screw \(\{ \cV_{\cG / \cF} \}\text{:}\)

\begin{equation*} \{ \cV_{\cG / \cF} \} = \left\{ \begin{array}{l} \dte \bv \\ L\dte \cos\te \bz \end{array} \right\}_B \end{equation*}

Then we find

\begin{equation*} \vel_{P \in \cG/ \cF} = L\dte \cos\te \bz + \dte \bv \times (-\rho \bi) = L\dte \cos\te \bz - \rho \dte \bj \end{equation*}

where \(\bj = \bv \times \bi\text{.}\) Finally we obtain

\begin{equation*} \boxed{ \vel_{P/\cF} = - \dro \bi + L\dte \cos\te \bz - \rho \dte \bj} \end{equation*}

Likewise to find \(\vel_{P/\cE}\) from \(\vel_{P/\cF}\text{,}\) we need transport velocity \(\vel_{P \in \cF/ \cE}\text{.}\) This term is found by using the expression of kinematic screw \(\{ \cV_{\cF / \cE} \}\text{:}\)

\begin{equation*} \{ \cV_{\cF / \cE} \} = \left\{ \begin{array}{l} \dpsi \bz \\ \bze \end{array} \right\}_O \end{equation*}

This gives

\begin{equation*} \vel_{P \in \cF/ \cE}=\dpsi \bz \times \br_{OP}= \rho \dpsi \cos\te \bv \end{equation*}

Finally, by using \(\vel_{P/\cE} = \vel_{P/\cF} + \vel_{P \in \cF/ \cE}\text{,}\) we find

\begin{equation*} \boxed{ \vel_{P/\cE} = \rho\dpsi \cos\te \bv - \dro \bi + L\dte \cos\te \bz - \rho \dte \bj } \end{equation*}

We apply formula \(\ba_{P/\cF} = \ba_{P/\cG}+ \ba_{P \in\cG /\cF}+2\bom_{\cG/\cF}\times \vel_{P/\cG}\) with \(\ba_{P/\cG} = -\ddro \bi\) and transport acceleration

\begin{equation*} \ba_{P\in\cG /\cF} = \ba_{B/\cF} + \ddte \bv \times (-\rho \bi) - \rho \dte^2 \bi = \ba_{B/\cF} - \rho \ddte \bj + \rho \dte^2 \bi \end{equation*}

with

\begin{equation*} \ba_{B/\cF} = \frac{d}{dt} (L\dte \cos\te \bz) \Big|_\cF = L (\ddte\cos\te - \dte^2 \sin\te) \bz \end{equation*}

to find

\begin{equation*} \boxed{\ba_{P/\cF} = - \ddro \bi + L (\ddte\cos\te - \dte^2 \sin\te) \bz - \rho \ddte \bj + \rho \dte^2 \bi -2 \dro\dte \bj} \end{equation*}

Then we apply formula \(\ba_{P/\cE} = \ba_{P/\cF}+ \ba_{P \in\cF /\cE}+ 2\bom_{\cF/\cE}\times \vel_{P/\cF}\) with transport acceleration

\begin{equation*} \ba_{P\in\cF/\cE} = \rho\cos\te (\ddpsi \bv - \dpsi^2 \bu) \end{equation*}

and Coriolis acceleration

\begin{equation*} 2\bom_{\cF/\cE}\times \vel_{P/\cF}= 2\dpsi (\dro \cos\te+ \ro \dte \sin\te) \bv \end{equation*}

leading to the final expression

\begin{equation*} \boxed{ \ba_{P/\cE}= \ba_{P/\cF} + \rho\cos\te (\ddpsi \bv - \dpsi^2 \bu) +2 \dpsi (\dro \cos\te +\rho \dte\sin\te)\bv } \end{equation*}

It is likely faster to find \(\vel_{P/\cE}\) and \(\ba_{P/\cE}\) by direct time-differentiation relative to \(\cE\text{.}\) For instance, velocity \(\vel_{P/\cE}\) is found by taking the time-derivative of \(\br_{OP}= L \sin\te \bz - \ro \bi\) relative to \(\cE\text{:}\)

\begin{equation*} \vel_{P/\cE}= L\dte \cos\te \bz -\dro \bi -\ro (\dpsi \bz+ \dte \bv) \times \bi \end{equation*}

with \(\bz\times \bi= -\cos\te \bv\) and \(\bv\times \bi =\bj\text{.}\) We recover the result found above.

2.

A small ring \(P\) slides along a circular frame \(\cC\) of center \(O\) and radius \(R\) at constant speed \(v\text{.}\) Frame \(\cC\) rotates about its diameter \(Oz\) at constant angular velocity \(\om \be_z\) relative to referential \(\cE\text{.}\) At time \(t=0\text{,}\) \(\cC\) lies in the plane \(Oxz\) of \(\cE\) and \(P\) is at point \(A\) on axis \(Ox\text{.}\)

Find the velocity and acceleration of \(P\) relative to \(\cC\text{.}\)
Using the previous quantities, find the velocity and acceleration of \(P\) relative to \(\cE\text{.}\)

3.

A sailboat \(P\) travels in a current so that a buoy located at a fixed point \(O\) always stays directly abeam to \(P\text{,}\) that is, the line connecting \(O\) and \(P\) stays perpendicular to the boat’s relative velocity. The boat’s speed in still water is the constant \(v\text{.}\) The current has a constant velocity \(\bf u\) (relative to the ground). Denote by \(\be_r\) the unit vector directed from \(O\) to \(P\text{,}\) and by \(r(t)\) the radial distance. With axis \(Ox\) defined as the fixed line perpendicular to \(\bf u\text{,}\) \(\theta (t)\) is the angle made by line \(OP\) with axis \(Ox\text{.}\)

Find the absolute velocity of \(P\text{,}\) that is, measured relative to the ground on the polar basis \((\be_r, \be_\theta )\text{.}\)
Find the trajectory \(r(\theta)\) of \(P\) assuming that at time \(t=0\text{,}\) \(\theta = 0\) and \(r= r_0\text{.}\) Characterize this trajectory according to the ratio \(e= u/v\text{.}\) Plot the trajectories of \(P\) for \(e=0.5, 1, 2\text{.}\)

4.

Consider the “grasshopper” carousel of Figure 6.7.4. The system is composed of three interconnected rigid bodies \(\cA\text{,}\) \(\cB\) and \(\cC\text{.}\) It is comprised of a central mast \(\cA (O, \bha_1, \bha_2, \bha_3)\text{,}\) a system of arms such as \(\cB (A, \bhb_1, \bhb_2, \bhb_3)\) extending radially from the mast and supporting a passenger-carrying seat \(\cC (B, \bhc_1, \bhc_2 , \bhc_3)\text{.}\)

The mast \(\cA\) is in rotation with angle \(\psi\) about a vertical axis \((O, \be_3 = \bha_3 )\) relative to a referential \(\cE (O, \be_1, \be_2,\be_3)\) attached to the ground.
The arm \(\cB\) can rotate about axis \((A, \bha_2 = \bhb_2 )\) of \(\cA\) with angle \(\theta\) relative to \(\cA\text{.}\) An actuator (pneumatic jack) mounted between \(\cA\) and \(\cB\) is able to create sudden lifting or lowering motions of arm \(\cB\text{.}\)
The seat \(\cC\) supported by arm \(\cB\) can rotate about an axis \((B, \bhc_3= \bhb_3)\) with angle \(\phi\) relative to arm \(\cB\text{.}\) The points \(A\) and \(B\) are defined by \(\br_{OA}= h \be_3 + R \bha_1\) and \(\br_{AB} = L \bhb_1\) (\(h\text{,}\) \(R\text{,}\) and \(L\) are constant).

Find the kinematic screws \(\{\cV_{\cC/\cB}\}\text{,}\) \(\{\cV_{\cB/\cA}\}\) and \(\{\cV_{\cA/\cE}\}\text{.}\)
Find the kinematic screw \(\{\cV_{\cC/\cE}\}\) by using the kinematic loop formula.

Solution.

The motion of \(\cC\) relative to \(\cB\) is a rotation of axis \((B,\bhb_3)\) and angle \(\phi\text{:}\)

\begin{equation*} \{ \cV_{\cC / \cB} \} = \left\{ \begin{array}{l} \dphi \bhb_3 \\ \bze \end{array} \right\}_B \end{equation*}

The motion of \(\cB\) relative to \(\cA\) is a rotation of axis \((A,\bha_2)\) and angle \(\te\text{:}\)

\begin{equation*} \{ \cV_{\cB / \cA} \} = \left\{ \begin{array}{l} \dte \bha_2 \\ \bze \end{array} \right\}_A \end{equation*}

Finally the motion of \(\cA\) relative to \(\cE\) is a rotation of axis \((O,\be_3)\) and angle \(\psi\text{:}\)

\begin{equation*} \{ \cV_{\cA / \cE} \} = \left\{ \begin{array}{l} \dpsi \be_3 \\ \bze \end{array} \right\}_O \end{equation*}

We can find the kinematic screw of body \(\cC\) relative to \(\cE\) by using the loop equation
\begin{align*} \{ \cV_{\cC / \cE} \} \amp = \{ \cV_{\cC / \cB} \}+\{ \cV_{\cB / \cA} \}+\{ \cV_{\cA / \cE} \} \\\\ \amp = \left\{ \begin{array}{l} \dphi \bhb_3 \\ \bze \end{array} \right\}_B + \left\{ \begin{array}{l} \dte \bha_2 \\ \bze \end{array} \right\}_A + \left\{ \begin{array}{l} \dpsi \be_3 \\ \bze \end{array} \right\}_O\\ \amp = \left\{ \begin{array}{l} \dpsi \be_3+ \dte \bha_2 +\dphi \bhb_3 \\ \dte \bha_2\times \br_{AB}+\dpsi \be_3\times \br_{OB} \end{array} \right\}_B\\ \amp = \left\{ \begin{array}{l} \dpsi \be_3+ \dte \bha_2 +\dphi \bhb_3 \\ \dte \bha_2\times L \bhb_1 + \dpsi \be_3\times (h\be_3+ R\bha_1+ L \bhb_1) \end{array} \right\}_B\\ \amp = \left\{ \begin{array}{l} \dpsi \be_3+ \dte \bha_2 +\dphi \bhb_3 \\ -L\dte\bhb_3 + \dpsi (R \bha_2 + L \cos\te \bha_2) \end{array} \right\}_B \end{align*}

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