Problems

Problems 4.10 Problems

1.

Consider an equiprojective field \(P\in \cE\mapsto \buu_P\text{,}\) that, a field satisfying \(\br_{PQ} \cdot \buu_P = \br_{PQ} \cdot \buu_Q\) for any two points \(P\) and \(Q\text{.}\) Let \(O\) be a given point of \(\cE\text{.}\) Consider the linear operator \(\cL: \bX = \br_{OP} \mapsto \buu_P- \buu_O\text{.}\)

Show that \(\cL\) is skew-symmetric, that is, \(\cL (\bX)\cdot \bY =- \bX\cdot \cL (\bY)\text{.}\)
Show that \(\cL\) is necessarily linear.
Conclude that there exists a vector \(\bU\) such that
\begin{equation*} \buu_P = \buu_O + \bU \times \br_{OP} \end{equation*}

Solution.

a. Let \(O\) be an arbitrary point of \(\cE\text{.}\) Define vectors \(\bxx = \br_{OP}\) and \(\byy= \br_{OQ}\) associated with points \(P\) and \(Q\text{,}\) respectively. Let us find \(\cL (\bxx)\cdot \byy\text{:}\)
\begin{equation*} \cL (\bxx)\cdot \byy = (\buu_P - \buu_O)\cdot \br_{OQ}= \buu_P\cdot \br_{OP}+ \buu_P\cdot \br_{PQ} - \buu_O\cdot \br_{OQ} \end{equation*}
We then use equiprojectivity:
\begin{equation*} \cL (\bxx)\cdot \byy = \buu_O\cdot \br_{OP}+ \buu_Q\cdot \br_{PQ} - \buu_Q\cdot \br_{OQ} \end{equation*}
This gives \(\cL (\bxx)\cdot \byy =\buu_O\cdot \br_{OP} +\buu_Q\cdot \br_{PO}\text{.}\) We compare this expression with \(\cL (\byy)\cdot \bxx\text{:}\)
\begin{equation*} \cL (\byy)\cdot \bxx = (\buu_Q - \buu_O) \cdot \br_{OP} \end{equation*}
and find that \(\boxed{ \cL (\byy)\cdot \bxx = - \cL (\bxx)\cdot \byy}\text{.}\) Operator \(\cL\) is skew-symmetric.

b. To show that \(\cL\) is linear, for any real number \(\la\text{,}\) and any vector \(\byy\text{,}\) we find
\begin{equation*} \cL (\la \bxx)\cdot \byy = - \la\bxx \cdot \cL(\byy) = \la \cL (\bxx)\cdot \byy \end{equation*}
which implies that \(\cL (\la \bxx)= \la \cL (\bxx)\text{.}\) We can also show that
\begin{equation*} \cL(\bxx_1+\bxx_2) = \cL(\bxx_1) + \cL(\bxx_2) \end{equation*}
for all \(\bxx_1\) and \(\bxx_2\text{.}\)

c. Now we know that all skew-symmetric linear operators can be written in the form \(\cL(\bxx) = \bU \times \bxx\text{.}\)

In conclusion,
\begin{equation*} \buu_P - \buu_O = \bU \times \br_{OP}\text{.} \end{equation*}
\(\blacksquare\)

2.

Consider the unit cube \(OABC\) \(O'A'B'C'\text{.}\) See Figure 4.10.1. Denote by \(D\) the midpoint of side \(OA\) and by \(E\) the midpoint of side \(A'B'\text{.}\) Consider the three sliders

\begin{equation*} \{\cS_1 \} = \begin{Bmatrix} \br_{O'E} \\ \bze \end{Bmatrix}_{O'} \end{equation*}

\begin{equation*} \{\cS_2 \} = \begin{Bmatrix} \br_{CD} \\ \bze \end{Bmatrix}_{C} \end{equation*}

\begin{equation*} \{\cS_3 \} = \begin{Bmatrix} \br_{C'B} \\ \bze \end{Bmatrix}_{C'} \end{equation*}

Find the expression of screw

\begin{equation*} \{\cV \}=\{\cS_1 \}+\{\cS_2 \}+\{\cS_3 \} \end{equation*}

resolved at \(O\text{.}\)

Solution.

Using the basis \((\bi,\bj,\bk)\) shown in Figure 4.10.2, we find
\begin{equation*} \{\cS_1 \} = \begin{Bmatrix} \frac{1}{2} \bi + \bj \\ \bze \end{Bmatrix}_{O'} = \begin{Bmatrix} \frac{1}{2} \bi + \bj \\ (\frac{1}{2} \bi + \bj)\times (-\bk) =\frac{1}{2} \bj - \bi \end{Bmatrix}_{O} \end{equation*}

Figure 4.10.2.
\begin{equation*} \{\cS_2 \} = \begin{Bmatrix} \br_{CD} \\ \bze \end{Bmatrix}_{C}= \begin{Bmatrix}- \bi + \frac{1}{2} \bj \\ (- \bi + \frac{1}{2} \bj)\times (-\bi) = \frac{1}{2} \bk \end{Bmatrix}_{O} \end{equation*}

\begin{equation*} \{\cS_3 \} = \begin{Bmatrix} \br_{C'B} \\ \bze \end{Bmatrix}_{C'}= \begin{Bmatrix} \bj -\bk \\(\bj -\bk)\times (-\bi -\bk) =\bk -\bi+\bj \end{Bmatrix}_{O} \end{equation*}

We can now express screw
\begin{equation*} \{\cV \}=\{\cS_1 \}+\{\cS_2 \}+\{\cS_3 \} =\begin{Bmatrix} - \frac{1}{2} \bi + \frac{5}{2} \bj -\bk \\ -2\bi+\frac{3}{2}\bj + \frac{3}{2} \bk \end{Bmatrix}_{O} \end{equation*}
resolved at \(O\text{.}\)

\(\blacksquare\)

3.

Consider four distinct points \(A\text{,}\) \(B\text{,}\) \(C\text{,}\) and \(D\) of a Euclidean space \(\cE\text{.}\)

Show that the vector field \(P\in\cE \mapsto \bvv_P = \br_{PA} \times \br_{PB} + \br_{PC} \times \br_{PD}\) defines a screw \(\{\cV\}\text{.}\) Find its axis.

4.

Consider four points \(A (1,0,0)\text{,}\) \(B (-1,0,0)\text{,}\) \(C (0,1,0)\text{,}\) and \(D(0,0,1)\) forming a tetrahedron in a Euclidean space \(\cE (O, \be_x, \be_y, \be_z)\text{.}\) Denote by \(\{ \cU \}\text{,}\) \(\{ \cV \}\text{,}\) \(\{\cW \}\) and \(\{ \cX \}\) the sliders corresponding to the bound vectors \((A, a\,\br_{AB})\text{,}\) \((B,b \,\br_{BC}) \text{,}\) \((C, c \,\br_{CD})\) and \((D, d \,\br_{DA})\text{,}\) respectively. The quantities \(a\text{,}\) \(b\text{,}\) \(c\) and \(d\) are non-zero scalars. Denote by \(\{\cS\} = \{ \cU \} + \{ \cV \} + \{\cW \} + \{ \cX \}\text{.}\) See Figure 4.10.3.

a. Find the condition satisfied by the scalars \((a, b, c, d)\) for screw \(\{ \cS \} \) to be a slider.

b. Find the condition satisfied by the scalars \((a, b, c, d)\) for screw \(\{ \cS \}\) to be a couple.

5.

Given four points \(A\text{,}\) \(B\text{,}\) \(C\text{,}\) and \(D\) forming a regular tetrahedron, consider the sliders \(\{ \cU \}\text{,}\) \(\{ \cV \}\text{,}\) and \(\{\cW \}\) corresponding to the bound vectors \((A, \br_{AB})\text{,}\) \((B, \br_{BC})\text{,}\) and \((C,\br_{CD})\text{,}\) respectively.

Find the axis of screw \(\{\cS\} = \{ \cU \} + \{ \cV \} + \{\cW \}\text{.}\)

6.

Consider the tetrahedron \(ABCD\) shown in Figure 4.10.5 and the 4 sliders

\begin{equation*} \{\cS_1 \}=\left\{ \begin{array}{ll} \bS_1 \\ \bze \end{array} \right\}_E \;\; \{\cS_2 \}=\left\{ \begin{array}{ll} \bS_2 \\ \bze \end{array} \right\}_F \;\; \{\cS_3 \}=\left\{ \begin{array}{ll} \bS_3 \\ \bze \end{array} \right\}_G \;\; \{\cS_4 \}=\left\{ \begin{array}{ll} \bS_4 \\ \bze \end{array} \right\}_H \end{equation*}

Axis \((\bS_1,E)\) is normal to the face \(ABC\) of centroid \(E\text{,}\) pointing outward. Furthermore, the magnitude of \(\bS_1\) is equal to the area of this face \(ABC\text{.}\) The remaining 3 axes \((\bS_2,F)\text{,}\) \((\bS_3,G)\text{,}\) and \((\bS_4,H)\) are defined in a similar manner (points \(F\text{,}\) \(G\text{,}\) and \(H\) are the centroids of faces \(ABD\text{,}\)\(ACD\text{,}\) and \(BCD\text{,}\) respectively). They are normal to the faces \(ABD\text{,}\)\(ACD\text{,}\) and \(BCD\text{,}\) respectively, and the magnitude of the vectors \(\bS_i\) is equal to the area of the corresponding face.

Show that \(\{\cS_1 \}+ \{\cS_2\}+ \{\cS_3\}+ \{\cS_4\} = \{0\}\text{.}\)

7.

Consider the screw

\begin{equation*} \{\cV\} = \begin{Bmatrix}\bV \\ \vel_P\end{Bmatrix} \end{equation*}

defined over \(\cE\) and the vector field \(P\in\cE \mapsto {\bf u}_P= \bn \times (\bn\times \vel_P)\) where \(\bn\) is a given unit vector.

Under what condition does this field define a screw?

8.

Let \(\cB\) be a rigid body in motion in a referential \(\cE\text{.}\)

Show that the velocity field \(P\in\cB \mapsto \vel_{P/\cE}\) defines a screw.
Show that the acceleration field \(P\in\cB \mapsto \ba_{P/\cE}\) does not define a screw.
Show that the field \(P\in\cB \mapsto \ba_{P/\cE} + \vel_{P/\cE} \times \bom_{\cB/\cE}\) defines a screw.

9. Lie Algebra.

Consider in a Euclidean space \(\cE\) two screws \(\{\cU \}\) and \(\{\cV \}\) defined as

\begin{equation*} \{\cU \}= \left\{ \begin{array}{c} \bU \\ {\bf u}_P \end{array} \right\} , \qquad \{\cV \}= \left\{ \begin{array}{c} \bV \\ {\bf v}_P \end{array} \right\} \end{equation*}

Consider the vector field

\begin{equation*} P \in \cE \mapsto \bU\times {\bf v}_P - \bV \times {\bf u}_P \end{equation*}

Show that this vector field is equiprojective, and hence defines a screw. Find the resultant of this screw. Denote this screw as \(\{\cU \} \times \{\cV \}\text{.}\)
Show that
\begin{equation*} \{\cU \} \times \{\cV \} \cdot \{\cU \} = \{\cU \} \times \{\cV \} \cdot \{\cV \} = 0 \end{equation*}
Then show that \(\{\cU \} \times \{\cV \}= - \{\cV \} \times \{\cU \}\text{.}\) Now justify the notation \(\{\cU \} \times \{\cV \}\text{.}\)
When \(\{\cU \} \times \{\cV \}\) is not a couple, show that its axis is the common perpendicular to the axis \(\Delta_U\) of \(\{\cU\}\) and the axis \(\Delta_V\) of \(\{\cV\}\text{.}\)
Show that
\begin{equation*} \{\cU \} \times (\{\cV \} \times \{\cW \} ) + \{\cV \} \times (\{\cW \} \times \{\cU \} ) + \{\cW \} \times (\{\cU \} \times \{\cV \} ) = \{ 0 \} \end{equation*}
Given two vectors \(\bA\) and \(\bB\) and two points \(A\) and \(B\text{,}\) show that the vector field defined as
\begin{equation*} P \in \cE \mapsto \bA\times(\bB\times \br_{BP}) + (\bA\times \br_{AP})\times \bB \end{equation*}
defines a screw of the type \(\{\cU \} \times \{\cV \}\text{.}\) Then show that the axis of this screw, when it exists, is the common perpendicular line to lines \((A, \bA)\) and \((B, \bB)\text{.}\)

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