Problems

Problems 3.7 Problems

1.

Given three referentials \(\cA\text{,}\) \(\cB\) and \(\cC\text{,}\) show that

\begin{equation*} \bal_{\cA/\cB}+ \bal_{\cB/\cC} +\bal_{\cC/\cA} = \bom_{\cA/\cC} \times \bom_{\cB/\cC} \end{equation*}

Then find conditions for the loop equation \(\bal_{\cA/\cB}+ \bal_{\cB/\cC} +\bal_{\cC/\cA} = \bze\) to be valid.

2.

Show that the angular velocity of rigid body \(\cB\) relative to referential \(\cA\) can be expressed as

\begin{equation*} \bom_{\cB/\cA}= \frac{1}{2} \left( \bhb_1 \times \frac{d\bhb_1}{dt} + \bhb_2 \times \frac{d\bhb_2}{dt} + \bhb_3 \times \frac{d\bhb_3}{dt} \right) \end{equation*}

where \((\bhb_1, \bhb_2, \bhb_3)\) is a basis of \(\cB\text{,}\) and where the time-derivatives are taken relative to \(\cA\text{.}\)

3.

Assume that the motion of body \(\cB\) relative to referential \(\cA\) is characterized by the rotation \(\cR_{\te,\bu}\) of angle \(\alpha\) about unit vector \(\bu\text{.}\) Recall that any vector \(\bV_0\) attached to \(\cA\) is mapped to a vector \(\bV\) attached to \(\cB\) according to Rodrigues formula:

\begin{equation*} \bV= \bV_0 + \sin\alpha \bu\times\bV_0 + (1- \cos\alpha) \bu\times(\bu\times\bV_0) \end{equation*}

Show that angular velocity \(\bom_{\cB/\cA}\) can be found according to

\begin{equation*} \bom_{\cB/\cA} = \dal \bu + \sin\al \frac{d\bu }{dt} + (1-\cos\al)\bu \times \frac{d\bu }{dt} \end{equation*}

4.

A rigid body in the shape of an isosceles triangle \(AOB\) (\(OA = OB\)). rotates about the origin \(O\) of a referential \(\cE\text{.}\) See Figure 3.7.1. Show that the velocities of vertices \(A\) and \(B\) satisfy the following property: \(\vel_A \cdot \br_{OB} = - \vel_B \cdot \br_{OA}\)

Solution.

Express the velocity of \(A\) and \(B\) relative to \(\cE\) in terms of the angular velocity of the body:
\begin{equation*} \vel_A = \bom \times \br_{OA}, \qquad \vel_B = \bom \times \br_{OB} \end{equation*}
taking into account that \(\vel_O = \bze\) (\(O\) is a point of the body). This gives
\begin{equation*} \vel_A \cdot \br_{OB} = \br_{OB} \cdot ( \bom \times \br_{OA}), \qquad \vel_B \cdot \br_{OA} = \br_{OA} \cdot ( \bom \times \br_{OB}) \end{equation*}
Then we use the property of the triple scalar product
\begin{equation*} \br_{OB} \cdot ( \bom \times \br_{OA}) = \bom \cdot (\br_{OA} \times \br_{OB}) , \qquad \br_{OA} \cdot ( \bom \times \br_{OB}) = \bom \cdot (\br_{OB} \times \br_{OA}) \end{equation*}
We then find
\begin{equation*} \boxed{\vel_A \cdot \br_{OB} = - \vel_B \cdot \br_{OA}} \end{equation*}
\(\blacksquare\)

5.

Consider a rigid body \(\cB\) in motion relative to a referential \(\cE\text{.}\) Given a unit vector \(\bz\) attached to \(\cB\text{,}\) show that the angular velocity of \(\cB\) relative to \(\cE\) can be written in the following form

\begin{equation*} \bom_{\cB /\cE} = \om_z \bz + \bz \times \left(\frac{d \bz}{dt}\right)_\cE \end{equation*}

where \(\om_z = \bz\cdot \bom_{\cB /\cE} \text{.}\)

6.

In the Eulerian formulation of continuum mechanics, the velocity field \(\vel (x_1, x_2, x_3 , t)\) of a material system is defined at every point \((x_1, x_2, x_3)\) of a referential \(\cE (O, \be_1, \be_2, \be_3)\text{.}\) Show that the velocity field \(\vel (x_1, x_2, x_3 , t)\) is that of a rigid body \(\cF\) if and only if \(\partial_i v_j + \partial_j v_i = 0\) \((i,j =1,2,3)\) where \(\partial_i = \partial / \partial x_i\) and \(v_i = \vel \cdot \be_i\text{.}\) When this condition is satisfied, find the corresponding angular velocity \(\bom_{\cF / \cE}\text{.}\)

Solution.

First assume that the field \(\vel\) is the velocity field of a rigid body. Then at any time, this field satisfies \(\vel({\bf x}, t ) = \vel (\bze, t) + \bom \times \br_{OP}\) which leads to the equations of \((\be_1, \be_2, \be_3)\text{:}\)
\begin{equation*} \begin{array}{l} v_1 = v_{O1} + (\om_2 x_3 - \om_3 x_2) \\ v_2 = v_{O2} + (\om_3 x_1 - \om_1 x_3) \\ v_3 = v_{O3} + (\om_1 x_2 - \om_2 x_1) \end{array} \end{equation*}
where \(\br_{OP} = x_i \be_i\text{,}\) \(\vel_P = v_i \be_i\) and \(\vel_O = v_{Oi} \be_i\text{.}\) Then by taking \(\partial_i\) for each coordinate \(v_j\) we obtain:
\begin{equation*} \begin{array}{lll} \partial_1 v_1 = 0 \amp \partial_2 v_1 = -\om_3 \amp \partial_3 v_1 = \om_2 \\ \partial_1 v_2 = \om_3 \amp \partial_2 v_2 = 0 \amp \partial_3 v_2 = - \om_1 \\ \partial_1 v_3 = - \om_2 \amp \partial_2 v_3 = \om_1 \amp \partial_3 v_3 = 0 \end{array} \end{equation*}
which shows that \(\partial_i v_i =0\) and \(\partial_i v_j = -\partial_j v_i\) for \(i\neq j\text{.}\) We also get
\begin{equation*} \om_1 = (\partial_2 v_3 - \partial_3 v_2)/2\text{,} \end{equation*}

\begin{equation*} \om_2 = (\partial_3 v_1 - \partial_1 v_3)/2\text{,} \end{equation*}

\begin{equation*} \om_3 = (\partial_1 v_2 - \partial_2 v_1)/2 \end{equation*}
leading to the expression
\begin{equation*} \boxed{\bom = {1\over 2} \grad \cdot \vel} \end{equation*}
with \(\grad = \be_i \partial_i\text{.}\)

The converse proof consists in showing that, if \(\partial_i v_j = -\partial_j v_i\text{,}\) then field \(\vel\) is the velocity field of a rigid body follows from the skew-symmetry of tensor \(\grad \vel\text{.}\) \(\blacksquare\)

7.

Consider at a given time the velocities \(\vel_P\text{,}\) \(\vel_Q\) and \(\vel_R\) of three points attached to the same rigid body \(\cB\) in motion relative to a referential \(\cA\text{.}\) Then consider the plane passing through the endpoints of the three vectors \(\vel_P\text{,}\) \(\vel_Q\) and \(\vel_R\) originating from a common arbitrary point \(O\text{.}\) See Figure 3.7.2.

Show that the normal to this plane is directed along the angular velocity \(\bom_{\cB /\cA}\) of \(\cB\text{.}\)

Solution.

Define the 3 fictitious points (in the velocity space) \(A\text{,}\) \(B\) and \(C\) by
\begin{equation*} \br_{OA}= \vel_P, \qquad \br_{OB}= \vel_Q, \qquad \br_{OC}= \vel_R \end{equation*}
The normal to plane \(ABC\) is directed along vector \(\br_{AB}\times \br_{AC}\text{.}\) The relationship between \(\vel_P\) and \(\vel_Q\) gives
\begin{equation*} \br_{AB}= \bom\times \br_{PQ} \qquad{(1)} \end{equation*}
This shows that \(\bom\) is normal to \(\br_{AB}\text{.}\) Similarly we have
\begin{equation*} \br_{AC}= \bom\times \br_{PR} \qquad{(2)} \end{equation*}
This shows that \(\bom\) is normal to \(\br_{AC}\text{.}\)

Using (1) and (2) we find
\begin{equation*} \br_{AB}\times \br_{AC}= (\bom\times \br_{PQ})\times (\bom\times \br_{PR}) = \lambda \bom \end{equation*}
assuming \(\lambda = (\bom\times \br_{PQ}) \cdot \br_{PR} \neq 0\text{.}\) Since vector \(\br_{AB}\times \br_{AC}\) is normal to plane \(ABC\text{,}\) we conclude that \(\bom\) is also normal to the plane \(ABC\text{.}\)

\(\blacksquare\)

8.

A rigid body \(\cB\) is in motion relative to a referential \(\cE\text{.}\) We are interested to study the field \(\vel_{P\in \cB /\cE}\) at a particular instant when the points \(P\) lie on a line \(\cal L\text{.}\) Call \(\bom\) the angular velocity of \(\cB\) at this instant. Also denote by \(\vel_A = \vel_{A\in\cB /\cE}\) the velocity of a particular point \(A\) of \(\cal L\text{.}\)

Show that in general the magnitude of \(\vel_{P\in \cB /\cE}\) must reach a minimum along line \(\cal L\text{.}\) Find the point \(P_*\) where this minimum is reached.
Is there a point of \(\cal L\) for which \(\vel_{P\in \cB /\cE}\) is collinear to a unit vector \(\hat{\boldsymbol{\delta}}\) directed along \(\cal L\text{?}\)

9.

A rigid body \(\cB\) of basis \((\bhb_1 , \bhb_2, \bhb_3)\) and origin \(B\) is in motion relative to a referential \(\cE\text{.}\) Consider the three points \(P\text{,}\) \(Q\) and \(R\) fixed in \(\cB\) defined by the position vectors \(\br_{BP}= a \bhb_1\text{,}\) \(\br_{BQ}= a \bhb_2\text{,}\) and \(\br_{BR}= a \bhb_3\) (\(a\) is a positive constant). The motion of \(\cB\) relative to \(\cE\) is characterized by the following conditions valid at all time:

\begin{equation*} \vel_{P}\cdot\bhb_2 =2a \om ,\; \vel_{P}\cdot\bhb_3 = - a\om \end{equation*}

\begin{equation*} \vel_Q \cdot \bhb_1 = a\om, \; \vel_Q \cdot \bhb_3 = 0 \end{equation*}

\begin{equation*} \vel_R \cdot \bhb_1 = a \om, \; \vel_R \cdot \bhb_2 = a\om \end{equation*}

where \(\om\) is a positive constant.

Find the angular velocity \(\bom_{\cB / \cE}\text{,}\) the velocity and acceleration of point \(B\text{.}\) Show that the motion of \(\cB\) relative to \(\cE\) cannot be a rotation.

Solution.

First, we find angular velocity \(\bom = \bom_{\cB/\cE} = \om_1 \bhb_1 +\om_2 \bhb_2 +\om_3 \bhb_3\) by relating velocities \(\vel_P\) and \(\vel_Q\) to each other: from equation \(\vel_Q = \vel_P + \bom \times \br_{PQ}\text{,}\) we obtain
\begin{equation*} a\om \bhb_1 + q \bhb_2 = p \bhb_1 + 2a\om \bhb_2 - a\om \bhb_3 + (\om_1 \bhb_1 +\om_2 \bhb_2 +\om_3 \bhb_3)\times a (\bhb_2 -\bhb_1) \end{equation*}
where we have denoted the components \(p= \vel_P \cdot \bhb_1\) and \(q= \vel_Q \cdot \bhb_2\text{,}\) yet to be found. This yields 3 equations
\begin{equation*} a\om = p - a \om_3, \quad q= 2a\om - a \om_3, \quad 0=-a \om + a\om_1 + a\om_2 \qquad{(1-3)} \end{equation*}

We repeat by relating \(\vel_R\) to \(\vel_P\) (denoting \(r= \vel_R \cdot \bhb_3\)): we impose \(\vel_R = \vel_P + \bom \times \br_{PR}\) to obtain
\begin{equation*} a\om \bhb_1 + a\om \bhb_2 +r \bhb_3 = p \bhb_1 + 2a\om \bhb_2 - a\om \bhb_3 + (\om_1 \bhb_1 +\om_2 \bhb_2 +\om_3 \bhb_3)\times a (\bhb_3 -\bhb_1) \end{equation*}
yielding 3 additional equations
\begin{equation*} a\om = p + a\om_2 , \quad a \om = 2a\om - a \om_1 -a \om_3 , \quad r = -a \om + a\om_2 \qquad{(4-6)} \end{equation*}

We now solve equations (1-6) to find \((\om_1,\om_2,\om_3)\) and \((p,q,r)\text{:}\)
\begin{equation*} \om_1 = \om, \quad \om_2= \om_3 = 0 \Longrightarrow \boxed{\bom= \om \bhb_1} \end{equation*}

\begin{equation*} \boxed{p= a\om, \quad q= 2a \om, \quad r= -a\om} \end{equation*}

We can now find the velocity of \(B\text{:}\) for instance we relate \(\vel_B\) to \(\vel_P\)
\begin{equation*} \boxed{ \vel_B = \vel_P + \om \bhb_1 \times \br_{PB} = \vel_P = a\om( \bhb_1 + 2 \bhb_2 -\bhb_3) } \end{equation*}
Time-differentiation of \(\vel_B\) gives the acceleration of \(B\text{:}\)
\begin{equation*} \boxed{ \ba_B = a\om ( 2\bom \times \bhb_2 - \bom \times \bhb_3 ) = a\om^2 ( 2\bhb_3 + \bhb_2 ) } \end{equation*}
Since \(\bom\) is directed along unit vector \(\bhb_1\text{,}\) \(\bhb_1\) must be fixed relative to referential \(\cE\text{:}\)
\begin{equation*} \frac{d\bhb_1}{dt} = \bom \times \bhb_1 = \bze \end{equation*}
It is tempting to conclude that the motion of \(\cB\) is a rotation along some axis directed along \(\be_1\text{:}\) to verify this assertion, we ask if we can find points \(Q\) of \(\cB\) such that \(\vel_Q =0\text{:}\) then, such points would satisfy
\begin{equation*} \om \bhb_1 \times \br_{BQ} = -a\om( \bhb_1 + 2 \bhb_2 -\bhb_3) \end{equation*}
We see that this equation does not have a solution since \(\bhb_1 \cdot ( \bhb_1 + 2 \bhb_2 -\bhb_3) \neq 0\text{.}\) The motion of \(\cB\) is not a rotation. \(\blacksquare\)

10.

Consider a linear operator \(\cL : \bV \mapsto \bW = \cL(\bV)\text{.}\) Show that

\begin{equation*} \frac{d}{dt}_{|_\cA}(\cL) =\frac{d}{dt}_{|_\cB} (\cL) + \Omega_{B/A} \circ \cL - \cL\circ\Omega_{B/A} \end{equation*}

where \(\Omega_{B/A}\) is the operator: \(\bV \mapsto \bom_{\cA/\cB}\times \bV\text{.}\)

11.

Human centrifuges are large ground-based centrifuges used to submit pilots to g-force environment they would encounter during the maneuvers of high-performance aircraft. They provide a testing and training platform for pilots subjected to the adverse effects (such as loss of consciousness) of large accelerations. A model of a 3-axis human centrifuge displayed in Figure 3.7.3 is comprised of the following rigid bodies:

(i) the arm \(\cA ( O, \bha_1 , \bha_2, \bha_3 = \be_3)\) is rotation about vertical axis \((O, \be_3)\) of referential \(\cE ( O, \be_1 , \be_2, \be_3)\text{.}\)

(ii) the gimbal \(\cB (A, \bhb_1 , \bhb_2 = \bha_2 , \bhb_3 )\) in rotation about axis \((A, \bha_2)\) of arm \(\cA\text{.}\) Position of point \(A\) is defined by \(\br_{OA}= L \bha_1\) where \(L\) is a constant.

(iii) the cab \(\cC (A, \bhc_1 = \bhb_1 , \bhc_2 , \bhc_3 )\) including its passenger in rotation about axis \((A, \bhb_1)\) of gimbal \(\cB\text{.}\)

Hence the cab is allowed to pitch and roll relative to the arm. From the point of view of the passenger, the axes \((A, \bhc_1)\text{,}\) \((A, \bhc_2)\) and \((A,\bhc_3)\) points from side to side, back to front, and toe to head, respectively. The point \(C\) whose position is defined by \(\br_{AC} = l \bhc_3\) is located at the level of the passenger’s head. The acceleration felt by the pilot is the vector \(\bG = \bg - \ba_{C\in \cC/\cE}\) where \(\bg = - g \be_3\) is the gravitational acceleration, and \(\ba_{C\in \cC/\cE}\) is the acceleration of point \(C\) attached to body \(\cC\text{.}\)

The centrifuge has three degrees of freedom modeled by the angles \(\psi\text{,}\) \(\theta\) and \(\phi\) which define the orientations of the bases of \(\cA\text{,}\) \(\cB\) and \(\cC\) as defined in Figure 3.7.4.

Find the velocity \(\vel_{C\in \cC /\cE}\) as a function of angles \(\psi\text{,}\) \(\te\) and \(\phi\) and their time-derivatives.
Assume that \(l \ll L\) and set \(l=0\text{.}\) Express vector \(\bG\) in the basis \((\bhc_1 , \bhc_2 , \bhc_3 )\) attached to \(\cC\text{.}\) It is desired to find control laws of the centrifuge so as to guarantee that the acceleration felt by the pilot is collinear to \(\bhc_3\text{,}\) that is, \(\bG\cdot\bhc_1 = \bG\cdot\bhc_2 =0\text{.}\) Find the corresponding angles \(\te\) and \(\phi\) as functions of \(g\text{,}\) \(L\text{,}\) \(\dpsi\) and \(\ddpsi\text{.}\)

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