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Underwater Robots - Gianluca Antonelli.pdf

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2.3 Rigid Body’s Dynamics 23<br />

Let us define the Inertia tensor of the body B relative to the pole O :<br />

�<br />

I O = S T ( p ) S ( p ) ρdV,<br />

V B<br />

where S is the skew-symmetric operator defined in (2.6). The matrix I O is<br />

symmetric and positive definite. The positive diagonal elements I Oxx , I Oyy ,<br />

I Ozz are the inertia moments with respect to the three coordinate axes of<br />

the reference frame. The off diagonal elements are the products ofinertia .<br />

The relationship between the inertia tensor in two different frames I O<br />

and I � O ,related by arotation matrix R ,with the same pole O ,isthe following:<br />

I O = RI � O R T .<br />

The change of pole is related by the Steiner’s theorem :<br />

I O = I C + m S T ( p C ) S ( p C ) ,<br />

where I C is the inertial tensor relative tothe center of mass, when expressed<br />

in aframe parallel to the frame in which I O is defined.<br />

Notice that O can be either afixed or moving pole. In case of afixed pole<br />

the elements of the inertia tensor are function of time. Asuitable choice of<br />

the pole might beapoint fixed to the rigid body in away to obtain aconstant<br />

inertia tensor. Moreover, since the inertiatensor is symmetric positivedefinite<br />

is always possible to find aframe in which the matrix attains adiagonal form,<br />

this frame is called principal frame, also, if the pole coincides with the center<br />

of mass, it is called central frame. This is true also ifthe body does not have<br />

asignificant geometric symmetry.<br />

Let Ω be any point inspace and p Ω the corresponding position vector.<br />

Ω can be either moving or fixed with respect to the reference frame. The<br />

angular momentum of the body B relative to the pole Ω is defined as the<br />

vector:<br />

�<br />

k Ω = ˙p × ( p Ω − p ) ρdV. (2.28)<br />

V B<br />

Taking into account the definition of center of mass, (2.28) can be rewritten<br />

in the form:<br />

k Ω = I C ω + m ˙p C × ( p Ω − p C ) , (2.29)<br />

where ω is the angular velocity.<br />

The resultant moment µ Ω with respect to the pole Ω of arigid body<br />

subject to n external forces f 1 ,...,f n is:<br />

n�<br />

µ Ω = f i × ( p Ω − p i ) .<br />

i =1<br />

In case of asystem with constant mass and rigid body, the angular part<br />

of the Newton’s law ofmotion

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