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Kinematics and dynamics of rigid bodies 57
The position of the centre of mass {R G2 } 1 is then found by integrating the
elemental first moments of mass about frame O 2 and dividing by the total
mass m 2 :
1
{ RG2} 1 { RP}
dm
(2.156)
m
∫vol
1
The linear momentum {L 2 } 1 of the body is the linear momenta of the elements
of mass that comprise the body. If the position of an element of mass is given
by {R P } 1 then the velocity {V P } 1 of dm is given using the triangle law of
vector addition by
{V P } 1 {V O2 } 1 {V PO2 } 1
{V P } 1 {V O2 } 1 { 2 } 1 {R P } 1 (2.157)
The linear momentum {L} 1 of the body is therefore found by integrating
the mass particles factored by their velocity vectors {V P } 1 over the volume
of the body:
{ L2} 1 ∫ { VP}
1 d m
vol
Using the expression for {V P } 1 given in (2.157) this leads to
{ L } { V } d m{ } { R } dm
2 1 O2 1 vol 2 1 vol P 1
(2.158)
(2.159)
Using the expressions given in (2.155) and (2.156) we get an expression for
the linear momentum of Body 2 in terms of the overall mass m 2 and the
velocity at the mass centre {V G2 } 1 :
{L 2 } 1 m 2 {{V O2 } 1 { 2 } 1 {R G2 } 1 } m 2 {V G2 } 1 (2.160)
It can be noted that problems are often set up in multibody dynamics so that
the body frame coincides with the mass centre. In this case for Body 2, O 2
and G 2 would be coincident so that {RG 2 } 1 would be zero and the linear
momentum would be obtained directly from
{L 2 } 1 m 2 {V G2 } 1 m 2 {V O2 } 1 (2.161)
2.9 Angular momentum
The angular momentum {H P } 1 of the particle of material with mass dm in
Figure 2.27 can be found by taking the moment of the linear momentum of
the particle about the frame O 2 as follows:
{H P } 1 {R P } 1 {{V O2 } 1 { 2 } 1 {R P } 1 } dm (2.162)
Integrating this over the volume of the body leads to the angular momentum
{H 2 } 1 of the rigid body about the frame O 2 as follows:
∫
2
∫
{ H } { R } {{ V } { } { R } } dm
2 1 vol P 1 O2 1 2 1 P 1
∫
{ VO } { RP} d m
{ RP}
vol
vol
{{ } { R } } dm
2 1 1 1
2 1 P 1
∫
(2.163)
Note that in the second line of (2.163) the vector {V O2 } 1 comes out of the
integral and hence the order of the cross product is reversed necessitating
∫