Question:

For a rigid slab in centroidal rotation, show that the system of the effective forces consists of vectors $-\left( \Delta { m }_{ i } \right) { \omega }^{ 2 }{ r }_{ i }^{ \prime }$ and $\left( \Delta { m }_{ i } \right) (\alpha \times { r }_{ i }^{ \prime })$ attached to the various particles ${ P }_{ i }$ of the slab, where \omega and α are the angular velocity and angular acceleration of the slab, and where ${ r }_{ i }^{ \prime }$ denotes the position vector of the particle ${ P }_{ i }$ relative to the mass center G of the slab. Further show, by computing their sum and the sum of their moments about G, that the effective forces reduce to a couple $\overline { I } \alpha$.

Step-by-step

For centroidal rotation: ${ a }_{ i }={ { a }_{ i } }_{ t }{ + }{ { a }_{ i } }_{ n }=\alpha \times { r }_{ i }^{ \prime }-{ \omega }^{ 2 }{ r }_{ i }^{ \prime }$

Effective forces are:$\left( \Delta { m }_{ i } \right) { a }_{ i }=\left( \Delta { m }_{ i } \right) \left( \alpha \times { r }_{ i }^{ \prime } \right) -\left( \Delta { m }_{ i } \right) { \omega }^{ 2 }{ r }_{ i }^{ \prime }\left( \Delta { m }_{ i } \right) \left( \alpha \times { r }_{ i }^{ \prime } \right)$

$\sum \left( \Delta { m }_{ i } \right) { a }_{ i }=\sum \left( \Delta { m }_{ i } \right) \left( \alpha \times { r }_{ i }^{ \prime } \right) -\sum \left( \Delta { m }_{ i } \right) { \omega }^{ 2 }{ r }_{ i }^{ \prime }=\alpha \times \sum { (\Delta mi } ){ r }_{ i }^{ \prime }-{ \omega }^{ 2 } \sum { (\Delta mi } ){ r }_{ i }^{ \prime }$

Since G is the mass center,$\sum { (\Delta mi } ){ r }_{ i }^{ \prime }$

effective forces reduce to a couple, Summing moments about G, $\sum { \left( { r }_{ i }^{ \prime }\times \Delta { m }_{ i }{ a }_{ i } \right) } =\sum { \left[ { r }_{ i }^{ \prime }\times \left( \Delta { m }_{ i } \right) \left( \alpha \times { r }_{ i }^{ \prime } \right) \right] } -\sum { { r }_{ i }^{ \prime } } \times \left( \Delta { m }_{ i } \right) { \omega }^{ 2 }{ r }_{ i }^{ \prime }$

But,${ r }_{ i }^{ \prime }\times \left( \Delta { m }_{ i } \right) { \omega }^{ 2 }{ r }_{ i }^{ \prime }={ \omega }^{ 2 }\left( \Delta { m }_{ i } \right) \left( { r }_{ i }^{ \prime }\times { r }_{ i }^{ \prime } \right) =0$

and, ${ r }_{ i }^{ \prime }\times \left( \Delta { m }_{ i } \right) \left( \alpha \times { r }_{ i }^{ \prime } \right) =\left( \Delta { m }_{ i } \right) { r }_{ 1 }^{ \prime 2 }\alpha$

Thus,$\sum { \left( { r }_{ i }^{ \prime }\times \Delta { m }_{ i }{ a }_{ i } \right) } =\sum { \left( \Delta { m }_{ i } \right) } { r }_{ 1 }^{ 2 }\alpha =\left[ \sum { \left( \Delta { m }_{ i } \right) { r }_{ i }^{ \prime 2 } } \right] \alpha$

Since${ \left( \Delta { m }_{ i } \right) { r }_{ i }^{ \prime 2 } }= \overline { I }$ the moment of the couple is $\overline { I } \alpha.$