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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.

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