Circular logic.

Classical Mechanics Level 3

A particle is exhibiting circular motion from an initial angular velocity ${ \omega '}$ at ${ T }=0$ .

Its radius of curvature is r . It is also know that the tangential acceleration of this particle is:

${ a }_{ T }=\ T\times \sqrt { { a }_{ N } } \times \sqrt { r }$

Find the correct relation for the time dependence of the angular velocity ( $\omega$ ) at the instant T( $\neq 0$ )

{ \omega ' }^{ \sqrt { 3 } }=\quad \sqrt [ 5 ]{ \ln { \frac { \omega }{ { T }^{ 5 } } } }

\ln { \cos { \omega ' } } ={ e }^{ \sqrt [ \omega ]{ { 3 }^{ T } } }

{ \ln { \left( \frac { \omega ' }{ \omega } \right) } }^{ 2 }={ T }^{ 2 }

{ e }^{ { T }^{ 2 } }={ (\frac { \omega }{ \omega ' } })^{ 2 }

1 solution

Navneel Mandal
Jul 20, 2015

Tangential Accelaration ${ a }_{ T }=\alpha *r$

where (alpha is the rate of change of angular velocity(dw/dt), and r is radius Centripetal accelaration formula=
${ a }_{ N }={ \omega }^{ 2 }*r$

It is know that, (aT)= t* {(sqrt) aN}*{(sqrt)r}

so, dw/dt= t* w

therefore, dw/w =t dt

Integrating with limits, $\int _{ \omega ' }^{ \omega }{ \frac { 1 }{ \omega } } =\int _{ 0 }^{ T }{ t\quad dt }$ , Which'll give $\log _{ e }{ \omega } -\log _{ e }{ \omega ' } =\frac { T }{ 2 } ^{ 2 }$ If we solve this equation, we'll get the answer.

Be sure to notice the difference b/w this answer and the other log answer

Circular logic.

1 solution

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