Find the tangential and normal components of the acceleration of a point

Tangential and Normal components of the acceleration

Find the tangential and normal components of the acceleration of a point 

describing the curve y² – 2x = 1 -2y with the uniform speed v when the 

particle is at (0, -1 ± `\sqrt 2 \)`.

Solution:

    The particle moving along a curve

    Tangential component = `\a_{T}` _­= `\frac{dv}{dt}`= 0

    And

    Normal component = `\a_{N}` = `\frac{v^2}{\rho }`

    We know that roh equation

            `\rho  = \left[ \frac{1 + \left( \frac{dy}{dx} \right)^2}\frac{d^2y}{dx^2} \right]^{\frac{3}{2}} -  -  -  -  -  -  -  -  - \left( A \right)`

    Given curve

            Y² – 2x = 1 – 2y

    Taking derivative w.r.t     x

            2y`\frac{dy}{dx}` - 2 =  - 2`\frac{dy}{dx}`

            2y`\frac{dy}{dx}` + 2`\frac{dy}{dx}` = 2

            2`\frac{dy}{dx}\left( y + 1 \right)` = 2

            `\frac{dy}{dx}\left( y + 1 \right)` = 1

            `\frac{dy}{dx}= \frac{1}{y + 1} - \ - \ - \ - \ - \ - \left( 1 \right)`

    At point ( 0 , - 1  ± `\sqrt 2 \)`

            `\frac{dy}{dx}= \frac{1}{- 1 \pm \sqrt 2 + 1}`

            `\frac{dy}{dx}= \frac{1}{\pm \sqrt 2 }`

    From equation (1)

             `\frac{dy}{dx}` = `\{y + 1}^{ - 1}`

    Taking derivative w.r.t x

            `\frac{d^2y}{dx^2}`=  `-\(y + 1)\^-2\frac{1}{\left(y + 1 )\}` 

            `\frac{d^2y}{dx^2}`=  `\frac{-1}{(y + 1)\^-2}.\frac{1}{\left(y + 1 )\}` 

            `\frac{d^2y}{dx^2}`=  `\frac{-1}{(y + 1)\^3}` 

    At point (0, -1 ± `\sqrt 2 \)`

            `\frac{d^2y}{dx^2}`=  `\frac{-1}{(- 1\ \pm \sqrt 2  + 1)\^3}`       

            `\frac{d^2y}{dx^2}`=  `\frac{-1}{(\ \pm \sqrt 2  )\^3}` 

    Now equation (A) becomes

            `\rho  = \left[ \frac{1 + \left(\frac{1}{\pm \sqrt 2})^2}\frac{-1}{(\ \pm \sqrt 2  )\^3}]^{\frac{3}{2}}`

            `\rho  = \left[ \frac{1 + \left( \frac{1}{ \pm \left( 2 \right)^{\frac{1}{2}}} \right)^2}\frac{ - 1}{\left(  \pm \\sqrt 2  \right)^3} \right]^{\frac{3}{2}}`

            `\rho  = \left[ \frac{1 + \frac{1}{2}}{\frac{ - 1}{\left(  \pm \\sqrt 2  \right)}^3} \right]^{\frac{3}{2}}`

            `\rho  = \left[ \frac{\frac{3}{2}}{\frac{ - 1}{\left(  \pm \sqrt 2  \right)^3}} \right]^{\frac{3}{2}}`

            `\rho  = \left[ 3\left(  \pm \\sqrt 2  \right) \right]^\frac{3}{2}`

            `\rho  = \left[  \pm \3\sqrt 2  \right]^\frac{3}{2}`

    Normal component = `\a_{N}` ₌ `\frac{v^2}{\left[  \pm \3\sqrt 2 \right]^{\frac{3}{2}}`