Tangential and Normal components of the acceleration
Question No. 1
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} `