MTH404 ASSIGNMENT NO. 1 FALL 2022

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MTH404 ASSIGNMENT NO. 1 FALL 2022 || 100% RIGHT SOLUTION || DYNAMICS || BY VuTech

Assignment # 01 MTH404 (Fall 2022)

Maximum Marks: 10 Due Date: November 20, 2022

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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} `

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

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