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

MTH404 ASSIGNMENT NO. 1 FALL 2022

For More Updates Visit Website

https://vutechofficial.blogspot.com/

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

DON’T MISS THESE: Important instructions before attempting the solution of this assignment:

 •      To solve this assignment, you should have good command on 1 - 5 lectures.

•       Upload assignments properly through LMS, No assignment will be accepted through email.

 •      Write your ID on the top of your solution file.

• Don’t use colorful back grounds in your solution files.
• Use Math Type or Equation Editor Etc. for mathematical symbols.
• You should remember that if we found the solution files of some students are same then we will reward zero marks to all those students.
• Try to make solution by yourself and protect your work from other students, otherwise you and the student who send same solution file as you will be given zero marks.
• Also remember that you are supposed to submit your assignment in Word format any other like scan images, etc. will not be accepted and we will give zero marks correspond to these assignments.

KINDLY, DON’T COPY PASTE

SUBSCRIBE, SHARE, LIKE AND COMMENTS FOR MORE UPDATES

SEND WHATSAPP OR E-MAIL FOR ANY QUERY

0325-6644800

kamranhameedvu@gmail.com

For More Updates Visit Website

https://vutechofficial.blogspot.com/

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}}`

 

KINDLY, DON’T COPY PASTE

SUBSCRIBE, SHARE, LIKE AND COMMENTS FOR MORE UPDATES

SEND WHATSAPP OR E-MAIL FOR ANY QUERY

0325-6644800

kamranhameedvu@gmail.com

For More Updates Visit Website

https://vutechofficial.blogspot.com/