Chapter 4, Problem 97P

To determine

The normal acceleration of particle passing through $x=1$ and $y=2$.

Answer to Problem 97P

The normal acceleration of particle passing through $x=1$ and $y=2$ is $\underset{\_}{3.996k^2}$.

Explanation of Solution

The vector field of flow is $\stackrel{\rightharpoonup }{V}=k\left(x^2-y^2\right)\stackrel{\rightharpoonup }{i}-2kxy\stackrel{\rightharpoonup }{j}$ and the radius of curvature of streamline is $R=\frac{{\left[1+{y^{\prime }}^2\right]}^{\frac{3}{2}}}{\left[y^{″}\right]}$

Write the streamline equation for velocity field.

  $\frac{dy}{dx}=\frac{v}{u}$.  ...... (I)

Here, the velocity in $y$ direction is $v$ and the velocity in $x$ direction is $u$

Write the expression for first derivative of $y$.

  $y^{\prime }=\frac{dy}{dx}$

Here, the first derivative of $y$ is $y^{\text{'}}$.

Substitute $y^{\text{'}}$ for $\frac{dy}{dx}$ in Equation (I).

  $y^{\prime }=\frac{v}{u}$  ...... (II)

Substitute $-2kxy$ for $v$ and $k\left(x^2-y^2\right)$ for $u$ in Equation (II).

  $y^{\prime }=\frac{-2kxy}{k\left(x^2-y^2\right)}$  ...... (III).

Differentiate Equation (III) with respect to $x$.

  $\begin{array}{l}\frac{d}{dx}{y}^{\prime }=\frac{d}{dx}\frac{-2kxy}{k\left(x^2-y^2\right)}\\ y^{″}=\frac{d}{dx}\frac{-2kxy}{k\left(x^2-y^2\right)}\\ y^{″}=\frac{\left(x^2-y^2\right)\left(-2y-2x{y}^{\text{'}}\right)+\left(2xy\right)\left(2x-2y{y}^{\text{'}}\right)}{{\left(x^2-y^2\right)}^2}\end{array}$   ...... (IV)

Substitute $\frac{-2kxy}{k\left(x^2-y^2\right)}$ for $y^{\text{'}}$ in Equation (IV).

  $\begin{array}{c}{y}^{″}=\frac{\left(x^2-y^2\right)\left(-2y-2x\frac{-2kxy}{k\left(x^2-y^2\right)}\right)+\left(2xy\right)\left(2x-2y\frac{-2kxy}{k\left(x^2-y^2\right)}\right)}{{\left(x^2-y^2\right)}^2}\\ =\frac{-2y{\left(x^2-y^2\right)}^2+8x^{2}y\left(x^2-y^2\right)+8x^2{y}^3}{{\left(x^2-y^2\right)}^3}\end{array}$   ...... (V)

Write the expression for the radius of curvature of streamline.

  $R=\frac{{\left[1+y^{\text{'}2}\right]}^{\frac{3}{2}}}{\left[y^{"}\right]}$..... (V)

Write the expression for the velocity field of flow.

  $\stackrel{\rightharpoonup }{V}=k\left(x^2-y^2\right)\stackrel{\rightharpoonup }{i}-2kxy\stackrel{\rightharpoonup }{j}$   ...... (VI)

Write the expression for the resultant velocity.

  $V=\sqrt{a^2+b^2}$  ...... (VII)

Write the expression for the normal acceleration.

  $a_n=\frac{V^2}{R}$   ...... (VIII).

Calculation:

Substitute $1$ for $x$ and $2$ for $y$ in Equation (V).

  $\begin{array}{c}{y}^{″}=\frac{-2\text{×2}{\left(1^2-2^2\right)}^2+8\text{×}{1}^2\text{×2}\left(1^2-2^2\right)+8\text{×}{1}^2\text{×}{2}^3}{{\left(1^2-2^2\right)}^3}\\ =\frac{-36-48+64}{-27}\\ =\frac{-20}{-27}\\ =0.74074\end{array}$

Substitute $1$ for $x$ and $2$ for $y$ in Equation (III).

  $\begin{array}{c}{y}^{\prime }=\frac{-2\text{×1×2}}{\left(1^2-2^2\right)}\\ =\frac{-4}{-3}\\ =1.334\end{array}$

Substitute $1.334$ for $y^{\text{'}}$ and $0.74074$ for $y^{″}$ in Equation (V).

  $\begin{array}{c}R=\frac{{\left[1+{\left(1.334\right)}^2\right]}^{\frac{3}{2}}}{\left[0.74074\right]}\\ =\frac{{\left[1+1.779556\right]}^{\frac{3}{2}}}{\left[0.74074\right]}\\ =6.256\end{array}$

Substitute $1$ for $x$ and $2$ for $y$ in Equation (III).

  $\begin{array}{c}\stackrel{\rightharpoonup }{V}=k\left(1^2-2^2\right)\stackrel{\rightharpoonup }{i}-2k\text{×1×2}\stackrel{\rightharpoonup }{j}\\ =-3k\stackrel{\rightharpoonup }{i}-4k\stackrel{\rightharpoonup }{j}\end{array}$

Substitute $-3k$ for $a$ and $-4k$ for $b$ in Equation (VII).

  $\begin{array}{c}V=\sqrt{{\left(-3k\right)}^2+{\left(-4k\right)}^2}\\ =\sqrt{9k+16k}\\ =5k\end{array}$

Substitute $5k$ for $V$ and $6.256$ for $R$ in Equation (VIII).

  $\begin{array}{c}{a}_n=\frac{{\left(5k\right)}^2}{6.256}\\ =\frac{25k^2}{6.256}\\ =3.996k^2\end{array}$

Conclusion:

The normal acceleration of particle passing through is calculated by the expression $a_n=\frac{V^2}{R}$