Chapter 11.5, Problem 11.180P

For the conic helix of Prob. 11.95, determine the angle that the osculating plane forms with the y axis.

To determine

The angle $\left(\beta \right)$ that the osculating plane forms with y axis.

Answer to Problem 11.180P

The angle $\left(\beta \right)$ that the osculating plane forms with y axis is $\underset{\_}{{\tan }^{-1}\left|\frac{R\left(2{\omega }_n^2{t}^2\right)}{c\sqrt{4+{\omega }_n^2{t}^2}}\right|}$.

Explanation of Solution

Given Information:

The three dimensional motion of a particle is defined by the position vector $r=\left(Rt\cos {\omega }_{n}t\right)i+ctj+\left(Rt\sin {\omega }_{n}t\right)k$.

Calculation:

Write three dimensional motion of particle position vector equation.

$r=\left(Rt\cos {\omega }_{n}t\right)i+ctj+\left(Rt\sin {\omega }_{n}t\right)k$ (1)

Write the expression for velocity using the relation:

$v=\frac{dr}{dt}$

Substitute $\left(Rt\cos {\omega }_{n}t\right)i+ctj+\left(Rt\sin {\omega }_{n}t\right)k$ for $r$.

$\begin{array}{c}v=\frac{d\left(Rt\cos {\omega }_{n}t\right)i+ctj+\left(Rt\sin {\omega }_{n}t\right)k}{dt}\\ =R\left(\cos {\omega }_{n}t-{\omega }_n\sin {\omega }_{n}t\right)i+cj+R\left(\sin {\omega }_{n}t+{\omega }_{n}t\cos {\omega }_{n}t\right)k\end{array}$ (2)

Here, $v_x$ is $R\left(\cos {\omega }_{n}t-{\omega }_n\sin {\omega }_{n}t\right)$, $v_y$ is $c$ and $v_z$ is $R\left(\sin {\omega }_{n}t+{\omega }_{n}t\cos {\omega }_{n}t\right)$.

Write the expression for acceleration using the relation:

$a=\frac{dv}{dt}$

Substitute $R\left(\cos {\omega }_{n}t-{\omega }_n\sin {\omega }_{n}t\right)i+cj+R\left(\sin {\omega }_{n}t+{\omega }_{n}t\cos {\omega }_{n}t\right)k$ for $v$.

$\begin{array}{c}a=\frac{d\left(R\left(\cos {\omega }_{n}t-{\omega }_n\sin {\omega }_{n}t\right)i+cj+R\left(\sin {\omega }_{n}t+{\omega }_{n}t\cos {\omega }_{n}t\right)k\right)}{dt}\\ ={\omega }_{n}R\left[\left(-2\sin {\omega }_{n}t{\omega }_{n}t\cos {\omega }_{n}t\right)i+\left(2\cos {\omega }_{n}t-{\omega }_{n}t\sin {\omega }_{n}t\right)k\right]\end{array}$ (3)

Show the vector $\left(v\times a\right)$ is perpendicular to the osculating plane as in Figure (1).

Calculate the vector $\left(v\times a\right)$:

Substitute $R\left(\cos {\omega }_{n}t-{\omega }_n\sin {\omega }_{n}t\right)i+cj+R\left(\sin {\omega }_{n}t+{\omega }_{n}t\cos {\omega }_{n}t\right)k$ for $v$ and ${\omega }_{n}R\left[\left(-2\sin {\omega }_{n}t{\omega }_{n}t\cos {\omega }_{n}t\right)i+\left(2\cos {\omega }_{n}t-{\omega }_{n}t\sin {\omega }_{n}t\right)k\right]$ for $a$.

$\begin{array}{c}\left(v\times a\right)={\omega }_{n}R\left|\begin{array}{ccc}i& j& k\\ R\left(\cos {\omega }_{n}t-{\omega }_n\sin {\omega }_{n}t\right)& c& R\left(\sin {\omega }_{n}t+{\omega }_{n}t\cos {\omega }_{n}t\right)\\ \left(-2\sin {\omega }_{n}t{\omega }_{n}t\cos {\omega }_{n}t\right)& 0& \left(2\cos {\omega }_{n}t-{\omega }_{n}t\sin {\omega }_{n}t\right)\end{array}\right|\\ ={\omega }_{n}R\left\{\begin{array}{l}\left(c\left(2\cos {\omega }_{n}t-{\omega }_{n}t\sin {\omega }_{n}t\right)i\right)\\ +R\left[\begin{array}{l}\left(-\sin {\omega }_{n}t+{\omega }_{n}t\cos {\omega }_{n}t\right)\left(2\sin {\omega }_{n}t+{\omega }_{n}t\cos {\omega }_{n}t\right)-\\ \left(\cos {\omega }_{n}t-{\omega }_{n}t\sin {\omega }_{n}t\right)\left(2\cos {\omega }_{n}t-{\omega }_{n}t\sin {\omega }_{n}t\right)\end{array}\right]j\\ +c\left(2\sin {\omega }_{n}t+{\omega }_{n}t\cos {\omega }_{n}t\right)k\end{array}\right\}\\ ={\omega }_{n}R\left[\begin{array}{l}c\left(2\cos {\omega }_{n}t-{\omega }_{n}t\sin {\omega }_{n}t\right)i-R\left(2+{\omega }_n^2{t}^2\right)j\\ +c\left(2\sin {\omega }_{n}t+{\omega }_{n}t\cos {\omega }_{n}t\right)k\end{array}\right]\end{array}$

Calculate the $\left(v\times a\right)j$:

$\begin{array}{c}\left(v\times a\right)j={\omega }_{n}R\left[-R\left(2+{\omega }_n^2{t}^2\right)\right]\\ =-{\omega }_n{R}^2\left(2+{\omega }_n^2{t}^2\right)\end{array}$

Calculate the $\left|\left(v\times a\right)\right|$:

$\begin{array}{c}\left|\left(v\times a\right)\right|={\omega }_{n}R{\left[c\left(2\cos {\omega }_{n}t-{\omega }_{n}t\sin {\omega }_{n}t\right)i-R\left(2+{\omega }_n^2{t}^2\right)j+c\left(2\sin {\omega }_{n}t+{\omega }_{n}t\cos {\omega }_{n}t\right)k\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ ={\omega }_{n}R{\left[c^2\left(4+{\omega }_n^2{t}^2\right)+R^2{\left(2+{\omega }_n^2{t}^2\right)}^2\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\end{array}$

Calculate the angle $\left(\cos \alpha \right)$ using the relation:

$\cos \alpha =\frac{\left(v\times a\right)j}{\left|\left(v\times a\right)\right|\left|j\right|}$

Substitute $-{\omega }_n{R}^2\left(2+{\omega }_n^2{t}^2\right)$ for $\left(v\times a\right)j$, 1 for $\left|j\right|$, and ${\omega }_{n}R{\left[c^2\left(4+{\omega }_n^2{t}^2\right)+R^2{\left(2+{\omega }_n^2{t}^2\right)}^2\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$ for $\left|\left(v\times a\right)\right|$.

$\begin{array}{c}\cos \alpha =\frac{-{\omega }_n{R}^2\left(2+{\omega }_n^2{t}^2\right)}{{\omega }_{n}R{\left[c^2\left(4+{\omega }_n^2{t}^2\right)+R^2{\left(2+{\omega }_n^2{t}^2\right)}^2\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\times 1}\\ =\frac{-R\left(2+{\omega }_n^2{t}^2\right)}{{\left[c^2\left(4+{\omega }_n^2{t}^2\right)+R^2{\left(2+{\omega }_n^2{t}^2\right)}^2\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\end{array}$

Write the expression for angle $\left(\alpha \right)$ using Figure 1:

$\begin{array}{c}\beta =\alpha -90°\\ \alpha =\beta +90°\end{array}$

Substitute $\beta +90°$ for $\alpha$ in $\cos \alpha$ :

$\begin{array}{c}\cos \alpha =\cos \left(\beta +90°\right)\left(\because \cos \theta +90°=-\sin \theta \right)\\ =-\sin \beta \end{array}$

Substitute $\frac{-R\left(2+{\omega }_n^2{t}^2\right)}{{\left[c^2\left(4+{\omega }_n^2{t}^2\right)+R^2{\left(2+{\omega }_n^2{t}^2\right)}^2\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}$ for $\cos \alpha$.

$\frac{-R\left(2+{\omega }_n^2{t}^2\right)}{{\left[c^2\left(4+{\omega }_n^2{t}^2\right)+R^2{\left(2+{\omega }_n^2{t}^2\right)}^2\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}=-\sin \beta$ . (4)

Show the representation of equation (4) as in Figure (2):

Calculate the angle $\left(\beta \right)$ that the osculating plane forms with y axis

$\begin{array}{c}\tan \beta =\frac{opp}{adj}\\ =\frac{R\left(2+{\omega }_n^2{t}^2\right)}{c\sqrt{4+{\omega }_n^2{t}^2}}\\ ={\tan }^{-1}\left(\frac{R\left(2+{\omega }_n^2{t}^2\right)}{c\sqrt{4+{\omega }_n^2{t}^2}}\right)\end{array}$

Therefore, the angle $\left(\beta \right)$ that the osculating plane forms with y axis is $\underset{\_}{{\tan }^{-1}\left|\frac{R\left(2{\omega }_n^2{t}^2\right)}{c\sqrt{4+{\omega }_n^2{t}^2}}\right|}$.