Chapter 11.5, Problem 11.180P

To determine

The angle between oscillating plane and y axis.

Answer to Problem 11.180P

$\beta =ta{n}^{-1}\frac{R(2+{\omega }_n{}^2{t}^2)}{c\sqrt{(4+{\omega }_n{}^2{t}^2)}}$

Explanation of Solution

Given information:

According to problem $11.95$ :

$r=(Rtcos{\omega }_{n}t)i+ctj+(Rtsin{\omega }_{n}t)k$

The first derivative of motion is equal to velocity,

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

The first derivative of velocity is equal to acceleration,

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

Calculation:

According to given information,

$r=(Rtcos{\omega }_{n}t)i+ctj+(Rtsin{\omega }_{n}t)k$

Differentiate to find velocity,

$v=\frac{dr}{dt}=R(cos{\omega }_{n}t-{\omega }_{n}tsin{\omega }_{n}t)i+cj+R(sin{\omega }_{n}t+{\omega }_{n}tcos{\omega }_n t)k$

Differentiate to find acceleration

$a=\frac{dv}{dt}=R(-{\omega }_{n}sin{\omega }_{n}t-{\omega }_{n}sin{\omega }_{n}t-{\omega }_n{}^{2}tcos{\omega }_{n}t)i+R({\omega }_{n}cos{\omega }_{n}t+{\omega }_{n}cos{\omega }_{n}t-{\omega }_n{}^{2}tsin{\omega }_{n}t)k$

Solve and rearrange,

$a={\omega }_{n}R[-(2sin{\omega }_{n}t+{\omega }_{n}tcos{\omega }_{n}t)i+(2cos{\omega }_{n}t-{\omega }_{n}tsin{\omega }_{n}t)k]$

Now, find the vector $(v\times a)$ which is perpendicular to the oscillating plane,

$\begin{array}{l}(v\times a)= {\omega }_{n}R\{c(2cos{\omega }_{n}t-{\omega }_{n}tsin{\omega }_{n}t)i+R[-(sin{\omega }_{n}t+{\omega }_{n}tcos{\omega }_{n}t)(2sin{\omega }_{n}t+{\omega }_{n}tcos{\omega }_{n}t)\\ -(cos{\omega }_{n}t-{\omega }_{n}tsin{\omega }_{n}t)(2cos{\omega }_{n}t-{\omega }_{n}tsin{\omega }_{n}t)]j+c(2sin{\omega }_{n}t+{\omega }_{n}tcos{\omega }_{n}t)k\end{array}$

Therefore,

$(v\times a)={\omega }_{n}R\left[c(2cos{\omega }_{n}t-{\omega }_{n}tsin{\omega }_{n}t)i-R(2+{\omega }_n{}^2{t}^2)j+c(2sin{\omega }_{n}t+{\omega }_{n}tcos{\omega }_{n}t)k\right]$

The angle $\alpha$ between $y$ axis and vector $(v\times a)$ is be defined as:

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

But, we know that,

$\left|j\right|=1$

$(v\times a).j=-{\omega }_n{R}^2(2+{\omega }_n{}^2{t}^2)$

$\left|(v\times a)\right|={\omega }_{n}R{\left[c^2(4+{\omega }_n{}^2{t}^2)+R^2{(2+{\omega }_n{}^2{t}^2)}^2\right]}^{1/2}$

Therefore,

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

According to above figure,

The angle $\beta$ between oscillating plane and y axis is equal to,

$\beta =\left(\alpha -90°\right)$

Then,

$\begin{array}{l}cos\alpha =cos\left(90°+\beta \right)=-sin\beta \\ -sin\beta =\frac{-R(2+{\omega }_n{}^2{t}^2)}{{\left[c^2(4+{\omega }_n{}^2{t}^2)+R^2{(2+{\omega }_n{}^2{t}^2)}^2\right]}^{1/2}}\end{array}$

Therefore,

$\begin{array}{l}tan\beta =\frac{R(2+{\omega }_n{}^2{t}^2)}{c\sqrt{(4+{\omega }_n{}^2{t}^2)}}\\ \beta =ta{n}^{-1}\frac{R(2+{\omega }_n{}^2{t}^2)}{c\sqrt{(4+{\omega }_n{}^2{t}^2)}}\end{array}$

Conclusion:

The angle $\beta$ between the oscillating plane and y axis is calculated using the derivative relation.