Chapter 11.5, Problem 11.152P

To determine

The radius $\left(\rho \right)$ of curvature of the path described by the particle when time (t) is 0 sec.

Answer to Problem 11.152P

The radius $\left(\rho \right)$ of curvature of the path described by the particle when time (t) is 0 sec is $\underset{\_}{2.50\text{ft}}$.

Explanation of Solution

Given Information:

The three dimensional motion of a particle is defined by the position vector is $r=\left(At\cos t\right)i+\left(A\sqrt{t^2+1}\right)j+\left(Bt\sin t\right)k$.

The curve described by the particle lies on the hyperboloid is ${\left(\frac{y}{A}\right)}^2-{\left(\frac{x}{A}\right)}^2-{\left(\frac{z}{B}\right)}^2=1$.

The value of A and B are 3 and 1 respectively.

Calculation:

Write the three dimensional motion of a particle position vector equation.

$r=\left(At\cos t\right)i+\left(A\sqrt{t^2+1}\right)j+\left(Bt\sin t\right)k$ (1)

Here, x is $\left(At\cos t\right)i$, y is $\left(A\sqrt{t^2+1}\right)j$ and z is $\left(Bt\sin t\right)k$.

Consider x:

$\begin{array}{l}x=\left(At\cos t\right)\\ \cos t=\frac{x}{At}\end{array}$ (2)

Consider y:

$\begin{array}{l}y=\left(A\sqrt{t^2+1}\right)\\ \sqrt{t^2+1}=\frac{y}{A}\\ t^2={\left(\frac{y}{A}\right)}^2-1\end{array}$ (3)

Consider z:

$\begin{array}{l}z=\left(Bt\sin t\right)\\ \sin t=\frac{z}{Bt}\end{array}$ (4)

Calculate the $t^2$ using the relation:

${\cos }^{2}t+{\sin }^{2}t=1$

Substitute $\frac{x}{At}$ for $\cos t$ and $\frac{z}{Bt}$ for $\sin t$.

$\begin{array}{l}{\left(\frac{x}{At}\right)}^2+{\left(\frac{z}{Bt}\right)}^2=1\\ t^2={\left(\frac{x}{A}\right)}^2+{\left(\frac{z}{B}\right)}^2\end{array}$

Check whether the position vector equation satisfied the curve equation or not.

Substitute ${\left(\frac{x}{A}\right)}^2+{\left(\frac{z}{B}\right)}^2$ for $t^2$ in equation (3).

$\begin{array}{l}{\left(\frac{x}{A}\right)}^2+{\left(\frac{z}{B}\right)}^2={\left(\frac{y}{A}\right)}^2-1\\ {\left(\frac{x}{A}\right)}^2+{\left(\frac{z}{B}\right)}^2-{\left(\frac{y}{A}\right)}^2=-1\\ -{\left(\frac{y}{A}\right)}^2+{\left(\frac{x}{A}\right)}^2+{\left(\frac{z}{B}\right)}^2=-1\\ {\left(\frac{y}{A}\right)}^2-{\left(\frac{x}{A}\right)}^2-{\left(\frac{z}{B}\right)}^2=1\end{array}$

Hence, the equation is satisfied.

Rewrite the Equation (1).

Substitute 3 for A and 1 for B in Equation (1).

$r=\left(3t\cos t\right)i+\left(3\sqrt{t^2+1}\right)j+\left(1t\sin t\right)k$

Write the expression for velocity using the relation:

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

Substitute $\left(At\cos t\right)i+\left(A\sqrt{t^2+1}\right)j+\left(Bt\sin t\right)k$ for $r$.

$\begin{array}{c}v=\frac{d\left(3t\cos t\right)i+\left(3\sqrt{t^2+1}\right)j+\left(1t\sin t\right)k}{dt}\\ =3\left(\cos t-t\sin t\right)i+3\frac{t}{\sqrt{t^2+1}}j+\left(\sin t+t\cos t\right)k\end{array}$ (5)

Calculate velocity vector $\left(v\right)$ when time is 0 sec:

Substitute 0 for t in Equation (5).

$\begin{array}{c}v=3\left(\cos \left(0\right)-\left(0\right)\sin \left(0\right)\right)i+3\frac{\left(0\right)}{\sqrt{{\left(0\right)}^2+1}}j+\left(\sin \left(0\right)+\left(0\right)\cos \left(0\right)\right)k\\ =3\left(1-0\right)i+0j+\left(0\right)k\\ =3i+0j+0k\end{array}$

Here, $v_x$ is $3$, $v_y$ is $0$ and $v_z$ is $0$.

Calculate the velocity $\left(v^2\right)$ using the relation:

$v^2=v_x^2+v_y^2+v_z^2$

Substitute $3$ for $v_x$, $0$ for $v_y$ and $0$ for $v_z$.

$\begin{array}{c}{v}^2=3^2+0^2+0^2\\ =9{\left(\text{ft/s}\right)}^2\end{array}$

Write the expression for acceleration vector $\left(a\right)$ using the relation:

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

Substitute $3\left(\cos t-t\sin t\right)i+3\frac{t}{\sqrt{t^2+1}}j+\left(\sin t+t\cos t\right)k$ for $v$.

$\begin{array}{c}a=\frac{d\left(3\left(\cos t-t\sin t\right)i+3\frac{t}{\sqrt{t^2+1}}j+\left(\sin t+t\cos t\right)k\right)}{dt}\\ =3\left(-\sin t-\sin t-t\cos t\right)i+3\frac{\sqrt{t^2+1}-t\left(\frac{t}{\sqrt{t^2+1}}\right)}{\left(t^2+1\right)}j+\left(\cos t+\cos t-t\sin t\right)k\\ =-3\left(2\sin t+t\cos t\right)i+3\frac{1}{{\left(t^2+1\right)}^{\left(\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}}j+\left(2\cos t-t\sin t\right)k\end{array}$

Substitute 0 sec for t.

$\begin{array}{c}a=-3\left(2\sin \left(0\right)+\left(0\right)\cos \left(0\right)\right)i+3\frac{1}{{\left({\left(0\right)}^2+1\right)}^{\left(\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}}j+\left(2\cos \left(0\right)-\left(0\right)\sin \left(0\right)\right)k\\ =-3\left(0\right)i+\left(3\right)j+\left(2-0\right)k\\ =0i+3j+2k\end{array}$

Here, $a_x$ is 0, $a_y$ is 3 and $a_z$ is 1.

Calculate the magnitude $\left(a\right)$ of acceleration at time 0 sec.

$a=\sqrt{a_x^2+a_y^2+a_z^2}$

Substitute 0 is $a_x$, 3 for $a_y$ and 2 for $a_z$.

$\begin{array}{c}a=\sqrt{0^2+3^2+2^2}\\ =\sqrt{0+9+4}\\ =3.61{\text{ft/s}}^{\text{2}}\end{array}$

When time (t) is zero the above equation become zero,

$\frac{dv}{dt}=0$

The tangential acceleration $\left(a_t\right)$ is the rate of change speed with respective to time,

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

Write the expression for normal component of acceleration $\left(a_n\right)$:

$a_n=\frac{v^2}{\rho }$

Calculate the radius $\left(\rho \right)$ using the total acceleration formula:

$a^2=a_t^2+a_n^2$

Substitute $\frac{dv}{dt}$ for $a_t$ and $\frac{v^2}{\rho }$ for $a_n$.

$a^2={\left(\frac{dv}{dt}\right)}^2+{\left(\frac{v^2}{\rho }\right)}^2$

Substitute 0 for $\frac{dv}{dt}$.

$\begin{array}{l}{a}^2=0^2+{\left(\frac{v^2}{\rho }\right)}^2\\ a=\sqrt{{\left(\frac{v^2}{\rho }\right)}^2}\\ a=\left(\frac{v^2}{\rho }\right)\end{array}$

Substitute $3.61{\text{ft/s}}^{\text{2}}$ for a and $9{\left(\text{ft/s}\right)}^2$ for $v^2$.

$\begin{array}{l}3.61=\left(\frac{9}{\rho }\right)\\ \rho =\left(\frac{9}{3.61}\right)\\ \rho =2.493\text{ft}\\ \rho \approx \text{2}\text{.50}\text{ft}\end{array}$

Therefore, the radius $\left(\rho \right)$ of curvature of the path described by the particle when time (t) is 0 sec is $\underset{\_}{2.50\text{ft}}$.