The equations of motion of a body are x = Acosω t , y =…

The equations of motion of a body are $x = A\cos\omega t$, $y = A\sin\omega t$, $z = Bt$, with $A = 2\ \mathrm{m}$, $B = 4\ \mathrm{m/s}$, $\omega = 5\ \mathrm{rad/s}$. Find the radius of curvature of its trajectory.

Explanation of Method

Radius of curvature of a helix.

Step 1: What the path looks like

In the $xy$-plane the body goes around a circle of radius $A$; at the same time it rises uniformly along $z$. The path is a helix (like a spring).

Step 2: Velocity and acceleration

\[ \vec v = -A\omega\sin\omega t\,\vec i + A\omega\cos\omega t\,\vec j + B\,\vec k, \]

\[ v^2 = A^2\omega^2 + B^2 = 100 + 16 = 116\ \mathrm{m^2/s^2}. \]

\[ \vec a = -A\omega^2\cos\omega t\,\vec i - A\omega^2\sin\omega t\,\vec j, \]

\[ a = A\omega^2 = 50\ \mathrm{m/s^2}. \]

The speed is constant, so there is no tangential acceleration: $\vec a$ points horizontally toward the $z$-axis and is perpendicular to $\vec v$ (check: $\vec v \cdot \vec a = A^2\omega^3\sin\cos - A^2\omega^3\cos\sin + 0 = 0$). All of it is normal acceleration.

$A$ $z$ $\vec v$ $\vec a$ $v^2 = A^2\omega^2 + B^2 = 116\ \mathrm{m^2/s^2}$ $a = A\omega^2 = 50\ \mathrm{m/s^2}\ \perp \vec v$ $\rho = v^2/a_n = 2.32\ \mathrm{m}$ $\text{(larger than } A = 2\ \mathrm{m)}$

velocity
acceleration

The body moves on a helix: around the $z$-axis on a cylinder of radius $A$ while rising steadily. The acceleration points horizontally toward the axis and is perpendicular to the velocity, so it is entirely normal.

Step 3: Radius of curvature

For any curved path, $a_n = v^2/\rho$: \[\rho = \frac{v^2}{a_n} = \frac{A^2\omega^2 + B^2}{A\omega^2} = \frac{116}{50}.\]

\[\rho = A + \frac{B^2}{A\omega^2} = 2.32\ \mathrm{m}\]

Check with the cross-product formula

$ |\vec v \times \vec a| = \sqrt{A^2\omega^4B^2 + A^4\omega^6} = \sqrt{40\,000 + 250\,000} \approx 538.5 $

$ \rho = \dfrac{v^3}{|\vec v \times \vec a|} = \dfrac{116^{3/2}}{538.5} = \dfrac{1249.4}{538.5} \approx 2.32\ \mathrm{m}\ \checkmark $

Result

\[\boxed{\,\rho = A + \frac{B^2}{A\omega^2} = 2.32\ \mathrm{m}\,}\]

Remark

About the original solution. Its formula $\rho = v^3/|\vec v \times \vec a|$ and all the algebra were correct, but the final number was wrong: substituting the values gives 2.32 m, not 2 m. The radius of curvature is larger than the cylinder radius $A$ because the rise along $z$ stretches the curve; with $B = 0$ the path would be a circle and $\rho = A$.