In one line
Split the acceleration into tangential and normal parts using the velocity direction, then use \( \rho = v^2/a_n \).
Explained simply
The radius of curvature tells how sharp a bend is at a point: a small radius means a tight turn. We cannot always see the circle, but we know how much of the acceleration is spent on turning, and that gives the radius.
Technically, \( a_n = |v_x a_y - v_y a_x| / v \) (the part of \( \vec{a} \) perpendicular to \( \vec{v} \)), so \( \rho = v^2/a_n = v^3/|v_x a_y - v_y a_x| \).
How it works
- From x(t), y(t), compute \( v_x, v_y, a_x, a_y \) at the instant.
- Speed \( v = \sqrt{v_x^2 + v_y^2} \).
- Tangential: \( a_t = (v_x a_x + v_y a_y)/v \) (projection of a on v).
- Normal: \( a_n = \sqrt{a^2 - a_t^2} \), or \( |v_x a_y - v_y a_x|/v \).
- \( \rho = v^2/a_n \).
- Projectile at the top of its path: velocity is horizontal (\( u\cos\theta \)) and the whole of g is normal, so \( \rho = u^2\cos^2\theta / g \).
Key formula
\( \rho = \dfrac{v^2}{a_n} = \dfrac{v^3}{|v_x a_y - v_y a_x|} \) (m). For a path y = f(x): \( \rho = \dfrac{[1 + (dy/dx)^2]^{3/2}}{|d^2y/dx^2|} \).
Quick example
\( x = 2t^2 \), \( y = t^3 \) at t = 1 s: \( v_x = 4, v_y = 3, a_x = 4, a_y = 6 \), v = 5 m/s. \( a_t = (16 + 18)/5 = 6.8 \) m/s\(^2\), \( a_n = |24 - 12|/5 = 2.4 \) m/s\(^2\) (check: \( 6.8^2 + 2.4^2 = 52 \)). \( \rho = 25/2.4 = 10.42 \) m. Projectile at 20 m/s and 30°: at the top, \( \rho = (20\cos30^\circ)^2/9.81 = 30.6 \) m.
Common mistakes
- Using the total acceleration in \( \rho = v^2/a \).
- Assuming the normal acceleration of a projectile is g at every point; that is true only at the top.
- Arithmetic slips in \( v^3 \); compute v first, then cube.
How MU asks it
Usually the final 2 to 3 marks of a 6 to 8-mark curvilinear motion problem: 'also find the radius of curvature at t = 2 s' or 'find the radius of curvature of a projectile at its highest point'. Show \( a_t \), \( a_n \) and \( \rho \) on separate lines.