A particle moves along a curved path. At a certain point its speed is 6 m/s and total acceleration is 5 m/s². Its speed varies with arc length s according to v² = 16 + 6s. If the radius of curvature at that point is N m, find N.
Physics · Circular Motion · relative angular velocity and radius of curvature
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Question

A particle moves along a curved path. At a certain point its speed is 6 m/s and total acceleration is 5 m/s². Its speed varies with arc length s according to v² = 16 + 6s. If the radius of curvature at that point is N m, find N.

Physics · Circular Motion · relative angular velocity and radius of curvature

G1 Physics · Accepted Answer

From v² = 16 + 6s, tangential acceleration is aₜ = v dv/ds = ½ d(v²)/ds = ½×6 = 3 m/s². Given total acceleration a = 5 m/s², the normal component is aₙ = √(a²−aₜ²) = √(25−9) = 4 m/s². Hence R = v²/aₙ = 36/4 = 9 m. Therefore N = 9.