Answer to Question #127881 in Mechanics | Relativity for Tad Medina

Question

A particle travels around a circle of radius 5 m, changing its speed at a constant rate. At a certain point A, the speed is 3m/s. After travelling another quarter revolution to point B, the speed has increased to 6 m/s. Determine the magnitude of the acceleration of the particle at B. *

Answer 1

$Solution:\\Find\ the\ tangential\ acceleration\\a_\tau=\frac{v^2-v_0^2}{s}=\frac{v^2-v_0^2}{0.5\pi R}=\\ =\frac{36-9}{0.5\times3.14\times5 }=3.44\frac{m}{s^2}\\then\ normal\ acceleration\\a_n=\frac{v^2}{R}=\frac{36}{5}=7.2\frac{m}{s^2}\\Acceleration\ at\ point\ B\ is \\the\ vector\ sum\ of\ two\ components \\a=\sqrt{a_\tau^2+a_n^2}=\sqrt{3.44^2+7.2^2}=7.98\frac{m}{s^2}$