Answer to Question #137409 in Electric Circuits for Sneha

Question

A point moves along an arc of a circle of radius R. Its velocity depends on the distance covered s as v = a sqrt(s), where a is a constant. Find the angle α between the vector of the total acceleration and the vector of velocity as a function of s.

Answer 1

Since $V=A\times \sqrt S$

$a_r=\frac {V^2}{R}=\frac {A^2\times S}{R}$ ,

$a_t=\frac {dV}{dt}=\frac{A}{2\times \sqrt S}$ $\times \frac{dS}{dt}=\frac {A}{2\times \sqrt S}$ $\times A \sqrt S= \frac {A^2}{2}$ .

Since $a_t$ is a positive constant, the particle velocity increases with time, and the tangential acceleration vector and the velocity vector coincide in direction. Therefore, the angle between $\vec V$ and $\vec a$ (total acceleration) is equal to, the angle between $a_t\times \hat v_t$ and $\vec a$ , and it can be found $α$ by the formula: $\tg {\alpha}=\frac {\lvert{a_n}\lvert}{\lvert a_t \lvert}$ $=\frac {\frac {a^2\times s}{R}}{0.5\times a^2}$$=\frac {2\times S}{R}$ , therefore $\alpha=\arctg \frac {2\times s}{R}$ .