A body moves along a circle of radius R so that its speed…

A body moves along a circle of radius $R$ so that its speed depends on the distance travelled as $v(s) = k\sqrt s$, where $k$ is a constant. Find the angle $\theta$ between the total acceleration and the velocity as a function of $s$.

Explanation of Method

Angle between acceleration and velocity when v = k√s.

Step 1: Tangential acceleration

Square the law: $v^2 = k^2 s$. This is $v^2 = 2a_t s$ (uniform acceleration from rest, as in Translational Kinematics 051) with

\[ a_t = \frac{k^2}{2} \ \ \text{(constant)}. \]

(The chain rule gives the same: $a_t = \dfrac{dv}{ds}\,v = \dfrac{k}{2\sqrt s}\cdot k\sqrt s = \dfrac{k^2}{2}$.)

Step 2: Normal acceleration

\[ a_n = \frac{v^2}{R} = \frac{k^2 s}{R}. \]

Step 3: The angle

The tangential part lies along the velocity and the normal part is perpendicular to it, so

\[ \tan\theta = \frac{a_n}{a_t} = \frac{k^2 s / R}{k^2/2}. \]

\[\tan\theta = \frac{2s}{R}, \qquad \theta = \arctan\frac{2s}{R}\]

$\theta\ (^\circ)$ $s/R$ $0$ $0.5$ $1$ $2$ $3$ $4$ $5$ $45$ $90$ $s = R/2:\ a_t = a_n,\ \theta = 45^\circ$ $\theta = \arctan\tfrac{2s}{R}$ $\to 90^\circ$

At the start the acceleration points along the velocity ($\theta = 0$). As the body speeds up, the normal part grows while the tangential part stays constant, so the acceleration turns toward the centre, approaching 90°.

Result

\[\boxed{\,\tan\theta = \frac{2s}{R}, \qquad \theta = \arctan\frac{2s}{R}\,}\]

Remark

Note. The constant $k$ cancels: the angle depends only on how far along the circle (in radii) the body has gone. After half a radius, $\theta = 45^\circ$; after one full lap ($s = 2\pi R$), $\tan\theta = 4\pi$, so $\theta \approx 85.5^\circ$.