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

A body moves along a circle of radius $R$ so that its tangential acceleration always has the same magnitude as its normal acceleration. At $t = 0$ its speed is $v_0$, and the motion is decelerating. Find (a) the speed as a function of time and of the distance travelled; (b) the total acceleration as a function of speed and of distance.

Explanation of Method

Braking with equal tangential and normal acceleration.

Step 1: The equation of motion

The motion slows down, so $dv/dt$ is negative and equal in size to $a_n = v^2/R$: $\frac{dv}{dt} = -\frac{v^2}{R}$.

Step 2: (a) Speed as a function of time

\[ \int_{v_0}^{v} \frac{dv}{v^2} = -\int_0^t \frac{dt}{R} \;\Longrightarrow\; \frac1v - \frac1{v_0} = \frac tR. \]

\[v(t) = \frac{v_0}{1 + v_0 t/R} = \frac{v_0 R}{R + v_0 t}\]

Step 3: (a) Speed as a function of distance

Use $\dfrac{dv}{dt} = v\dfrac{dv}{ds}$: \[v\,\frac{dv}{ds} = -\frac{v^2}{R} \;\Longrightarrow\; \frac{dv}{v} = -\frac{ds}{R} \;\Longrightarrow\; \ln\frac{v}{v_0} = -\frac sR.\]

\[v(s) = v_0\,e^{-s/R}\]

$\text{fraction}$ $s/R$ $0$ $1$ $2$ $3$ $0.5$ $1$ $v/v_0 = e^{-s/R}$ $a/a_0 = e^{-2s/R}$

speed
total acceleration

Along the path, the speed falls by the same factor $e$ for every radius travelled, and the acceleration falls twice as fast. ($a_0 = \sqrt2\,v_0^2/R$ is the initial acceleration.) The body never stops completely.

Step 4: (b) Total acceleration

The two parts are equal and perpendicular: $a = \sqrt{a_t^2 + a_n^2} = \sqrt2\,a_n$.

\[a(v) = \frac{\sqrt2\,v^2}{R}, \qquad a(s) = \frac{\sqrt2\,v_0^2}{R}\,e^{-2s/R}\]

The total acceleration always makes 45° with the velocity (and points backward, toward the centre).

Result

\[\boxed{\,v(t) = \frac{v_0}{1 + v_0 t/R} = \frac{v_0 R}{R + v_0 t}\,}\]

\[\boxed{\,v(s) = v_0\,e^{-s/R}\,}\]

\[\boxed{\,a(v) = \frac{\sqrt2\,v^2}{R}, \qquad a(s) = \frac{\sqrt2\,v_0^2}{R}\,e^{-2s/R}\,}\]

Remark

About the original solution. It wrote $dv/dt = +v^2/R$ (speeding up) but then used a form of the result that belongs to slowing down; the two sign slips cancel, so its final formulas are right. With the correct minus sign the steps above lead there directly. Going $s$ directly from $v(s)$ with $v\,dv/ds$ also avoids solving for $t$.