A material point moves along a circle with speed v(t) = bt…

A material point moves along a circle with speed $v(t) = bt$, where $b = 1\ \mathrm{m/s^2}$. What is its total acceleration at the moment when it has turned through $\theta = 0.6\pi\ \mathrm{rad}$?

Explanation of Method

Total acceleration after turning through 0.6π.

Step 1: Tangential acceleration

\[ a_t = \frac{dv}{dt} = b = 1\ \mathrm{m/s^2}. \]

Step 2: Speed after turning through $\theta$

The point starts from rest with constant tangential acceleration $b$. After travelling the arc $s = \theta R$: $v^2 = 2bs = 2b\,\theta R$.

Step 3: Normal acceleration

\[ a_n = \frac{v^2}{R} = 2b\theta = 2 \cdot 1 \cdot 0.6\pi = 1.2\pi \approx 3.77\ \mathrm{m/s^2}. \]

The radius cancels, so it is not needed.

$0.6\pi$ $\text{start}$ $a_t = b = 1$ $a_n = 2b\theta \approx 3.77$ $a \approx 3.90\ \mathrm{m/s^2}$ $v^2 = 2bs = 2b\,\theta R$ $a_n = \tfrac{v^2}{R} = 2b\theta$ $(R \text{ cancels})$

tangential
normal
total

The point starts from rest and speeds up along the circle; after sweeping $0.6\pi$ rad (108°, blue arc) its accelerations are as shown, drawn to scale. The answer does not depend on the size of the circle.

Step 4: Total acceleration

\[ a = \sqrt{a_t^2 + a_n^2} = b\sqrt{1 + 4\theta^2} = \sqrt{1 + 14.21}. \]

\[a \approx 3.90\ \mathrm{m/s^2}\]

Result

\[\boxed{\,a \approx 3.90\ \mathrm{m/s^2}\,}\]

Remark

Note. The original solution found the same result by integrating $\omega = bt/R$; the relation $v^2 = 2bs$ from uniform acceleration gives it in one line. The total acceleration makes an angle $\arctan(2\theta) \approx 75^\circ$ with the direction of motion.