The normal acceleration of a body moving along a circle of…

The normal acceleration of a body moving along a circle of radius $R$ changes with time as $a_n(t) = At + Bt^2$, where $A$ and $B$ are constants. Find (a) the angular velocity as a function of time; (b) the tangential and total acceleration.

Explanation of Method

Angular velocity and accelerations from a given normal acceleration.

Step 1: Speed from the normal acceleration

\[ a_n = \frac{v^2}{R} \;\Longrightarrow\; v(t) = \sqrt{R\,(At + Bt^2)}. \]

Step 2: (a) Angular velocity

\[ \omega = \frac{v}{R} = \frac{\sqrt{R(At + Bt^2)}}{R}. \]

\[\omega(t) = \sqrt{\frac{At + Bt^2}{R}}\]

$a_n = At + Bt^2$ $a_t = \tfrac{dv}{dt}$ $a = \sqrt{a_t^2 + a_n^2}$ $\vec v$ $1)\ a_n \to v = \sqrt{R\,a_n}$ $2)\ \omega = v/R$ $3)\ a_t = dv/dt$

The given normal acceleration fixes the speed at every moment through $a_n = v^2/R$. From the speed follow the angular velocity and, by differentiating, the tangential acceleration. (Arrow lengths are only illustrative.)

Step 3: (b) Tangential acceleration

Differentiate $v^2 = R(At + Bt^2)$ with respect to time: $2v\,\dfrac{dv}{dt} = R(A + 2Bt)$. So

\[ a_t = \frac{dv}{dt} = \frac{R(A + 2Bt)}{2v} = \frac{A + 2Bt}{2}\sqrt{\frac{R}{At + Bt^2}}. \]

\[a_t = \frac{(A + 2Bt)\sqrt R}{2\sqrt{At + Bt^2}}\]

Step 4: (b) Total acceleration

\[a = \sqrt{a_t^2 + a_n^2} = \sqrt{\frac{R(A + 2Bt)^2}{4(At + Bt^2)} + (At + Bt^2)^2}\]

Result

\[\boxed{\,\omega(t) = \sqrt{\frac{At + Bt^2}{R}}\,}\]

\[\boxed{\,a_t = \frac{(A + 2Bt)\sqrt R}{2\sqrt{At + Bt^2}}\,}\]

\[\boxed{\,a = \sqrt{a_t^2 + a_n^2} = \sqrt{\frac{R(A + 2Bt)^2}{4(At + Bt^2)} + (At + Bt^2)^2}\,}\]

Remark

Note. Differentiating $v^2$ instead of $v$ (step 3) avoids the chain rule on a square root and gives the same result as the original solution. At $t \to 0$ the speed is zero but $a_t$ becomes very large (it contains $1/\sqrt t$): such a motion is an idealisation near the start.