$\let\oldhat\hat \renewcommand{\vec}[1]{\mathbf{#1}} \renewcommand{\hat}[1]{\oldhat{\mathbf{#1}}}$

Suppose we have a car moving on a circular track of radius $b$ and speed $v=ct$, where $t$ is time and $c\in\mathbb{R^+}$.

Let the velocity vector be written as: $$\vec v = v\vec u$$ where $|\vec v|=v=ct$ and $\vec u$ is a unit vector in the direction of motion. Thus by the Chain Rule, $$\vec{\dot v}=\vec{a}=v'\vec u+v\vec{\dot u}\tag1$$

But in general for nonuniform motion in a circle, $$\vec{a}=a_T\vec{u}+a_R\vec r\tag2$$ where $\vec{r}$ is a unit radial vector always pointing inwards to the center of the circle, $a_T$ is the magnitude of the tangential acceleration in the direction of $\vec v$, and $a_R$ is the magnitude of the radial acceleration.

Reconciling $(1)$ and $(2)$, one finds $$a_T=v'$$ but its not so clear for $a_R$. I would like to somehow equate $v\vec{\dot u }=a_R\vec r$ by using the more general formula that $$a_R=\frac{v^2}{r}$$ But for this particular problem $$v\vec{\dot u}=ct\vec{\dot u}=^? \frac{c^2t^2}{b}\vec r$$ So it must be then that since $\vec{u}$ and $\vec{\dot u}$ are perpendicular that $$\frac{t}{b}\vec{\dot u}=\vec r$$ right? How can I proceed further?

I'm hoping to utilize the above work to help me show the $\theta$ such that $$\vec v\cdot \vec a=va\cos\theta$$ will be equal to $45^{\circ}$ at $t=\sqrt{\frac{b}{c}}$.


1 Answer 1


You had the correct tangential, $c$, and radial, $\dfrac {c^2t^2}{b}$, accelerations

If the radius is constant then in polar coordinates $\vec v = b\,\dot \theta \hat \theta = c\,t\, \hat \theta \Rightarrow \dot \theta = \dfrac{ct}{b} \Rightarrow \ddot \theta = \dfrac cb$ and $\vec a = -b\,\dot \theta^2\, \hat r + b\,\ddot\theta \,\hat \theta =-\dfrac{c^2t^2}{b}\hat r +c\,\hat \theta = - c\,\hat r + c\,\hat \theta $ at $t=\sqrt{\dfrac bc}$ so the angle between $\vec v$ and $\vec a$ is $45^\circ$.


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