(a) Find the position \(\mathbf{r}=\mathbf{r}( t)\) of a particle that moves counterclockwise along a circle of radius \(R\) with a constant speed \(v_{0}\).

(b) Find the velocity and acceleration of the particle.

(c) Find the magnitude of the acceleration.

Solution For convenience, we place the circle of radius \(R\) in the \(xy\)-plane, with its center at the origin, and assume that at time \(t=0\) the particle is on the positive \(x\)-axis.

(a) The particle is moving counterclockwise along the circle, as shown in Figure 28. If \(\theta (t)\) is the angle between the positive \(x\)-axis and the position vector of the particle \(\mathbf{r}=\) \(\mathbf{r}(t)\), then the vector \(\mathbf{r=r}( t) \) is \begin{equation*} \mathbf{r}(t)=R \cos [\theta (t)]{\bf i}+R\;\sin [\theta (t)]{\bf j}\tag{1} \end{equation*}

787

Notice that \(\mathbf{r}( 0)\;=\;R\mathbf{i,}\) as required. Also for the motion to be counterclockwise, the function \(\theta\;=\;\theta ( t) \) must be increasing.

(b) The velocity \(\mathbf{v}\) of the particle is \begin{equation*} \mathbf{v}(t)=\frac{d\mathbf{r}}{dt}=-R\;\sin [\theta (t)]\frac{d\theta }{dt} \mathbf{i}+R\;\cos [\theta (t)]\frac{d\theta }{dt}\mathbf{j}\qquad \color{#0066A7}{\hbox{Use the Chain Rule.}} \end{equation*}

and the speed \(v\) of the particle is \begin{eqnarray*} v(t)&=&\left\Vert \mathbf{v}(t)\right\Vert\;=\;\sqrt{R^{2}\sin ^{2}[\theta (t)]\left( \frac{ d\theta }{dt}\right) ^{2}+R^{2}\cos ^{2}[\theta (t)]\left( \frac{d\theta }{dt} \right) ^{2}}\\[6pt] &=&\sqrt{R^{2}\!\left( \frac{d\theta }{dt}\right) ^{2}}=R\left\vert \frac{d\theta }{dt}\right\vert \end{eqnarray*}

Since we know the speed is constant, \(v(t)=v_{0}\). Also, since the function \(\theta\;=\;\theta ( t) \) is increasing, \(\dfrac{d\theta }{dt}>0\). As a result, \begin{eqnarray*} v_{0}&=&R\frac{d\theta }{dt} \\[4pt] \frac{d\theta }{dt}&=&\frac{v_{0}}{R} \end{eqnarray*}

Since \(\dfrac{d\theta }{dt}\) is the rate at which the angle \(\theta \) is changing, the quantity \(\dfrac{v_{0}}{R}\) is the angular speed \(\omega \) of the particle. That is, \begin{equation*} \frac{d\theta }{dt}=\omega \end{equation*}

We solve this differential equation, using \(\theta ( 0)\;=\;0\) as the initial condition. \begin{eqnarray*} d\theta &=&\omega dt \\[3pt] \theta ( t) &=&\omega t+k \\[3pt] \theta ( 0) &=&k=0 \end{eqnarray*}

So, \(\theta (t)=\omega t\). Now substitute \(\theta (t)=\omega t\) into the vector function \(\mathbf{r=r}( t)\) [statement (1)] to obtain \begin{equation*} \mathbf{r}(t)=R\;\cos ( \omega t) \mathbf{i}+R\;\sin ( \omega t) \mathbf{j} \end{equation*}

Then the velocity \(\mathbf{v}\) and acceleration \(\mathbf{a}\) of the particle are \begin{eqnarray*} \mathbf{v}\;=\;\mathbf{v}(t) &=& -R\omega \sin ( \omega t) \mathbf{i} +R\omega \cos ( \omega t) \mathbf{j} \\[3pt] \mathbf{a}\;=\;\mathbf{a}(t) &=& -R\omega ^{2}\cos ( \omega t) \mathbf{i }-R\omega ^{2}\sin ( \omega t) \mathbf{j}=-\omega ^{2}\left[ R\;\cos ( \omega t) \mathbf{i}+R\;\sin ( \omega t) \mathbf{ j}\right]\\[3pt] &=&-\omega ^{2}\mathbf{r}(t) \end{eqnarray*}

(c) Since \(\omega=\dfrac{v_{0}}{R}\), the magnitude of the acceleration is \begin{equation*} \boxed {\left\Vert \mathbf{a}(t)\right\Vert\;=\;\omega ^{2}\left\Vert \mathbf{r}(t)\right\Vert\;=\;\omega ^{2}R=\dfrac{v_{0}^{2}}{R}}\tag{2} \end{equation*}