An equilateral triangle ABC with side length $l$ moves in a plane. Let's establish a coordinate system with origin at A and the x-axis aligned with the vector $\vec{AC}$. The coordinates are: A(0, 0), C($l$, 0), and B($l\cos(60^\circ)$, $l\sin(60^\circ)$) = B($l/2$, $l\sqrt{3}/2$). The position vectors relative to A are $\vec{r}_{C/A} = l\hat{i}$ and $\vec{r}_{B/A} = (l/2)\hat{i} + (l\sqrt{3}/2)\hat{j}$. The motion of point A is given by $\vec{v}_A = \omega_0 \vec{AC} = l\omega_0 \hat{i}$ and $\vec{a}_A = 2\omega_0^2 \vec{AC} = 2l\omega_0^2 \hat{i}$. For planar motion, the angular velocity and acceleration are $\vec{\omega} = \omega\hat{k}$ and $\vec{\alpha} = \alpha\hat{k}$.

[Q1] Magnitude of the velocity of point C

We first determine the angular velocity $\omega$ of the triangle using the relative velocity equation between points A and B: $\vec{v}_B = \vec{v}_A + \vec{\omega} \times \vec{r}_{B/A}$.

$$ \vec{v}_B = l\omega_0 \hat{i} + (\omega\hat{k}) \times \left(\frac{l}{2}\hat{i} + \frac{l\sqrt{3}}{2}\hat{j}\right) = l\omega_0 \hat{i} - \frac{\omega l\sqrt{3}}{2}\hat{i} + \frac{\omega l}{2}\hat{j} $$ $$ \vec{v}_B = \left(l\omega_0 - \frac{\omega l\sqrt{3}}{2}\right)\hat{i} + \left(\frac{\omega l}{2}\right)\hat{j} $$

The magnitude of $\vec{v}_B$ is given as $v_B = l\omega_0/2$. So, $v_B^2 = l^2\omega_0^2/4$.

$$ \left(l\omega_0 - \frac{\omega l\sqrt{3}}{2}\right)^2 + \left(\frac{\omega l}{2}\right)^2 = \frac{l^2\omega_0^2}{4} $$

Expanding and simplifying by dividing by $l^2$:

$$ \omega_0^2 - \sqrt{3}\omega\omega_0 + \frac{3}{4}\omega^2 + \frac{1}{4}\omega^2 = \frac{\omega_0^2}{4} $$ $$ \omega^2 - \sqrt{3}\omega_0\omega + \frac{3}{4}\omega_0^2 = 0 $$

This is a perfect square: $(\omega - \frac{\sqrt{3}}{2}\omega_0)^2 = 0$, which gives a unique solution for the angular velocity:

$$ \omega = \frac{\sqrt{3}}{2}\omega_0 $$

Now, we find the velocity of point C using the relative velocity equation $\vec{v}_C = \vec{v}_A + \vec{\omega} \times \vec{r}_{C/A}$.

$$ \vec{v}_C = l\omega_0 \hat{i} + \left(\frac{\sqrt{3}}{2}\omega_0\hat{k}\right) \times (l\hat{i}) = l\omega_0 \hat{i} + \frac{\sqrt{3}}{2}l\omega_0 \hat{j} $$

The magnitude of the velocity of C is:

$$ v_C = |\vec{v}_C| = \sqrt{(l\omega_0)^2 + \left(\frac{\sqrt{3}}{2}l\omega_0\right)^2} = \sqrt{l^2\omega_0^2 + \frac{3}{4}l^2\omega_0^2} = \sqrt{\frac{7}{4}l^2\omega_0^2} $$ $$ v_C = \frac{l\omega_0\sqrt{7}}{2} $$

[Q2] Magnitude of the acceleration of point C

First, determine the angular acceleration $\alpha$ using the relative acceleration equation between A and B: $\vec{a}_B = \vec{a}_A + \vec{\alpha} \times \vec{r}_{B/A} - \omega^2 \vec{r}_{B/A}$. We have $\omega^2 = ( \frac{\sqrt{3}}{2}\omega_0 )^2 = \frac{3}{4}\omega_0^2$.

$$ \vec{a}_B = 2l\omega_0^2\hat{i} + (\alpha\hat{k})\times\left(\frac{l}{2}\hat{i} + \frac{l\sqrt{3}}{2}\hat{j}\right) - \frac{3}{4}\omega_0^2\left(\frac{l}{2}\hat{i} + \frac{l\sqrt{3}}{2}\hat{j}\right) $$ $$ \vec{a}_B = \left(2l\omega_0^2 - \frac{\alpha l\sqrt{3}}{2} - \frac{3l\omega_0^2}{8}\right)\hat{i} + \left(\frac{\alpha l}{2} - \frac{3\sqrt{3}l\omega_0^2}{8}\right)\hat{j} $$ $$ \vec{a}_B = l\left(\frac{13\omega_0^2}{8} - \frac{\alpha\sqrt{3}}{2}\right)\hat{i} + l\left(\frac{\alpha}{2} - \frac{3\sqrt{3}\omega_0^2}{8}\right)\hat{j} $$

The magnitude is given as $a_B = l\omega_0^2/2$, so $a_B^2 = l^2\omega_0^4/4$.

$$ l^2\left[\left(\frac{13\omega_0^2}{8} - \frac{\alpha\sqrt{3}}{2}\right)^2 + \left(\frac{\alpha}{2} - \frac{3\sqrt{3}\omega_0^2}{8}\right)^2\right] = \frac{l^2\omega_0^4}{4} $$

Expanding and grouping terms by powers of $\alpha$:

$$ \alpha^2 - (2\sqrt{3}\omega_0^2)\alpha + \frac{45}{16}\omega_0^4 = 0 $$

This quadratic equation for $\alpha$ yields two solutions:

$$ \alpha = \frac{2\sqrt{3}\omega_0^2 \pm \sqrt{12\omega_0^4 - 4(45/16)\omega_0^4}}{2} = \frac{2\sqrt{3}\omega_0^2 \pm \sqrt{3/4}\omega_0^2}{2} = \frac{2\sqrt{3} \pm \sqrt{3}/2}{2}\omega_0^2 $$ $$ \alpha_1 = \frac{5\sqrt{3}}{4}\omega_0^2 \quad \text{and} \quad \alpha_2 = \frac{3\sqrt{3}}{4}\omega_0^2 $$

Now we find the acceleration of C, $\vec{a}_C = \vec{a}_A + \vec{\alpha} \times \vec{r}_{C/A} - \omega^2 \vec{r}_{C/A}$.

$$ \vec{a}_C = 2l\omega_0^2\hat{i} + (\alpha\hat{k})\times(l\hat{i}) - \frac{3}{4}\omega_0^2(l\hat{i}) = \left(2l\omega_0^2 - \frac{3l\omega_0^2}{4}\right)\hat{i} + \alpha l\hat{j} $$ $$ \vec{a}_C = \frac{5l\omega_0^2}{4}\hat{i} + \alpha l\hat{j} $$

We calculate the magnitude of $\vec{a}_C$ for each value of $\alpha$.

Case 1: $\alpha = \alpha_1 = \frac{5\sqrt{3}}{4}\omega_0^2$

$$ a_{C1} = \sqrt{\left(\frac{5l\omega_0^2}{4}\right)^2 + \left(\frac{5\sqrt{3}}{4}l\omega_0^2\right)^2} = \frac{l\omega_0^2}{4}\sqrt{25 + 25(3)} = \frac{l\omega_0^2}{4}\sqrt{100} $$ $$ a_{C1} = \frac{5l\omega_0^2}{2} $$

Case 2: $\alpha = \alpha_2 = \frac{3\sqrt{3}}{4}\omega_0^2$

$$ a_{C2} = \sqrt{\left(\frac{5l\omega_0^2}{4}\right)^2 + \left(\frac{3\sqrt{3}}{4}l\omega_0^2\right)^2} = \frac{l\omega_0^2}{4}\sqrt{25 + 9(3)} = \frac{l\omega_0^2}{4}\sqrt{52} $$ $$ a_{C2} = \frac{l\omega_0^2\sqrt{13}}{2} $$