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(C) Let the radius of the circle be $R = 1$ and the mass of each particle be $m = 1$. Let the angular speed be $\omega$.
At any time $t$, the angular

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(C) Let the radius of the circle be $R = 1$ and the mass of each particle be $m = 1$. Let the angular speed be $\omega$.
At any time $t$, the angular position of the first particle is $	heta_1 = \pi/2 - \omega t$ and the second particle is $	heta_2 = \pi/2 + \omega t$.
The velocity vectors of the two particles are $\vec{v}_1 = v(-\sin	heta_1 \hat{i} + \cos	heta_1 \hat{j})$ and $\vec{v}_2 = v(\sin	heta_2 \hat{i} + \cos	heta_2 \hat{j})$, where $v = R\omega = \omega$.
The total linear momentum is $\vec{P} = m\vec{v}_1 + m\vec{v}_2 = \omega[(-\sin	heta_1 + \sin	heta_2)\hat{i} + (\cos	heta_1 + \cos	heta_2)\hat{j}]$.
Using the sum-to-product identities:
$-\sin	heta_1 + \sin	heta_2 = 2\sin(\frac{	heta_2 - 	heta_1}{2})\cos(\frac{	heta_2 + 	heta_1}{2}) = 2\sin(\omega t)\cos(\pi/2) = 0$.
$\cos	heta_1 + \cos	heta_2 = 2\cos(\frac{	heta_1 + 	heta_2}{2})\cos(\frac{	heta_1 - 	heta_2}{2}) = 2\cos(\pi/2)\cos(-\omega t) = 0$.
Wait, let's re-evaluate the angle $	heta$. The angle $	heta$ shown in the figure is the angle with the horizontal.
Let the particles be at angles $\phi_1 = \pi/2 - \omega t$ and $\phi_2 = \pi/2 + \omega t$ from the positive $x$-axis.
The velocity vectors are $\vec{v}_1 = v(-\sin\phi_1 \hat{i} + \cos\phi_1 \hat{j})$ and $\vec{v}_2 = v(\sin\phi_2 \hat{i} + \cos\phi_2 \hat{j})$.
$\vec{P} = m\vec{v}_1 + m\vec{v}_2 = v[(\sin\phi_2 - \sin\phi_1)\hat{i} + (\cos\phi_1 + \cos\phi_2)\hat{j}]$.
$\sin\phi_2 - \sin\phi_1 = 2\sin(\frac{\phi_2 - \phi_1}{2})\cos(\frac{\phi_2 + \phi_1}{2}) = 2\sin(\omega t)\cos(\pi/2) = 0$.
$\cos\phi_1 + \cos\phi_2 = 2\cos(\frac{\phi_1 + \phi_2}{2})\cos(\frac{\phi_1 - \phi_2}{2}) = 2\cos(\pi/2)\cos(-\omega t) = 0$.
Actually, the magnitude of momentum is $P = |\vec{P}| = 2mv \cos(\frac{\phi_2 - \phi_1}{2}) = 2mv \cos(\omega t)$.
Since $	heta = \pi/2 - \omega t$, we have $\omega t = \pi/2 - 	heta$.
Thus, $P = 2mv \cos(\pi/2 - 	heta) = 2mv \sin	heta$.
As $	heta$ goes from $\pi/2$ to $-\pi/2$, $\sin	heta$ goes from $1$ to $-1$. The magnitude $P = |2mv \sin	heta|$ goes from $2mv$ to $0$ and back to $2mv$.

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