A particle moves in a circular path of radius R with constant angular velocity ω. If the particle's position vector makes an angle θ with the positive x-axis at time t=0, what is the magnitude of its acceleration vector at time t?

When a particle moves in a circle with constant angular velocity $\omega$, its speed is constant but its direction changes continuously. This change in direction causes acceleration called centripetal acceleration, which always points towards the center of the circle.

The magnitude of the centripetal acceleration is given by:

$a = R \omega^2$

where $R$ is the radius of the circle and $\omega$ is the angular velocity.

The initial angle $\theta$ at time $t=0$ does not affect the magnitude of acceleration because acceleration depends only on the radius and angular velocity, not on the position on the circle.

So, at any time $t$, the magnitude of acceleration remains constant and equals $R \omega^2$.

Imagine you are running around a circular track at a steady speed. Even though you run at the same speed, you keep changing direction. This change in direction means you are accelerating, even if your speed doesn't change. The question asks: if you start at some angle on the circle and keep moving with the same angular speed, what is the size (magnitude) of your acceleration at any time? To find this, we need to understand how acceleration works when moving in a circle.

Given:

• Radius of circle, $R$
• Angular velocity, $\omega$
• Initial angle $\theta$ at $t=0$ (does not affect magnitude)

Step 1: Understand that acceleration in uniform circular motion is centripetal acceleration.

Step 2: Formula for centripetal acceleration magnitude:

$a = R \omega^2$

Step 3: Since $\theta$ does not affect magnitude, acceleration magnitude at any time $t$ is:

$\boxed{a = R \omega^2}$

This is the final answer.