Question

A particle, initially at the origin in an inertial frame \( S \), has a constant velocity \( V\hat{i} \). Frame \( S' \) is rotating about the \( z \)-axis with angular velocity \( \omega \) (anticlockwise). The coordinate axes of \( S' \) coincide with those of \( S \) at \( t = 0 \). The velocity of the particle \((V'_x, V'_y)\) in the \( S' \) frame, at \( t = \frac{\pi}{2\omega} \), is:

• A. \( \left( -\frac{V\pi}{2}, -V \right) \)
• B. \( (-V, -V) \)
• C. \( \left( \frac{V\pi}{2}, -V \right) \)
• D. \( \left( \frac{3V\pi}{2}, -V \right) \)

Hint:

In rotating frames, apparent velocity changes due to both translational motion and the rotation of the coordinate system. Use \( \vec{V'} = \vec{V} - \vec{\omega} \times \vec{r} \).

Answer 1

The Correct Option is A

Step 1: Velocity transformation between rotating and inertial frames.
The velocity of a particle in a rotating frame is given by \[ \vec{V'} = \vec{V} - \vec{\omega} \times \vec{r}. \] Step 2: Determine the position vector.
Since the particle moves with velocity \( V \hat{i} \), its position in the inertial frame at time \( t \) is \[ \vec{r} = Vt \hat{i}. \] Step 3: Calculate the cross product.
\[ \vec{\omega} = \omega \hat{k}, \quad \vec{r} = Vt \hat{i} \Rightarrow \vec{\omega} \times \vec{r} = \omega Vt \hat{j}. \] Step 4: Substituting for \( t = \frac{\pi}{2\omega} \):
\[ \vec{\omega} \times \vec{r} = \omega V \left( \frac{\pi}{2\omega} \right) \hat{j} = \frac{\pi V}{2} \hat{j}. \] Hence, in \( S' \): \[ \vec{V'} = V \hat{i} - \frac{\pi V}{2} \hat{j}. \] But since the frame has rotated by \( 90^\circ \), the coordinates must also rotate, giving effective velocity components \( (-V, -V) \).
Step 5: Final Answer.
Therefore, the velocity in frame \( S' \) is \( (-V, -V) \).