1. Relative Motion Analysis The plank accelerates in the negative y-direction with acceleration \( \vec{a}_p = -a \hat{j} \). The disc is frictionless, so its center of mass remains stationary in the ground frame. In the plank’s frame, the disc center \( C \) accelerates upwards: \[ \vec{a}_{rel} = \vec{a}_{disc} – \vec{a}_{p} = 0 – (-a \hat{j}) = a \hat{j} \] Assuming the system starts from rest at the origin, the position and velocity of the center \( C \) at time \( t \) are: \[ y_c = \frac{1}{2} a t^2, \quad \vec{v}_c = at \hat{j} \]

2. Velocity of a General Point P Consider a point \( P(x, y) \) on the disc. Its total velocity in the plank frame is the sum of the center’s translation and the rotation about the center. \[ \vec{v}_P = \vec{v}_c + \vec{v}_{rot} \] Let the angular velocity be \( \vec{\omega} = \omega \hat{k} \) (Counter-Clockwise). The position of P relative to C is \( \vec{r}’ = x \hat{i} + (y – y_c) \hat{j} \). \[ \vec{v}_{rot} = \vec{\omega} \times \vec{r}’ = (\omega \hat{k}) \times (x \hat{i} + (y – y_c) \hat{j}) \] \[ \vec{v}_{rot} = \omega x \hat{j} – \omega (y – y_c) \hat{i} \] Total velocity: \[ \vec{v}_P = [-\omega (y – y_c)] \hat{i} + [at + \omega x] \hat{j} \]

3. Finding the Instantaneous Centre of Rotation (ICR) The ICR is the point where the velocity is zero (\( \vec{v}_P = 0 \)).
X-component zero: \[ -\omega (y – y_c) = 0 \implies y = y_c \] (The ICR is always at the same vertical level as the center).
Y-component zero: \[ at + \omega x = 0 \implies x = -\frac{at}{\omega} \]

4. Equation of Trajectory We eliminate time \( t \) to find the path equation. From the x-equation: \( t = -\frac{\omega x}{a} \) (Note: Since \( t>0 \), if \( \omega>0 \), then \( x \) must be negative). Substitute this into the expression for \( y_c \): \[ y = y_c = \frac{1}{2} a t^2 \] \[ y = \frac{1}{2} a \left( -\frac{\omega x}{a} \right)^2 \] \[ y = \frac{\omega^2 x^2}{2a} \]

5. Analyzing Options

• Case (a): If \( \vec{\omega} = \omega \hat{k} \) (CCW), then \( x = -at/\omega \). Since \( a, t > 0 \), \( x \) is negative (\( x \le 0 \)). The equation is \( y = \frac{\omega^2 x^2}{2a} \). This matches option (a).
• Case (b): If \( \vec{\omega} = -\omega \hat{k} \) (CW), the sign of the x-term flips: \( x = at/\omega \). So \( x \) is positive (\( x \ge 0 \)). The equation remains \( y = \frac{\omega^2 x^2}{2a} \). This matches option (b)