A horizontal disk has radial frictionless slot in which small block is confined to slide. The disk turns anti clockwise about its center with constant angular velocity of 3 rad/s. If the block slides along the slot with a constant speed of 0.2 m/s relative to slot, the magnitude of Coriolis acceleration in m/s², is

The correct option is 1.2.

To find the magnitude of the Coriolis acceleration, we can use the formula for Coriolis acceleration of a particle moving relative to a rotating reference frame.

The Coriolis acceleration vector ${\mathbf{a}}_c$ is defined as:
${\mathbf{a}}_c=2(\mathbf{\omega }\times {\mathbf{v}}_{\mathrm{rel}})$
where:
• $\mathbf{\omega }$ represents the angular velocity vector of the rotating frame (the disk).
• ${\mathbf{v}}_{\mathrm{rel}}$ represents the velocity vector of the block relative to the rotating frame (along the radial slot).

Since the disk is rotating horizontally in an anticlockwise direction, the angular velocity vector $\mathbf{\omega }$ points vertically upwards, perpendicular to the plane of the disk. The relative velocity vector ${\mathbf{v}}_{\mathrm{rel}}$ lies in the horizontal plane along the radial slot. Because these two vectors are perpendicular to each other, the angle between them is 90°.

Thus, the magnitude of the Coriolis acceleration is given by:
$a_c=2\omega v_{\mathrm{rel}}sin(90°)=2\omega v_{\mathrm{rel}}$

Given the values from the problem:
• Angular velocity of the disk, $\omega =3\text{rad/s}$
• Constant relative speed, $v_{\mathrm{rel}}=0.2\text{m/s}$

Substituting these values into our equation:
$a_c=2\cdot 3\cdot 0.2$
$a_c=1.2{\text{m/s}}^2$

Consequently, the magnitude of the Coriolis acceleration is 1.2 m/s².