Question

A horizontal disk has a radial frictionless slot in which a small block is confined to slide. The disk turns anticlockwise about its centre with a 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 the slot, then the magnitude of Coriolis acceleration in \( \text{m/s}^2 \) is

• A. \( 1.2 \)
• B. \( 0.6 \)
• C. \( 2.4 \)
• D. \( 0.3 \)

Hint:

Use the Coriolis acceleration formula, twice the angular velocity times the relative sliding speed.

Answer 1

The Correct Option is A

Coriolis acceleration shows up whenever something moves relative to a rotating frame, and it can be derived instead of just recalled from a formula.

Write the block's position along the slot as $r$, measured from the disk centre, with the block moving outward at relative speed $v_r = dr/dt = 0.2$ m/s. In the fixed, non-rotating frame, the block's velocity has a radial part $v_r$ and a tangential part $\omega r$ coming from the disk's own spin.

Differentiating the tangential part $\omega r$ with respect to time picks up two pieces: one from $r$ changing, giving $\omega \, dr/dt = \omega v_r$, and one from the tangential direction itself turning at rate $\omega$, giving another $\omega v_r$ term. Adding these two equal pieces gives a total of $2\omega v_r$ in the tangential direction, which is the Coriolis acceleration.

Putting in the numbers, $\omega = 3$ rad/s and $v_r = 0.2$ m/s, gives $a_c = 2(3)(0.2) = 1.2 \text{ m/s}^2$, confirming option (A) and ruling out the other three values, which do not match either $\omega v_r$ alone (0.6) or any other combination of the given numbers.