Question

If a slider moves at a velocity \( v \) on a link rotating at speed \( \omega \,\text{rad/s} \), the Coriolis component of its acceleration is

• A. \( \omega v \)
• B. \( 2\omega v \)
• C. \( 2\omega^2 v \)
• D. \( 2\omega v^2 \)

Hint:

Coriolis acceleration always acts perpendicular to the direction of relative velocity.

Answer 1

The Correct Option is B

Step 1: Consider the slider's absolute velocity as the vector sum of its velocity along the link (\( v \)) and the velocity due to the link's rotation (\( \omega r \), perpendicular to the link).
Step 2: As the link rotates through a small angle \( d\theta = \omega\,dt \), the sliding-velocity vector \( v \) (fixed in direction along the link) rotates with the link, contributing an acceleration component \( \omega v \) perpendicular to the link; simultaneously, the point's own outward motion at rate \( v \) changes the radius, contributing an equal additional component \( \omega v \) from the changing tangential velocity \( \omega r \).
Step 3: Adding these two equal contributions together:
\[ a_{cor} = \omega v + \omega v = 2\omega v \]

Answer 2

A quick sanity check uses specific numbers. Suppose the link rotates at \( \omega = 2\,\text{rad/s} \) and the slider moves along it at \( v = 3\,\text{m/s} \). The well-established mechanism-analysis result for this configuration gives a Coriolis component of \( 12\,\text{m/s}^2 \) for these values. Testing each option:

1. \( \omega v \): Gives \( 2 \times 3 = 6\,\text{m/s}^2 \), only half of the expected \( 12\,\text{m/s}^2 \), confirming this option is missing a factor of 2.
2. \( 2\omega v \): Gives \( 2 \times 2 \times 3 = 12\,\text{m/s}^2 \), matching the expected result exactly.
3. \( 2\omega^2 v \): Gives \( 2 \times 4 \times 3 = 24\,\text{m/s}^2 \), double the expected value, confirming the extra factor of \( \omega \) is not physically present.
4. \( 2\omega v^2 \): Gives \( 2 \times 2 \times 9 = 36\,\text{m/s}^2 \), triple the expected value, confirming the extra factor of \( v \) is not physically present.

The numeric check confirms that only \( 2\omega v \) reproduces the standard Coriolis acceleration result.

Therefore, the correct answer is \( 2\omega v \).