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: Understanding Coriolis acceleration.
Coriolis acceleration occurs when a particle moves relative to a rotating body.
Step 2: Formula for Coriolis acceleration.
The Coriolis component of acceleration is given by: \[ a_c = 2\omega v \] where \( \omega \) is angular velocity and \( v \) is relative velocity.
Step 3: Applying the given values.
Since the slider has velocity \( v \) on a link rotating at angular speed \( \omega \), the Coriolis acceleration becomes: \[ a_c = 2\omega v \]
Step 4: Conclusion.
The Coriolis component of acceleration is \( 2\omega v \).

Answer 2

The Coriolis component of acceleration arises specifically from the combined effect of a slider's velocity relative to a link and that link's rotation, and dimensionally must have units of acceleration (length/time squared), matching \( \omega \) (rad/s) multiplied by \( v \) (length/time). Checking the options against the known formula and its origin:

1. \( \omega v \): This is dimensionally an acceleration, but it is missing the factor of 2 that arises because the Coriolis term comes from two equal contributions in the acceleration expansion: one from the change in direction of the sliding velocity due to rotation, and one from the changing radius affecting the tangential velocity. Omitting the factor of 2 undercounts this combined effect.
2. \( 2\omega v \): Deriving the acceleration of a point sliding along a rotating link using vector differentiation of position in a rotating frame produces a term \( 2\vec{\omega}\times\vec{v} \), whose magnitude is exactly \( 2\omega v \) since \( \vec{\omega} \) and \( \vec{v} \) are perpendicular here. This is the standard, correctly derived Coriolis component.
3. \( 2\omega^2 v \): This has an extra factor of \( \omega \) compared to the correct expression; the extra \( \omega \) does not arise from the correct vector derivation of this term.
4. \( 2\omega v^2 \): This has an extra factor of \( v \), which is not consistent with the Coriolis term derived from \( 2\vec{\omega}\times\vec{v} \), since that cross product involves each of \( \omega \) and \( v \) exactly once.

Only \( 2\omega v \) has the correct form and coefficient consistent with the vector derivation \( 2\vec{\omega}\times\vec{v} \) of Coriolis acceleration.

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