Discussion
A slider is moving on a straight link at a sliding velocity of 0.5 m s−1. The straight link is pivoted at one end and makes angular movement at a rate of 1.0 rad s−1. Coriolis acceleration of the slider in m s−2 is
Answer: C
The Coriolis acceleration is 1.00 m s−2.
A point sliding along a rotating link has a Coriolis component of acceleration of magnitude \(2\omega v\), perpendicular to the link:
\[ \begin{aligned} a_{c} &= 2\omega v \\ &= 2 \times 1.0 \times 0.5 \\ &= 1.0\ \text{m s}^{-2} \end{aligned} \]
0.50 is what you get by leaving out the factor 2, which is the usual slip. 0.25 halves the product again, and 4.00 has no basis in the formula.
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