In a rotary engine the angular velocity of the cylinder center line is 25 rad/sec and the relative velocity of a point on the cylinder center line w.r.t. cylinder is 10 m/sec. Corioli’s acceleration will be

In a rotary engine the angular velocity of the cylinder center line is 25 rad/sec and the relative velocity of a point on the cylinder center line w.r.t. cylinder is 10 m/sec. Corioli’s acceleration will be Correct Answer 500m/sec2

Corioli’s component = 2Vω = 2 x 10 x 25 = 500500m/sec2.
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