+1 vote
in Machine Kinematics by (38.0k points)
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

(a) 500m/sec^2

(b) 250m/sec^2

(c) 1000m/sec^2

(d) 2000m/sec^2

I have been asked this question in exam.

My doubt stems from Method of Locating Instantaneous Centres in a Mechanism topic in section Velocity in Mechanisms of Machine Kinematics

1 Answer

0 votes
by (106k points)
Correct choice is (a) 500m/sec^2

To elaborate: Corioli’s component = 2Vω

                                 = 2 x 10 x 25 = 500500m/sec^2.

Related questions

We welcome you to Carrieradda QnA with open heart. Our small community of enthusiastic learners are very helpful and supportive. Here on this platform you can ask questions and receive answers from other members of the community. We also monitor posted questions and answers periodically to maintain the quality and integrity of the platform. Hope you will join our beautiful community
...