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The angular velocity of the pinion is 100 rpm and that of the gear is 30 rpm. The path of recess is equal to 50 mm and the path of contact = 100 mm. Find the velocity of sliding at the beginning of the contact.

The angular velocity of the pinion is 100 rpm and that of the gear is 30 rpm. The path of recess is equal to 50 mm and the path of contact = 100 mm. Find the velocity of sliding at the beginning of the contact. Correct Answer 680.58 mm/s

Np = 100 rpm. Therefore, ωp= 10.47 rad/s Ng = 30 rpm. Therefore, ωg= 3.14 rad/s Path of approach = Path of contact – Path of recess = 100 mm – 50 mm = 50 mm. Velocity of sliding = (ωp + ωg) x Path of approach = (10.47 + 3.14) x 50 = 680.58 mm/s.

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