Question

The constant 40-N force is applied to the 36-kg stepped cylinder as shown. The centroidal radius of gyration of the cylinder is k = 200 mm, and it rolls on the incline without slipping. If the cylinder is at rest when the force is first ap- plied, determine its angular velocity weight seconds later.

Answer 1

The angular velocity of the stepped cylinder is determined as 2.35 rad/s.

Centripetal acceleration of the cylinder

The centripetal acceleration of the cylinder is calculated as follows;

F = ma

a = F/m

a = 40/ 36

a = 1.11 m/s²

Velocity of the cylinder

a = v²/r

v² = ar

v² = 1.11 x 0.2

v² = 0.222

v = 0.47 m/s

Angular velocity of the cylinder

v = ωr

ω = v/r

ω = 0.47/0.2

ω = 2.35 rad/s

Learn more about angular velocity here: brainly.com/question/6860269

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