Question

A disk with a rotational inertia of 5.0 kg. m2 and a radius of 0.25 m rotates on a frictionless fixed axis perpendicu-
lar to the disk and through its center. A force of 8.0 N is applied tangentially to the rim. If the disk starts at rest,
then after it has turned through half a revolution its angular velocity is :
(1) 0.57 rad/s (2) 0.64 rad/s
(3) 1.6 rad/s (4) 3.2 rad/s

Answer 1

Answer:

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Answer 2

Answer:

           c)1.6rad/s

Explanation:

In the disk rational force and orbit motion using torque formula,

Torque = F*r = I*α

F*r = I*α

8.0*0.25 = 5.0*α

α = 0.4 rad/s^2

ω^2 = ωo^2 + 2θα

ω^2 = 0 + 2π(0.4) --------(1 revolution = 2π, half revolution = π)

ω^2 = 2.5133

ω = 1.6 rad/s