39. A disk with a rotational inertia of 5.0 kg m2 and aradius of 0.25 m rotates

Question

39. A disk with a rotational inertia of 5.0 kg m2 and aradius of 0.25 m rotates on a frictionless fixed axis perpendicular to the disk and throughits center. A force of 2.0 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: 39. A disk with a rotational inertia of 5.0 kg m2 and aradius of 0.25 m rotates on a frictionless fixed axis perpendicular to the disk and throughits center. A force of 2.0 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:

Explanation / Answer

= rxF = 2.0N*0.25m = 0.50 N.m since = I => = /I = 0.50N.m/5.0kg.m2 = 0.1rad/s2 since = (1/2) t2, when = (half rev) we have: = (1/2)*0.1 rad/s2 t2 =>t = (2/0.1 rad/s2) = 7.92 s the angule velocity = t = 0.1rad/s2 *7.92 s =0.792 rad /s the angular velocity at half rev is 0.792 rad/s. . =>t = (2/0.1 rad/s2) = 7.92 s the angule velocity = t = 0.1rad/s2 *7.92 s =0.792 rad /s the angular velocity at half rev is 0.792 rad/s. .

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