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A solid cylinder has a moment of inertia of 0.08 kg- m^2. A torque of 5.0...

A solid cylinder has a moment of inertia of 0.08 kg- m^2. A torque of 5.0 N-m is applied to the disk so it begins to spin on an axis going through its center( assume the disk was originally at rest). what is the angular acceleration of the disk? What is the rotational kinetic energy of the disk after 2 seconds?

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Answer #1

here,

the moment of inertia , I = 0.08 kg.m^2

torque applied , T = 5 N.m

the angular accelration , alpha = T/I

alpha = 5/0.08 rad/s^2

alpha = 62.5 rad/s^2

time taken , t = 2 s

final speed , w = 0 + alpha * t

w = 125 rad/s

the rotational kinetic energy of the disk after 2 seconds , KE = 0.5 * I * w^2

KE = 0.5 * 0.08 * 125^2 J

KE = 625 J

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