A constant tangential force of 30 N is applied to the rim of a homogeneous solid disc weighing 10 kg, mounted on an axle.

A constant tangential force of 30 N is applied to the rim of a homogeneous solid disc weighing 10 kg, mounted on an axle. Determine the kinetic energy of the disk in 4 s after the start of the force.

m = 10 kg.
F = 30 N.
t = 4 s.
w0 = 0
Ek -?
The kinetic energy of a homogeneous disk Ek is determined by the formula: Ek = J * w ^ 2/2, where J is the moment of inertia of the disk, w is the angular velocity.
The moment of inertia of the disk J is determined by the formula: J = m * r ^ 2/2, where m is the mass of the disk, r is the radius of the disk.
Let us write 2 Newton’s law for rotational motion: ε = M / J, where ε is the angular acceleration, M is the moment of force.
M = F * r.
ε = (w – w0) / t.
Since w0 = 0, then ε = w / t.
w / t = 2 * F * r / m * r ^ 2.
w = 2 * F * t / m * r.
Ek = m * r ^ 2 * (2 * F * t) ^ 2/2 * (m * r) ^ 2 = 2 * F ^ 2 * t ^ 2 / m.
Ek = 2 * (30 N) ^ 2 * (4 s) ^ 2/10 kg = 2880 J.
Answer: the disk has kinetic energy Ek = 2880 J.