A constant force of 100N is applied to the rim of a homogeneous disc with a radius of R = 0.2m and a mass

A constant force of 100N is applied to the rim of a homogeneous disc with a radius of R = 0.2m and a mass of 1.2kg. When rotating, a frictional moment of force equal to 5N * m acts on the disk. What is the angular acceleration of the disc?

R = 0.2 m.

m = 1.2 kg.

F = 100 N.

Mtr = 5 N * m.

ε -?

Let us write Newton’s 2 law for rotational motion: I * ε = M – Mfr, where I is the moment of inertia of the disk, ε is the angular acceleration of the disk, M is the moment of force that rotates the disk, Mtr is the moment of friction force.

ε = (M – Mtr) / I.

We find the moment of inertia of a homogeneous disk by the formula: I = m * R ^ 2/2.

The moment of force M, which rotates the disk, we express by the formula: M = F * R.

ε = 2 * (F * R – Mtr) / m * R ^ 2.

ε = 2 * (100 N * 0.2 m – 5 N * m) / 1.2 kg * (0.2 m) ^ 2 = 625 rad / s ^ 2.

Answer: the angular acceleration of the disk is ε = 625 rad / s ^ 2.