Consider a mechanical clutch that consists of two heavy disks that can engage or disengage. At the beginning disk 1 with mass m1 = 12kg and diameter d1 = 60cm is at rest (f1 = 0min−1 ) and disengaged from disk 2 with mass m2 = 8kg and diameter d2 = 40cm that is rotating with a frequency of f2 = 200min−1 . When we engage the clutch both disks become connected and disk 1 is accelerated while disk 2 decelerates due to a portion of its rotational energy being used to accelerate disk 1. In full contact both disks are rotating with the same angular velocity. Calculate this final angular velocity ω and the corresponding frequency f. We are neglecting further loss of energy due to heat as a result of friction. This friction and heat is the reason why real clutches wear out over time. The moment of inertia for a solid disk can be found in the textbook.

The system is formed by the two discs, during the impact the system remains isolated, we have the forces are internal, this implies that the external torque is zero and the angular momentum is conserved

Initial Before sticking

L₀ = 0 + I₂ w₂

Final after coupling

[tex]L_{f}[/tex] = (I₁ + I₂) w

The moments of inertia of a disk with an axis of rotation in its center are

I = ½ M R²

How the moment is preserved

L₀ = [tex]L_{f}[/tex]

I₂ w₂ = (I₁ + I₂) w

w = w₂ I₂ / (I₁ + I₂)

Let's reduce the units to the SI System

d₁ = 60 cm = 0.60 m

d₂ = 40 cm = 0.40 m

f₂ = 200 min-1 (1 min / 60 s) = 3.33 Hz

Angular velocity and frequency are related.

w₂ = 2 π f₂

w₂ = 2π 3.33

w₂ = 20.94 rad / s

Let's replace

w = w₂ (½ M₂ R₂²) / (½ M₁ R₁² + ½ M₂ R₂²)

w = w₂ M₂ R₂² / (M₁ R₁² + M₂ R₂²)

Let's calculate

w = 20.94 8 0.40² / (12 0.60² + 8 0.40²)

w = 20.94 1.28 / 5.6

w = 4,786 rad / s

Angular velocity and frequency are related.

w = 2π f

f = w / 2π

f = 4.786 / 2π

f = 0.76176 Hz