Question

A circular disc rotates on a thin air film with a period of 0.3 s. Its moment of inertia about its axis
of rotation is 0.06 kg m2. A small mass is dropped onto the disc and rotates with it. The moment
of inertia of the mass about the axis of rotation is 0.04 kg m2. Determine the final period of the
rotating disc and mass.

Answer 1

A circular disc rotates on a thin air film with a period of 0.3 s. Its moment of inertia about its axis of rotation is 0.06 kg m². A small mass is dropped onto the disc and rotates with it. The moment of inertia of the mass about the axis of rotation is 0.04 kg m². Determine the final period of the rotating disc and mass.

Solution.

$$T_1 = 0.3\,s, I_1 = 0.06\,kg \cdot m^2, I = 0.04\,kg \cdot m^2;$$ $$T_1 - ?$$

The angular momentum of the disc is:

$$L_1 = I_1 \omega_1;$$

$I_1$ - the moment of inertia of the disc;

$\omega_1$ - the angular velocity of the disc.

$$\omega_1 = \frac{2\pi}{T_1};$$

$T_1$ - the time period of the disc.

$$L_1 = I_1 \frac{2\pi}{T_1};$$

The angular momentum of the disc with the small mass is:

$$L_2 = I_2 \frac{2\pi}{T_2};$$ $$I_2 = I_1 + I;$$

$I_2$ - the moment of inertia of the system: the disc with the small mass.

$I$ - the moment of inertia of the small mass.

$$L_2 = (I_1 + I) \frac{2\pi}{T_2}.$$

The law of conservation of angular momentum acts in this situation, because no external torque acts on a disc then:

$$L_1 = L_2;$$ $$I_1 \frac{2\pi}{T_1} = (I_1 + I) \frac{2\pi}{T_2};$$ $$\frac{I_1}{T_1} = \frac{I_1 + I}{T_2};$$ $$T_2 = \frac{I_1 + I}{I_1} T_1;$$ $$T_2 = \left(1 + \frac{I}{I_1}\right) T_1.$$ $$T_2 = \left(1 + \frac{0.06}{0.04}\right) 0.3 = 0.75(s).$$

Answer: $T_2 = 0.75s$.