Question

a circular disc rotates on a thin air film with a period of 0.3second. its moment of inertia about its axis of rotation is 0.06kg/m*2.a small mass is dropped onto the disc and rotates with it. the moment of inertia of the mass about axis of rotation is 0.04kg/m*2.determine final period of rotating disc and mass?

Answer 1

Answer on Question #39930 – Physics – Mechanics | Kinamatics | Dynamics

1. A circular disc rotates on a thin air film with a period of 0.3 second. Its moment of inertia about its axis of rotation is $0.06\,\mathrm{kg/m^2}$. A small mass is dropped onto the disc and rotates with it. The moment of inertia of the mass about axis of rotation is $0.04\,\mathrm{kg/m^2}$. Determine final period of rotating disc and mass.

The momentum of the conservative system keeps the constant. So, $L = \frac{2\pi I}{T} = (I + I_1)\omega_1$ and new angular velocity becomes $\omega_1 = \frac{2\pi I}{T(I + I_1)}$.

Thus, new period equals to $T_1 = \frac{2\pi}{\omega_1}$, $\boxed{T_1 = \left(1 + \frac{I_1}{I}\right)T}$.

The moment of inertia of the mass about axis of rotation is $I_1 = m r^2$, where $r$ is distance from the mass to the axis. So, $\boxed{m = I_1 / r^2}$. As there is no info about this distance, we cannot evaluate it and this quantity remains in symbolic form.

Let check the dimensions.

$$\boxed{T_1} = s, \quad \boxed{m} = \frac{kg \cdot m^2}{m^2} = kg.$$

Let evaluate the quantities.

$$T_1 = \left(1 + \frac{0.04}{0.06}\right) \cdot 0.3 = 0.5(s).$$

Answer: $0.5\,s$, $m = \frac{I_1}{r^2}$.