Question

A disk with moment of inertia I1 rotates about a frictionless, vertical axle with angular speed$\mathit{\omega }$. A second disk, this one having moment of inertia I₂ and initially not rotating, drops onto the first disk (Fig. ). Because of friction between the surfaces, the two eventually reach the same angular speed ${\mathit{\omega }}_{\mathbf{f}}$.

(a) Calculate${\mathit{\omega }}_{\mathbf{f}}$.

(b) Calculate the ratio of the final to the initial rotational energy.

Short answer

a). The answer is, $\frac{I_1}{(I_1+I_2)}{\omega }_1$.

b). The ratio of the final to the initial rotational energy $\frac{K{E}_f}{K{E}_i}=\frac{I_1}{\left(I_1+I_2\right)}$

Step-by-step solution

Law of Conservation of Angular Momentum.

The Law of Conservation of Angular Momentum is states that the Total Initial angular momentum is equal to the total Final angular momentum

Total Initial angular momentum = Total Final angular momentum

Calculation(a) 

Let,Moment of Inertia of Disk one $I_1$.

Moment of Inertia of Disk two .$I_2$

Initial Angular speed of disk one ${\omega }_1$.

Initial angular speed of disk two iszero.

Using law of conservation of Angular momentum,

$$\begin{gathered} I_1{\omega }_1+I_2\times 0=(I_1+I_2)\omega \\ \omega =\frac{I_1}{(I_1+I_2)}{\omega }_1 \end{gathered}$$

The ratio of the final to the initial rotational energy.

Initial Rotational Energy,

$$\begin{gathered} K{E}_i=\frac{1}{2}{I}_1{\omega }_1^2+\frac{1}{2}{I}_2(0{)}^2 \\ K{E}_i=\frac{1}{2}{I}_1{\omega }_1^2 \end{gathered}$$

Final Rotational Energy,

$$\begin{gathered} K{E}_f=\frac{1}{2}(I_1+I_2){\omega }^2 \\ K{E}_f=\frac{1}{2}(I_1+I_2)\{\frac{I_1}{(I_1+I_2)}{\omega }_1{\}}^2 \\ K{E}_f=\frac{1}{2}.\frac{I_1^2}{(I_1+I_2)}{\omega }_1^2 \\ K{E}_f=\frac{\frac{1}{2}.\frac{I_1^2}{(I_1+I_2)}{\omega }_1^2}{\frac{1}{2}{I}_1{\omega }_1^2} \end{gathered}$$

$\frac{K{E}_f}{K{E}_i}=\frac{I_1}{\left(I_1+I_2\right)}$