91Ó°ÊÓ

1. The rotational inertia of electric motor is,${\text{I}}_{\text{m}}=2.0×{10}^{−3}\text{kg}.{\text{m}}^2$
2. The rotational inertia of probe is,${\text{I}}_{\text{p}}=12\text{kg}.{\text{m}}^2.$
3. The angle through which the probe is rotated is,${\text{θ}}_{\mathrm{p}}=30°$

Using the conservation law of the angular momentum we can find the angle through which the motor is rotated. As in one rotation the motor is rotated through $\mathbf{360}$°, we can find the number of revolutions for the angle through which the motor is rotated.

Formula:

The law of conservation of angular momentum,${\mathbf{L}}_{\mathbf{i}}\mathbf{}\mathbf{=}{\mathbf{L}}_{\mathbf{f}}$

$\mathbf{}\mathbf{\pm õθ}\mathbf{=}\mathbf{∫}\mathbf{\pm õÓ\neg }$

The law of conservation of angular momentum gives

$\begin{array}{c}{\text{L}}_{\text{i}}={\text{L}}_{\text{f}}\\ {\text{I}}_{\text{m}}{\text{Ӭ}}_{\text{m}}={\text{I}}_{\text{p}}{\text{Ӭ}}_{\text{p}}\\ {\text{I}}_{\text{m}}{\text{θ}}_{\text{m}}={\text{I}}_{\text{p}}{\text{θ}}_{\mathrm{p}}\end{array}$

$(\mathrm{As}{\mathrm{I}}_{\mathrm{m}}{\mathrm{θ}}_{\mathrm{m}}=∫{\text{I}}_{\text{m}}{\text{Ӭ}}_{\text{m}}\mathrm{dt}\mathrm{and}{\text{I}}_{\text{p}}{\text{θ}}_{\mathrm{p}}=∫{\text{I}}_{\text{p}}{\text{Ӭ}}_{\text{p}}\mathrm{dt})$

${\text{θ}}_{\text{m}}=\frac{{\text{I}}_{\text{p}}{\text{θ}}_{\mathrm{p}}}{{\text{I}}_{\text{m}}}$

$\begin{array}{l}{\text{θ}}_{\text{m}}=\frac{12×30}{2×{10}^{−3}}\\ {\text{θ}}_{\text{m}}=180000°\end{array}$

So, the no. of revolutions of the rotor is

$\begin{array}{l}\mathrm{N}={\text{θ}}_{\text{m}}/(360°/\mathrm{rev})\\ \mathrm{N}=180000°/(360°/\mathrm{rev})\\ \mathrm{N}=500\mathrm{rev}\end{array}$

Therefore, the no. of revolutions of the rotor is$5.0×{10}^2$.