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Q50P

Expert-verifiedFound in: Page 324

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**The rotor of an electric motor has rotational inertia ${{\mathbf{I}}}_{{\mathbf{m}}}{\mathbf{=}}{\mathbf{}}{\mathbf{2}}{\mathbf{.0}}{\mathbf{}}{\mathbf{\times}}{\mathbf{}}{{\mathbf{10}}}^{\mathbf{-}\mathbf{3}}{\mathbf{}}{\mathbf{kg}}{\mathbf{}}{{\mathbf{m}}}^{{\mathbf{2}}}{\mathbf{}}$about its central axis. The motor is used to change the orientation of the space probe in which it is mounted. The motor axis is mounted along the central axis of the probe; the probe has rotational inertia role="math" localid="1660985808865" ${{\mathbf{I}}}_{{\mathbf{p}}}{\mathbf{=}}{\mathbf{}}{\mathbf{12}}{\mathbf{}}{\mathbf{kg}}{\mathbf{.}}{\mathbf{}}{{\mathbf{m}}}^{{\mathbf{2}}}$**** about this axis. Calculate the number of revolutions of the rotor required to turn the probe through ${\mathbf{30}}{\mathbf{\xb0}}$**** about its central axis. **

The number of revolutions of the rotor required to turn the probe through $30\xb0$ about its central axis are $5.0\times {10}^{2}$.

- The rotational inertia of electric motor is, ${\text{I}}_{\text{m}}=2.0\times {10}^{-3}\text{kg}.{\text{m}}^{2}$
- The rotational inertia of probe is,${\text{I}}_{\text{p}}=12\text{kg}.{\text{m}}^{2}.$
- The angle through which the probe is rotated is,${\text{\theta}}_{\mathrm{p}}=30\xb0$

**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,** ${{\mathbf{L}}}_{{\mathbf{i}}}{\mathbf{}}{\mathbf{=}}{{\mathbf{L}}}_{{\mathbf{f}}}$

${\mathbf{}}{\mathbf{I\theta}}{\mathbf{=}}{\mathbf{\int}}{\mathbf{I\omega}}$

The law of conservation of angular momentum gives

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

$(\mathrm{As}{\mathrm{I}}_{\mathrm{m}}{\mathrm{\theta}}_{\mathrm{m}}=\int {\text{I}}_{\text{m}}{\text{\omega}}_{\text{m}}\mathrm{dt}\mathrm{and}{\text{I}}_{\text{p}}{\text{\theta}}_{\mathrm{p}}=\int {\text{I}}_{\text{p}}{\text{\omega}}_{\text{p}}\mathrm{dt})$

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

$\begin{array}{l}{\text{\theta}}_{\text{m}}=\frac{12\times 30}{2\times {10}^{-3}}\\ {\text{\theta}}_{\text{m}}=180000\xb0\end{array}$

So, the no. of revolutions of the rotor is

$\begin{array}{l}\mathrm{N}={\text{\theta}}_{\text{m}}/(360\xb0/\mathrm{rev})\\ \mathrm{N}=180000\xb0/(360\xb0/\mathrm{rev})\\ \mathrm{N}=500\mathrm{rev}\end{array}$

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

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