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Rotational Dynamics — Physics STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 SciencePhysicsRotational Dynamics4 Marks
Question
A flywheel rotating about an axis through its centre and perpendicular to its plane loses $100 \mathrm{~J}$ of energy on slowing down from $60 \mathrm{rpm}$ to $30 \mathrm{rpm}$. Find its moment of inertia about the given axis and the change in its angular momentum.
✓

Answer

Data : $\mathrm{f}_1=60 \mathrm{rpm}=60 / 60 \mathrm{rot} / \mathrm{s}=1 \mathrm{rot} / \mathrm{s}, \mathrm{f}_2=30 \mathrm{rpm}=30 / 60 \mathrm{rot} / \mathrm{s}=\frac{1}{2} \mathrm{rot} / \mathrm{s}, \Delta \mathrm{E}=-100 \mathrm{~J}$
(i) Rotational KE, $E=\frac{1}{2} I \omega^2=\frac{1}{2} I(2 \pi f)^2=2 \pi^2 I f^2$
$
\begin{aligned}
& \text { The change in KE, } \Delta E=E_2-E_1 \\
& =2 \pi^2 I\left(f_2^2-f_1^2\right) \\
& \therefore I=\frac{\Delta E}{2 \pi^2\left(f_2{ }^2-f_1{ }^2\right)} \\
& =\frac{-100}{2(3.142)^2\left[\left(\frac{1}{2}\right)^2-1^2\right]} \\
& =\frac{-100}{2(3.142)^2\left(-\frac{3}{4}\right)}=\frac{200}{3(3.142)^2} \\
& =6.753 \mathrm{~kg} \cdot \mathrm{m}^2 \\
&
\end{aligned}
$
This gives the $\mathrm{Ml}$ of the flywheel about the given axis.
(ii) Angular momentum, $L=l \omega=I(2 \pi f) 2 \pi l f$
The change in angular momentum, $\Delta \mathrm{L}$
$
\begin{aligned}
& =L_2-L_1=2 \pi \mid\left(f_2-f_1\right) \\
& =2 \times 3.142 \times 6.753\left(\frac{1}{2}-1\right) \\
& =-3.142 \times 6.753=-21.22 \mathrm{kg.m} / \mathrm{s}
\end{aligned}
$

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