A uniform horizontal disc is freely rotating about a vertical axis passing through its centre at the rate of $180 \mathrm{rpm}$. A blob of wax of mass $1.9 \mathrm{~g}$ falls on it and sticks to it at $25 \mathrm{~cm}$ from the axis. If the frequency of rotation is reduced by $60 \mathrm{rpm}$, calculate the moment of inertia of the disc.
Q 114.1
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Data : $\mathrm{f}_1=180 \mathrm{rpm}=180 / 60 \mathrm{rot} / \mathrm{s}=3 \mathrm{rot} / \mathrm{s}, \mathrm{f}_2=(180-60) \mathrm{rpm}=120 / 60 \mathrm{rot} / \mathrm{s}=2 \mathrm{rot} / \mathrm{s}, \mathrm{m}$ $=1.9 \mathrm{~g}=1.9 \times 10^{-3} \mathrm{~kg}, \mathrm{r}=25 \mathrm{~cm}=0.25 \mathrm{~m}$
Let $I_1$ be the $M l$ of the disc. Let $I_2$ be the Ml of the disc and the blob.
$
\therefore I_2=I_1+m r^2
$
According to the principle of conservation of angular momentum,
$
\begin{aligned}
& I_1 \omega_1=I_2 \omega_2 \quad \\
& \therefore I_1\left(2 \pi f_1\right)=\left(I_1+m r^2\right)\left(2 \pi f_2\right) \\
& \therefore I_1 f_1=\left(I_1+m r^2\right) f_2 \\
& \therefore I_1\left(f_1-f_2\right)=m r^2 f_2 \\
& \therefore I_1=\frac{m r^2 f_2}{f_1-f_2}=\frac{1.9 \times 10^{-3} \times(0.25)^2 \times 2}{3-2} \\
& =3.8 \times 10^{-3} \times 6.25 \times 10^{-2} \\
& =2.375 \times 10^{-4} \mathrm{~kg} \cdot \mathrm{m}^2 \\
&
\end{aligned}
$
Let $I_1$ be the $M l$ of the disc. Let $I_2$ be the Ml of the disc and the blob.
$
\therefore I_2=I_1+m r^2
$
According to the principle of conservation of angular momentum,
$
\begin{aligned}
& I_1 \omega_1=I_2 \omega_2 \quad \\
& \therefore I_1\left(2 \pi f_1\right)=\left(I_1+m r^2\right)\left(2 \pi f_2\right) \\
& \therefore I_1 f_1=\left(I_1+m r^2\right) f_2 \\
& \therefore I_1\left(f_1-f_2\right)=m r^2 f_2 \\
& \therefore I_1=\frac{m r^2 f_2}{f_1-f_2}=\frac{1.9 \times 10^{-3} \times(0.25)^2 \times 2}{3-2} \\
& =3.8 \times 10^{-3} \times 6.25 \times 10^{-2} \\
& =2.375 \times 10^{-4} \mathrm{~kg} \cdot \mathrm{m}^2 \\
&
\end{aligned}
$
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