A horizontal disc is rotating about a transverse axis through its centre at 100 rpm. A 20 gram blob of wax falls on the disc and sticks to it at 5 ~cm from its axis. The moment of inertia of the disc about its axis passing through its centre is 2 10^ -4 ~kg m^2. Calculate the new frequency of rotation of the disc.

Data : f_1=100 rpm, m=20 ~g=20 10^ -3 ~kg, r=5 ~cm=5 10^ -2 ~m, l_1=I_ disc =2 10^ -4 kg m^2 The Ml of the disc and blob of wax is l_2=l_1 & +m r^2 & = (2 10^ -4 )+ (20 10^ -3 ) (5 10^ -2 )^2 & = (2 10^ -4 )+ (20 10^ -3 ) (25 10^ -4 ) & =(2+0.5) 10^ -4 =2.5 10^ -4 ~kg m^2 By the principle of conservation of angular & momentum, I_1 _1=I_2 _2 & I_1 (2 f_1 )=I_2 (2 f_2 ) & f_2=I_1 f_1 I_2 = (2 10^ -4 )(100) 2.5 10^ -4 =200 5 / 2 =80 rpm This is the new frequency of rotation.

A horizontal disc is rotating about a transverse axis through its…

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A horizontal disc is rotating about a transverse axis through its centre at $100 \mathrm{rpm}$. A $20 \mathrm{gram}$ blob of wax falls on the disc and sticks to it at $5 \mathrm{~cm}$ from its axis. The moment of inertia of the disc about its axis passing through its centre is $2 \times 10^{-4} \mathrm{~kg} \cdot \mathrm{m}^2$. Calculate the new frequency of rotation of the disc.

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Answer

$
\text { Data }: \mathrm{f}_1=100 \mathrm{rpm}, \mathrm{m}=20 \mathrm{~g}=20 \times 10^{-3} \mathrm{~kg}, \mathrm{r}=5 \mathrm{~cm}=5 \times 10^{-2} \mathrm{~m}, \mathrm{l}_1=\mathrm{I}_{\text {disc }}=2 \times 10^{-4}
$
$\mathrm{kg} \cdot \mathrm{m}^2$
The $\mathrm{Ml}$ of the disc and blob of wax is
$
\begin{aligned}
l_2=l_1 & +m r^2 \\
& =\left(2 \times 10^{-4}\right)+\left(20 \times 10^{-3}\right)\left(5 \times 10^{-2}\right)^2 \\
& =\left(2 \times 10^{-4}\right)+\left(20 \times 10^{-3}\right)\left(25 \times 10^{-4}\right) \\
& =(2+0.5) \times 10^{-4}=2.5 \times 10^{-4} \mathrm{~kg} \cdot \mathrm{m}^2
\end{aligned}
$
By the principle of conservation of angular
$
\begin{aligned}
& \text { momentum, } I_1 \omega_1=I_2 \omega_2 \\
& \therefore I_1\left(2 \pi f_1\right)=I_2\left(2 \pi f_2\right) \\
& \therefore f_2=\frac{I_1 f_1}{I_2}=\frac{\left(2 \times 10^{-4}\right)(100)}{2.5 \times 10^{-4}}=\frac{200}{5 / 2}=80 \mathrm{rpm}
\end{aligned}
$
This is the new frequency of rotation.

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