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If $2 \mathrm{~kg}$ mass is rotating on a circular path of radius $0.8 \mathrm{~m}$ with angular velocity of $44 \mathrm{rad} / \mathrm{sec}$. If radius of the path becomes $1 \mathrm{~m}$, then what will be the value of angular velocity?
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The correct answer is:
$28.16 \mathrm{rad} / \mathrm{sec}$
Mass $(m)=2 \mathrm{~kg}$; initial radius of the path
$\left(r_1\right)=0.8 \mathrm{~m}$; initial angular velocity
$\left(\omega_1\right)=44 \mathrm{rad} / \mathrm{sec}$ and final radius of the path $\left(r_2\right)=1 \mathrm{~m}$
Moment of inertia, $I_1=m r_1^2=2 \times(0.8)^2$ $=1.28 \mathrm{~kg} \mathrm{~m}^2$ and $I_2=m r_2^2=2 \times(1)^2=2 \mathrm{~kg} \mathrm{\textrm {m } ^ { 2 }}$
Therefore from the law of conservation of angular momentum $I_1 \omega_1=I_2 \omega_2$
or $\omega_2=\frac{I_1 \times \omega_1}{I_2}=\frac{1.28 \times 44}{2}$
$\omega_2=28.16 \mathrm{rad} / \mathrm{sec}$
$\left(r_1\right)=0.8 \mathrm{~m}$; initial angular velocity
$\left(\omega_1\right)=44 \mathrm{rad} / \mathrm{sec}$ and final radius of the path $\left(r_2\right)=1 \mathrm{~m}$
Moment of inertia, $I_1=m r_1^2=2 \times(0.8)^2$ $=1.28 \mathrm{~kg} \mathrm{~m}^2$ and $I_2=m r_2^2=2 \times(1)^2=2 \mathrm{~kg} \mathrm{\textrm {m } ^ { 2 }}$
Therefore from the law of conservation of angular momentum $I_1 \omega_1=I_2 \omega_2$
or $\omega_2=\frac{I_1 \times \omega_1}{I_2}=\frac{1.28 \times 44}{2}$
$\omega_2=28.16 \mathrm{rad} / \mathrm{sec}$
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