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A hoop of radius $2 \;m$ weighs $100\; kg$. It rolls along a horizontal floor so that its centre of mass has a speed of $20\; cm/s$. How much work has to be done to stop it?
Solution
Radius of the hoop, $r=2 m$
Mass of the hoop, $m=100 kg$
Velocity of the hoop, $v=20 cm / s =0.2 m / s$
Total energy of the hoop $=$ Translational $KE +$ Rotational $KE$
$E_{ T }=\frac{1}{2} m v^{2}+\frac{1}{2} I \omega^{2}$
Moment of inertia of the hoop about its centre, $I=m r^{2}$
$E_{ T }=\frac{1}{2} m v^{2}+\frac{1}{2}\left(m r^{2}\right) \omega^{2}$
But we have the relation, $v=r \omega$
$\therefore E_{ r }=\frac{1}{2} m v^{2}+\frac{1}{2} m r^{2} \omega^{2}$
$=\frac{1}{2} m v^{2}+\frac{1}{2} m v^{2}=m v^{2}$
The work required to be done for stopping the hoop is equal to the total energy of the hoop.
$\therefore$ Required work to be done, $W=m v^{2}=100 \times(0.2)^{2}=4 J$
