A hoop of radius 2m weighs 100kg. It rolls along a horizontal floor so that its centre of mass has a speed of 20cm/s. How much work has to be done to stop it?
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Radius of the hoop, $r=2 \mathrm{~m}$ Mass of the hoop, $\mathrm{m}=100 \mathrm{~kg}$ Velocity of the hoop, $\mathrm{v}=20 \mathrm{~cm} / \mathrm{s}=0.2 \mathrm{~m} / \mathrm{s}$ Total energy of the hoop $=$ Translational KE + Rotational KE $\mathrm{E}_{\mathrm{r}}=\frac{1}{2} \mathrm{mv}^2+\frac{1}{2} \mathrm{I} \omega^2$ Moment of inertia of the hoop about its centre, I $=m r^2 E_{\mathrm{r}}=\frac{1}{2} \mathrm{mv}^2+\frac{1}{2} \mathrm{mr}^2 \omega^2$ But we have the relation, $\mathrm{v}=\mathrm{r} \omega \therefore \mathrm{E}_{\mathrm{r}}=\frac{1}{2} \mathrm{mv}^2+\frac{1}{2} \mathrm{mr}^2 \omega^2$
$\therefore \mathrm{E}_{\mathrm{r}}=\frac{1}{2} \mathrm{mv}^2+\frac{1}{2} \mathrm{mv}^2 \therefore \mathrm{E}_{\mathrm{r}}=\mathrm{mv}^2$ 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, $W=m v^2=100 \times(0.2)^2=4 \mathrm{~J}$
$\therefore \mathrm{E}_{\mathrm{r}}=\frac{1}{2} \mathrm{mv}^2+\frac{1}{2} \mathrm{mv}^2 \therefore \mathrm{E}_{\mathrm{r}}=\mathrm{mv}^2$ 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, $W=m v^2=100 \times(0.2)^2=4 \mathrm{~J}$
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