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?
Exercise
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Radius of the hoop, r = 2m Mass of the hoop, m = 100kg Velocity of the hoop, v = 20cm/s = 0.2m/s Total energy of the hoop = Translational KE + Rotational KE $\text{E}_{\text{r}}=\frac{1}{2}\text{mv}^{2}+\frac{1}{2}\text{I}\omega^2$ Moment of inertia of the hoop about its centre, $I = mr^2 \text{E}_{\text{r}}=\frac{1}{2}\text{mv}^{2}+\frac{1}{2}\text{mr}^2\omega^2$ But we have the relation, $\text{v}=\text{r}\omega$
$\therefore\ \text{E}_{\text{r}}=\frac{1}{2}\text{mv}^{2}+\frac{1}{2}\text{mr}^2\omega^2$
$\therefore\ \text{E}_{\text{r}}=\frac{1}{2}\text{mv}^{2}+\frac{1}{2}\text{mv}^2$
$\therefore\ \text{E}_{\text{r}}=\text{mv}^2$ 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, $W = mv^2 = 100 \times (0.2)^2 = 4J$
$\therefore\ \text{E}_{\text{r}}=\frac{1}{2}\text{mv}^{2}+\frac{1}{2}\text{mr}^2\omega^2$
$\therefore\ \text{E}_{\text{r}}=\frac{1}{2}\text{mv}^{2}+\frac{1}{2}\text{mv}^2$
$\therefore\ \text{E}_{\text{r}}=\text{mv}^2$ 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, $W = mv^2 = 100 \times (0.2)^2 = 4J$
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