A hoop of radius 2 m weighs 100 kg.

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

$E_{\mathrm{T}}=\frac{1}{2} m v^{2}+\frac{1}{2} I \omega^{2}$

Moment of inertia of the hoop about its centre, $I=m r^{2}$

$E_{\mathrm{T}}=\frac{1}{2} m v^{2}+\frac{1}{2}\left(m r^{2}\right) \omega^{2}$

But we have the relation, $v=r \omega$

$\therefore E_{\mathrm{T}}=\frac{1}{2} m v^{2}+\frac{1}{2} m r^{2} \omega^{2}$

$=\frac{1}{2} m v^{2}+\frac{1}{2} m v^{2}=m v^{2}$

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, $W=m v^{2}=100 \times(0.2)^{2}=4 \mathrm{~J}$