Answer

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**Hint**From the given value of the speed of the centre of mass, we can calculate the translational kinetic energy if the hoop by using its formula. Then, using the relation between the angular and the translational velocity, we can determine the angular velocity. Also, since the hoop is circular in shape, we can use the formula for the moment of inertia of the circle to get the MI of the hoop. Then using the formula for the rotational kinetic energy we can determine its value for the hoop. Finally, using the work-energy theorem, we can calculate the required work done.

Formula used: The formulae used for solving this question are

${K_T} = \dfrac{1}{2}m{v^2}$

${K_R} = \dfrac{1}{2}I{\omega ^2}$

\[\omega = \dfrac{v}{R}\]

**Complete step-by-step solution:**

We know that the translational kinetic of a moving body is given by

${K_T} = \dfrac{1}{2}m{v^2}$

The mass of the hoop is $m = 100kg$ and the velocity of the centre of mass is $v = 20cm/s = 0.2m/s$. Substituting these above, we get

${K_T} = \dfrac{1}{2}\left( {100} \right){\left( {0.2} \right)^2}$

$ \Rightarrow {K_T} = \dfrac{1}{2}\left( {100} \right){\left( {0.2} \right)^2}$

On solving, we get

${K_T} = 2J$.................(1)

Now, we know that the rotational kinetic energy is given by

${K_R} = \dfrac{1}{2}I{\omega ^2}$ ……….(2)

Since the hoop is circular in shape, so its moment of inertia is given by

$I = m{r^2}$

Substituting $m = 100kg$ and $r = 2m$ we get

$I = 100 \times {2^2}$

$ \Rightarrow I = 400kg{m^2}$ …….(3)

Also, the angular velocity is given by

$\omega = \dfrac{v}{r}$

Substituting $v = 0.2m/s$ and $r = 2m$ we get

$\omega = \dfrac{{0.2}}{2}$

$ \Rightarrow \omega = 0.1m/s$ ………………...(4)

Substituting (3) and (4) in (1) we get

${K_R} = \dfrac{1}{2}\left( {400} \right){\left( {0.1} \right)^2}$

$ \Rightarrow {K_R} = 2J$

So the total kinetic energy of the hoop is given by

$K = {K_R} + {K_T}$

Substituting ${K_T} = 2J$ and ${K_R} = 2J$ we get

$K = 2 + 2$

$ \Rightarrow K = 4J$

This is the initial kinetic energy of the hoop. When the hoop stops, its final kinetic energy becomes zero, that is,

$K' = 0$

Now, we know by the work energy theorem that the work done is equal to the change in kinetic energy, that is,

$W = \Delta K$

$ \Rightarrow W = K' - K$

Substituting $K' = 0$ and $K = 4J$ we get

$W = 0 - 4$

$ \Rightarrow W = - 4J$

**Hence, the magnitude of the required work done is equal to $4J$.**

**Note:**The negative sign of the work done is because the stopping force is opposite to the motion of the hoop. Do not forget to include the calculation for the rotational kinetic energy, along with translational kinetic energy of the hoop. Also, do not forget to write the velocity in the SI unit of meters per second, which is given in the CGS unit of centimeters per second.

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