A force is applied to the rim of a disk that can rotate like a merry-go-round, so as to change its angular velocity. Its initial and final angular velocities, respectively, for four situations are: (a) $\mathbf{\text{-2rad/s,\hspace{0.17em}5rad/s}}$ ; (b)$\mathbf{\text{2rad/s,\hspace{0.17em}5rad/s}}$ ; (c)$\mathbf{\text{-2rad/s,\hspace{0.17em}-5rad/s}}$ ; and (d)$\mathbf{\text{2rad/s,\hspace{0.17em}-5rad/s}}$. Rank the situations according to the work done by the torque due to the force, greatest first.

The values of initial and final angular velocity for point,

1. (-2 rad/sec, 5 rad/sec)
2. (2 rad/sec, 5 rad/sec)
3. (-2 rad/sec, -5 rad/sec)
4. (2 rad/sec, -5 rad/sec)

The work-energy theorem states that the net work done by the forces on an object equals the change in its kinetic energy.

Find the K.E of the disk in each situation from given angular velocities. Then using the work-energy theorem we can find the corresponding work done and rank the situations accordingly.

Formulae are as follows:

Work energy theorem,$\mathrm{W}=\mathrm{\Delta K}={\mathrm{I\omega }}_{\mathrm{f}}^2-{\mathrm{I\omega }}_{\mathrm{i}}^2$

Where, k is kinetic energy, w is work done 𝜔_i is initial angular velocity and𝜔_f is final angular velocity.

According to the work-energy theorem,

$\mathrm{W}=\mathrm{\Delta K}={\mathrm{I\omega }}_{\mathrm{f}}^2-{\mathrm{I\omega }}_{\mathrm{i}}^2$

In case of given situations,

localid="1663073785540" $\begin{array}{c}\mathrm{W}=\mathrm{\Delta K}\\ =\mathrm{I}{\left(5\right)}^2-\mathrm{I}{(-2)}^2\\ =21I\text{\hspace{0.17em}J}\end{array}$

Therefore, the work done is localid="1663073843365" $21I\text{\hspace{0.17em}J}$.

According to the work-energy theorem,

$\mathrm{W}=\mathrm{\Delta K}={\mathrm{I\omega }}_{\mathrm{f}}^2-{\mathrm{I\omega }}_{\mathrm{i}}^2$

localid="1663073893164" $\begin{array}{c}\mathrm{W}=\mathrm{\Delta K}\\ =\text{I}{\left(5\right)}^2-\text{I}{\left(2\right)}^2\\ =21I\text{\hspace{0.17em}J}\end{array}$

Therefore, the work done is localid="1663073932184" $21I\text{\hspace{0.17em}J}$ .

According to the work-energy theorem,

$\mathrm{W}=\mathrm{\Delta K}={\mathrm{I\omega }}_{\mathrm{f}}^2-{\mathrm{I\omega }}_{\mathrm{i}}^2$

localid="1663074118667" $\begin{array}{c}\mathrm{W}=\mathrm{\Delta K}\\ =\mathrm{I}{(-5)}^2-\mathrm{I}{(-2)}^2\\ =21I\text{\hspace{0.17em}J}\end{array}$

Therefore, the work done is localid="1663074194722" $21I\text{\hspace{0.17em}J}$.

According to the work-energy theorem,

$\mathrm{W}=\mathrm{\Delta K}={\mathrm{I\omega }}_{\mathrm{f}}^2-{\mathrm{I\omega }}_{\mathrm{i}}^2$

$\begin{array}{c}\mathrm{W}=\mathrm{\Delta K}\\ =\text{I}{(-5)}^2-\text{I}{\left(2\right)}^2\\ =21I\text{\hspace{0.17em}J}\end{array}$

Therefore, the work done is$21I\text{\hspace{0.17em}J}$.

From the answers of part (a), (b), (c), and (d), we can conclude that the work done in each situation is the same.