91影视

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,$\mathrm{W}=\mathrm{螖碍}={\mathrm{滨蝇}}_{\mathrm{f}}^2鈭�{\mathrm{滨蝇}}_{\mathrm{i}}^2$

Where, k is kinetic energy, w is work done 饾湐_i is initial angular velocity and饾湐_f is final angular velocity.

According to the work-energy theorem,

$\mathrm{W}=\mathrm{螖碍}={\mathrm{滨蝇}}_{\mathrm{f}}^2鈭�{\mathrm{滨蝇}}_{\mathrm{i}}^2$

In case of given situations,

localid="1663073785540" $\begin{array}{c}\mathrm{W}=\mathrm{螖碍}\\ =\mathrm{I}{\left(5\right)}^2鈭�\mathrm{I}{(鈭�2)}^2\\ =21\text{鈥�}I\text{鈥塉}\end{array}$

Therefore, the work done is localid="1663073843365" $21\text{鈥�}I\text{鈥塉}$.

According to the work-energy theorem,

$\mathrm{W}=\mathrm{螖碍}={\mathrm{滨蝇}}_{\mathrm{f}}^2鈭�{\mathrm{滨蝇}}_{\mathrm{i}}^2$

localid="1663073893164" $\begin{array}{c}\mathrm{W}=\mathrm{螖碍}\\ =\text{I}{\left(5\right)}^2鈭�\text{I}{\left(2\right)}^2\\ =21\text{鈥�}I\text{鈥塉}\end{array}$

Therefore, the work done is localid="1663073932184" $21\text{鈥�}I\text{鈥塉}$ .

According to the work-energy theorem,

$\mathrm{W}=\mathrm{螖碍}={\mathrm{滨蝇}}_{\mathrm{f}}^2鈭�{\mathrm{滨蝇}}_{\mathrm{i}}^2$

localid="1663074118667" $\begin{array}{c}\mathrm{W}=\mathrm{螖碍}\\ =\mathrm{I}{(鈭�5)}^2鈭�\mathrm{I}{(鈭�2)}^2\\ =21\text{鈥�}I\text{鈥塉}\end{array}$

Therefore, the work done is localid="1663074194722" $21\text{鈥�}I\text{鈥塉}$.

According to the work-energy theorem,

$\mathrm{W}=\mathrm{螖碍}={\mathrm{滨蝇}}_{\mathrm{f}}^2鈭�{\mathrm{滨蝇}}_{\mathrm{i}}^2$

$\begin{array}{c}\mathrm{W}=\mathrm{螖碍}\\ =\text{I}{(鈭�5)}^2鈭�\text{I}{\left(2\right)}^2\\ =21I\text{鈥塉}\end{array}$

Therefore, the work done is$21\text{鈥�}I\text{鈥塉}$.

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