a)
If the length is 2 meters and the rod rotates about its center, so 2 meters is the diameter, and the radius of rotation is 1 meter.
Then, if the force is 5 N, let's calculate the torque:
[tex]\begin{gathered} \tau=F\cdot r \\ \tau=5\cdot1=5\text{ Nm} \end{gathered}[/tex]
Then, calculating the rotational inertia and the angular acceleration, we have:
[tex]\begin{gathered} I=\frac{1}{2}mr^2 \\ I=\frac{1}{2}\cdot1\cdot1^2=0.5 \\ \\ \alpha=\frac{\tau}{I}=\frac{5}{0.5}=10 \end{gathered}[/tex]
The angular acceleration is 10 rad/s², so to reach an angular velocity of 30 rad/s, it takes 3 seconds.
b)
The rotational energy can be calculated with the formula below:
[tex]\begin{gathered} E_k=\frac{1}{2}I\cdot\omega^2 \\ E_k=\frac{1}{2}\cdot0.5\cdot30^2 \\ E_k=225\text{ J} \end{gathered}[/tex]
c)
The angular momentum is given by:
[tex]\begin{gathered} L=I\cdot\omega \\ L=0.5\cdot30 \\ L=15 \end{gathered}[/tex]
Then, since the rod turns into a sphere, the new rotational inertia is:
[tex]\begin{gathered} I=\frac{2}{5}mr^2 \\ I=\frac{2}{5}\cdot1\cdot1^2 \\ I=\frac{2}{5}=0.4 \end{gathered}[/tex]
So the new angular velocity is:
[tex]\begin{gathered} L=I\cdot\omega \\ 15=0.4\cdot\omega \\ \omega=\frac{15}{0.4}=37.5\text{ rad/s} \end{gathered}[/tex]
