Solution
We need to find the work done by the external agency to keep the angular velocity $\omega$ constant while the sleeve moves radially outward. This work equals the change in kinetic energy of the sleeve plus the work done against any conservative potentials (none here).
1. Velocity Components:
The sleeve has two velocity components:
Using the work-energy theorem in the rotating frame: $$ \int_{r_1}^{r_2} (m \omega^2 r) dr = \frac{1}{2} m v_r^2 – 0 $$ $$ m \omega^2 \left[ \frac{r^2}{2} \right]_{r_1}^{r_2} = \frac{1}{2} m v_r^2 $$ $$ m \omega^2 (r_2^2 – r_1^2) = m v_r^2 $$ $$ v_r^2 = \omega^2 (r_2^2 – r_1^2) $$
The sleeve has two velocity components:
- Tangential velocity: $v_{\theta} = r \omega$
- Radial velocity: $v_r = \frac{dr}{dt}$
Using the work-energy theorem in the rotating frame: $$ \int_{r_1}^{r_2} (m \omega^2 r) dr = \frac{1}{2} m v_r^2 – 0 $$ $$ m \omega^2 \left[ \frac{r^2}{2} \right]_{r_1}^{r_2} = \frac{1}{2} m v_r^2 $$ $$ m \omega^2 (r_2^2 – r_1^2) = m v_r^2 $$ $$ v_r^2 = \omega^2 (r_2^2 – r_1^2) $$
2. Total Kinetic Energy:
The total kinetic energy in the lab frame is the sum of radial and tangential kinetic energies. $$ K = \frac{1}{2} m (v_r^2 + v_{\theta}^2) $$ At position $r_2$: $$ K_f = \frac{1}{2} m \left[ \omega^2 (r_2^2 – r_1^2) + (r_2 \omega)^2 \right] $$ $$ K_f = \frac{1}{2} m \omega^2 (2r_2^2 – r_1^2) $$ At position $r_1$ (initially at rest relative to rod): $$ K_i = \frac{1}{2} m (0 + (r_1 \omega)^2) = \frac{1}{2} m \omega^2 r_1^2 $$
The total kinetic energy in the lab frame is the sum of radial and tangential kinetic energies. $$ K = \frac{1}{2} m (v_r^2 + v_{\theta}^2) $$ At position $r_2$: $$ K_f = \frac{1}{2} m \left[ \omega^2 (r_2^2 – r_1^2) + (r_2 \omega)^2 \right] $$ $$ K_f = \frac{1}{2} m \omega^2 (2r_2^2 – r_1^2) $$ At position $r_1$ (initially at rest relative to rod): $$ K_i = \frac{1}{2} m (0 + (r_1 \omega)^2) = \frac{1}{2} m \omega^2 r_1^2 $$
3. Work Done by External Agency:
According to the Work-Energy Theorem, the work done by the external agency ($W_{ext}$) is equal to the change in kinetic energy of the system. $$ W_{ext} = K_f – K_i $$ $$ W_{ext} = \frac{1}{2} m \omega^2 (2r_2^2 – r_1^2) – \frac{1}{2} m \omega^2 r_1^2 $$ $$ W_{ext} = \frac{1}{2} m \omega^2 (2r_2^2 – 2r_1^2) $$ $$ W_{ext} = m \omega^2 (r_2^2 – r_1^2) $$ The work done by the external agency is **$m \omega^2 (r_2^2 – r_1^2)$**.
According to the Work-Energy Theorem, the work done by the external agency ($W_{ext}$) is equal to the change in kinetic energy of the system. $$ W_{ext} = K_f – K_i $$ $$ W_{ext} = \frac{1}{2} m \omega^2 (2r_2^2 – r_1^2) – \frac{1}{2} m \omega^2 r_1^2 $$ $$ W_{ext} = \frac{1}{2} m \omega^2 (2r_2^2 – 2r_1^2) $$ $$ W_{ext} = m \omega^2 (r_2^2 – r_1^2) $$ The work done by the external agency is **$m \omega^2 (r_2^2 – r_1^2)$**.