1. Diagram & System Analysis

The system is a differential pulley. The pulling thread P is continuous with the winding on the upper drum. Based on the winding logic:

• Input: Thread P is pulled to the right with velocity $v$, causing the upper pulley to rotate clockwise.
• Left String: Attached to the Outer Drum ($R$). As the pulley rotates clockwise, this string winds up.
• Right String: Attached to the Inner Drum ($r$). As the pulley rotates clockwise, this string unwinds down.

2. Kinematics

Angular Velocity ($\omega$):
Since string P unwinds from the outer radius $R$ at velocity $v$:

$$ \omega = \frac{v}{R} \quad (\text{Clockwise}) $$

String Velocities:

• Left String ($v_L$): Connected to the outer radius $R$. Clockwise rotation pulls this point upward. $$ v_L = \omega R = \left( \frac{v}{R} \right) R = v \quad (\uparrow) $$
• Right String ($v_R$): Connected to the inner radius $r$. Clockwise rotation moves this point downward. $$ v_R = \omega r = \left( \frac{v}{R} \right) r = v \frac{r}{R} \quad (\downarrow) $$

Load Velocity ($v_m$):
The movable pulley velocity is the average of the string velocities. Taking upward as positive:

$$ v_m = \frac{v_L + (-v_R)}{2} $$ $$ v_m = \frac{v – v \frac{r}{R}}{2} = \frac{v}{2} \left( 1 – \frac{r}{R} \right) $$

Since $R > r$, $v_m$ is positive, meaning the load moves upward.

3. Power Calculation

The power delivered by the pulling agency goes into increasing the gravitational potential energy of the load (since kinetic energy is constant due to constant velocity).

$$ P = F \cdot v = m g \cdot v_m $$

Substituting $v_m$:

$$ P = m g \left[ \frac{v}{2} \left( 1 – \frac{r}{R} \right) \right] $$

4. Final Answer

The power delivered by the pulling agency is:

$$ P = \frac{mg v}{2} \left( 1 – \frac{r}{R} \right) $$