Solution: Rotating Pulley and Thermal Expansion
Figure 2: The pulley D serves as an electrical node connected to terminal C. The wire splits into left (AD) and right (BD) sections.
Part (a): Mechanism of Rotation
The pulley rotates because of differential thermal expansion caused by unequal power dissipation in the two segments of the wire.
1. Potential Analysis:
The pulley D is electrically connected to terminal C.
- The left portion of the wire is connected between A and D (effectively A and C). The potential difference is $V_{AC} \approx V_{\text{battery}}$.
- The right portion of the wire is connected between B and D (effectively B and C). The potential difference is $V_{BC}$.
2. Unequal Heating:
Since current flows from A to C, there is a voltage drop across resistor $R$. Therefore, $V_{AC} > V_{BC}$.
Power dissipated is $P = V^2/r$. Since the left wire has a higher voltage across it, it dissipates more heat ($P_{\text{left}} > P_{\text{right}}$).
3. Dynamics:
Higher heat leads to greater thermal expansion in the left wire. The spring keeps the wire taut. To accommodate the excess length generated on the left, the pulley rotates to shift wire from the left side to the right side. The rotation stops when the system reaches a steady thermal state where heat generation equals heat dissipation.
Part (b): Derivation of Angular Displacement
We aim to show that the angular displacement $\Delta \theta$ is proportional to the power consumed by load resistance $R_L$.
1. Relation between Angle and Expansion:
Let $\delta l_1$ and $\delta l_2$ be the thermal expansions of the left and right wire segments, respectively.
For the pulley to compensate for the difference in length and keep the system symmetric/taut, the arc length rotated $s$ must satisfy:
$$ s = a \Delta \theta \propto (\delta l_1 – \delta l_2) $$
Since thermal expansion is proportional to the change in temperature ($\delta l \propto \Delta T$), and in steady state $\Delta T$ is proportional to Power ($P$), we have:
$$ \Delta \theta \propto (P_1 – P_2) $$
2. Power Difference Calculation:
Let $V$ be the battery voltage.
$$ P_1 \propto V^2 \quad \text{(Voltage across A-C)} $$
$$ P_2 \propto V_{BC}^2 \quad \text{(Voltage across B-C)} $$
Therefore, $\Delta \theta \propto (V^2 – V_{BC}^2)$.
Using difference of squares: $V^2 – V_{BC}^2 = (V – V_{BC})(V + V_{BC})$.
Note that $V – V_{BC} = V_{AB}$ (Voltage drop across resistor $R$).
So, $\Delta \theta \propto V_{AB} (V + V_{BC})$.
3. Approximations ($R_L \gg R$):
Since $R_L \gg R$, the voltage drop across $R$ is small, so $V_{BC} \approx V$.
The term $(V + V_{BC}) \approx 2V$ (a constant factor).
Thus, the rotation is largely proportional to $V_{AB}$:
$$ \Delta \theta \propto V_{AB} $$
Using Ohm’s Law for the resistor $R$: $V_{AB} = I R$.
Since $R_L \gg R$, the current $I \approx V/R_L$.
$$ V_{AB} \approx \frac{V}{R_L} R $$
4. Linking to Load Power:
The power consumed by the load $R_L$ is:
$$ P_{R_L} = \frac{V_{BC}^2}{R_L} \approx \frac{V^2}{R_L} $$
Comparing our result for rotation:
$$ \Delta \theta \propto \frac{V}{R_L} \propto \frac{V^2}{R_L} \frac{1}{V} $$
Since $V$ is constant, we see that:
$$ \Delta \theta \propto \frac{V^2}{R_L} $$
$$ \Delta \theta \propto P_{R_L} $$
Hence, the angular displacement is proportional to the power consumed by the resistance $R_L$.
Figure 2: The pulley D serves as an electrical node connected to terminal C. The wire splits into left (AD) and right (BD) sections.
The pulley rotates because of differential thermal expansion caused by unequal power dissipation in the two segments of the wire.
1. Potential Analysis:
The pulley D is electrically connected to terminal C.
- The left portion of the wire is connected between A and D (effectively A and C). The potential difference is $V_{AC} \approx V_{\text{battery}}$.
- The right portion of the wire is connected between B and D (effectively B and C). The potential difference is $V_{BC}$.
2. Unequal Heating:
Since current flows from A to C, there is a voltage drop across resistor $R$. Therefore, $V_{AC} > V_{BC}$.
Power dissipated is $P = V^2/r$. Since the left wire has a higher voltage across it, it dissipates more heat ($P_{\text{left}} > P_{\text{right}}$).
3. Dynamics:
Higher heat leads to greater thermal expansion in the left wire. The spring keeps the wire taut. To accommodate the excess length generated on the left, the pulley rotates to shift wire from the left side to the right side. The rotation stops when the system reaches a steady thermal state where heat generation equals heat dissipation.
We aim to show that the angular displacement $\Delta \theta$ is proportional to the power consumed by load resistance $R_L$.
1. Relation between Angle and Expansion:
Let $\delta l_1$ and $\delta l_2$ be the thermal expansions of the left and right wire segments, respectively. For the pulley to compensate for the difference in length and keep the system symmetric/taut, the arc length rotated $s$ must satisfy: $$ s = a \Delta \theta \propto (\delta l_1 – \delta l_2) $$ Since thermal expansion is proportional to the change in temperature ($\delta l \propto \Delta T$), and in steady state $\Delta T$ is proportional to Power ($P$), we have: $$ \Delta \theta \propto (P_1 – P_2) $$
2. Power Difference Calculation:
Let $V$ be the battery voltage. $$ P_1 \propto V^2 \quad \text{(Voltage across A-C)} $$ $$ P_2 \propto V_{BC}^2 \quad \text{(Voltage across B-C)} $$ Therefore, $\Delta \theta \propto (V^2 – V_{BC}^2)$.
Using difference of squares: $V^2 – V_{BC}^2 = (V – V_{BC})(V + V_{BC})$. Note that $V – V_{BC} = V_{AB}$ (Voltage drop across resistor $R$). So, $\Delta \theta \propto V_{AB} (V + V_{BC})$.
3. Approximations ($R_L \gg R$):
Since $R_L \gg R$, the voltage drop across $R$ is small, so $V_{BC} \approx V$. The term $(V + V_{BC}) \approx 2V$ (a constant factor). Thus, the rotation is largely proportional to $V_{AB}$: $$ \Delta \theta \propto V_{AB} $$ Using Ohm’s Law for the resistor $R$: $V_{AB} = I R$. Since $R_L \gg R$, the current $I \approx V/R_L$. $$ V_{AB} \approx \frac{V}{R_L} R $$
4. Linking to Load Power:
The power consumed by the load $R_L$ is: $$ P_{R_L} = \frac{V_{BC}^2}{R_L} \approx \frac{V^2}{R_L} $$ Comparing our result for rotation: $$ \Delta \theta \propto \frac{V}{R_L} \propto \frac{V^2}{R_L} \frac{1}{V} $$ Since $V$ is constant, we see that: $$ \Delta \theta \propto \frac{V^2}{R_L} $$ $$ \Delta \theta \propto P_{R_L} $$ Hence, the angular displacement is proportional to the power consumed by the resistance $R_L$.