Solution: Kinematics of Beetles on an Elastic Cord
1. Force Analysis
Let the total relaxed length of the cord be $l_0$ and stiffness be $k$.
We divide the cord into two segments at the position of the climbing beetle (Beetle 1):
- Top Segment: Natural length $l_{0,1}$.
- Bottom Segment: Natural length $l_{0,2}$, where $l_{0,1} + l_{0,2} = l_0$.
Tension in Bottom Segment ($T_2$): Supports only Beetle 2.
$$ T_2 = mg $$
Tension in Top Segment ($T_1$): Supports Beetle 1 and the tension from the bottom segment.
$$ T_1 = mg + T_2 = 2mg $$
2. Relating Natural Lengths to Current Lengths
Using Hooke’s Law, the stiffness of a segment is inversely proportional to its natural length ($k_{seg} = \frac{k l_0}{l_{seg}}$).
Length of Top Segment ($y_1$):
$$ y_1 = l_{0,1} + \text{Extension} = l_{0,1} + \frac{T_1}{k_{top}} = l_{0,1} + \frac{2mg}{k l_0 / l_{0,1}} = l_{0,1} \left( 1 + \frac{2mg}{k l_0} \right) $$
Let $\lambda_1 = 1 + \frac{2mg}{k l_0}$. Then $y_1 = l_{0,1} \lambda_1$.
Length of Bottom Segment ($y_2$):
$$ y_2 = l_{0,2} + \frac{T_2}{k_{bot}} = l_{0,2} + \frac{mg}{k l_0 / l_{0,2}} = l_{0,2} \left( 1 + \frac{mg}{k l_0} \right) $$
Let $\lambda_2 = 1 + \frac{mg}{k l_0}$. Then $y_2 = l_{0,2} \lambda_2$.
3. Kinematics
Beetle 1 climbs with velocity $u$ relative to the ceiling. Taking the downward direction as positive $y$:
$$ \frac{dy_1}{dt} = -u $$
Differentiating the expression for $y_1$:
$$ \frac{dy_1}{dt} = \frac{d l_{0,1}}{dt} \lambda_1 \implies -u = \dot{l}_{0,1} \lambda_1 \implies \dot{l}_{0,1} = \frac{-u}{\lambda_1} $$
Since the total natural length $l_0$ is constant, $\dot{l}_{0,2} = – \dot{l}_{0,1} = \frac{u}{\lambda_1}$.
4. Velocity of the Lower Beetle
The position of the lower beetle is $y = y_1 + y_2$. Its velocity $v$ is $\frac{dy}{dt}$.
$$ v = \frac{dy_1}{dt} + \frac{dy_2}{dt} = -u + \dot{l}_{0,2} \lambda_2 $$
Substituting $\dot{l}_{0,2}$:
$$ v = -u + \left( \frac{u}{\lambda_1} \right) \lambda_2 = u \left( \frac{\lambda_2}{\lambda_1} – 1 \right) = u \left( \frac{\lambda_2 – \lambda_1}{\lambda_1} \right) $$
Substituting back $\lambda_1$ and $\lambda_2$:
$$ \lambda_2 – \lambda_1 = \left( 1 + \frac{mg}{k l_0} \right) – \left( 1 + \frac{2mg}{k l_0} \right) = -\frac{mg}{k l_0} $$
$$ v = u \left( \frac{-mg / k l_0}{1 + 2mg / k l_0} \right) = -u \frac{mg}{k l_0 + 2mg} $$
The negative sign indicates the velocity is directed upwards.
Final Answer: The lower beetle moves with velocity $\frac{mgu}{kl_0 + 2mg}$ upwards ($\uparrow$).
Let the total relaxed length of the cord be $l_0$ and stiffness be $k$. We divide the cord into two segments at the position of the climbing beetle (Beetle 1):
- Top Segment: Natural length $l_{0,1}$.
- Bottom Segment: Natural length $l_{0,2}$, where $l_{0,1} + l_{0,2} = l_0$.
Tension in Bottom Segment ($T_2$): Supports only Beetle 2.
$$ T_2 = mg $$
Tension in Top Segment ($T_1$): Supports Beetle 1 and the tension from the bottom segment.
$$ T_1 = mg + T_2 = 2mg $$
Using Hooke’s Law, the stiffness of a segment is inversely proportional to its natural length ($k_{seg} = \frac{k l_0}{l_{seg}}$).
Length of Top Segment ($y_1$): $$ y_1 = l_{0,1} + \text{Extension} = l_{0,1} + \frac{T_1}{k_{top}} = l_{0,1} + \frac{2mg}{k l_0 / l_{0,1}} = l_{0,1} \left( 1 + \frac{2mg}{k l_0} \right) $$ Let $\lambda_1 = 1 + \frac{2mg}{k l_0}$. Then $y_1 = l_{0,1} \lambda_1$.
Length of Bottom Segment ($y_2$): $$ y_2 = l_{0,2} + \frac{T_2}{k_{bot}} = l_{0,2} + \frac{mg}{k l_0 / l_{0,2}} = l_{0,2} \left( 1 + \frac{mg}{k l_0} \right) $$ Let $\lambda_2 = 1 + \frac{mg}{k l_0}$. Then $y_2 = l_{0,2} \lambda_2$.
Beetle 1 climbs with velocity $u$ relative to the ceiling. Taking the downward direction as positive $y$: $$ \frac{dy_1}{dt} = -u $$ Differentiating the expression for $y_1$: $$ \frac{dy_1}{dt} = \frac{d l_{0,1}}{dt} \lambda_1 \implies -u = \dot{l}_{0,1} \lambda_1 \implies \dot{l}_{0,1} = \frac{-u}{\lambda_1} $$ Since the total natural length $l_0$ is constant, $\dot{l}_{0,2} = – \dot{l}_{0,1} = \frac{u}{\lambda_1}$.
The position of the lower beetle is $y = y_1 + y_2$. Its velocity $v$ is $\frac{dy}{dt}$. $$ v = \frac{dy_1}{dt} + \frac{dy_2}{dt} = -u + \dot{l}_{0,2} \lambda_2 $$ Substituting $\dot{l}_{0,2}$: $$ v = -u + \left( \frac{u}{\lambda_1} \right) \lambda_2 = u \left( \frac{\lambda_2}{\lambda_1} – 1 \right) = u \left( \frac{\lambda_2 – \lambda_1}{\lambda_1} \right) $$ Substituting back $\lambda_1$ and $\lambda_2$: $$ \lambda_2 – \lambda_1 = \left( 1 + \frac{mg}{k l_0} \right) – \left( 1 + \frac{2mg}{k l_0} \right) = -\frac{mg}{k l_0} $$ $$ v = u \left( \frac{-mg / k l_0}{1 + 2mg / k l_0} \right) = -u \frac{mg}{k l_0 + 2mg} $$
The negative sign indicates the velocity is directed upwards.