NLM CYU 10

Kinematics of Beetles on an Elastic Cord | Complete Solution

JEE Physics · Elasticity + Kinematics

Kinematics of Beetles on an Elastic Cord

Two complementary derivations: a segment-length method and an elegant infinitesimal-transfer method.

Result preview
The lower beetle moves upward with speed \( \displaystyle \frac{mgu}{k l_0+2mg} \).

Two beetles on a hanging elastic cord The upper beetle climbs upward. The cord above it has tension 2mg, while the cord below it has tension mg. Beetle 1 climbs at speed u Beetle 2 Tension = 2mg Tension = mg
Free-body information for the two cord segments.

Notation and model

The light cord has natural length \(l_0\) and whole-cord stiffness \(k\). Each beetle has mass \(m\). The upper beetle climbs with constant speed \(u\). Motion is sufficiently slow for each segment to remain in instantaneous equilibrium.

\(l_0\)natural cord length
\(k\)whole-cord stiffness
\(m\)mass of each beetle
\(u\)upper beetle’s speed
Solution 1

Segment-length method

Track how much natural cord belongs to the regions above and below the climbing beetle.

1

Find the tensions

The lower segment supports Beetle 2, so \(T_2=mg\). At Beetle 1, the upper segment must balance its weight together with \(T_2\). Hence

\[T_1=mg+T_2=2mg.\]
2

Write the stretched segment lengths

Let \(l_{01}\) and \(l_{02}\) be the natural lengths above and below Beetle 1. Then \(l_{01}+l_{02}=l_0\).

\[ y_1=l_{01}\left(1+\frac{2mg}{kl_0}\right) =l_{01}\lambda_1, \] \[ y_2=l_{02}\left(1+\frac{mg}{kl_0}\right) =l_{02}\lambda_2, \] where \[ \lambda_1=1+\frac{2mg}{kl_0}, \qquad \lambda_2=1+\frac{mg}{kl_0}. \]
3

Use the prescribed speed of Beetle 1

Take downward as positive. Because Beetle 1 climbs upward, \(dy_1/dt=-u\). Since \(\lambda_1\) is constant,

\[ -u=\lambda_1\frac{dl_{01}}{dt} \quad\Longrightarrow\quad \frac{dl_{01}}{dt}=-\frac{u}{\lambda_1}. \]

The total natural length is fixed, so

\[ \frac{dl_{02}}{dt} =-\frac{dl_{01}}{dt} =\frac{u}{\lambda_1}. \]
4

Differentiate the position of Beetle 2

Its downward coordinate is \(y=y_1+y_2\). Therefore,

\[ \begin{aligned} v &=\frac{dy_1}{dt}+\frac{dy_2}{dt}\\ &=-u+\frac{u}{\lambda_1}\lambda_2\\ &=u\left(\frac{\lambda_2-\lambda_1}{\lambda_1}\right)\\ &=-\frac{mgu}{kl_0+2mg}. \end{aligned} \]

The negative sign is relative to our downward-positive axis. Thus, Beetle 2 moves upward.

Solution 2

Infinitesimal-transfer method

Alternate solution contributed by Sayan Chatterjee. 📄 Download Solution PDF

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