NLM CYU 3

Solution to Question 3

1. Finding Radius of Curvature ($\rho$)

Consider a small element of the rope of length $ds$ at the lowest point. The forces acting on it are Tension $T_0$ tangentially and the normal force required to sustain the curvature.

$2T \sin(d\phi/2) \approx T d\phi$. This inward force balances the weight of the element $dm \cdot g$.

$$ T_0 d\phi = (\lambda ds) g $$ $$ T_0 \frac{d\phi}{ds} = \lambda g $$

Since curvature $\frac{1}{\rho} = \frac{d\phi}{ds}$:

$$ \frac{T_0}{\rho} = \lambda g \implies \rho = \frac{T_0}{\lambda g} $$

Now, find $T_0$. From the equilibrium of the half-rope:

• Vertical Equilibrium: $T \sin \theta = \frac{\lambda l g}{2}$
• Horizontal Equilibrium: $T \cos \theta = T_0$

From Vertical eq: $T = \frac{\lambda l g}{2 \sin \theta}$. Substitute this into Horizontal eq:

$$ T_0 = \left( \frac{\lambda l g}{2 \sin \theta} \right) \cos \theta = \frac{\lambda l g}{2 \tan \theta} $$

Substituting $T_0$ back into the radius formula:

$$ \rho = \frac{1}{\lambda g} \left( \frac{\lambda l g}{2 \tan \theta} \right) = \frac{l}{2 \tan \theta} $$

2. Finding Depth $h$

We consider the tangential forces on a small element of the rope. The net force along the tangent balances the component of gravity along the tangent.

$$ dT = (dm)g \sin \phi $$

Where $\phi$ is the angle with the horizontal. Integrating from the lowest point (where $\phi=0$, tension=$T_0$) to the top support (where $\phi=\theta$, tension=$T$):

$$ \int_{T_0}^{T} dT = \int \lambda g \sin \phi \, dl $$

We know from geometry that $dy = dl \sin \phi$. Substituting this into the integral:

$$ T – T_0 = \lambda g \int dy $$ $$ T – T_0 = \lambda g h $$

Now we substitute the values of $T$ and $T_0$ derived from equilibrium:

• $T = \frac{\lambda l g}{2 \sin \theta}$
• $T_0 = T \cos \theta$

Substituting these into the energy equation:

$$ T (1 – \cos \theta) = \lambda g h $$ $$ \frac{\lambda l g}{2 \sin \theta} (1 – \cos \theta) = \lambda g h $$

Canceling $\lambda g$:

$$ h = \frac{l}{2} \frac{(1 – \cos \theta)}{\sin \theta} $$

Using the half-angle identity $\frac{1 – \cos \theta}{\sin \theta} = \tan(\theta/2)$:

$$ h = \frac{l}{2} \tan\left(\frac{\theta}{2}\right) $$