Solution: Maximum Tension in Sliding Rope

Maximum Tension Calculation

Uniform Rope Sliding on a Hemisphere

Figure 1: Free body diagram of the rope segment.

1. System Variables

Consider a uniform rope of mass $M$ and length $L$. The linear mass density is $\lambda = M/L$. The rope subtends a total angle $\phi_0 = L/R$ at the center. We measure the angle $\theta$ from the vertical axis.

2. Calculating Acceleration ($a_t$)

Since the rope is inextensible, every point moves with the same tangential acceleration $a_t$. The net driving force is the sum of the tangential components of gravity acting on all elements $dm$.

The force on a small element at angle $\alpha$ is $dF = (dm)g \sin\alpha = (\lambda R d\alpha) g \sin\alpha$.

$$ F_{net} = \int_{0}^{\phi_0} \lambda R g \sin\alpha \, d\alpha = \lambda R g [-\cos\alpha]_0^{\phi_0} = \lambda R g (1 – \cos\phi_0) $$

Applying Newton’s Second Law ($F_{net} = M a_t$):

$$ a_t = \frac{\lambda R g (1 – \cos\phi_0)}{\lambda L} = \frac{Rg}{L}(1 – \cos\phi_0) $$

3. Finding Tension $T(\theta)$

Consider the lower segment of the rope, extending from angle $\theta$ to the end $\phi_0$. This segment has mass $m’ = \lambda R (\phi_0 – \theta)$.

The forces acting along the tangent are:

Gravity ($F_{seg}$): Pulls the segment down the slope.
Tension ($T$): Pulls the segment up the slope (as shown in the diagram).

The equation of motion is $F_{seg} – T = m’ a_t$. Calculating the gravitational pull on this segment:

$$ F_{seg} = \int_{\theta}^{\phi_0} \lambda R g \sin\alpha \, d\alpha = \lambda R g (\cos\theta – \cos\phi_0) $$

Solving for Tension:

$$ T(\theta) = \underbrace{\lambda R g (\cos\theta – \cos\phi_0)}_{\text{Gravity Pull}} – \underbrace{\lambda R (\phi_0 – \theta) a_t}_{\text{Inertial Term}} $$

4. Maximizing Tension

To find the maximum tension, we differentiate $T$ with respect to $\theta$ and set it to zero: $\frac{dT}{d\theta} = 0$.

$$ \frac{dT}{d\theta} = \lambda R g (-\sin\theta) – \lambda R a_t (-1) = 0 $$ $$ -\lambda R g \sin\theta + \lambda R a_t = 0 \implies g \sin\theta = a_t $$

Physically, this means tension is maximum at the point where the local component of gravity equals the overall acceleration of the rope.

Final Result

Substituting the value of $a_t$ derived in Step 2:

$$ g \sin\theta = \frac{Rg}{L}(1 – \cos\phi_0) $$

Using $\phi_0 = L/R$, we get:

$$ \theta = \sin^{-1} \left[ \frac{R}{L} \left( 1 – \cos \frac{L}{R} \right) \right] $$