Question

A rope of length $(\frac{\pi}{2}+1)R$ has been placed on a smooth sphere of radius $R$ as shown in figure. End A of the rope is at the top of the sphere and end B is overhanging. Mass per unit length of the rope is $\lambda$. The horizontal string holding this rope in place can tolerate tension equal to weight of the rope. Find the maximum mass $(M_0)$ of a block that can be tied to the end $B$ of the rope so that the string does not break.

Answer 1

$\lambda R (\frac{\pi}{2} - 1)$

Answer 2

The problem asks us to find the maximum mass $M_0$ that can be tied to the end B of the rope such that the horizontal string holding the rope in place does not break.

1. Analyze the Rope Segments: The total length of the rope is $L = (\frac{\pi}{2}+1)R$. The rope is placed on a smooth sphere of radius $R$. End A is at the top of the sphere. The rope curves down along the sphere and then hangs vertically. The curved part of the rope covers a quarter circle (from the top of the sphere to the point where it becomes vertical). Length of the curved part ($L_{curved}$) = $\frac{1}{4} (2\pi R) = \frac{\pi}{2}R$. Length of the overhanging straight part ($L_{overhanging}$) = Total length - Length of curved part $L_{overhanging} = (\frac{\pi}{2}+1)R - \frac{\pi}{2}R = R$.

2. Calculate the Force Exerted by the Rope and Block on the String: The horizontal string at point A provides the tension $T_{string}$ needed to prevent the rope from sliding off the sphere. This tension must balance the effective downward pull due to gravity on the rope segments and the block.

• Force from the curved part of the rope: Consider a small element of the rope of length $ds = R d\theta$ at an angle $\theta$ from the vertical (measured from the top, so $\theta=0$ at A). The mass of this element is $dm = \lambda ds = \lambda R d\theta$. The component of the weight of this element that pulls the rope tangentially along the sphere's surface is $dw_{tangential} = dm g \sin\theta = \lambda R g \sin\theta d\theta$. The total tangential force from the curved part (from $\theta=0$ to $\theta=\pi/2$) is: $F_{curved} = \int_{0}^{\pi/2} \lambda R g \sin\theta d\theta = \lambda R g [-\cos\theta]_{0}^{\pi/2}$ $F_{curved} = \lambda R g (-\cos(\pi/2) - (-\cos(0))) = \lambda R g (0 - (-1)) = \lambda R g$

• Force from the overhanging part of the rope: The mass of the overhanging part is $M_{overhanging} = \lambda L_{overhanging} = \lambda R$. The weight of the overhanging part is $W_{overhanging} = M_{overhanging} g = \lambda R g$. This force acts vertically downwards.

• Force from the block: The weight of the block is $W_{block} = M_0 g$. This force acts vertically downwards.

The total force pulling the rope down and to the right, which must be balanced by the tension in the horizontal string, is the sum of these forces: $T_{string} = F_{curved} + W_{overhanging} + W_{block}$ $T_{string} = \lambda R g + \lambda R g + M_0 g$ $T_{string} = (2\lambda R + M_0) g$

3. Determine the Maximum Tolerable Tension: The problem states that the horizontal string can tolerate tension equal to the weight of the rope. The total mass of the rope is $M_{total\_rope} = \lambda \times (\text{total length of rope}) = \lambda (\frac{\pi}{2}+1)R$. The weight of the total rope is $W_{total\_rope} = M_{total\_rope} g = \lambda (\frac{\pi}{2}+1)R g$. So, the maximum tension the string can tolerate is: $T_{max} = \lambda (\frac{\pi}{2}+1)R g$

4. Find the Maximum Mass $M_0$: For the string not to break, the tension in the string must be less than or equal to the maximum tolerable tension ($T_{string} \le T_{max}$). To find the maximum mass $M_0$, we set $T_{string} = T_{max}$: $(2\lambda R + M_0) g = \lambda (\frac{\pi}{2}+1)R g$ Divide both sides by $g$: $2\lambda R + M_0 = \lambda (\frac{\pi}{2}+1)R$ Solve for $M_0$: $M_0 = \lambda (\frac{\pi}{2}+1)R - 2\lambda R$ Factor out $\lambda R$: $M_0 = \lambda R \left(\frac{\pi}{2}+1 - 2\right)$ $M_0 = \lambda R \left(\frac{\pi}{2} - 1\right)$

The value of $\frac{\pi}{2} - 1 \approx 1.5708 - 1 = 0.5708$, which is positive, so $M_0$ is a positive mass.

The final answer is $\boxed{\lambda R (\frac{\pi}{2} - 1)}$.

Explanation of the solution:

1. Identify rope segments: The rope consists of a curved part on the sphere (length $\frac{\pi}{2}R$) and an overhanging straight part (length $R$).
2. Calculate forces causing tension:
   • The curved part contributes an effective downward pull of $\lambda R g$ (obtained by integrating tangential components of gravity).
   • The overhanging part contributes its full weight, $\lambda R g$.
   • The block contributes its weight, $M_0 g$.
3. Total tension: The total tension in the horizontal string is the sum of these pulls: $T_{string} = (2\lambda R + M_0)g$.
4. Maximum tolerable tension: The string can tolerate tension equal to the total weight of the rope: $T_{max} = \lambda (\frac{\pi}{2}+1)R g$.
5. Equate tensions: For the string not to break, $T_{string} = T_{max}$, which leads to $M_0 = \lambda R (\frac{\pi}{2} - 1)$.