Source: Physics Competition - Mechanics
Problem Sets:
Problem
On a smooth circular tabletop of radius $R$ with its center at O, a vertical post is fixed at the center. The intersection of the post with the tabletop is a convex, smooth, closed curve C. An inextensible, flexible, light string has one end fixed at a point on the curve C, and the other end is attached to a small block of mass $m$. The block is placed on the tabletop, and the string is pulled taut. The block is then given an initial velocity of magnitude $v_0$ in a direction perpendicular to the string. As the block moves on the tabletop, the string wraps around the post. The string breaks when its tension reaches $T_0$. It is assumed that the initial tension $T$ is less than $T_0$, and the block always remains on the tabletop before the string breaks.
The length of the taut part of the string at the moment it is about to break is:
$$l_f = \frac{mv_0^2}{T_0}$$The block of mass $m$ moves on a smooth horizontal tabletop, so the only horizontal force acting on it is the tension $\vec{T}$ from the string. The string wraps around the post, and the block's velocity $\vec{v}$ is always perpendicular to the taut part of the string.
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Conservation of Energy: The work done by the tension force on the block is given by $W = \int \vec{T} \cdot d\vec{s}$. Since the velocity vector $\vec{v} = d\vec{s}/dt$ is always perpendicular to the force vector $\vec{T}$, the dot product $\vec{T} \cdot \vec{v}$ is always zero. Consequently, the tension force does no work on the block.
$$W_T = 0$$By the work-energy theorem, the change in kinetic energy of the block is equal to the net work done on it. Since no work is done, the kinetic energy is conserved.
$$\Delta K = 0 \implies \frac{1}{2}mv^2 = \frac{1}{2}mv_0^2$$This implies that the speed of the block remains constant throughout its motion:
$$v = v_0$$ -
Centripetal Force and Tension: At any instant, the block is undergoing instantaneous circular motion. The radius of this motion is the current length of the taut part of the string, which we denote by $l$. The tension in the string provides the necessary centripetal force to maintain this motion.
$$T = F_{\text{centripetal}} = \frac{mv^2}{l}$$ -
Breaking Condition: We can substitute the constant speed $v=v_0$ into the tension equation:
$$T = \frac{mv_0^2}{l}$$The string breaks when its tension reaches the maximum value $T_0$. Let $l_f$ be the length of the taut part of the string at this moment.
$$T_0 = \frac{mv_0^2}{l_f}$$ -
Final Length: Solving for the final length $l_f$, we get:
$$l_f = \frac{mv_0^2}{T_0}$$This is the length of the taut part of the string at the moment it is about to break. The specific shape of the post C is not needed for this calculation.