91Ó°ÊÓ

In the study of chain dynamics, calculating acceleration is crucial to understanding the motion of the system. When a chain is released from rest, the acceleration can be determined by assessing the physical forces acting on it. For our scenario, the chain has a mass distribution given by its mass per unit length, represented as \(\rho\), and a gravitational force \(g\). As a small overhanging section of the chain, of length \(x\), causes the motion, the force of gravity acting specifically on this overhang is expressed as \(F = \rho x g\).
However, since the entire chain is prompted to move, we consider the balance of forces in terms of the total chain mass \(M\), which is \(\rho L\) because \(L\) is the total chain length. By applying Newton’s second law, \(F = M a\), which reconfigures to \(\rho x g = \rho L a\), we solve to find acceleration \(a\):

• The resultant formula for acceleration is: \(a = \frac{x}{L} g\).

This formula informs us that the acceleration is a function of the chain length \(L\), the distance overhung \(x\), and gravitational pull \(g\). The equation reveals that as more of the chain hangs over, the acceleration increases, emphasizing the role of the chain's changing mass distribution on its dynamic behavior.

Understanding tension within the chain during motion is another vital part of chain dynamics. Tension is essentially the force within the chain that tries to prevent it from accelerating too quickly. For each point of the chain, this tension adjusts depending on how much of the chain is still on the surface versus how much hangs over the edge.
In our specific case, the tension \(T\) at the smooth corner where the chain starts to descends is a bit complex. As the chain overhangs, the horizontal portion is initially at rest but must now accelerate to maintain connection with the rest of the moving chain. This requires a tension force that both balances the gravitational pull on the hanging section and assists the horizontal chain's acceleration.

• The tension in the chain is expressed as: \(T = \frac{\rho x g (L - x)}{L} \).

This equation shows how tension diminishes as more of the chain falls off the edge, given \(T\) is directly proportional to \(L - x\). Such analysis aids in deducing how forces interplay when an extended body like a chain undergoes motion over a pivot.

Finally, determining the velocity at which the chain's last link reaches the corner involves applying conservation of energy principles. When the chain is initially released, it possesses potential energy due to its height. As it falls, this energy converts to kinetic energy.

• The potential energy can be represented by \(\rho g \frac{L^2}{2}\).
• When the entire chain is vertical, just as the last link reaches the edge, its kinetic energy is \(\frac{1}{2} \rho L v^2\).

Equating these expressions, as energy is conserved, we solve the equation: \(\rho g \frac{L^2}{2} = \frac{1}{2} \rho L v^2\). This leads to solving for velocity \(v\):
\( v = \sqrt{gL} \).
This formula demonstrates that the velocity of the last link is dependent solely on the gravitational field strength \(g\) and the chain's total length \(L\). Hence, every time the chain is released from such a position, it will always reach this velocity by the time the last link arrives at the corner.