Calculating Kinetic Energy of a Slipping Chain

This problem involves a uniform chain slipping from a smooth horizontal table. We need to find the kinetic energy of the chain when half of its length is hanging, given that it started slipping when one-fourth of its length was hanging. Since the table is smooth, there is no friction, and we can apply the principle of conservation of mechanical energy or the work-energy theorem.

Understanding Potential Energy of a Hanging Chain

For a uniform chain hanging vertically, its potential energy can be calculated by considering the potential energy of its center of mass. If a length \(y\) of a uniform chain of linear mass density \(\lambda = \frac{m}{l}\) is hanging below a reference point (like the edge of the table), its mass is \(\lambda y\). The center of mass of this hanging part is at a depth of \(\frac{y}{2}\) below the reference point. Taking the edge of the table as the zero potential energy level, the potential energy of the hanging part is:

Potential Energy \(U = (\lambda y) \times g \times \left(-\frac{y}{2}\right) = -\frac{1}{2} \lambda g y^2\)

Substituting \(\lambda = \frac{m}{l}\), the potential energy of the hanging part is \(U = -\frac{1}{2} \frac{m}{l} g y^2\).

The part of the chain lying on the table has zero potential energy as it is at the reference level.

Initial State Analysis

Initially, a length of \(y_1 = \frac{l}{4}\) is hanging from the edge of the table. The rest of the chain (\(l - \frac{l}{4} = \frac{3l}{4}\)) is on the table.

• Initial hanging length, \(y_1 = \frac{l}{4}\).
• Mass of initial hanging part = \(\frac{m}{l} \times \frac{l}{4} = \frac{m}{4}\).
• Center of mass of initial hanging part is at depth \(\frac{l/4}{2} = \frac{l}{8}\) below the table edge.
• Initial potential energy, \(U_1 = -\frac{1}{2} \frac{m}{l} g \left(\frac{l}{4}\right)^2 = -\frac{1}{2} \frac{m}{l} g \frac{l^2}{16} = -\frac{mgl}{32}\).
• Initially, the chain is at rest, so initial kinetic energy, \(K_1 = 0\).
• Total initial mechanical energy, \(E_1 = U_1 + K_1 = -\frac{mgl}{32} + 0 = -\frac{mgl}{32}\).

Final State Analysis

Finally, a length of \(y_2 = \frac{l}{2}\) is hanging from the edge of the table. The rest of the chain (\(l - \frac{l}{2} = \frac{l}{2}\)) is on the table.

• Final hanging length, \(y_2 = \frac{l}{2}\).
• Mass of final hanging part = \(\frac{m}{l} \times \frac{l}{2} = \frac{m}{2}\).
• Center of mass of final hanging part is at depth \(\frac{l/2}{2} = \frac{l}{4}\) below the table edge.
• Final potential energy, \(U_2 = -\frac{1}{2} \frac{m}{l} g \left(\frac{l}{2}\right)^2 = -\frac{1}{2} \frac{m}{l} g \frac{l^2}{4} = -\frac{mgl}{8}\).
• Let the final kinetic energy be \(K_2\).
• Total final mechanical energy, \(E_2 = U_2 + K_2 = -\frac{mgl}{8} + K_2\).

Applying Conservation of Mechanical Energy

Since the table is smooth and only gravity is doing work on the hanging part (the table supports the horizontal part, but this force does no work as there is no vertical displacement), the total mechanical energy of the chain is conserved.

\(E_1 = E_2\)

\(U_1 + K_1 = U_2 + K_2\)

\(-\frac{mgl}{32} + 0 = -\frac{mgl}{8} + K_2\)

Now, we solve for \(K_2\):

\(K_2 = -\frac{mgl}{32} + \frac{mgl}{8}\)

\(K_2 = mgl \left(\frac{1}{8} - \frac{1}{32}\right)\)

To subtract the fractions, find a common denominator, which is 32:

\(K_2 = mgl \left(\frac{4}{32} - \frac{1}{32}\right)\)

\(K_2 = mgl \left(\frac{4-1}{32}\right)\)

\(K_2 = mgl \left(\frac{3}{32}\right)\)

\(K_2 = \frac{3}{32} mgl\)

Final Answer

The kinetic energy of the chain when half of its length is hanging is \(\frac{3}{32}mgl\).

Revision Table: Slipping Chain Kinetics

Key concepts used in solving this problem:

Additional Information: Work-Energy Theorem Approach

Alternatively, one could use the Work-Energy theorem, which states that the net work done on a system equals its change in kinetic energy.

• Work done by gravity as the chain slips from \(y_1 = l/4\) hanging to \(y_2 = l/2\) hanging is equal to the negative change in potential energy: \(W_{gravity} = -\Delta U = -(U_2 - U_1) = U_1 - U_2\).
• \(W_{gravity} = -\frac{mgl}{32} - \left(-\frac{mgl}{8}\right) = -\frac{mgl}{32} + \frac{mgl}{8} = \frac{3mgl}{32}\).
• Since the table is smooth, there's no work done by friction. The work done by the normal force from the table is also zero as it's perpendicular to the motion.
• The net work done is \(W_{net} = W_{gravity} = \frac{3mgl}{32}\).
• According to the Work-Energy Theorem, \(W_{net} = \Delta K = K_2 - K_1\).
• \(\frac{3mgl}{32} = K_2 - 0\).
• \(K_2 = \frac{3mgl}{32}\).

Both the conservation of energy and the work-energy theorem methods yield the same result, confirming the solution. This problem highlights how potential energy stored in the hanging part of the chain is converted into kinetic energy as more of the chain hangs and accelerates.