Question

A thin uniform rod \(A B\) of mass \(M\) and length \(L\) is hinged at one end \(\mathrm{A}\) to the horizontal floor. Initially it stands vertically. It is allowed to fall freely on the floor in the vertical plane. The angular velocity of the rod when its end \(B\) strikes the floor is \(\\{\mathrm{A}\\} \sqrt{(\mathrm{g} / \mathrm{L})}\) \(\\{\mathrm{B}\\} \sqrt{(2 \mathrm{~g} / \mathrm{L})}\) \(\\{C\\} \sqrt{(3 g / L)}\) \(\\{\mathrm{D}\\} 2 \sqrt{(\mathrm{g} / \mathrm{L})}\)

Short answer

The angular velocity of the rod when its end B strikes the floor is \(\sqrt{\frac{g}{L}}\).

Step-by-step solution

Define the initial and final positions

Initially, the rod is standing vertically, and when it falls, end B will strike the floor. We will call these two positions the initial and final positions, respectively.

Calculate the moment of inertia for the rod about point A

The moment of inertia (I) for a thin rod rotating about its end is given by the formula: \[I = \frac{1}{3}mL^2\] Where m is the mass of the rod and L is its length.

Apply conservation of mechanical energy

Since no external torques are acting on the rod, the conservation of mechanical energy applies. This means that the sum of the initial potential energy and kinetic energy equals the sum of the final potential energy and kinetic energy: \(U_{initial} + K_{initial} = U_{final} + K_{final}\) Initially, the rod has potential energy due to its height from the ground but no kinetic energy, as it is not moving. When the rod falls, its potential energy will decrease and it will gain kinetic energy.

Calculate the initial potential energy

At the initial position, the rod's center of mass is at a distance L/2 from the ground. Thus, we can calculate the initial potential energy (U) as: \(U_{initial} = mgh\) Where m is the mass of the rod, g is the acceleration due to gravity, and h is the height of the center of mass from the ground. Since h = L/2, the potential energy becomes: \(U_{initial} = mg(L/2)\)

Calculate the final potential energy and kinetic energy

At the final position, the center of mass of the rod is at the ground level, so U_final = 0. The kinetic energy (K) of the rod is given as: \(K_{final} = \frac{1}{2}I\omega^2\) Where ω is the angular velocity which we need to find.

Implement conservation of mechanical energy

Using the conservation of mechanical energy equation from Step 3, we can now substitute our calculations for the potential and kinetic energies: \(mg(L/2) = 0 + \frac{1}{2}(\frac{1}{3}mL^2)\omega^2\)

Solve for the angular velocity ω

Simplify the equation: \(mgL = mL^2\omega^2\) Now solve for ω: \(\omega = \sqrt{\frac{g}{L}}\) The angular velocity of the rod when its end B strikes the floor is \(\sqrt{\frac{g}{L}}\), which corresponds to option A.