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.

Key concepts

Moment of Inertia

Moment of inertia is an important concept in the study of rotational motion. It can be thought of as the rotational analog to mass in linear motion. Specifically, it measures an object's resistance to changes in its rotational motion.
For different shapes and axes of rotation, the moment of inertia will vary. In the case of a thin uniform rod rotating about one end, the moment of inertia is given by the formula: \[I = \frac{1}{3}mL^2\]Where:

• \(m\) is the mass of the rod
• \(L\) is the length of the rod

Using this formula, we can calculate how much torque (rotational force) is needed to cause the rod to rotate about its hinge. It also helps in determining how quickly the rod can change its rotational speed, which is crucial for solving rotational dynamics problems.

Conservation of Mechanical Energy

The conservation of mechanical energy is a fundamental principle in physics, especially useful in solving problems involving energy conversion between kinetic and potential forms. This principle states that in a closed system with no external forces, the total mechanical energy remains constant.
Mathematically, it is expressed as:\[U_{initial} + K_{initial} = U_{final} + K_{final}\] In the context of the falling rod, initially, there is potential energy due to the rod's height and no kinetic energy since the rod isn't moving. As the rod begins to fall, some of this potential energy converts into kinetic energy, causing the rod to rotate. By applying this principle, we can balance the energy equation and solve for unknown quantities such as angular velocity.

Thin Uniform Rod

A thin uniform rod is a common model in physics for simplifying problems about rigid bodies. It assumes the rod has a uniform mass distribution across its length, which is critical in calculating the moment of inertia and analyzing motion.
In our exercise, the rod's uniformity allows the center of mass to be conveniently located at the midpoint, \(L/2\) from the pivot point (hinged end). This property simplifies many calculations, making it easier to understand complex rotational dynamics. It allows us to predict the behavior of the rod as it rotates about the hinge and eventually strikes the ground.

Rotational Motion

Rotational motion occurs when an object spins around a fixed point or axis. This type of motion is described by angular variables like angular velocity, which tells us how fast the object is rotating.
For the falling rod problem, as it rotates from the vertical to the horizontal, each point on the rod traces out a circular path about the hinge point. The entire rod's motion can be described using concepts such as:

• Angular displacement: the angle through which the rod rotates
• Angular velocity \(\omega\): the rate of rotation at any given moment
• Angular acceleration: the rate at which angular velocity changes

Understanding rotational motion is crucial to solving problems like the one presented, where the rod's movement is described not by how fast each point moves, but by how quickly it spins around its pivot point.