Question

Statement \(-1-\) A thin uniform rod \(A B\) of mass \(M\) and length \(\mathrm{L}\) 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 ends \(B\) strikes the floor \(\sqrt{(3 g / L)}\) Statement \(-2\) - The angular momentum of the rod about the hinge remains constant throughout its fall to the floor. \(\\{\mathrm{A}\\}\) Statement \(-1\) is correct (true), Statement \(-2\) is true and Statement- 2 is correct explanation for Statement - 1 \\{B \\} Statement \(-1\) is true, statement \(-2\) is true but statement- 2 is not the correct explanation four statement \(-1\). \\{C\\} Statement \(-1\) is true, statement- 2 is false \\{D \(\\}\) Statement- 2 is false, statement \(-2\) is true

Short answer

Option B is the correct answer: Statement 1 is true, statement 2 is true, but statement 2 is not the correct explanation for statement 1. The main principle that led us to Statement 1 is the conservation of mechanical energy, not the conservation of angular momentum.

Step-by-step solution

(Step 1: Set up conservation of mechanical energy)

As the rod is allowed to fall freely from an initial vertical position, we can analyze the problem using the conservation of mechanical energy. At the initial vertical position, the rod's gravitational potential energy is at its highest, and it has no kinetic energy. At the instant just before the rod strikes the floor, all of its gravitational potential energy will have been converted to kinetic energy. Let's denote the initial gravitational potential energy as \(U_i\) and the final gravitational potential energy as \(U_f\). Similarly, we denote the initial kinetic energy as \(K_i\) and the final kinetic energy as \(K_f\). The conservation of mechanical energy can be expressed as: \[U_i + K_i = U_f + K_f\] Since the rod has no initial kinetic energy, \(K_i = 0\).

(Step 2: Calculate gravitational potential energy)

We can calculate the initial and final gravitational potential energies (\(U_i\) and \(U_f\)) using the following formula: \[U = mgh\] Where: - \(m\) is the mass of the rod (which is uniformly distributed); - \(g\) is the gravitational acceleration; and - \(h\) is the vertical height of the center of mass. Initially, the rod's center of mass is at a vertical height of \(\frac{L}{2}\), so: \[U_i = \frac{Mg\cdot L}{2}\] When the rod is about to strike the floor, the center of mass is at a height of \(0\), so: \[U_f = 0\] Substituting into the conservation of mechanical energy equation, we get: \[\frac{Mg\cdot L}{2} = K_f\]

(Step 3: Calculate the final kinetic energy)

The rod's final kinetic energy is related to its angular velocity, which we can denote as \(\omega\). We can express the final kinetic energy as: \[K_f = \frac{1}{2}I\omega^2\] Where: - \(I\) is the rod's moment of inertia about the hinge (end A); and - \(\omega\) is the angular velocity of the rod, which we want to find. The moment of inertia of a thin uniform rod about one of its ends is given by: \[I = \frac{M\cdot L^2}{3}\] Now, we can substitute this into our expression for \(K_f\): \[K_f = \frac{1}{2}\cdot\frac{M\cdot L^2}{3}\cdot\omega^2\]

(Step 4: Solve for the angular velocity \(\omega\))

Equating our expressions for \(K_f\), we can now solve for the angular velocity \(\omega\): \[\frac{Mg\cdot L}{2} = \frac{1}{2}\cdot\frac{M\cdot L^2}{3}\cdot\omega^2\] Canceling the common factors and solving for \(\omega\), we get: \[\omega = \sqrt{\frac{3g}{L}}\] So, Statement 1 is true.

(Step 5: Analyzing the angular momentum of the rod)

Let's now examine Statement 2, which claims that the angular momentum of the rod remains constant throughout its fall. The angular momentum of an object can be expressed as: \[L = I\omega\] Where: - \(L\) is the angular momentum; - \(I\) is the moment of inertia; and - \(\omega\) is the angular velocity. Since there are no external torques acting on the rod about the hinge, its angular momentum about the hinge will indeed remain constant during the fall, according to the conservation of angular momentum. Therefore, Statement 2 is also true. Now, we need to determine if Statement 2 is a correct explanation for Statement 1.

(Step 6: Relate statement 2 to statement 1)

From our previous analysis, we know that the angular velocity is \(\omega = \sqrt{\frac{3g}{L}}\), and this result came from the conservation of mechanical energy, which in turn was related to the constant gravitational potential energy and kinetic energy of the rod during its fall. Since Statement 2 is about conservation of angular momentum and confirmed the conservation of angular momentum throughout the fall, it is also in accordance with the result of Statement 1. However, Statement 2 does not directly explain why the angular velocity is \(\sqrt{\frac{3g}{L}}\) when the rod strikes the floor. The main principle that led us to Statement 1 is the conservation of mechanical energy, not the conservation of angular momentum. Thus, both statements are true, but Statement 2 is not a direct explanation for Statement 1. The correct answer is option B.

Key concepts

Angular Momentum

Angular momentum is an important concept in rotational mechanics, similar to linear momentum in translational motion. It helps us understand how rotational motion is conserved in a system with no external torques. For a rotating object, angular momentum (\( L \)) is given by the formula:

• \( L = I\omega \)
• Where:
• \( I \) is the moment of inertia
• \( \omega \) is the angular velocity

If no external torque acts on the system, the angular momentum remains constant. In our exercise, the rod attached to the hinge keeps its angular momentum constant throughout its fall. This helps verify that the rotational motion remains unchanged due to the absence of an external force affecting it.
In summary, whenever you know no external torques are present, you can be sure that angular momentum won't change, making it easier to solve rotational problems as it was in our example.

Moment of Inertia

The moment of inertia (\( I \)) plays a role similar to mass in linear motion but for rotational motion. It's a measure of how much resistance an object has against its change in angular motion. It depends on the object's distribution of mass relative to its axis of rotation.

• For a thin rod hinged at one end, the moment of inertia is:
• \( I = \frac{M\cdot L^2}{3} \)
• Where:
• \( M \) is mass
• \( L \) is length

Knowing the moment of inertia is crucial for calculating the kinetic energy and angular momentum of a rotating body. In the exercise, determining the moment of inertia helps us find the final angular velocity, which is necessary to determine how quickly the rod rotates when it strikes the floor.

Conservation of Energy

Conservation of energy is a fundamental principle in physics stating that energy within a closed system remains constant. In rotational mechanics, this means that the total mechanical energy (potential plus kinetic) of a system is conserved if no external forces like friction are present.
In our rod's fall:

• Initial Potential Energy: \( \frac{Mg\cdot L}{2} \)
• Final Potential Energy: \( 0 \)
• Initial Kinetic Energy: \( 0 \)
• Final Kinetic Energy: \( \frac{1}{2}I\omega^2 \)

By setting the initial energy equal to the final energy, the exercise demonstrates how to use conservation of energy to solve for unknown quantities. This principle allowed us to find the angular velocity of the rod, showcasing the pivotal role that energy conservation plays in understanding and predicting motion.

Kinetic Energy

Kinetic energy in rotational motion is the energy due to the rotation of a body. It’s linked to how fast an object rotates and its mass distribution. The formula for rotational kinetic energy (\( K \)) is:

• \( K = \frac{1}{2}I\omega^2 \)
• Where:
• \( I \) is moment of inertia
• \( \omega \) is angular velocity

In the exercise, as the rod falls, its gravitational potential energy is transformed into kinetic energy. By knowing the rod's moment of inertia and using the conservation of mechanical energy, we calculated the rod's final angular velocity. This approach shows how changes in height affect the rotational speed through energy conversion, a cornerstone of rotational dynamics.

Gravitational Potential Energy

Gravitational potential energy (\( U \)) is the energy stored in an object due to its height relative to a reference point. For objects in uniform gravitational fields, it is given by:

• \( U = mgh \)
• Where:
• \( m \) is mass
• \( g \) is acceleration due to gravity
• \( h \) is height

For the uniform rod initially standing vertically, the center of mass is at height \( \frac{L}{2} \), giving it initial gravitational potential energy. As the rod falls, this potential energy is converted to kinetic energy, powering the rod's rotation. Understanding potential energy's conversion into kinetic energy is crucial for analyzing systems in motion, predicting outcomes, and applying conservation principles effectively.