Question

A uniform rod of length $\mathrm{L}$ is suspended from one end such that it is free to rotate about an axis passing through that end and perpendicular to the length, what maximum speed must be imparted to the lower end so that the rod completes one full revolution? $$\begin{gathered} \mathrm{A} \\ \sqrt{(2\mathrm{g}\mathrm{L})} \end{gathered}$$ $$\begin{gathered} \mathrm{B} \\ 2\sqrt{(\mathrm{gL})} \end{gathered}$$ $$\begin{gathered} C \\ \sqrt{(6gL)} \end{gathered}$$ $$\begin{gathered} \mathrm{D} \\ 2\sqrt{(2g\mathrm{L})} \end{gathered}$$

Short answer

The maximum speed that should be imparted to the lower end of the rod to complete one full revolution is $\sqrt{6gL}$. Hence, the correct answer is $$\begin{gathered} C \\ \sqrt{(6gL)} \end{gathered}$$.

Step-by-step solution

Calculate the initial gravitational potential energy

To find the initial gravitational potential energy, we will take the potential energy at the lowest point (when the rod is horizontal) as zero. The center of mass (COM) of the rod is at a distance L/2 from the rotational axis, so the initial height of the center of mass is (L/2)*sin(90), which is L/2. The initial gravitational potential energy (PE_initial) is given by: $$P{E}_{initial}=mgh=mg(\frac{L}{2})$$ where m is the mass of the rod, g is the acceleration due to gravity, h is the initial height of the COM, and L is the length of the rod.

Calculate the final gravitational potential energy

When the rod completes a full revolution, the center of mass will reach the highest point directly above the rotational axis. The final height of the COM will be equal to the length of the rod. The final gravitational potential energy (PE_final) is given by: $$P{E}_{final}=mgh=mgL$$

Calculate the change in potential energy and use conservation of energy to find the change in kinetic energy

The change in potential energy (PE_change) is the difference between the initial and final potential energies. Conservation of mechanical energy states that the change in kinetic energy (KE_change) will be equal to the change in potential energy. Therefore: $$\mathrm{\Delta }PE=P{E}_{final}-P{E}_{initial}=mgL-mg(\frac{L}{2})$$ $$\mathrm{\Delta }KE=\mathrm{\Delta }PE=\frac{mgL}{2}$$

Calculate the initial and final angular velocities

At the lowest point (initial position), the rod is given an initial speed, denoted as v_initial. The initial angular velocity (ω_initial) can be calculated as: $${\omega }_{initial}=\frac{v_{initial}}{L}$$ When the rod reaches the highest point (final position), its linear speed at the lower end, denoted as v_final, is zero, and the final angular velocity (ω_final) is: $${\omega }_{final}=\frac{v_{final}}{L}=0$$

Calculate the change in kinetic energy using the rotational inertia

The rotational inertia (I) of a rod rotating about one end is given by: $$I=\frac{1}{3}m{L}^2$$ The change in kinetic energy (KE_change) can also be calculated using the initial and final angular velocities and rotational inertia: $$\mathrm{\Delta }KE=\frac{1}{2}(I{\omega }_{final}^2-I{\omega }_{initial}^2)=\frac{1}{2}(0-\frac{1}{3}m{L}^2{\omega }_{initial}^2)$$ Now, we will equate this expression of KE_change to the one obtained in Step 3: $$\frac{1}{2}(-\frac{1}{3}m{L}^2{\omega }_{initial}^2)=\frac{mgL}{2}$$

Solve for the initial angular velocity and then find the initial linear speed

Divide both sides by m and L: $$-\frac{1}{3}L{\omega }_{initial}^2=g$$ Multiply both sides by -3: $$L{\omega }_{initial}^2=3g$$ Now, we find the initial angular velocity: $${\omega }_{initial}=\sqrt{\frac{3g}{L}}$$ Finally, we will find the initial linear speed (v_initial) by substituting the initial angular velocity back into the equation from Step 4: $$v_{initial}=L{\omega }_{initial}=L\sqrt{\frac{3g}{L}}$$ $$v_{initial}=\sqrt{3gL}$$ So, the maximum speed that should be imparted to the lower end of the rod to complete one full revolution is $\sqrt{3gL}$. Therefore, the correct answer is: $$\begin{gathered} C \\ \sqrt{(6gL)} \end{gathered}$$

Key concepts

Gravitational Potential Energy

Gravitational potential energy is the energy an object possesses because of its position in a gravitational field, usually related to its height above a reference point. In the exercise, this concept is key to understanding how energy is stored and converted as the rod moves from its lowest point to the highest point in its rotation.

When the rod is horizontal, its center of mass is at its initial position, and its gravitational potential energy is calculated as $$P{E}_{initial}=mg\left(\frac{L}{2}\right)$$
where

• $m$ is the mass of the rod,
• $g$ is the acceleration due to gravity,
• $L$ is the length of the rod.

As the rod completes a full circle, its center of mass ascends to the highest point, increasing its potential energy to $$P{E}_{final}=mgL$$. Thus, the change in potential energy, which equals the gain in gravitational potential energy, is vital in the energy conservation analysis.

Angular Velocity

Angular velocity represents how fast an object rotates or revolves, indicating the angle an object, such as the rod in the exercise, sweeps per unit time. It's crucial for determining how the speed imparted to the lower end translates to rotational motion.

At the rod's lowest point, we calculate its initial angular velocity using the relationship $${\omega }_{initial}=\frac{v_{initial}}{L}$$
Here, $v_{initial}$ is the linear speed given to the rod's end, and ${\omega }_{initial}$ is its angular velocity when the rod starts moving.

• This calculation involves converting linear speed at the tip to rotational speed around the axis.
• As the rod rotates and reaches its upright position, its final angular velocity drops to zero since the linear speed at the lower end halts.

Understanding angular velocity allows you to link linear speed to the rod’s rotation and predict its movement through different points in its trajectory.

Conservation of Energy

The principle of conservation of energy is foundational in physics and plays a crucial role in solving the exercise. It states that the total mechanical energy in a closed system remains constant if only conservative forces, like gravity, are acting.

In the context of the rotating rod, we need to account for both gravitational potential energy and kinetic energy. Initially, the rod's energy is a combination of these two forms. As it rotates upwards, potential energy increases while kinetic energy decreases, maintaining the balance as expressed by: $$\mathrm{\Delta }KE=\mathrm{\Delta }PE$$

• Here, $\mathrm{\Delta }KE$ is the change in kinetic energy, and $\mathrm{\Delta }PE$ is the change in potential energy.
• The decrease in kinetic energy exactly offsets the increase in potential energy, and vice versa.

This conservation principle underlies the calculation of initial speed, ensuring the rod’s rotation remains feasible and energy is appropriately redistributed as it swings up to its highest point.

Rotational Inertia

Rotational inertia, or moment of inertia, quantifies an object's resistance to changes in its rotational motion. It depends on the mass distribution relative to the axis of rotation. For the rod in the exercise, rotational inertia critically affects how bodily movements affect the energy conversion processes.

The rotational inertia of a rod about one end is given by $$I=\frac{1}{3}m{L}^2$$
In calculations, this inertial value helps determine the rod's angular motion dynamics. It's incorporated into the energy equations as $$\mathrm{\Delta }KE=\frac{1}{2}(I{\omega }_{final}^2-I{\omega }_{initial}^2)$$

• This relation shows how inertia and angular velocities determine the rotational kinetic energy changes.
• Larger inertia implies more energy is needed to change the angular velocity, affecting the initial speed required for a complete revolution.

Understanding rotational inertia aids in systematically analyzing how the rod's mass distribution influences its rotational kinematics and energy considerations.