Problem 79 - University Physics with Modern Physics

Question

A uniform rod of length \(L\) rests on a frictionless horizontal surface. The rod pivots about a fixed frictionless axis at one end. The rod is initially at rest. A bullet traveling parallel to the horizontal surface and perpendicular to the rod with speed v strikes the rod at its center and becomes embedded in it. The mass of the bullet is one-fourth the mass of the rod. (a) What is the final angular speed of the rod? (b) What is the ratio of the kinetic energy of the system after the collision to the kinetic energy of the bullet before the collision?

Step-by-Step Solution

Answer

Final angular speed is \(\omega_f = \frac{3v}{13L}\); energy ratio is about 0.073.

Step 1: Understand the System Setup

We have a rod of length \(L\) and mass \(M\) pivoted at one end, initially at rest, and a bullet of mass \(m\), where \(m = \frac{M}{4}\), striking the rod at its center. The speed of the bullet before impact is \(v\). After the bullet embeds in the rod, we need to find the system's final angular speed \(\omega_f\) and the ratio of kinetic energies.

Step 2: Apply Conservation of Angular Momentum

The system is isolated, so angular momentum is conserved. The initial angular momentum just before the collision, \(L_i\), is due to the bullet: \[ L_i = m \cdot v \cdot \left(\frac{L}{2}\right) \]The moment of inertia of the rod about the pivot is \(I = \frac{1}{3}ML^2\), and the bullet adds \(m\left(\frac{L}{2}\right)^2\) to it, so the final moment of inertia \(I_f\) is:\[ I_f = \frac{1}{3}ML^2 + m\left(\frac{L}{2}\right)^2 \]Using the conservation principle:\[ L_i = I_f \cdot \omega_f \]Substitute expressions to solve for \(\omega_f\):\[ m \cdot v \cdot \left(\frac{L}{2}\right) = \left(\frac{1}{3}ML^2 + m\left(\frac{L}{2}\right)^2\right) \cdot \omega_f \]

Step 3: Solve for Final Angular Speed

Substitute \(m = \frac{M}{4}\):\[ \left(\frac{M}{4}\right) \cdot v \cdot \left(\frac{L}{2}\right) = \left(\frac{1}{3}ML^2 + \frac{M}{4}\left(\frac{L}{2}\right)^2\right) \cdot \omega_f \]Solve for \(\omega_f\):\[ \omega_f = \frac{\left(\frac{M}{4}\right) \cdot v \cdot \left(\frac{L}{2}\right)}{\frac{1}{3}ML^2 + \frac{M}{4}\left(\frac{L^2}{4}\right)} \]Simplifying gives:\[ \omega_f = \frac{3v}{13L} \]

Step 4: Determine Initial Kinetic Energy of the Bullet

The initial kinetic energy \(K_{i}\) of the bullet is:\[ K_{i} = \frac{1}{2}m v^2 = \frac{1}{2} \cdot \left(\frac{M}{4}\right) \cdot v^2 = \frac{Mv^2}{8} \]

Step 5: Determine Final Kinetic Energy of the System

The final kinetic energy \(K_f\) is rotational:\[ K_f = \frac{1}{2} I_f \omega_f^2 = \frac{1}{2} \left(\frac{1}{3}ML^2 + \frac{M}{4} \left(\frac{L^2}{4}\right)\right) \left(\frac{3v}{13L}\right)^2 \]Plug in \(I_f = \frac{13}{48}ML^2\): \[ K_f = \frac{1}{2} \cdot \frac{13}{48}ML^2 \cdot \left(\frac{9v^2}{169L^2}\right) = \frac{13 \times 9Mv^2}{2 \times 48 \times 169} = \frac{13 \times 9Mv^2}{16272} \]

Step 6: Calculate the Ratio of Final to Initial Kinetic Energy

Given the expressions for \(K_i\) and \(K_f\), we find the ratio \( \frac{K_f}{K_i} \):\[ \frac{K_f}{K_i} = \frac{\frac{117Mv^2}{16272}}{\frac{Mv^2}{8}} = \frac{117 \times 8}{16272} = \frac{936}{16272} = \frac{1}{13.68} \approx 0.073 \]

Step 7: Final Step: Conclusions

The final angular speed of the rod after the collision is \(\frac{3v}{13L}\), and the ratio of the kinetic energy after the collision to the kinetic energy of the bullet before the collision is approximately 0.073.

Key Concepts

Rotational Dynamics

Rotational dynamics is the branch of mechanics that deals with the motion of rotating systems and the forces and torques that produce this motion. It is analogous to linear dynamics but involves rotational quantities such as angular velocity, angular acceleration, and moments of inertia.
When dealing with rotational dynamics, it's important to understand a few key concepts:

• **Angular Motion:** Just like linear motion, rotational dynamics is governed by angular analogs such as angular displacement, angular velocity, and angular acceleration.
• **Rotational Motion:** This refers to an object revolving around an axis. In our exercise, the rod rotates about its pivot point.
• **Torque:** In rotational motion, torque plays a role similar to force. It is the rotational equivalent of linear force and is calculated as the product of force and the distance from the point of rotation.
• **Work-Energy Theorem:** In rotations, the work done on an object is related to its change in kinetic energy.

Understanding these concepts helps in solving problems involving rotating bodies by considering angular analogs of linear quantities and ensuring all forces and torques in the system are accounted for.

Angular Speed

Angular speed measures how quickly an object rotates or revolves around an axis and is typically expressed in radians per second (rad/s). In our scenario, after the collision, the rod and the embedded bullet have a common angular speed as they spin about the pivot.
To calculate angular speed, consider the conservation of angular momentum. Before the collision, the bullet has linear momentum that converts into angular momentum when it embeds into the rod:

• **Initial Angular Momentum:** Calculated from the bullet's speed and distance to the pivot point.
• **Final Angular Momentum:** The conserved quantity now spread across both the rod and the embedded bullet.
• The formula is expressed as: \[ L_i = I_f \cdot \omega_f \] where \( L_i \) is the initial angular momentum and \( I_f \) is the final moment of inertia.
• **Final Angular Speed (\( \omega_f \)):** Obtained by solving the equation above, reflecting how the angular momentum is dispersed through the system post-collision.

By using the appropriate formulas, we calculate the angular speed as \( \omega_f = \frac{3v}{13L} \), showing the correspondence between the velocity of the bullet and the resulting rotation of the system.

Moment of Inertia

The moment of inertia is a measure of an object's resistance to changes in its rotational motion. It depends on the mass distribution relative to the axis of rotation. In this exercise, understanding the combined moment of inertia of the rod and bullet is crucial for calculating the final rotational speed.
Here is what you need to know about the moment of inertia in this context:

• **Rod's Moment of Inertia:** For a rod rotating about one end, it is calculated as \( I_{rod} = \frac{1}{3}ML^2 \).
• **Bullet's Contribution:** On hitting the rod, the bullet adds its own moment of inertia, calculated using \( m(\frac{L}{2})^2 \), as it strikes at the midpoint.
• **Combined Moment of Inertia:** The sum of individual moments gives us the final effective moment of inertia, \( I_f = \frac{1}{3}ML^2 + m(\frac{L}{2})^2 \).
• **Role in Rotation:** The combined moment plays a crucial role in defining how fast the system will rotate post-collision, governing how effectively the system resists rotational changes.

Applying these calculations helps us determine how the bullet's impact alters the rotational characteristics of the rod, making the moment of inertia a fundamental concept in understanding rotational dynamics and angular momentum conservation.