Problem 91 - 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 angeed 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

(a) Final angular speed \( \omega_f = \frac{6v}{19L} \); (b) Energy ratio is \( \frac{19}{12} \).

Step 1: Determine the Masses

Let the mass of the rod be \( M \), and the mass of the bullet be \( \frac{1}{4}M \). The bullet strikes the center of the rod, embedding itself to become a part of the system.

Step 2: Apply Conservation of Linear Momentum

Before the collision, the linear momentum of the system is only due to the bullet, given by \( \frac{1}{4}Mv \) (where \( M \) is the mass of the rod and \( v \) is the speed of the bullet). After the collision, the bullet and rod move together. The combined mass is \( M + \frac{1}{4}M = \frac{5}{4}M \). Let \( v_f \) be the velocity of the system's center of mass immediately after the collision.Using conservation of linear momentum:\[ \frac{1}{4}Mv = \frac{5}{4}M v_f \]Solving for \( v_f \):\[ v_f = \frac{1}{5}v \]

Step 3: Apply Conservation of Angular Momentum

For angular momentum conservation, calculate the initial angular momentum provided by the bullet impact. Let the pivot be point \( O \), and the bullet strikes at the center of rod at a distance \( \frac{L}{2} \) from \( O \).Initial angular momentum is \( \frac{1}{4}Mv \cdot \frac{L}{2} = \frac{1}{8}MLv \).After the collision, the system rotates with angular velocity \( \omega_f \). The moment of inertia of the rod about the pivot is \( \frac{1}{3}ML^2 \) (for the rod) and for the bullet turned part of rod motion after collision is \( \frac{1}{16}ML^2 \).Thus, the total moment of inertia \( I \) is:\[ I = \frac{1}{3}ML^2 + \frac{1}{16}ML^2 = \frac{19}{48}ML^2 \]Using angular momentum conservation:\[ \frac{1}{8}MLv = \left( \frac{19}{48}ML^2 \right)\omega_f \]Solving for \( \omega_f \):\[ \omega_f = \frac{6v}{19L} \]

Step 4: Compute Kinetic Energy Before and After

Before the collision, the kinetic energy (KE) is due to the bullet:\[ KE_{\text{initial}} = \frac{1}{2}\left( \frac{1}{4}M \right)v^2 = \frac{1}{8}Mv^2 \]After the collision, the kinetic energy is rotational kinetic energy:\[ KE_{\text{final}} = \frac{1}{2}\left( \frac{19}{48}ML^2 \right) \left( \frac{6v}{19L} \right)^2 = \frac{1}{2} \times \frac{19}{48}ML^2 \times \frac{36v^2}{361L^2} \]Simplifying gives:\[ KE_{\text{final}} = \frac{19 \times 18}{48 \times 361}Mv^2 = \frac{19}{96}Mv^2 \]

Step 5: Calculate the Ratio of Kinetic Energies

The ratio of the kinetic energy after the collision to that before is given by:\[ \text{Ratio} = \frac{KE_{\text{final}}}{KE_{\text{initial}}} = \frac{\frac{19}{96}Mv^2}{\frac{1}{8}Mv^2} \]Simplifying the expression:\[ \frac{19}{96} \times \frac{8}{1} = \frac{19}{12} \]

Key Concepts

Moment of Inertia

The moment of inertia is a fundamental concept in rotational dynamics. It measures an object's resistance to changes in its rotational motion. Consider it the rotational equivalent of mass in linear motion. For any rotating body, the moment of inertia depends on how the mass is distributed relative to the axis of rotation.

In our rod and bullet system, the original problem calculates the total moment of inertia after the bullet becomes embedded in the rod. The rod, pivoting around one end, has a moment of inertia given by \( \frac{1}{3}ML^2 \). However, once the bullet embeds itself at the rod's center, it alters the system's moment of inertia. The bullet adds its own contribution, \( \frac{1}{16}ML^2 \), bringing the total to \( \frac{19}{48}ML^2 \). For clarity, notice how the moment of inertia is higher when mass, like the bullet, is away from the axis, affecting how easy or hard it is to initiate rotation.

Kinetic Energy Transformation

Kinetic energy transformation occurs during the collision between the bullet and the rod. Initially, the kinetic energy exists entirely as linear kinetic energy in the moving bullet. This can be calculated as\( KE_{\text{initial}} = \frac{1}{8}Mv^2 \).

After the collision, as the bullet becomes part of the rod, the system gains rotational kinetic energy. The final rotational kinetic energy of the newly formed bullet-rod system is given by \( KE_{\text{final}} = \frac{19}{96}Mv^2 \). This conversion showcases the laws of conservation: the energy isn't lost but transformed from linear motion of the bullet alone to rotational motion of the rod-bullet system.

Understanding these transformations helps us see how energy is conserved and redistributed in mechanical processes, a vital part of physics that explains many real-world systems beyond this textbook exercise.

Rotational Dynamics

Rotational dynamics provides insight into systems experiencing rotational motion. It's governed by angular analogues to linear dynamics principles, such as Newton's laws. Conservation of angular momentum plays a critical role. It states that in a closed system with no external torques, the total angular momentum remains constant.

Initially, the angular momentum comes solely from the bullet, hitting the rod's center at a perpendicular angle. It is calculated as \( \frac{1}{8}MLv \). During the collision, the system's angular momentum post-collision remains the same, and calculating the final angular velocity \( \omega_f = \frac{6v}{19L} \), it ensures this conservation law holds true.

• Rotational motion is more complex than linear motion due to the influence of torque, moment of inertia, and angular velocity.
• The final rotational state depends heavily on where and how the colliding object interacts with another.
• Understanding these dynamics provides insight into how real-world systems like machinery, wheels, or even rollercoasters work.