Problem 93 - University Physics with Modern Physics

Question

A target in a shooting gallery consists of a vertical square wooden board, 0.250 \(\mathrm{m}\) on a side and with mass 0.750 \(\mathrm{kg}\) , that pivots on a horizontal axis along its top edge. The board is struck face-on at its center by a bullet with mass 1.90 \(\mathrm{g}\) that is traveling at 360 \(\mathrm{m} / \mathrm{s}\) and that remains embedded in the board. (a) What is the angular speed of the board just after the bullet's impact? (b) What maximum height above the equilibrium position does the center of the board reach before starting to swing down again? (c) What minimum bullet speed would be required for the board to swing all the way over after impact?

Step-by-Step Solution

Answer

(a) \(5.47 \text{ rad/s}\); (b) \(0.0319 \text{ m}\); (c) \(31.15 \text{ m/s}\)

Step 1: Calculate Moment of Inertia

First, calculate the moment of inertia of the board about the pivot point using the formula for a rectangle pivoting about one edge. The formula is \( I = \frac{1}{3} m l^2 \), where \( m = 0.750 \, \text{kg} \) and \( l = 0.250 \, \text{m} \).\[ I = \frac{1}{3} \times 0.750 \times (0.250)^2 = 0.01563 \, \text{kg} \cdot \text{m}^2 \]

Step 2: Use Conservation of Angular Momentum

Use conservation of angular momentum to find the angular speed \( \omega \) just after the collision. The initial angular momentum is the linear momentum of the bullet \( mv \) times the distance to the pivot.\[ L_{\text{initial}} = mvx = 0.0019 \times 360 \times 0.125 = 0.0855 \, \text{kg} \cdot \text{m}^2/\text{s} \]The angular momentum is conserved, so \( I \omega = 0.0855 \). Solving for \( \omega \):\[ \omega = \frac{0.0855}{0.01563} = 5.47 \, \text{rad/s} \]

Step 3: Calculate Maximum Height Using Energy Conservation

Convert the rotational kinetic energy to gravitational potential energy to find the maximum height. The initial kinetic energy is \( \frac{1}{2} I \omega^2 \). The potential energy at maximum height is \( mgh \). Equate them to solve for \( h \).\[ \frac{1}{2} \times 0.01563 \times (5.47)^2 = 0.750 \times 9.8 \times h \]\[ h = \frac{0.2344}{7.35} \approx 0.0319 \, \text{m} \]

Step 4: Find Minimum Bullet Speed for Full Rotation

For the board to complete a full swing over, it must have enough energy to reach the point where its center of mass is at the top of the circle. This requires the kinetic energy to be equal to the gravitational potential energy at the height of \( l \).\[ \frac{1}{2} I \omega^2 = mgh \to mv^2 = mgl \, \text{for center mass at the top:} \]\[ 1.90 \cdot 10^{-3} \cdot v^2 = 0.750 \cdot 9.8 \cdot 0.250 \]Solve for \( v \):\[ v^2 = \frac{0.750 \cdot 9.8 \cdot 0.250}{1.90 \cdot 10^{-3}} \approx 971.05 \]\[ v = 31.15 \, \text{m/s} \]

Step 5: Final Calculation: Minimum Bullet Speed

The minimum bullet speed required for the board to swing over is \( 31.15 \, \text{m/s} \).

Key Concepts

Angular Momentum Conservation

Angular momentum conservation is a core principle in physics. It states that the total angular momentum of an isolated system remains constant if no external torque acts on it. In our exercise, this principle helps determine the angular speed of the board after it is struck by the bullet. The bullet contributes its linear momentum, which translates into angular momentum because it hits the board off-center.

• Before the impact, the bullet's linear momentum is calculated as the product of its mass and velocity, resulting in a certain amount of angular momentum.
• At the moment of impact, the bullet embeds itself in the board. Hence, the total angular momentum remains the same before and just after the collision.
• This conservation allows us to calculate the board's angular speed by equating the initial angular momentum with the final angular momentum of the system.

Understanding this principle is vital, as it shows how the motion and interaction of objects can be analyzed when forces like impacts occur.

Moment of Inertia

The moment of inertia is akin to the rotational analog of mass in linear motion. It tells us how much an object resists angular acceleration when a torque is applied. For a rectangular board pivoting at an edge, the moment of inertia depends on the mass and size of the board. To compute the moment of inertia, the formula used is \[I = \frac{1}{3} m l^2\]where:

• \( m \) is the mass of the board, and
• \( l \) is the length of the board's side.

This formula stems from how mass is distributed relative to the axis of rotation. In essence, the farther the mass is distributed from the axis, the greater the moment of inertia, requiring more effort to change its rotational state. Spanning from simple shapes to complex bodies, understanding the moment of inertia assists in predicting how objects will rotate under various conditions.

Energy Conservation

Energy conservation is a vital concept that helps us track how energy flows in a system. In rotational dynamics, kinetic and potential energy transformations occur during motion. After the bullet impacts the board, energy conservation principles help to find the maximum height the board reaches as it swings up. Initially, the system possesses rotational kinetic energy obtained by the bullet’s impact, calculated using:\[\text{Kinetic Energy} = \frac{1}{2} I \omega^2\]Then, as the board rises, this kinetic energy converts into gravitational potential energy at the height of its swing:\[\text{Potential Energy} = mgh\]Setting these energies equal allows for calculating the maximum height:

• The higher the board swings, the more kinetic energy is converted to potential energy and vice versa when descending.

This transformation showcases how energy is conserved and transposed between forms, vital in predicting outcomes in various physical phenomena, like swings or pendulums.