Problem 60 - Fundamentals of Physics

Question

A \(1.0 \mathrm{~g}\) bullet is fired into a \(0.50 \mathrm{~kg}\) block attached to the end of a \(0.60 \mathrm{~m}\) nonuniform rod of mass \(0.50 \mathrm{~kg} .\) The block-rod-bullet system then rotates in the plane of the figure, about a fixed axis at \(A .\) The rotational inertia of the rod alone about that axis at \(A\) is \(0.060 \mathrm{~kg} \cdot \mathrm{m}^{2}\). Treat the block as a particle. (a) What then is the rotational inertia of the block-rod-bullet system about point \(A ?\) (b) If the angular speed of the system about \(A\) just after impact is \(4.5 \mathrm{rad} / \mathrm{s},\) what is the bullet's speed just before impact?

Step-by-Step Solution

Answer

The rotational inertia of the block-rod-bullet system is 0.2436 kg·m², and the bullet's speed just before impact is 109.62 m/s.

Step 1: Calculate the Moment of Inertia of the Block

The block can be treated as a point mass. The moment of inertia of a point mass is calculated using the formula \( I = mr^2 \), where \( m \) is the mass of the block, and \( r \) is the distance to the pivot point. Here, \( m = 0.50 \text{ kg} \) and \( r = 0.60 \text{ m} \). Thus, the moment of inertia is \( I_{block} = 0.50 \cdot (0.60)^2 \).

Step 2: Calculate the Moment of Inertia of the Bullet

The bullet is also treated as a point mass. Thus, the moment of inertia is given by \( I = mr^2 \), where \( m = 0.001 \text{ kg} \) (1.0 g converted to kg) and \( r = 0.60 \text{ m} \). Thus, the moment of inertia is \( I_{bullet} = 0.001 \cdot (0.60)^2 \).

Step 3: Sum the Moments of Inertia

The total rotational inertia of the system is the sum of the rotational inertia of the rod, block, and bullet. Thus, \( I_{total} = I_{rod} + I_{block} + I_{bullet} \), where \( I_{rod} = 0.060 \text{ kg} \cdot \text{m}^2 \). Calculate \( I_{total} \) using the results from Steps 1 and 2.

Step 4: Use Conservation of Angular Momentum

The angular momentum just before and just after the collision is conserved. Before impact, only the bullet has linear momentum which becomes rotational momentum after the collision. Set the initial angular momentum (\( I_{bullet} v_{bullet} \)) equal to the final angular momentum (\( I_{total} \omega \)), where \( \omega = 4.5 \text{ rad/s} \). Solve for \( v_{bullet} \).

Key Concepts

Angular Momentum

Angular momentum is a fundamental property in rotational dynamics, akin to linear momentum in linear dynamics. It indicates the quantity of rotation of an object and is conserved in an isolated system. Just like linear momentum is mass multiplied by velocity, angular momentum (\( L \) is the product of the moment of inertia (\( I \) and angular velocity (\( \omega \): \( L = I \omega \).

• In rotational scenarios, angular momentum accounts for the rotational motion of objects.
• It measures how much effort is needed to rotate an object around an axis.

In the original problem, the bullet's linear motion contributes to the angular momentum of the block-rod-bullet system once it embeds itself into the block.
Upon impact, the kinetic energy of the bullet is transformed into rotational motion, making angular momentum a crucial concept for solving how fast the bullet was initially moving.

Moment of Inertia

Moment of inertia is a measure of an object's resistance to changes in its rotational motion. In simpler terms, it tells us how difficult it is to change the state of rotation of an object. The moment of inertia depends on the mass distribution relative to the axis of rotation.

• For point masses, the moment of inertia (\( I \) is calculated as the product of the mass (\( m \) and the square of the distance (\( r \) from the axis of rotation: \( I = mr^2 \).
• The farther the mass is from the axis, the higher the moment of inertia, and vice versa.

In the given exercise, calculating the moment of inertia involves summing the contributions from each component: the rod, the block, and the bullet. This cumulative moment of inertia is essential in determining how the system will respond to torques and angular velocities.

Conservation of Angular Momentum

The principle of conservation of angular momentum states that if no external torque acts on a system, its total angular momentum remains constant. This concept is critical in understanding systems with colliding or interacting objects, such as in the problem at hand.

• Before impact, the bullet alone possesses linear momentum, which, upon collision, is converted to angular momentum in the block-rod-bullet system.
• After the collision, the total angular momentum remains the same, but is distributed over the entire system.

By applying the conservation of angular momentum, we can equate the initial angular momentum (from the bullet) to the final angular momentum (of the entire system). This relationship allows us to deduce the bullet's initial speed before the impact, knowing the final angular velocity and the system's moment of inertia after the collision.