Question

A thin, uniform metal bar, 2.00 m long and weighing 90.0 N, is hanging vertically from the ceiling by a frictionless pivot. Suddenly it is struck 1.50 m below the ceiling by a small 3.00-kg ball, initially traveling horizontally at 10.0 m/s. The ball rebounds in the opposite direction with a speed of 6.00 m/s. (a) Find the angular speed of the bar just after the collision. (b) During the collision, why is the angular momentum conserved but not the linear momentum?

Short answer

The angular speed of the bar is approximately 5.89 rad/s. Angular momentum is conserved due to the absence of external torques.

Step-by-step solution

Identify the Initial and Final States

This is an inelastic collision problem involving a ball and a bar. Before the collision, only the ball has linear momentum, moving towards the bar. After the collision, the ball has rebounded, and the bar gains angular velocity.

Calculate Initial Angular Momentum

The initial linear momentum of the ball is given by \( p_{i} = m \cdot v_{i} \), where \( m = 3.00 \text{ kg} \) and \( v_{i} = 10.0 \text{ m/s} \). Hence, \( p_{i} = 3.00 \cdot 10.0 = 30.0 \text{ kg m/s} \). The initial angular momentum \( L_{i} \) about the pivot is \( L_{i} = r \cdot p_{i} = 1.50 \cdot 30.0 = 45.0 \text{ kg m}^2/\text{s} \).

Calculate Final Angular Momentum

After the collision, the final linear momentum of the ball is \( p_{f} = m \cdot (-v_{f}) \), where \( v_{f} = 6.00 \text{ m/s} \) in the opposite direction. Thus, \( p_{f} = 3.00 \cdot (-6.00) = -18.0 \text{ kg m/s} \). The final angular momentum of the ball about the pivot is \( L_{f, ball} = r \cdot p_{f} = 1.50 \cdot (-18.0) = -27.0 \text{ kg m}^2/\text{s} \).

Apply the Conservation of Angular Momentum

The total initial angular momentum \( L_{i} \) must equal the total final angular momentum \( L_{f} = L_{f, ball} + L_{f, bar} \). Hence, \( 45.0 = -27.0 + L_{f, bar} \). Thus, \( L_{f, bar} = 72.0 \text{ kg m}^2/\text{s} \).

Calculate Angular Speed of the Bar

The moment of inertia \( I \) of the bar about the pivot is given by \( I = \frac{1}{3} mL^2 \). Here, \( m = \frac{90.0}{9.81} \approx 9.17 \text{ kg} \) and \( L = 2.00 \text{ m} \). Thus, \( I = \frac{1}{3} \times 9.17 \times 2^2 = 12.22 \text{ kg m}^2 \).\( L_{f, bar} = I \cdot \omega \), therefore, \( \omega = \frac{L_{f, bar}}{I} = \frac{72.0}{12.22} \approx 5.89 \text{ rad/s} \).

Reason for Conservation of Angular Momentum

Angular momentum is conserved because no external torques act on the system. Linear momentum is not conserved because there is an external force from the pivot, which applies a torque affecting linear momentum but not angular momentum.

Key concepts

Inelastic Collision

An inelastic collision is a type of collision in which the colliding objects stick together or move with some different velocities after impact. Although the total kinetic energy is not conserved due to the dissipation of energy in the form of heat, sound, or deformation, the total momentum is conserved.
In our exercise, a 3.00-kg ball hits a hanging metal bar. During the collision, the ball changes direction, indicating an inelastic property since kinetic energy is not conserved.
This conclusively demonstrates a real-world example of an inelastic collision, where motion and some energy transform within the structures involved.

Moment of Inertia

The moment of inertia is an essential concept in rotational mechanics. It measures an object's resistance to changes in its rotation rate. You can think of it as the rotational equivalent of mass in linear motion.
In the exercise, the moment of inertia of the bar is calculated to see how it impacts the bar's angular velocity post-collision. Using the formula: \[ I = \frac{1}{3} m L^2 \] where \( m \) is the mass of the bar and \( L \) is its length, we find the bar's moment of inertia is 12.22 kg m².
This is crucial because it helps determine how the impact from the ball affects the bar’s rotation speed.

Conservation of Momentum

In every collision, like the one between the ball and the bar, the principle of conserving momentum plays a pivotal role. This principle states that the total momentum before the collision must equal the total momentum after, as long as no external forces act upon the system.
Even though the linear momentum of our system isn't conserved due to the external pivot force, the angular momentum is conserved. This is because no external torque is acting on the system during the collision.

• Total initial angular momentum: 45.0 kg m²/s
• Final angular momentum shared by the rebounding ball and spinning bar: 45.0 kg m²/s

This conservation ensures we can accurately solve for post-collision angular velocity.

Angular Velocity

Angular velocity refers to how fast an object spins around an axis. This is a direct outcome of the angular momentum for a rotating system like our metal bar.
In our case, post-collision, the bar has gained angular velocity, which we calculate using:\[ \omega = \frac{L_{f, \text{bar}}}{I} \]where \( L_{f, \text{bar}} \) is the final angular momentum of the bar and \( I \) is the moment of inertia. This results in an angular velocity of approximately 5.89 rad/s.
Understanding this calculation shows how the change in form of momentum manifests as a pace in the object's rotation after being struck by another moving object.