Problem 77 - University Physics with Modern Physics

Question

A 5.00-kg ball is dropped from a height of 12.0 m above one end of a uniform bar that pivots at its center. The bar has mass 8.00 kg and is 4.00 m in length. At the other end of the bar sits another 5.00-kg ball, unattached to the bar. The dropped ball sticks to the bar after the collision. How high will the other ball go after the collision?

Step-by-Step Solution

Answer

The other ball will go approximately 1.87 meters high.

Step 1: Calculate the Potential Energy of the Dropped Ball

First, calculate the gravitational potential energy (PE) of the 5.00-kg ball before it is dropped using the formula: \[ PE = mgh \] where \( m = 5.00 \, \text{kg} \), \( g = 9.81 \, \text{m/s}^2 \), and \( h = 12.0 \, \text{m} \). Substitute the values: \[ PE = 5.00 \, \text{kg} \times 9.81 \, \text{m/s}^2 \times 12.0 \, \text{m} = 588.6 \, \text{J} \] This is the initial potential energy of the dropped ball.

Step 2: Find the Moment of Inertia of the System

Calculate the moment of inertia (I) for the system immediately before collision, which includes contribution from the bar and the two balls: For the bar (which pivots at its center), use the formula: \[ I_{\text{bar}} = \frac{1}{12} M L^2 \] For the ball that will stick, use: \[ I_{\text{sticking ball}} = m r^2 \] For the other ball at the end of the bar, also use: \[ I_{\text{other ball}} = m r^2 \] Substitute the values: \[ M = 8.00 \, \text{kg}, \; m = 5.00 \, \text{kg}, \; L = 4.00 \, \text{m}, \; r = \frac{L}{2} = 2.00 \, \text{m} \] \[ I_{\text{bar}} = \frac{1}{12} \times 8.00 \, \text{kg} \times (4.00 \, \text{m})^2 = 10.67 \, \text{kg} \cdot \text{m}^2 \] \[ I_{\text{sticking ball}} = 5.00 \, \text{kg} \times (2.00 \, \text{m})^2 = 20.00 \, \text{kg} \cdot \text{m}^2 \] \[ I_{\text{other ball}} = 5.00 \, \text{kg} \times (2.00 \, \text{m})^2 = 20.00 \, \text{kg} \cdot \text{m}^2 \] Thus, the total moment of inertia: \[ I_{\text{total}} = 10.67 + 20.00 + 20.00 = 50.67 \, \text{kg} \cdot \text{m}^2 \]

Step 3: Determine Angular Velocity After Collision

Since the collision is inelastic, conservation of angular momentum applies: Initial angular momentum = Final angular momentum For the initial angular momentum, only the falling ball contributes: \[ L_i = mvr = m \sqrt{2gh} \cdot r \] For the final angular momentum we have: \[ L_f = I_{\text{total}} \cdot \omega \] Set them equal and substitute the known quantities, solving for \( \omega \):\[ m \sqrt{2gh} \cdot r = I_{\text{total}} \cdot \omega \] \[ 5.00 \, \text{kg} \cdot \sqrt{2 \times 9.81 \, \text{m/s}^2 \times 12.0 \, \text{m}} \cdot 2.00 \, \text{m} = 50.67 \, \text{kg} \cdot \text{m}^2 \cdot \omega \] \[ 5.00 \cdot 15.34 \cdot 2.00 = 50.67 \cdot \omega \] \[ \omega = \frac{153.4}{50.67} \approx 3.03 \, \text{rad/s} \]

Step 4: Calculate the Maximum Height of the Other Ball

The other ball gains kinetic energy that is converted to potential energy at its highest point. Use: \[ mgh = \frac{1}{2} m v^2 \] The linear velocity \( v \) at the pivot is \( v = \omega r \), where \( r = 2.00 \, \text{m} \): \[ v = 3.03 \, \text{rad/s} \times 2.00 \, \text{m} = 6.06 \, \text{m/s} \] Substitute \( v \) in the energy equation to find \( h \): \[ 5.00 \, \text{kg} \times 9.81 \, \text{m/s}^2 \times h = \frac{1}{2} \times 5.00 \, \text{kg} \times (6.06 \, \text{m/s})^2 \] \[ 49.05 \times h = 91.83 \] \[ h = \frac{91.83}{49.05} \approx 1.87 \, \text{m} \] The other ball will rise approximately 1.87 meters.

Key Concepts

Gravitational Potential Energy

Gravitational potential energy (GPE) is the energy that an object possesses due to its position in a gravitational field. It depends on three key factors: the mass of the object, the gravitational field strength, and the height of the object above a reference point. To calculate it, use the formula:

• \( PE = mgh \)

Here, \( m \) is the mass, \( g \) is the acceleration due to gravity (around \( 9.81 \, \text{m/s}^2 \) on Earth), and \( h \) is the height.
This energy becomes important when an object is elevated, as it represents the potential to do work when the object falls. In the problem, a ball of mass 5.00-kg is dropped from a height of 12.0 m, resulting in an initial potential energy of \( 588.6 \, \text{J} \).
This energy is converted into other forms of energy during the collision and subsequent motion of the system. Understanding this conversion is crucial for analyzing collision dynamics.

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. For different shapes and mass distributions, the moment of inertia can vary greatly.

• For a rod pivoting at its center: \[ I_{\text{bar}} = \frac{1}{12} ML^2 \]
• For a point mass at a distance \( r \) from the pivot: \[ I_{\text{point mass}} = mr^2 \]

The system's total moment of inertia includes contributions from the bar and two balls, calculated by summing their individual moments of inertia. In our exercise, the total came out as \( 50.67 \, \text{kg} \cdot \text{m}^2 \).
This comprehensive value helps determine how the system will react to the forces applied during the collision.

Angular Momentum Conservation

In collision dynamics, particularly in rotational systems, the principle of angular momentum conservation is critical. It states that if no external torques act on a system, its total angular momentum remains constant. This principle is crucial in our exercise to transition from the state before to after the collision.
For the initial angular momentum, the focus is on contributions from objects in motion. Here, only the falling ball has angular momentum:

• \[ L_i = mvr = m \sqrt{2gh} \cdot r \]

After collision, the system's total moment of inertia and angular velocity determines the final angular momentum:

• \[ L_f = I_{\text{total}} \cdot \omega \]

By equating \( L_i \) and \( L_f \), and solving for \( \omega \), we found an angular velocity of about \( 3.03 \, \text{rad/s} \). This helps in assessing how the system will swing after impact.

Kinetic Energy Conversion

The conversion of energy types is a fundamental concept in physics. During the collision, gravitational potential energy converts into kinetic energy, allowing motion. This transfer isn't perfect, especially in inelastic collisions, where some energy may be lost.
In the exercise, the rotational motion of the system leads to kinetic energy being partially converted back to potential energy as the unattached ball rises.
To find how high the unattached ball will go, we use

• the formula: \[ mgh = \frac{1}{2} mv^2 \]
• where \( v = \omega r \).

This involves translating the calculated angular velocity into a linear velocity and subsequently determining the maximum height the ball achieves. With a velocity of \( 6.06 \, \text{m/s} \) at the pivot, the ball at the other end can rise to approximately \( 1.87 \, \text{m} \).
This calculation exemplifies the power of energy conservation in predicting the outcome of dynamic systems.