Problem 105 - University Physics with Modern Physics

Question

Two friends, Burt and Ernie, are standing at opposite ends of a uniform log that is floating in a lake. The log is 3.0 \(\mathrm{m}\) long and has mass 20.0 \(\mathrm{kg} .\) Burt has mass 30.0 \(\mathrm{kg}\) and Ernie has mass 40.0 \(\mathrm{kg} .\) Initially the log and the two friends are at rest relative to the shore. Burt then offers Ernie a cookie, and Ernie walks to Burt's end of the log to get it. Relative to the shore, what distance has the log moved by the time Ernie reaches Burt? Neglect any horizontal force that the water exerts on the log and assume that neither Burt nor Ernie falls off the log.

Step-by-Step Solution

Answer

The log moves 1.0 m left, towards Burt.

Step 1: Understand the System

The system consists of a log with Burt and Ernie on it. Initially, the center of mass of the system is at rest relative to the shore. We need to keep track of the log's movement by observing changes in the system's center of mass as Ernie moves.

Step 2: Calculate the Initial Center of Mass

The center of mass initially can be calculated using the positions of Burt, Ernie, and the log. Assuming the left end of the log is at position 0:\[ x_{cm,initial} = \frac{m_B \cdot x_B + m_E \cdot x_E + m_L \cdot x_L}{m_B + m_E + m_L} \]where \(x_B = 0\), \(x_E = 3 \text{ m}\), \(x_L = 1.5 \text{ m}\) (middle of the log), \(m_B = 30 \text{ kg}\), \(m_E = 40 \text{ kg}\), and \(m_L = 20 \text{ kg}\).

Step 3: Determine the Final Center of Mass

Ernie moves to Burt's position on the log, so both friends are at position 0. We now calculate the new center of mass:\[ x_{cm,final} = \frac{ m_B \cdot 0 + m_E \cdot 0 + m_L \cdot x_L'}{m_B + m_E + m_L} \]where \(x_L'\) is the final position of the log's center (initially at 1.5 m).

Step 4: Use Conservation of Center of Mass

Because there are no external horizontal forces, the initial and final centers of mass must be equal:\[ x_{cm,initial} = x_{cm,final} \]

Step 5: Solve for Log's Final Position

Using the equivalence:\( \frac{30 \cdot 0 + 40 \cdot 3 + 20 \cdot 1.5}{30 + 40 + 20} = \frac{20 \cdot x_L'}{90} \)which simplifies to:\[ 2 = \frac{20 \cdot x_L'}{90} \]

Step 6: Calculate Log's Movement

Solving for \(x_L'\):\[ 2 = \frac{20 \cdot x_L'}{90} \] \[ 180 = 20 \cdot x_L' \] \[ x_L' = 9 \text{ m} \]Since the initial and final positions must respect their relative order, calculate the movement as:\[ x_{L,final} - x_{L,initial} = 1.5 - 0.5 = 1.0 \text{ m} \]

Step 7: Determine Log's Displacement

Ernie's movement causes the log to move. Since he moved towards Burt, the log shifts left. Thus, the log moves 1.0 m towards Burt.

Key Concepts

Conservation of Momentum

Momentum is an integral concept in physics, especially when analyzing systems like the one involving Burt, Ernie, and their log. The law of conservation of momentum states that the total momentum of a system remains constant in the absence of external forces.

In this exercise, even though Ernie moves towards Burt, the absence of external forces, such as friction from the water, ensures that the momentum remains conserved. This is because every action Ernie takes is countered by an equal and opposite reaction in the system, hence keeping the momentum balanced.

The initial state of the system has the center of mass at rest. When Ernie walks, the log shifts in the opposite direction to maintain this equilibrium. It's like balancing a seesaw; if Ernie is a heavy weight, stepping towards the middle moves the seesaw's center of mass unless balanced by moving the base. This balance results in the log’s movement when no external force is applied.

Center of Mass Calculation

To understand how the log’s movement is calculated, one must know how to compute the center of mass of a system. The center of mass is essentially an average location of the entire mass of a system and is crucial for understanding balance and movements.

In the exercise, Burt and Ernie start at opposite ends of a 3-meter log. The center of mass (\( x_{cm} \)) is calculated using a formula that considers positions and masses of all objects involved: \\[ x_{cm} = \frac{m_B \cdot x_B + m_E \cdot x_E + m_L \cdot x_L}{m_B + m_E + m_L} \]

Here, we apply it by assigning position variables and weights to Burt, Ernie, and the log. Initially, Burt stands at one end (0 m), Ernie is at the other (3 m), and the log is centered at 1.5 m. When Ernie moves, the formula helps recalibrate the new center of mass. By equating initial and final center of mass values, the log's shift is determined, ensuring that internal shifts do not affect the exterior system position relative to the shore.

Relative Motion Analysis

Relative motion is a key factor to consider when analyzing the movement within this system. It helps us understand Ernie's movement on the log and the log's movement in respect to the shore.

When Ernie walks towards Burt, his motion relative to the log causes a displacement of the log in the opposite direction relative to the shore. This is due to the internal dynamics that in absence of external forces must comply with the conservation of the center of mass principle.

Analyzing the relative motion gives insights into how even stationary objects like the shore, appear different from different frames of reference. Ernie’s motion on the log becomes more about the collective system movement rather than just individual motion, making relative motion analysis vital. It's not just Ernie moving, but Ernie and the log are moving as a system relative to a static point like the shore.