Problem 48 - University Physics with Modern Physics

Question

A thin uniform rod has a length of 0.500 m and is rotating in a circle on a frictionless table. The axis of rotation is perpendicular to the length of the rod at one end and is stationary. The rod has an angular velocity of 0.400 rad/s and a moment of inertia about the axis of 3.00 \(\times 10^{-3}\) kg \(\cdot\) m\(^2\). A bug initially standing on the rod at the axis of rotation decides to crawl out to the other end of the rod. When the bug has reached the end of the rod and sits there, its tangential speed is 0.160 m/s. The bug can be treated as a point mass. What is the mass of (a) the rod; (b) the bug?

Step-by-Step Solution

Answer

(a) Mass of rod: 0.036 kg; (b) Mass of bug: 0.003 kg.

Step 1: Understand Conservation of Angular Momentum

The angular momentum of the system (rod + bug) remains constant because there is no external torque acting on it. The initial angular momentum is equal to the final angular momentum. Let the initial angular velocity be \(\omega_0 = 0.400\, \text{rad/s}\) and the final angular velocity be \(\omega_f\).

Step 2: Establish the Initial Conditions

Initially, the bug is sitting at the axis, so the system's angular momentum \(L_i\) is determined by the rod alone: \[ L_i = I_\text{rod} \cdot \omega_0 = (3.00 \times 10^{-3}) \cdot 0.400 = 1.20 \times 10^{-3} \, \text{kg} \cdot \text{m}^2/\text{s} \]

Step 3: Establish the Final Conditions

When the bug reaches the end of the rod, the total moment of inertia is the sum of the rod's and the bug's contributions:\[ I_f = I_\text{rod} + m_\text{bug} \cdot R^2 \]where \(R = 0.500\, \text{m}\) is the length of the rod. The final angular momentum \(L_f\) is:\[ L_f = I_f \cdot \omega_f \]

Step 4: Determine the Mass of the Rod

First, solve for \(m_\text{rod}\). The rod's moment of inertia about one end is \( I_\text{rod} = \frac{1}{3}m_\text{rod}L^2 \):\[ 3.00 \times 10^{-3} = \frac{1}{3}m_\text{rod} \cdot (0.500)^2 \]Solving for \(m_\text{rod}\), yields:\[ m_\text{rod} = \frac{3.00 \times 10^{-3} \times 3}{0.500^2} = 0.036 \, \text{kg} \]

Step 5: Solve for the Final Angular Velocity

Using the bug's tangential speed \(v_t = 0.160\, \text{m/s}\) and \(v_t = \omega_f \times R\):\[ \omega_f = \frac{0.160}{0.500} = 0.320 \, \text{rad/s} \]

Step 6: Solve for the Mass of the Bug

Using the conservation of angular momentum, equate initial and final angular momentums:\[ 1.20 \times 10^{-3} = (3.00 \times 10^{-3} + m_\text{bug} \cdot 0.250) \cdot 0.320 \, \text{kg} \cdot \text{m}^2/\text{s} \]Solve for \(m_\text{bug}\):\[ 1.20 \times 10^{-3} = 0.960 \times 10^{-3} + 0.320 \cdot m_\text{bug} \cdot 0.250 \]\[ 0.24 \times 10^{-3} = 0.080 \cdot m_\text{bug} \]\[ m_\text{bug} = \frac{0.24 \times 10^{-3}}{0.080} = 3.00 \times 10^{-3} \, \text{kg} \]

Step 7: Final Thoughts

We computed \(m_\text{rod}\) using its moment of inertia definition and solved for \(\omega_f\) and subsequently \(m_\text{bug}\) using conservation of angular momentum.

Key Concepts

Moment of Inertia

The moment of inertia is a concept used to quantify how much an object resists rotational motion. Think of it like rotational mass, a property that determines how easily an object begins to spin or stops spinning. For our rod, its moment of inertia about one end is given by the formula:

• For a thin rod rotating about an end: \[ I_\text{rod} = \frac{1}{3} m_\text{rod} L^2 \]

Here, \(m_\text{rod}\) is the mass of the rod, and \(L\) is its length. This formula tells us that the ease with which the rod rotates is proportional to both its mass and the square of its length.

In the original exercise, the rod's moment of inertia was given as \(3.00 \times 10^{-3}\, \text{kg} \cdot \text{m}^2\). We used this information to solve for the rod's mass \(m_\text{rod}\) using moment of inertia and leveraging the known length of the rod.

Angular Velocity

Angular velocity depicts how fast something is spinning around an axis and is measured in radians per second (rad/s). If you picture a merry-go-round, the angular velocity tells you how quickly it's turning. In the given problem, the initial angular velocity of the rod is.

• \(\omega_0 = 0.400 \, \text{rad/s}\)

When the bug moves to the rod's end, it affects the system's angular velocity due to conservation of angular momentum. The final angular velocity \(\omega_f\) can be found using the relation between tangential speed \(v_t\) of the bug and the radius \(R\) of the rod:

• \[ \omega_f = \frac{v_t}{R} \]

Substituting \(v_t = 0.160 \, \text{m/s}\) and \(R = 0.500\, \text{m}\), we determined that \(\omega_f = 0.320 \, \text{rad/s}\). Understanding angular velocity helps to realize how motion changes, especially when external conditions like mass distribution change as with the movement of the bug.

Rotational Motion

Rotational motion was demonstrated through the rod moving in a circle around the stationary axis at one end. Here, Conservation of Angular Momentum is a key principle which states that in the absence of external torque, the total angular momentum of a system remains constant. This principle is represented by the equation:

• \(L_i = L_f\)

Let's break this down:

• The initial angular momentum \(L_i\) is based only on the rod: \[L_i = I_\text{rod} \cdot \omega_0\]
• The final angular momentum \(L_f\) involves both the rod and the bug: \[L_f = I_f \cdot \omega_f\]

As the bug crawls to the end of the rod, it changes the distribution of mass and thereby the total moment of inertia \(I_f\). This change affects the entire system's angular velocity. By using the formula for angular momentum and solving equations that include these variables, we can understand the intricate dance of rotational motion. This same principle helps solve the mass of the bug and the rod within the context of this rotating system.