Problem 97 - University Physics with Modern Physics

Question

A horizontal plywood disk with mass 7.00 \(\mathrm{kg}\) and diameter 1.00 \(\mathrm{m}\) pivots on frictionless bearings about a vertical axis through its center. You attach a circular model-railroad track of negligible mass and average diameter 0.95 m to the disk. A 1.20 -kg, battery-driven model train rests on the tracks. To demonstrate conservation of angular momentum, you switch on the train's engine. The train moves counterclockwise, soon attaining a constant speed of 0.600 \(\mathrm{m} / \mathrm{s}\) relative to the tracks. Find the magnitude and direction of the angular velocity of the disk relative to the earth.

Step-by-Step Solution

Answer

The disk rotates clockwise with an angular velocity of 0.387 rad/s.

Step 1: Understand the Initial Conditions

We start with a horizontal plywood disk of mass \(7.00 \, \text{kg}\) and diameter \(1.00 \, \text{m}\), thus leading to a radius \(R_d = 0.50 \, \text{m}\). A train of mass \(1.20 \, \text{kg}\) is on a track with average diameter \(0.95 \, \text{m}\), so its radius \(R_t = 0.475 \, \text{m}\). Initially, the system is at rest.

Step 2: Calculate the Moment of Inertia

The moment of inertia of the disk \(I_d\) is given by the formula for a circular disk: \[ I_d = \frac{1}{2} M R_d^2 \]Plugging in the values: \[ I_d = \frac{1}{2} \times 7.00 \, \text{kg} \times (0.50 \, \text{m})^2 = 0.875 \, \text{kg} \, \text{m}^2 \]

Step 3: Angular Momentum of the Train

When the train reaches a speed of \(0.600 \, \text{m/s}\), its linear velocity \(v_t\) is related to its angular velocity \(\omega_t\) by the relation \(v_t = R_t \omega_t\). Solving for \(\omega_t\):\[ \omega_t = \frac{v_t}{R_t} = \frac{0.600 \, \text{m/s}}{0.475 \, \text{m}} = 1.263 \, \text{rad/s} \]The angular momentum of the train \(L_t\) is given by:\[ L_t = m_t R_t^2 \omega_t = 1.20 \, \text{kg} \times (0.475 \, \text{m})^2 \times 1.263 \, \text{rad/s} = 0.339 \, \text{kg} \, \text{m}^2/\text{s} \]

Step 4: Apply Conservation of Angular Momentum

Since the system was initially at rest and is closed, the conservation of angular momentum states that the sum of angular momentum is zero:\[ L_d + L_t = 0 \]Where \(L_d\) is the angular momentum of the disk with angular velocity \(\omega_d\). This gives us:\[ I_d \omega_d + L_t = 0 \]Solving for \(\omega_d\):\[ \omega_d = -\frac{L_t}{I_d} = -\frac{0.339 \, \text{kg} \, \text{m}^2/\text{s}}{0.875 \, \text{kg} \, \text{m}^2} = -0.387 \, \text{rad/s} \]

Step 5: Determine the Direction

The negative sign indicates that the direction of the disk's angular velocity is opposite to that of the train's motion. Thus, if the train moves counterclockwise, the disk must rotate clockwise relative to the Earth.

Key Concepts

Moment of Inertia

The moment of inertia is an essential concept in understanding rotation. It tells us how difficult it is to change an object's rotation rate. Think of it as the rotational equivalent of mass in linear motion.
It depends on how the mass is distributed in relation to the axis of rotation. For a circular disk, like our plywood disk, the formula is:

• \( I_d = \frac{1}{2} M R_d^2 \)

Here, \(M\) is the mass and \(R_d\) is the radius of the disk.
In our example, substituting the mass and radius gives \(I_d = 0.875 \, \text{kg m}^2\). This tells us how the disk would resist changes to its rotational speed.
When multiple objects interact, like our disk and the train on a circular track, each has its moment of inertia. This value is critical for calculating angular momentum and understanding how they influence each other in rotational motion.

Angular Velocity

Angular velocity is a measure of how fast an object rotates or revolves relative to another point. It's the rotational equivalent of linear velocity. In our exercise, it's used to describe how quickly the train and the disk rotate.
The train's angular velocity, \( \omega_t \), is determined from its linear speed \( v_t \) and track radius \( R_t \):

• \( v_t = R_t \omega_t \)
• \( \omega_t = \frac{v_t}{R_t} \)

For the train moving at \(0.600 \, \text{m/s}\) on a \(0.475 \, \text{m}\) radius track, \(\omega_t\) evaluated to \(1.263 \, \text{rad/s}\).
The disk's angular velocity, \( \omega_d \), was found using conservation laws, calculated as \(-0.387 \, \text{rad/s}\), indicating it rotates opposite to the train. Angular velocities are crucial as they help understand how different parts of a system move relative to each other.

Angular Momentum

Angular momentum is a pivotal concept reflecting the rotational motion's quantity, similar to how linear momentum reflects straight motion. It's the product of an object's moment of inertia and its angular velocity.
In rotational systems, like our disk-and-train setup, it's essential for understanding equilibrium and dynamics. The angular momentum of the train was:

• \( L_t = m_t R_t^2 \omega_t \)

Solving gives \(0.339 \, \text{kg} \, \text{m}^2/\text{s}\) for the train. According to the conservation of angular momentum:

• \( L_d + L_t = 0 \)

This means no external torques act, so the initial and final angular momentum is balanced. As the train moves counterclockwise, it gains positive angular momentum, making the disk rotate clockwise to maintain the balance of the system. This exchange and conservation highlight how connected components of a system react to each other's motions. Understanding this can help solve complex problems involving rotational motion in physics.