Problem 47 - Fundamentals of Physics

Question

A track is mounted on a large wheel that is free to turn with negligible friction about a vertical axis (Fig. \(11-48\) ). A toy train of mass \(m\) is placed on the track and, with the system initially at rest,the train's electrical power is turned on. The train reaches speed \(0.15 \mathrm{~m} / \mathrm{s}\) with respect to the track. What is the wheel's angular speed if its mass is \(1.1 \mathrm{~m}\) and its radius is \(0.43 \mathrm{~m} ?\) (Treat it as a hoop, and neglect the mass of the spokes and hub.)

Step-by-Step Solution

Answer

The wheel's angular speed is approximately 0.317 rad/s.

Step 1: Identify Relevant Physical Concepts

The law of conservation of angular momentum states that the initial angular momentum of a system is equal to its final angular momentum if no external torques are acting on it. Here, we will apply this principle since the system starts from rest and remains isolated.

Step 2: Define Variables and Known Values

Let the mass of the train be \( m \), the mass of the wheel be \( M = 1.1m \), the radius of the wheel be \( R = 0.43 \ \mathrm{m} \), and the speed of the train relative to the track be \( v = 0.15 \ \mathrm{m/s} \). We need to find the angular speed \( \omega \) of the wheel.

Step 3: Express Initial Angular Momentum

Initially, the system is at rest, so initial angular momentum \( L_i = 0 \).

Step 4: Express Final Angular Momentum of the System

After the train starts moving, it has linear momentum and hence contributes to the system's angular momentum. The angular momentum due to the train is given by \( L_{train} = m v R \). The wheel turns with angular speed \( \omega \) and has moment of inertia \( I \) for a hoop, which is \( I = MR^2 \). The angular momentum of the wheel is \( L_{wheel} = I \omega = MR^2 \omega \).

Step 5: Apply Conservation of Angular Momentum

Since the initial angular momentum was zero, the total final angular momentum must also be zero. Therefore, set up the equation: \( L_{train} + L_{wheel} = 0 \). This gives us: \[ m v R - M R^2 \omega = 0 \]

Step 6: Solve for Angular Speed \( \omega \)

Re-arranging the equation gives: \( m v R = M R^2 \omega \). Solving for \( \omega \) gives: \[ \omega = \frac{m v}{MR} \]. Substituting \( M = 1.1m \) and simplifying, we have: \[ \omega = \frac{v}{1.1 R} \]. Substituting \( v = 0.15 \, \mathrm{m/s} \) and \( R = 0.43 \, \mathrm{m} \), we find: \[ \omega = \frac{0.15}{1.1 \times 0.43} \, \mathrm{rad/s} \]. Calculate to find \( \omega \).

Step 7: Calculate Numerical Value of \( \omega \)

Perform the calculation to find \( \omega \): \( \omega = \frac{0.15}{0.473} \approx 0.317 \) rad/s.

Key Concepts

Moment of Inertia

The moment of inertia is a measure of how difficult it is to change an object's rotational motion. It's often compared to mass in linear motion. For rotating systems, the moment of inertia depends not just on the mass of the object, but also how that mass is distributed relative to the axis of rotation. In our example, the wheel is modeled as a hoop.

For a hoop, the mass is distributed along the edge at a distance R from the center. This gives it a moment of inertia \( I = MR^2 \).

• \( M \) is the mass of the hoop (in this problem, 1.1 times the train's mass)
• \( R \) is the radius (0.43 m in this problem)

The larger the value of \( I \), the harder it is to increase or decrease its rotational speed. The moment of inertia plays a crucial role when forces initiate rotation, as seen in the train and wheel system.

Angular Speed

Angular speed represents how quickly an object is rotating. It's the angle an object turns through in a certain amount of time and is measured in radians per second (rad/s).

In our train problem, after the train speeds up, the wheel also starts spinning. Their angular velocity compensates for each other to satisfy the conservation of angular momentum.
The formula to find the angular speed \( \omega \) of the wheel is derived from conservation principles:
\[ \omega = \frac{v}{1.1R} \]

• \( v \) is the linear speed of the train (0.15 m/s)
• \( R \) is the wheel's radius (0.43 m)

Plugging in these values, you calculate \( \omega \) to see how fast the wheel turns due to the train's motion.

Linear Momentum

Linear momentum is a quantity expressing the motion of an object and is calculated by the product of an object's mass and velocity. It's a vector quantity, which means it has both magnitude and direction.
In the scenario with the train and the wheel, linear momentum helps us understand the initial motion of the train.When the train speeds up to 0.15 m/s, it has linear momentum calculated as:

• \( p = m \times v \)

This momentum causes the train to gain angular momentum as it traverses on the wheel, which needs to be countered by the wheel's own rotation to maintain the conservation of angular momentum, hence initiating the wheel's angular speed. This interplay of motion between the train and the wheel exemplifies the relationship between linear and angular momentum in rotational dynamics.