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A steel bicycle wheel (without the rubber tire) is rotating freely with an angular speed of 18.00 rad/s. The temperature of the wheel changes from \(-100.0\) to \(+300.0^{\circ} \mathrm{C}\) . No net external torque acts on the wheel, and the mass of the spokes is negligible. (a) Does the angular speed increase or decrease as the wheel heats up? Why? (b) What is the angular speed at the higher temperature?

Short Answer

Expert verified
(a) The angular speed decreases; (b) It is approximately 17.86 rad/s.

Step by step solution

01

Understand the Physical Principles

As the wheel heats up, it expands due to thermal expansion. Since no external torque acts on the wheel, angular momentum is conserved. This means the product of the moment of inertia and angular velocity remains constant. We'll use this principle to find the new angular speed.
02

Apply Conservation of Angular Momentum

The initial angular momentum is given by \( I_i \omega_i \), and the final angular momentum is \( I_f \omega_f \). Since they are equal, we have:\[I_i \omega_i = I_f \omega_f\]Due to thermal expansion, \( I_f = kI_i \), where \( k \) represents the increased moment of inertia at the higher temperature. Thus, the equation becomes:\[\omega_f = \frac{\omega_i}{k}\].
03

Determine the Change in Moment of Inertia

The moment of inertia of a circular wheel changes with temperature, as it depends on the square of the radius, \( I \propto r^2 \). For linear expansion, \( r_f = r_i(1 + \alpha \Delta T) \), where \( \alpha \) is the coefficient of linear expansion, and \( \Delta T \) is the temperature change. Therefore,\( I_f \approx I_i(1 + 2\alpha \Delta T) \).
04

Calculate the Angular Speed at Higher Temperature

Given that \( \alpha \approx 11 \times 10^{-6} / ^{\circ}C \) for steel and \( \Delta T = 400^{\circ}C \), find \( k \):\[k = 1 + 2 \times 11 \times 10^{-6} \times 400 = 1.0088 \approx 1.008\]Then, calculate \( \omega_f \):\[\omega_f = \frac{18.00}{1.008} \approx 17.86 \text{ rad/s}\]
05

Interpret the Result

The wheel's angular speed decreases slightly because its moment of inertia increases due to thermal expansion, and angular momentum is conserved.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Thermal Expansion
Thermal expansion is the process where materials expand as they increase in temperature. This occurs because as atoms vibrate faster with heat, they tend to take up more space. For metals, this is defined by a constant known as the coefficient of linear expansion, denoted by \( \alpha \). In our exercise, steel is the material in question with \( \alpha \approx 11 \times 10^{-6} / ^{\circ}C \).

As the temperature of the bicycle wheel moves from \(-100.0^{\circ}C\) to \(+300.0^{\circ}C\), we notice a direct temperature change, \( \Delta T = 400^{\circ}C \).
  • In thermal expansion, any length of the object changes in line with the formula: \( L_f = L_i(1 + \alpha \Delta T) \)
  • For circular objects like the bicycle wheel, this affected dimension is the radius, which in turn changes the moment of inertia.
These changes, though tiny on a per-degree scale, are compounded in large temperature differences, affecting other physical properties, such as rotational motion.
Angular Speed
Angular speed is a measure of how fast an object rotates or spins. It is expressed as radians per second (rad/s) and is crucial in determining how fast a rotation or cycle is completed.

In this exercise, the wheel initially rotates at 18.00 rad/s. When the wheel undergoes thermal expansion, it changes some dynamics. Without external interference, this is accompanied by a change in angular speed. Why? Because angular momentum is conserved.

  • Without external torque, the conservation law implies \( I_i \omega_i = I_f \omega_f \)
  • Since \( I_f = kI_i \), we rearrange to get the new angular speed: \( \omega_f = \omega_i / k \)
Upon heating, the wheel has a slightly slower angular speed: about 17.86 rad/s, indicating the system's balance between increasing inertia and diminishing speed.
Moment of Inertia
Moment of inertia helps us understand how mass distribution affects a rotating object. It is crucial for calculations involving rotational motion. For a wheel, the moment of inertia relies heavily on the radius and mass distribution along that radius.

In our scenario, as the wheel expands, the radius increases and thereby the moment of inertia increases too. We denote the initial and final moments of inertia as \( I_i \) and \( I_f \) respectively. With thermal expansion, the relationship modifies as:

  • \( I \propto r^2 \), means a small change in radius induces a larger change in inertia.
  • Linear expansion adds the formula: \( I_f = I_i(1 + 2\alpha \Delta T) \)
By this relationship, we calculate an increase factor \( k \sim 1.008 \), confirming that when \( r \) changes even slightly, inertia adopts a consequential change.

This understanding is crucial as it informs why the wheel slows down, keeping angular momentum constant despite thermal changes.

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