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Chapter 40: Problem 73

Do we need to consider quantum effects for everyday rotating objects? Estimate the differences between rotational energy levels for a spinning baton compared to the energy of the baton. Assume the baton consists of a uniform \(32-\mathrm{cm}\) -long bar with a mass of \(260 \mathrm{~g}\) and two small end masses, each of mass \(380 \mathrm{~g},\) and that it rotates at \(1.6 \mathrm{rev} / \mathrm{s}\)

Short Answer

Expert verified
Quantum effects are negligible for everyday rotating objects as quantum energy differences are extremely small compared to classical rotational energy.

Step by step solution

01

Calculate the Moment of Inertia

The moment of inertia for a rotating object is a measure of its resistance to changes in its rotation.For the baton, which consists of a uniform bar and two end masses, the total moment of inertia is the sum of individual moments:- Moment of inertia for the rod (uniform bar):\[ I_{rod} = \frac{1}{12} M L^2 \]where \( M = 260 \, \text{g} = 0.26 \, \text{kg} \) and \( L = 0.32 \, \text{m} \).- Moment of inertia for the point masses (at the ends of the rod):\[ I_{masses} = 2(m r^2) \]where \( m = 380 \, \text{g} = 0.38 \, \text{kg} \) and \( r = 0.16 \, \text{m} \).Calculating each:\[ I_{rod} = \frac{1}{12} \times 0.26 \times (0.32)^2 = 0.0022 \, \text{kg} \cdot \text{m}^2 \]\[ I_{masses} = 2(0.38 \times (0.16)^2) = 0.0195 \, \text{kg} \cdot \text{m}^2 \]\[ I_{total} = I_{rod} + I_{masses} = 0.0022 + 0.0195 = 0.0217 \, \text{kg} \cdot \text{m}^2 \].
02

Calculate Rotational Energy

The rotational kinetic energy of an object is given by the formula:\[ E_r = \frac{1}{2} I \omega^2 \]where \( I \) is the moment of inertia and \( \omega \) is the angular velocity in rad/s.Given the object rotates at \( 1.6 \, \text{rev/s} \), convert this to rad/s by multiplying by \( 2\pi \):\[ \omega = 1.6 \times 2\pi = 10.05 \, \text{rad/s} \]Now, calculate the rotational energy:\[ E_r = \frac{1}{2} \times 0.0217 \times (10.05)^2 = 1.0958 \, \text{J} \].
03

Calculate Quantum Rotational Energy Level Differences

Quantum mechanics tells us that for rotating systems, the energy difference between quantum levels is given by:\[ \Delta E = \frac{\hbar^2}{2I} \times (J+1) \]\( J \) is the quantum rotational number, and \( \hbar \) (reduced Planck's constant) is approximately \( 1.055 \times 10^{-34} \, \text{J} \cdot \text{s} \).Calculate the difference between the first two levels \( (J=0 \rightarrow J=1) \):\[ \Delta E = \frac{(1.055 \times 10^{-34})^2}{2 \times 0.0217} \times (1+1) \approx 5.13 \times 10^{-68} \, \text{J} \].
04

Compare Quantum Effects with Everyday Energy

Now that we have both the classical rotational energy and the energy difference between quantum levels:- Classical rotational energy: \( 1.0958 \, \text{J} \)- Difference between quantum energy levels: \( 5.13 \times 10^{-68} \, \text{J} \)Compare the magnitudes:The difference in energy between quantum levels is extremely small compared to the classical energy, indicating that quantum effects are insignificant for such macroscopic rotating objects.

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Key Concepts

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

Rotational Energy
Rotational energy is a crucial concept in physics when studying objects in motion, particularly those that spin or rotate. It is the kinetic energy due to the rotation of an object. Its quantity depends on two key factors: the moment of inertia and the angular velocity.
  • Moment of Inertia: This measures how much an object resists rotational changes. Think of it as rotational mass.
  • Angular Velocity: This describes how fast the object spins, measured in radians per second (rad/s).
The formula for calculating rotational kinetic energy is: \[ E_r = \frac{1}{2} I \omega^2 \]Here, \( E_r \) is the rotational energy, \( I \) is the moment of inertia, and \( \omega \) is the angular velocity. In the case of the baton, its rotational energy is primarily driven by these two factors.
Moment of Inertia
The moment of inertia is essentially how resistant an object is to changes in its angular motion. It is analogous to mass in linear motion but applies to rotation about an axis.
  • Formula for a Uniform Rod: The moment of inertia \( I \) for a rod rotating about its center is calculated as \( I_{rod} = \frac{1}{12} M L^2 \), where \( M \) is mass, and \( L \) is length.
  • Point Masses: For point masses located at a distance from the rotation axis, the moment of inertia is \( I_{masses} = 2(m r^2) \), where \( m \) is mass and \( r \) is the distance to the axis.
Combining these for the baton (a rod with two end masses), the total moment of inertia is the sum: \[ I_{total} = I_{rod} + I_{masses} \].
It's important to correctly identify and add the contributions from different parts when calculating.
Angular Velocity
Angular velocity gives us the rate at which an object rotates, expressed in radians per second (rad/s). It is a vector quantity, meaning it has both magnitude and direction.
  • Conversion from Revolutions to Radians: To convert revolutions per second to radians per second, multiply by \( 2\pi \).
    So, for the baton spinning at 1.6 rev/s, its angular velocity \( \omega \) becomes \( 1.6 \times 2\pi \).
  • Role in Rotational Energy: This angular velocity directly affects the rotational energy of the object, as seen in the formula \( E_r = \frac{1}{2} I \omega^2 \).
Understanding angular velocity is vital for calculating how energetic a rotating system is.
Quantum Rotational Levels
In the realm of quantum mechanics, rotation is quantized, meaning that an object can only possess specific rotational energies, described by quantum rotational levels.
  • Quantum Number \( J \): This number denotes the rotational level, where energy differences between levels are determined.
  • Energy Differences: Quantum energy level differences are calculated by:\[ \Delta E = \frac{\hbar^2}{2I} \times (J+1) \] where \( \hbar \) is the reduced Planck's constant.
For objects in everyday situations, like our baton, differences between quantum rotational levels \( J = 0 \rightarrow J = 1 \) are minuscule compared to the total energy. While quantum effects might seem fascinating, they reveal themselves only in atomic and subatomic scales.
This comparison shows why classical descriptions suffice for most macroscopic objects.

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Most popular questions from this chapter

(III) A \(120-\mathrm{V}\) rms \(60-\mathrm{Hz}\) voltage is to be rectified with a full-wave rectifier as in Fig. \(40,\) where \(R=28 \mathrm{k} \Omega\) , and \(C=35 \mu \mathrm{F}\) (a) Make a rough estimate of the average current. \((b)\) What happens if \(C=0.10 \mu \mathrm{F} ?\)

Calculate the longest-wavelength photon that can cause an electron in silicon \(\left(E_{\mathrm{g}}=1.14 \mathrm{eV}\right)\) to jump from the valence band to the conduction band.

(II) What is the occupancy probability for a conduction electron in copper at \(T=295 \mathrm{K}\) for an energy \(E=1.015 E_{\mathrm{F}} ?\)

(II) The energy gap \(E_{\text { g }}\) germanium is 0.72 eV. When used as a photon detector, roughly how many electrons can be made to jump from the valence to the conduction band by the passage of a \(730-\mathrm{keV}\) photon that loses all its energy in this fashion?

Common salt, \(\mathrm{NaCl}\), has a density of \(2.165 \mathrm{~g} / \mathrm{cm}^{3} .\) The molecular weight of \(\mathrm{NaCl}\) is \(58.44 .\) Estimate the distance between nearest neighbor \(\mathrm{Na}\) and \(\mathrm{Cl}\) ions. [Hint: Each ion can be considered to have one "cube" or "cell" of side \(s\) (our unknown) extending out from it.
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