/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 47 Two square reflectors, each \(1.... [FREE SOLUTION] | 91Ó°ÊÓ

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Two square reflectors, each \(1.50 \mathrm{~cm}\) on a side and of mass \(4.00 \mathrm{~g}\), are located at opposite ends of a thin, extremely light, \(1.00-\mathrm{m}\) rod that can rotate without friction and in vacuum about an axle perpendicular to it through its center (Fig. P12.47). These reflectors are small enough to be treated as point masses in moment-of-inertia calculations. Both reflectors are illuminated on one face by a sinusoidal light wave having an electric field of amplitude \(1.25 \mathrm{~N} / \mathrm{C}\) that falls uniformly on both surfaces and always strikes them perpendicular to the plane of their surfaces. One reflector is covered with a perfectly absorbing coating, and the other is covered with a perfectly reflecting coating. What is the angular acceleration of this device?

Short Answer

Expert verified
The angular acceleration is found using torque and moment of inertia relations, calculated from radiation pressures on the reflectors.

Step by step solution

01

Calculate Force due to Absorbing Coating

The force on the perfectly absorbing reflector due to radiation pressure can be computed using the formula: \[ F = \frac{1}{c} \times (\text{Power}) = \frac{PA}{c}, \]where \(P\) is the power of the light wave, and \(A\) is the area. First, calculate the intensity (power per unit area) using: \[ I = \frac{1}{2} \epsilon_0 c E^2, \] where \(\epsilon_0 = 8.85 \times 10^{-12} \text{ C}^2/\text{N} \cdot \text{m}^2\) and \(E = 1.25 \text{ N/C}\). Then find the force: Using \(A = (1.50 \times 10^{-2})^2 \text{ m}^2\), and simplifying gives an intensity \(I\). Compute the power \(P = I\cdot A\), and then the force \(F = \frac{P}{c}\).
02

Calculate Force due to Reflecting Coating

The force on the perfectly reflecting reflector is twice that on the absorbing one, because the momentum transfer is doubled upon reflection: \[ F_r = 2 \times \frac{P}{c}, \]where \(P\) is the same power calculated previously. Compute this force using the same steps as for the absorbing reflector, but multiply the resulting force by 2 for the momentum reflection.
03

Calculate Net Force Exerted on the System

The net force exerted on the system is the difference between the forces on the reflecting and absorbing reflectors, as they act in opposite directions on the system: \[ F_{\text{net}} = F_r - F_a, \]where \(F_r\) is the force on the reflecting surface and \(F_a\) is on the absorbing surface. This net force will provide a torque on the rod.
04

Calculate Torque due to Net Force

Torque \(\tau\) is related to force and the distance from the pivot: \[ \tau = F_{\text{net}} \times d, \]where \(d = 0.5 \text{ m}\) is half the length of the rod, the distance from the center to where each force acts. Use the previously calculated net force to find the torque.
05

Calculate Moment of Inertia

The moment of inertia \(I\) for the system of point masses is: \[ I = 2 \times m \times d^2, \]where each mass \(m = 4.00 \times 10^{-3} \text{ kg}\) and \(d = 0.5 \text{ m}\). Calculate this to find the total moment of inertia.
06

Calculate Angular Acceleration

The angular acceleration \(\alpha\) is related to torque and moment of inertia by the equation: \[ \alpha = \frac{\tau}{I}. \]Use the values calculated for \(\tau\) and \(I\) from the previous steps to find \(\alpha\).

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Key Concepts

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

Moment of Inertia
The moment of inertia is a fundamental concept when discussing rotational motion, as it determines how easily an object can be rotated. Think of it like the rotational equivalent of mass in linear motion. When we say a body has a large moment of inertia, it means more torque is required to change its rotational speed.
In the context of our exercise, each point mass is located at a fixed distance from the pivot point, which greatly influences the moment of inertia. The formula used is simple: \[ I = \sum m_i r_i^2 \] Here, every mass (\(m_i\)) is multiplied by the square of its distance (\(r_i\)) from the axis of rotation. Since we have two identical point masses located equidistant from the center, we calculate the system's moment of inertia by considering both point masses.
This calculation helps us to determine how the applied torque will affect the system's angular acceleration.
Radiation Pressure
Radiation pressure refers to the pressure exerted upon any surface due to the exchange of momentum between the light wave and the surface it strikes. In simpler terms, light, even though it has no mass, has momentum and can exert a force. This concept plays a critical role in our problem since it defines how light can exert force on the reflectors at the ends of the rod.
The formula for radiation pressure on a perfectly absorbing surface is: \[ P = \frac{I}{c} \] where \(I\) is the intensity of the light and \(c\) is the speed of light. For a perfectly reflecting surface, the radiation pressure is doubled, because the light wave bounces back, transferring additional momentum.
Understanding this is key to computing the forces acting on each reflector, as they contribute to the net torque affecting the rod's rotation.
Torque Calculation
Torque is the rotational equivalent of force. In our problem, the net force acting on the rod due to radiation pressure results in a torque that causes the rod to rotate. Torque depends on the magnitude of the force, the distance from the rotation axis, and the angle at which the force is applied. However, since our forces are acting perpendicular to the rod, the angle simplifies our calculation.
We can calculate torque using: \[ \tau = F_{\text{net}} \times d \] where \(F_{\text{net}}\) is the net force and \(d\) is the distance from the pivot to where the force acts. In this scenario, \(d\) is half the length of the rod.
The net torque will ultimately allow us to find the angular acceleration, illustrating how effectively the forces cause the rod to rotate.
Point Masses
Point masses are an essential simplification when calculating the moment of inertia in this exercise. When we deal with rotational systems, it's often ideal to consider objects as point masses if the actual dimensions of the masses are negligible compared to their distances from the pivot.
This simplification allows us to ignore detailed distributions of mass within the reflectors, thus making our calculations more straightforward and manageable. Instead of calculating with complex shapes, we assume all mass located at a single point, simplifying the application of the inertia formula.
By treating them as point masses, we focus only on the mass and distance, helping to efficiently compute the values needed for subsequent steps like torque calculations and determining angular acceleration.

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