91Ó°ÊÓ

An insulating rigid rod of length \(2 a\) and negligible mass is attached at its center to a pivot at the origin and is free to rotate in the \(x y\) -plane. A small ball with mass \(M\) and charge \(Q\) is attached to one end of the rod. A second small ball with mass \(M\) and no charge is attached to the other end. A constant electric field \(\vec{E}=-E \hat{\imath}\) is present in the region \(y>0\) while the region \(y<0\) has a vanishing electric field. Define \(\vec{r}\) as the vector that points from the center of the rod to the charged end of the rod, and \(\theta\) as the angle between \(\vec{r}\) and the positive \(x\) -axis. The rod is oriented so that \(\theta=0\) and is given an infinitesimal nudge in the direction of increasing \(\theta\). (a) Write an expression for the vector \(\vec{r}\). (b) Determine the torque \(\vec{\tau}\) about the center of the rod when \(0 \leq \theta \leq \pi\). (c) Determine the torque on the rod about its center when \(\pi \leq \theta \leq 2 \pi\). (d) What is the moment of inertia \(I\) of the system about the \(z\) -axis? (e) The potential energy \(U(\theta)\) is determined by \(\tau=-d U / d \theta .\) Use this equation to write an expression for \(U(\theta)\) over the range \(0 \leq \theta \leq 4 \pi\) using the convention that \(U(0)=0 .\) Make sure that \(U(\theta)\) is continuous. (f) The angular velocity of the rod is \(\omega=\omega(\theta) .\) Using \(\tau=I d^{2} \theta / d t^{2}\) show that the energy \(\frac{1}{2} I \omega^{2}+U(\theta)\) is conserved. (g) Using energy conservation, determine an expression for the angular velocity at the \(n\) th time the positive charge crosses the negative \(y\) -axis.

Step by step solution

01

Write an expression for the vector \(r\)

From the problem's information, the vector \(r\) will point from the origin (midway point of the rod) to the positive end of the rod. Since the rod rotates in the xy-plane, and the angle \(\theta\) between \(r\) and the positive x-direction is given, express \(r\) in terms of \(\theta\) and the rod's length. So, \(\vec{r} = a(\cos(\theta) \hat{\imath} + \sin(\theta) \hat{\jmath})\).

02

Determine the torque \(\tau\) for \(0 \leq \theta \leq \pi\)

The torque produced by the electric field on the positively charged particle is \(\tau = r F_{\perp}\), where \(F_{\perp}\) is the force perpendicular to \(r\). Since the electric field \(\vec{E}=-E\hat{\imath}\) is horizontal, the force \(F=Q E\) on the charged particle also is horizontal. When \(0 \leq \theta \leq \pi\), the angle between \(\vec{r}\) and \(\vec{F}\) is \(180 - \theta\). Hence, \(\tau = r F \sin(180 - \theta) = a Q E \sin(\pi - \theta) = a Q E \sin(\theta)\). Note that the area \(y<0\) has no electric field, so \(\tau = 0\).

03

Determine the torque \(\tau\) for \(\pi \leq \theta \leq 2\pi\)

For this part of the motion, the positively charged particle is still in the area \(y>0\), so \(\tau = a Q E \sin(\theta)\) there. On the other hand, for the part of \(\theta\) where the charged particle is in the area \(y<0\), \(\tau = 0\). Therefore, we can write that \(\tau = a Q E \sin(\theta)\) for \(\pi \leq \theta < 3\pi / 2\) and \(\tau = 0\) for \(3\pi / 2 \leq \theta \leq 2\pi\).

04

Determine the moment of inertia \(I\) of the system

The moment of inertia for a system of point particles is given by \(I = \sum m_i r_i^2\), where \(m_i\) and \(r_i\) are the mass and distance from the rotation axis of the \(i\)th particle. Here, \(I = M a^2 + M a^2 = 2 M a^2\).

05

Write an expression for the potential energy \(U(\theta)\)

Integrating the equation \(\tau = -dU/d\theta\) gives \(U(\theta) = - \int_0^{\theta} \tau d\theta\). Using the expressions for the torque for \(0 \leq \theta < 3\pi / 2\) and \(\tau = 0\) for \(3\pi / 2 \leq \theta \leq 2\pi\), and fitting these results together to create a continuous function, gives \(U(\theta) = - a Q E (1-\cos(\theta))\) for \(0 \leq \theta < 2\pi\) and \(U(\theta) = - 2 a Q E (1-\cos(\theta))\) for \(2\pi \leq \theta < 4\pi\). Thus, \(U(\theta)\) is continuous for \(0 \leq \theta < 4\pi\)

06

Show that energy is conserved

Starting from the equation \(\tau = I d^2\theta / dt^2\), and using that \(\tau = -dU(\theta) / d\theta\) and that the kinetic energy is \(\frac{1}{2} I \omega^2 = \frac{1}{2} I (d\theta / dt)^2\), one can write that \(d / dt (\frac{1}{2} I (d\theta / dt)^2 + U(\theta)) = 0\). Thus, the energy \(\frac{1}{2} I \omega^2 + U(\theta)\) is a constant of motion - it is conserved.

07

Expression for angular velocity

Using the conservation of energy, write that \(\frac{1}{2} I \omega_n^2 + U(\theta = \pi n) = \frac{1}{2} I \omega_0^2 + U(\theta = 0)\). This leads to \(\omega_n = \sqrt{\omega_0^2 - \frac{2}{I}[U(\pi n) - U(0)]}\). Then, using the expression for \(U(\theta)\) from step 5 and \(U(0) = 0\), the result becomes \(\omega_n = \sqrt{\omega_0^2 - 2\frac{Q E a}{I}(1-\cos(\pi n))}\), where \(n\) is a positive integer. This is the angular velocity at the \(n\)th time the positive charge crosses the negative \(y\) -axis.

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Key Concepts

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

Rotational Motion

When an object rotates around a fixed point, it is undergoing rotational motion. This kind of motion is common in atoms to celestial bodies and is quantified by variables such as angular velocity and angular acceleration. The most basic example is a particle moving in a circle at a constant speed. While the speed might not change, its direction does, which implies an acceleration towards the center of the circle. In the context of the given exercise, the rod with balls represents a simple rotational system where after an initial nudge, the rod exhibits rotational motion around the pivot at the origin.

The pivot point is crucial as it acts as the center of rotation, akin to the sun being the center around which planets revolve. For objects extending in space (like our rod with masses), the rotational motion is influenced by the distribution of mass around the pivot; a concept closely tied to the moment of inertia, which we will discuss further in another section.

Electric Fields

Electric fields are a fundamental concept in electromagnetism, representing the force fields around charged particles. Imagine them as the influence a charge exerts on the space around it. In our exercise, we deal with a constant electric field, denoted by \, which impacts the charged ball on the rod. The force due to the electric field on a charge is given by \, where \ is the force and \(Q\) is the charge of the particle.

This force induces torque, which then affects the rotational motion. It's critical to understand that electric fields can do work on charged particles, changing their kinetic energy, which is directly linked to the conservation of energy that we will also delve into more deeply.

Moment of Inertia

A pivotal concept in understanding rotational motion is the moment of inertia (\(I\)), often considered the rotational equivalent of mass in linear motion. It quantifies an object's resistance to changes in its rotational motion and depends on the mass distribution relative to the axis of rotation. Mathematically, for a system of point masses, it can be expressed as \(I = \sum m_i r_i^2\), where \(m_i\) is the mass and \(r_i\) is the distance of the \(i\)th point from the axis of rotation.

In the exercise, we calculate the moment of inertia for the rod and balls system about the pivot point. Understanding that the two balls are equidistant from the pivot, we can easily calculate this property which will later help us determine the rod’s rotational motion since it directly affects the angular acceleration and the energy within the system.

Energy Conservation

The principle of energy conservation is a cornerstone of physics, stating that the total energy in an isolated system remains constant over time. Energy can be transformed from one form to another but never created or destroyed. In rotational systems, energy conservation involves kinetic and potential energies. Kinetic energy in rotational motion is expressed as \(\frac{1}{2}I\omega^2\), where \(\omega\) is the angular velocity.

In our exercise, the conservation of mechanical energy implies that any work done by external torques (like those induced by the electric field) will result in a change in the total mechanical energy – kinetic plus potential – of the system. Following the calculation laid out in the textbook, we can confirm that the sum of kinetic and potential energy remains constant, proving the principle of energy conservation. This allows us to solve for unknown variables such as angular velocity at different points in the motion.