An insulating rigid rod of length 2 a and negligible mass is...

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.

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.

Key Concepts

Vector Representation in Rigid Body Kinematics

This concept deals with describing the position of a point on a rigid body in space by using vectors. Here, one expresses the position of an object (such as the charged ball) in terms of its distance from the pivot and an angle measured from a fixed direction (often using polar coordinates), which is fundamental for setting up the analysis of the system’s configuration.

Torque and the Cross Product in Rotational Dynamics

Torque is the rotational analog of force and is calculated as the cross product of the position vector (from the pivot to the point where the force is applied) and the force vector. Understanding this concept is essential for determining how forces (such as electric forces on a charged particle) produce rotation about an axis.

Force on a Charged Particle in an Electric Field

A charged particle in an electric field experiences a force that is proportional to both its charge and the electric field strength. This concept is important for analyzing how the electric field exerts a force on the charged end of the rod, which in turn generates torque and influences the system’s dynamics.

Moment of Inertia for Systems of Point Masses

The moment of inertia quantifies how the mass of a system is distributed relative to an axis of rotation, influencing the angular acceleration produced by a given torque. For systems consisting of point masses, it is calculated as the sum of each mass multiplied by the square of its distance from the axis, a key factor in rotational energy computations.

Potential Energy in Conservative Rotational Systems

For conservative forces, the potential energy can be found by integrating the torque with respect to the angle, with appropriate boundary conditions to ensure continuity. This concept helps in characterizing the energy landscape of the system as a function of the angle, providing insights into equilibrium and stability.

Conservation of Mechanical Energy in Rotational Motion

This principle states that in the absence of non-conservative forces, the total mechanical energy—comprising rotational kinetic energy and potential energy—remains constant over time. It is a powerful tool for analyzing the dynamics of the system when the forces and torques change with the angle of rotation.

Transcript

0:00 Greetings.

00:02 This problem has an insulated rod of link 2a, and it can rotate in the x, y, plane.

00:09 It has this charge q at one end and has a mass of m, and on the other end it has a mass m but doesn't have a charge.

00:20 And above the x axis, there is this field that points in the negative x direction, and below the x axis, we do not have a field.

00:31 The charge q, the charge q, the is the location is indicated by this vector r.

00:41 And so we can write the vector r as its x and y components.

00:47 So that's what this is here, the x component and its y component.

00:51 And there is our vector r.

00:54 The force is our coulomb force, so it's just equal to the charge cube and the field that is in e.

01:03 And so here is our force.

01:05 In the negative x direction and so the torque on that charge is equal to r cross f and so we have this we want to do the cross product we have this torque a q e sine theta and send the z direction are the clock or counterclockwise direction and when we we have no force below the x -axis so we have no torque below the x -axis so we have no torque below the x -axis let's see.

01:43 Our moment of inertia for each of those mass is the moment of inertia is just m times a squared and so we have two of them so the moment of inertia for the entire rod is 2 m a squared.

02:04 We're asked to find a function of the potential energy given that the potential energy differentiated with the respect to theta is equal to our torque.

02:19 So we just need to do the integral.

02:23 We do that, and we go on the left side, and on the right side we'll have our torque d theta.

02:41 When we do this integral, we'll get you on that side.

02:48 On this side we'll get a qe, a cosine theta, plus our constant of integration.

02:59 We're told that you evaluated at 0 must be 0.

03:08 So for pitt into 0, we'll have a quee and cosine of 0 is 1, and we have our constant, and so that must equal 0, and that forces our constant must be equal to a negative a queue.

03:30 So finally we can write our potential energy as aquee, cosine, as a qe, cosine, and that forces our constant, not plus but minus a queue and if you pull out the aquee you're left with cosine minus cosine minus cosine beta minus one so that's what i have here they asked you to find the potential energy between zero and four pi so that's just what i have here it's this is your potential energy in between zero and and in between 2 pi and 3 pi and your potential energy is 0 in between pi and 2 pi and in between 3 pi and 4 pi so that's just what all this stuff says here okay part f is where we do i've got a few tricks we're told to start with our definition of torque here and we already know this one from earlier and so basically i'm just going to have these two equal each other.

04:59 But i'm going to write this guy as d -o -mega -d -t because we have this relationship.

05:13 And remember the definition of omega is just d -theta -d -t.

05:18 We'll use that in a minute.

05:22 So i can write this equation here like this.

05:29 So first thing i'm going to do is multiply through by omega.

05:35 And on this side, i'm going to write it down as omega.

05:37 On this side, i'm going to write it down as its definition d -theta -d -t...

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