- #1

Hak

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- Homework Statement
- A small ball of mass ##m##, and carrying a positive charge ##q##, is suspended

by an insulating string of length ##\ell##. The pendulum so formed is placed, at rest, in a

homogeneous, vertical magnetic field of strength ##B##. Experiment shows that, if the

ball is initially knocked slightly sideways, then it swings back and forth, with the

plane of its swing slowly rotating.

1. Find the final angular frequency ##\omega_0## of the pendulum.

2. How long does it take for the plane to make one complete revolution?

- Relevant Equations
- \

Consider a pendulum swinging at the North Pole in an inertial frame of reference [tex]xyz[/tex] with the [tex]x[/tex] and [tex]y[/tex] axes in the plane tangent to the pole and the suspension placed at elevation [tex]z = l[/tex]. There is no induction [tex]\mathbf{B}[/tex] and in the chosen system the plane of oscillation [tex]xz[/tex] of the pendulum is constant. The acting forces are the weight [tex]m \vec g[/tex] and the tension[tex]\vec T[/tex]of the wire. Given [tex]\theta[/tex] the angle of oscillation we have [tex]T_x =-mg \sin \theta[/tex]and [tex]T_z = mg \cos\theta[/tex] and the equations of motion result in [tex]m\frac{ d^2 x}{dt^2} = - mg \sin\theta[/tex] and [tex]m \frac{d^2 z}{dt^2}=mg \cos\theta - mg[/tex]. Now we assume that the pendulum oscillation is small, i.e., [tex]\sin\theta \sim \theta[/tex] and [tex]cos\theta \sim 1[/tex], whereby the second equation (and the related motion) is negligible while the first equation can be posed [tex]\frac{d^2\theta}{dt^2} = -\frac{g}{l}{\theta}[/tex] and represents a harmonic motion of pulsation [tex]\omega_1 = \sqrt{\frac{g}{l}}[/tex], hence the well-known formula [tex]\tau_1= 2\pi \sqrt{\frac{l}{g}}[/tex]. Now suppose we enter the vector [tex]\vec B[/tex] directed as the [tex]z[/tex] axis. The ball comes to be subject to the Lorentz force [tex]\vec F_L = q \vec v \times \vec B[/tex], which in the [tex]xy[/tex] plane results in a centripetal force relative to a circular trajectory of radius [tex]R[/tex]. The ball comes to be subjected to the Lorentz force [tex]\vec F_L = q \vec v \times \vec B[/tex], which in the [tex]xy[/tex] plane results in a centripetal force relative to a circular trajectory of radius [tex]R[/tex]. The force is perpendicular to the velocity v (thus at zero work on the ball) but tends to move it around the center of the circle. The observer thus observes a rotation of the pendulum-ball's plane of oscillation m. Similar to what the noninertial observer of the Foucault pendulum sees, which participates at the north pole in the rotation of the Earth, due to the Coriolis force [tex]F_C = -2m \vec \omega_\tau \times \vec v_r[/tex]. If we refer to the center of the circle, the Lorentz force is opposite to the verse of [tex]\vec R[/tex] as the Coriolis force. But the thing to note is that it (is easily demonstrated) is much less than the weight force [tex]mg[/tex], about [tex]10^{-5}[/tex] of the same. This means that the angular frequency [tex]\omega_0[/tex], to which it will give rise with the same velocity [tex]v[/tex] of pendulum oscillation, will be much less than the frequency [tex]\omega_1= \sqrt{\frac{g}{l}}[/tex] of pendulum oscillation. Therefore, the rotation of the pendulum-ball oscillation plane appears very slow to the observer. Also, having made the premise about the signs, we can refer to magnitude. It will be [tex]m\frac{v^2}{R}=|qv|B[/tex] to obtain, simplifying and dividing by [tex]m[/tex], [tex]\omega_0 = \frac{|qB|}{m}[/tex]. Then the relative period would be [tex]\tau = \frac{2\pi m}{|qB|}[/tex]. In reality, only in one half-period does the pendulum swing so as to travel clockwise around the circumference, while in the other half-period the swing of the pendulum-ball is in the opposite direction. Therefore, the actual period, time taken to carry out one revolution around the center in a clockwise direction, will be twice as long as [tex]\tau[/tex], that is, [tex]\tau_0= 2 \tau = \frac{4 \pi m}{|qB|}[/tex].

Below, instead, is the official solution:

The ball, at position [tex]\mathbf{r}[/tex] relative to its rest position, and moving with velocity [tex]\mathbf{v}[/tex], is acted upon by the tension force in the string, the gravitational force and the Lorentz force caused by the magnetic field. Using the usual approximations for small-amplitude motions of pendulums, the horizontal projection of the tension inthe string is [tex]-m \omega^2 \mathbf{r}[/tex], where [tex]\omega = \sqrt {\frac{g}{l}}[/tex] is the angular frequency of the pendulum without any magnetic field. The ball’s equation of motion, in the presence of the magnetic field, is [tex]m \dot v = q \mathbf{v} \times \mathbf {B} - m \omega^2 \mathbf{r}[/tex].

This quite sophisticated (vector) differential equation can be solved without using calculus, if we notice that it is very similar to the equation of motion of a simple pendulum in a rotating frame of reference (a Foucault pendulum at the North Pole). To describe the small-amplitude swinging of a planar pendulum with angular frequency [tex]\omega_0[/tex], in a frame of reference rotating with angular velocity [tex]\mathbf{\Omega}[/tex] relative to the inertial reference frame ([tex]\mathbf{\Omega}[/tex] is a vertical vector), we need an equation of motionthat contains the Coriolis force as well as the centrifugal one: [tex]m \dot v = 2m \mathbf{v} \times \mathbf {\Omega} - m \omega_0^2 \mathbf{r} + m \Omega^2 \mathbf{r}[/tex].

The solution to this equation is well known. Since the oscillation plane is fixed in the inertial frame of reference, in a reference frame rotating with angular velocity [tex]\mathbf{\Omega}[/tex], the plane rotates with angular velocity [tex]- \mathbf{\Omega}[/tex]; the period of a whole revolution is [tex]T = \frac{2 \pi}{\Omega}[/tex].

Comparing equations (1) and (2), it can be seen that the observed rotation rate is [tex]\mathbf{\Omega} = \frac{qB}{2m}[/tex] and that the angular velocity vector of the rotating plane has a direction opposed to that of [tex]B[/tex]. The actual angular frequency [tex]\omega_0[/tex] of the pendulum can be expressed entirely in terms of the other parameters of the problem:

[tex]\omega_0 = \sqrt{\frac{g}{\ell} + \left(\frac{qB}{2m}\right)^2}[/tex],

but this has no bearing on the rotation rate of the plane of oscillation.

So, the time needed for the plane containing the pendulum’s motion to complete one revolution is: [tex]T = \frac{4\pi m}{qB}[/tex].

2. The quantity [tex]qB/m[/tex] is called the

Where am I mistaken?

Below, instead, is the official solution:

The ball, at position [tex]\mathbf{r}[/tex] relative to its rest position, and moving with velocity [tex]\mathbf{v}[/tex], is acted upon by the tension force in the string, the gravitational force and the Lorentz force caused by the magnetic field. Using the usual approximations for small-amplitude motions of pendulums, the horizontal projection of the tension inthe string is [tex]-m \omega^2 \mathbf{r}[/tex], where [tex]\omega = \sqrt {\frac{g}{l}}[/tex] is the angular frequency of the pendulum without any magnetic field. The ball’s equation of motion, in the presence of the magnetic field, is [tex]m \dot v = q \mathbf{v} \times \mathbf {B} - m \omega^2 \mathbf{r}[/tex].

This quite sophisticated (vector) differential equation can be solved without using calculus, if we notice that it is very similar to the equation of motion of a simple pendulum in a rotating frame of reference (a Foucault pendulum at the North Pole). To describe the small-amplitude swinging of a planar pendulum with angular frequency [tex]\omega_0[/tex], in a frame of reference rotating with angular velocity [tex]\mathbf{\Omega}[/tex] relative to the inertial reference frame ([tex]\mathbf{\Omega}[/tex] is a vertical vector), we need an equation of motionthat contains the Coriolis force as well as the centrifugal one: [tex]m \dot v = 2m \mathbf{v} \times \mathbf {\Omega} - m \omega_0^2 \mathbf{r} + m \Omega^2 \mathbf{r}[/tex].

The solution to this equation is well known. Since the oscillation plane is fixed in the inertial frame of reference, in a reference frame rotating with angular velocity [tex]\mathbf{\Omega}[/tex], the plane rotates with angular velocity [tex]- \mathbf{\Omega}[/tex]; the period of a whole revolution is [tex]T = \frac{2 \pi}{\Omega}[/tex].

Comparing equations (1) and (2), it can be seen that the observed rotation rate is [tex]\mathbf{\Omega} = \frac{qB}{2m}[/tex] and that the angular velocity vector of the rotating plane has a direction opposed to that of [tex]B[/tex]. The actual angular frequency [tex]\omega_0[/tex] of the pendulum can be expressed entirely in terms of the other parameters of the problem:

[tex]\omega_0 = \sqrt{\frac{g}{\ell} + \left(\frac{qB}{2m}\right)^2}[/tex],

but this has no bearing on the rotation rate of the plane of oscillation.

So, the time needed for the plane containing the pendulum’s motion to complete one revolution is: [tex]T = \frac{4\pi m}{qB}[/tex].

*Notes*. 1. It is interesting that the rotational period of the pendulum’s plane depends only on the strength of the magnetic field and the data associated withthe small ball, and that it does not depend on the natural period of the pendulum (i.e. it is independent of [tex]\ell[/tex] and [tex]g[/tex]).2. The quantity [tex]qB/m[/tex] is called the

*cyclotron angular frequency*associated with the given magnetic field; it is the angular velocity of a particle, with charge-to-mass ratio [tex]q/m[/tex], that will orbit in a homogeneous magnetic field of strength [tex]B[/tex]. The plane of the pendulum in the problem rotates with one-half of the cyclotron angular frequency.Where am I mistaken?