Question 8.24: A circular disk of radius R and mass M carries n point charg...

A circular disk of radius R and mass M carries n point charges (q), attached at regular intervals around its rim. At time t = 0 the disk lies in the xy plane, with its center at the origin, and is rotating about the z axis with angular velocity \omega_{0}, when it is released. The disk is immersed in a (time-independent) external magnetic field

B (s, z)=k(-s \hat{ s }+2 z \hat{ z }) .

where k is a constant.

(a) Find the position of the center if the ring, z(t), and its angular velocity, ω(t), as functions of time. (Ignore gravity.)

(b) Describe the motion, and check that the total (kinetic) energy—translational plus rotational—is constant, confirming that the magnetic force does no wor k \cdot^{26}

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(a) Initially, the disk will rise like a helicopter. The force on one charge (velocity v =\omega R \hat{ \phi }+v_{z} \hat{ z } ) is

F _{i}=q( v \times B )=q k\left|\begin{array}{ccc}\hat{ s } & \hat{\phi} & \hat{ z } \\0 & \omega R & v_{z} \\-R & 0 & 2 z\end{array}\right|=q k\left(2 \omega R z \hat{ s }-R v_{z} \hat{\phi}+\omega R^{2} \hat{ z }\right) .

The net force on all the charges is

F =\sum_{i=1}^{n} F _{i}=n q k R^{2} \omega \hat{ z }=M \frac{d^{2} z}{d t^{2}} \hat{ z } ; \quad \frac{d^{2} z}{d t^{2}}=\left(\frac{n q k R^{2}}{M}\right) \omega.                 [1]

The net torque on the disk is

N =\sum_{i=1}^{n}\left( r _{i} \times F _{i}\right)=n(R \hat{ s }) \times\left(-q k R v_{z} \hat{ \phi }\right)=-n q k R^{2} v_{z} \hat{ z }=I \frac{d \omega}{d t} \hat{ z }

where I is the moment of inertia of the disk. So

\frac{d \omega}{d t}=-\left(\frac{n q k R^{2}}{I}\right) \frac{d z}{d t}.                                                                                                 [2]

Differentiate [2], and combine with [1]:

\frac{d^{2} z}{d t^{2}}=-\left(\frac{I}{n q k R^{2}}\right) \frac{d^{2} \omega}{d t^{2}}=\left(\frac{n q k R^{2}}{M}\right) \omega \Rightarrow \frac{d^{2} \omega}{d t^{2}}=\alpha \omega, \quad \text { where } \quad \alpha \equiv \frac{n q k R^{2}}{\sqrt{M I}}

The solution (with initial angular velocity \omega_{0} and initial angular acceleration 0) is

\omega(t)=\omega_{0} \cos \alpha t .

Meanwhile,

\frac{d z}{d t}=-\left(\frac{I}{n q k R^{2}}\right) \frac{d \omega}{d t}=\left(\frac{I}{n q k R^{2}}\right) \omega_{0} \alpha \sin \alpha t=\omega_{0} \sqrt{\frac{I}{M}} \sin \alpha t.

z(t)=\omega_{0} \sqrt{\frac{I}{M}} \int_{0}^{t} \sin \alpha t d t=\frac{\omega_{0}}{\alpha} \sqrt{\frac{I}{M}}(1-\cos \alpha t).

[The problem is a little cleaner if you make the disk massless, and assign a mass m to each of the charges. Then \left.M \rightarrow n m \text { and } I \rightarrow n m R^{2}, \text { so } \alpha=q k R / m \text { and } z(t) \rightarrow\left(\omega_{0} R / \alpha\right)(1-\cos \alpha t) .\right]

(b) The disk rises and falls harmonically, as its rotation slows down and speeds up. The total energy is

E=\frac{1}{2} M v_{z}^{2}+\frac{1}{2} I \omega^{2}=\frac{1}{2} M \omega_{0}^{2} \frac{I}{M} \sin ^{2} \alpha t+\frac{1}{2} I \omega_{0}^{2} \cos ^{2} \alpha t=\frac{1}{2} I \omega_{0}^{2}\left(\sin ^{2} \alpha t+\cos ^{2} \alpha t\right)=\frac{1}{2} I \omega_{0}^{2} ,

which is constant (and equal to the initial energy). [Of course, if you didn’t notice that the rotation rate is changing, you might think the magnetic force is doing work, as the disk oscillates up and down.]

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