Question

Question: 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 role="math" localid="1653403772759" $B\left(s,z\right)=k\left(-s\overbrace{s}+2z\overbrace{z}\right)$, where k is a constant.

(a) Find the position of the center if the ring, $z\left(t\right)$, and its angular velocity, $\omega \left(t\right)$, 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 work.

Short answer

(a) The position of the center is $z\left(t\right)=\frac{{\omega }_0}{\alpha }\sqrt{\frac{1}{M}}\left(1-\cos \alpha t\right)$.

(b) The total kinetic energy is $\frac{1}{2}I{{\omega }_0}^2$, and the disk rises and falls harmonically as its rotation slows down and speeds up.

Step-by-step solution

Expression for the force on one charge:

Suppose, initially, the disk will rise like a helicopter.

Write the expression for the force on one charge.

$F_i=q\left(v\times B\right)$ …… (1)

Here, q is the charge, v is the velocity, and B is the magnetic field.

The velocity is given as:

$v=\omega R\overbrace{\varnothing }+v_2\overbrace{z}$

Determine the position of the center:

(a)

From equation (1) write the expression for the force.

$$\begin{gathered} F_i=qk\left|\begin{array}{ccc}\overbrace{s}& \overbrace{\varnothing }& \overbrace{z}\\ 0& \omega R& v_z\\ -R& 0& 2z\end{array}\right| \\ {F}_i=qk\left(2\omega Rz\overbrace{s}-R{v}_z\overbrace{\varnothing }+\omega R^2\overbrace{z}\right) \end{gathered}$$

Write the expression for the net force on all the charges.

$F=\sum _{i=1}^n{F}_i$

Substitute the value of in the above expression.

$$\begin{gathered} F=qk\left(2\omega Rz\overbrace{s}-R{v}_z\overbrace{\varnothing }+\omega R^2\overbrace{z}\right) \\ F=nqk{R}^2\omega \overbrace{Z} \\ F=M\frac{d^{2}z}{d{t}^2}\overbrace{z} \\ \frac{d^{2}z}{d{t}^2}=\left(\frac{nqk{R}^2}{M}\right)\omega \end{gathered}$$ …… (2)

Write the expression for the net torque on the disk.;

$N=\sum _{i=1}^n\left(r_i\times F_i\right)$

Substitute the known values in the above expression.

$$\begin{gathered} N=n\left(R\overbrace{s}\right)\times \left(-qkR{v}_z\overbrace{\varnothing }\right) \\ N=-nqk{R}^2{v}_z\overbrace{z} \\ N=I\frac{d\omega }{dt}\overbrace{z} \end{gathered}$$

Here, I is the moment of inertia of the disk, which is give as:

$\frac{d\omega }{dt}=-\left(\frac{nqk{R}^2}{I}\right)\frac{dz}{dt}$ …… (3)

Differentiate equation (3) and combine with equation (2).

$$\begin{gathered} \frac{d^{2}z}{d{t}^2}=-\left(\frac{1}{nqk{R}^2}\right)\frac{d^2\omega }{d{t}^2}\left(\frac{d^2\omega }{d{t}^2}=-\left(\frac{nqk{R}^2}{MI}\omega \right)\right) \\ \frac{d^{2}z}{d{t}^2}=\left(\frac{nqk{R}^2}{M}\omega \right) \end{gathered}$$

Write the solution with initial angular velocity and initial angular acceleration 0.

$\omega \left(t\right)={\omega }_0\cos \alpha t$

Hence, the equation becomes,

$$\begin{gathered} \frac{dz}{dt}=-\left(\frac{1}{nqk{R}^2}\right)\frac{d\omega }{dt} \\ \frac{dz}{dt}=\left(\frac{1}{nqk{R}^2}\right){\omega }_0\alpha \sin \alpha t \\ \frac{dz}{dt}={\omega }_0\sqrt{\frac{1}{M}}\sin \alpha t \end{gathered}$$

On further solving,

$$\begin{gathered} z\left(t\right)={\omega }_0\sqrt{\frac{I}{M}}{\int }_0^t\sin \alpha tdt \\ z\left(t\right)=\frac{{\omega }_0}{\alpha }\sqrt{\frac{I}{M}}\left(1-\cos \alpha t\right) \end{gathered}$$

Therefore, the position of the center is $z\left(t\right)=\frac{{\omega }_0}{\alpha }\sqrt{\frac{I}{M}}\left(1-\cos \alpha t\right)$.

Determine the total kinetic energy:

(b)

The disk rises and falls harmonically as its rotation slows down and speeds up. Hence, the total energy will be,

$$\begin{gathered} E=\frac{1}{2}m{v_z}^2+\frac{1}{2}I{\omega }^2 \\ E=\frac{1}{2}I{{\omega }_0}^2{\sin }^2\alpha t+\frac{1}{2}I{{\omega }_0}^2{\cos }^2\alpha t \\ E=\frac{1}{2}I{{\omega }_0}^2\left({\sin }^2\alpha t+{\cos }^2\alpha t\right) \\ E=\frac{1}{2}I{{\omega }_0}^2 \end{gathered}$$

Therefore, the total kinetic energy is $\frac{1}{2}I{{\omega }_0}^2$and the disk rises and falls harmonically as its rotation slows down and speeds up.