Q 22. Consider a hollow cylinder of mass \(m\), radius \(R\) and length \(L\) placed on a smooth frictionless surface as shown in figure. Let its curved surface area be charged with a uniform surface charge density, \(\sigma\). It is placed in a region of time varying magnetic field, \(B = B_0\ln(1+\Omega t)\). At time, \(t=0\), sand starts to pour into the cylinder at a rate of \(\lambda\) (kg \(\text{s}^{-1}\)). The density of the sand is \(\rho\). Assume that when the cylinder is completely filled, the magnetic field becomes constant at whatever value it is then. Further assume that as the sand falls, it always forms a perfect cylinder.

1. Find its angular velocity, \(\omega\) as a function of time.

2. Further, consider the special case when \(\frac{m \Omega}{\lambda} = \frac{\eta + 2}{\eta + 4}\), where \(\eta\) is the ratio \({R \over L}\). Now, find the maximum angular velocity obtained and at what time it is obtained.

3. (Assume the previous condition still holds) If we want the final angular velocity to be the maximum angular velocity achieved, what is the ratio of mass of total sand in the cylinder at that time to the mass of the cylinder. i.e. \({\rho \pi R^2 L \over m} = \; ?\)

Solution:

First, we find the moment of inertia of only the cylinder

\[I = \frac{mR^2}{2} \left( \frac{\eta + 4}{\eta + 2} \right)\]

Then, we can find the moment of inertia of the sand as a function of time,

\[I_{sand} = \frac{\lambda R^2}{2} t\]

For convenience, let \(b = \frac{\lambda R^2}{2}\)

From Faraday’s law, we can write:

\[\oint \vec{E}_{in} \cdot \vec{dl} = \frac{-d \phi_B }{dt}\]

Integrating over the closed loop, (\(\vec{E}_{in}\) is constant for the loop), we get,

\[E_{in} = \frac{- B_0 R \Omega }{2 (1 + \Omega t)}\]

\[\implies \tau = \frac{ B_0 R \Omega \pi L \sigma}{(1 + \Omega t)}\]

Integrating,

\[\omega = B_0 \pi R L \sigma {\ln \left| 1+ \Omega t \right| \over (I + bt)}\]

In general,

\[\omega_{max} = \frac{\exp\left[ W\left( {I \Omega - b \over eb } \right) + 1 \right] - 1}{\Omega}\]

Where \(W\) is the product log function(Or Lambert W function).

Under the given conditions,

\[\omega_{max} = \frac{2 B_0 \pi L \sigma \Omega}{\lambda e} = \frac{B_0 \pi R L \sigma}{Ie}\]

From this, we get:

\[\frac{\rho \pi R^2 L}{m} = \left( \frac{\eta + 4}{\eta + 2} \right) (e-1)\]