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Viscous Force

As per the sh...

Question

As per the shown figure the central solid cylinder starts with initial angular velocity $ω_{0}$ . Find out the time after which the angular velocity becomes half (velocity gradient uniform).

A

\frac{m (R_{2} - R_{1}) l n 2}{4 π η l R_{2}}

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B

\frac{m (R_{2} - R_{1}) l n 2}{4 π η l R_{1}}

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C

\frac{m (R_{2} + R_{1}) l n 2}{4 π η l R_{2}}

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D

\frac{m (R_{2} + R_{1}) l n 2}{4 π η l R_{1}}

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Solution

The correct option is B $\frac{m (R_{2} - R_{1}) l n 2}{4 π η l R_{1}}$
We know that $F_{v} = η A \frac{d v}{d z}$

Given, initial angular velocity, $= ω_{0}$

Thickness of the liquid, $= R_{2} - R_{1}$

Viscous force on the cylinder,
$F_{v} = η A \frac{d v}{d z}$
So, velocity gradient,
$\frac{d v}{d z} = \frac{ω R_{1} - 0}{R_{2} - R_{1}}$
$F = η (2 π R_{1} l) \frac{ω R_{1}}{R_{2} - R_{1}}$

Torque due to viscous force about central axis of the cylinder,
$τ = F R_{1} = \frac{2 π η R_{1}^{3} ω l}{R_{2} - R_{1}}$
$I α = \frac{2 π η R_{1}^{3} ω l}{R_{2} - R_{1}}$
$\frac{m R_{1}^{2}}{2} (- \frac{d ω}{d t}) = \frac{2 π η R_{1}^{3} ω l}{R_{2} - R_{1}}$

$- \int_{w_{0}}^{w_{0} / 2} \frac{d ω}{ω} = \frac{4 π η R_{1} l}{m (R_{2} - R_{1})} \int_{0}^{t} d t$
$t = \frac{m (R_{2} - R_{1}) l η 2}{4 π η l R_{1}}$

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