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Q.

A plastic circular disc of radius 10 cm is placed on a thin oil film of thickness 2 mm, spread over a flat horizontal surface. The torque (in $10^{- 2} N - m$ ) required to spin the disc about its central vertical axis with a constant angular velocity 8 rad/sec is (coefficient of viscosity of oil is $1.0 k g / m - s$ ) (take $π = 3.14$ )

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answer is 63.

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$τ = \int F . r = \int_{0}^{R} η (2 π r d r) (\frac{r w}{t}) . r$

$\begin{array}{l} = \frac{2 π η W R^{4}}{4 t} = \frac{2 \times 3.14 \times 1 \times 8 \times 10^{- 4}}{4 \times 2 \times 10^{3}} \\ = 0.625 N - m \approx 0.63 N - m \end{array}$

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A plastic circular disc of radius 10 cm is placed on a thin oil film of thickness 2 mm, spread over a flat horizontal surface. The torque (in 10−2N−m ) required to spin the disc about its central vertical axis with a constant angular velocity 8 rad/sec is (coefficient of viscosity of oil is 1.0 kg/m−s ) (take π = 3.14 )