Question

A circular disc of diameter d is slowly rotated in liquid viscosity at small distance h from fixed flat surface. Find the torque required to maintain an angular velocity .

Answer 1

Answer:

$\tau = \dfrac{\pi\eta\omega d^4}{32h}$

Explanation:

Given,

A circular disc of diameter d is slowly rotated in liquid viscosity at small distance h from fixed flat surface

To find: the torque required to maintain an angular velocity

Solution:

$\displaystyle\sf d\tau = r\eta \left(2\pi r dr\dfrac{\omega r}{h}\right)\\\\\tau = \int_0^r \dfrac{\eta}{h} 2\pi\omega r^3dr\\\\\tau = \int_0^{d/2} \dfrac{\eta\omega}{h} 2\pi r^3 dr\\\\\tau = \left[\dfrac{2\pi\eta\omega}{h}\dfrac{r^4}{4}\right]_0^{d/2}\\\\\boxed{\tau = \dfrac{\pi\eta\omega d^4}{32h}}$