A thin horizontal disc of radius $R = 10 ext{ cm}$ is located within a cylindrical cavity filled w

Question

A thin horizontal disc of radius $R = 10 	ext{ cm}$ is located within a cylindrical cavity filled with oil of viscosity $\eta = 0.08 	ext{ P}$ ($1 	ext{ Poise} = 0.1 	ext{ Pa}\cdot	ext{s}$). The clearance between the disc and the top horizontal plane is $h = 1.0 	ext{ mm}$, and the clearance between the disc and the bottom horizontal plane is also $h = 1.0 	ext{ mm}$. The disc rotates with a constant angular velocity $\omega = 60 	ext{ rad/s}$. Neglecting end effects, which of the following statements is/are correct?

Professor PC Thomas Classes · Answer

Let's analyze the forces acting on an elementary annular area of the disc. Consider a ring of radius $r$ and radial thickness $dr$. Its area on one side is $dA = 2\pi r dr$.

The linear velocity of the fluid at this radius (sticking to the disc) is $v = \omega r$. Since the clearance $h$ is small, the velocity profile in the gap between the disc and the stationary planes is linear. The velocity gradient is $\frac{dv}{dz} = \frac{\omega r}{h}$.

According to Newton's law of viscosity, the viscous shear stress is $ au_{ ext{shear}} = \eta \frac{dv}{dz} = \frac{\eta \omega r}{h}$.
The viscous force on the elementary ring on ONE side of the disc is $dF = au_{ ext{shear}} dA = \left(\frac{\eta \omega r}{h} ight)(2\pi r dr) = \frac{2\pi \eta \omega r^2}{h} dr$.
The elementary viscous torque on this ring is $d au = r \cdot dF = \frac{2\pi \eta \omega r^3}{h} dr$ (Option C is correct).

Integrating from $r = 0$ to $r = R$ gives the total torque on ONE side of the disc: $ au_1 = \int_0^R \frac{2\pi \eta \omega r^3}{h} dr = \frac{\pi \eta \omega R^4}{2h}$.
Since the oil is present on both the top and bottom surfaces, the total torque is $ au = 2 au_1 = \frac{\pi \eta \omega R^4}{h}$ (Option A is correct).

The power required to maintain this rotation against the viscous drag is $P = au \omega = \frac{\pi \eta \omega^2 R^4}{h}$. Note that $P \propto R^4$ (so Option D is incorrect).

Now, substitute the given numerical values:
$\eta = 0.08 ext{ P} = 0.008 ext{ Pa}\cdot ext{s}$
$R = 10 ext{ cm} = 0.1 ext{ m}$
$\omega = 60 ext{ rad/s}$
$h = 1.0 ext{ mm} = 10^{-3} ext{ m}$

$P = \frac{\pi (0.008 ext{ Pa}\cdot ext{s}) (60 ext{ s}^{-1})^2 (0.1 ext{ m})^4}{10^{-3} ext{ m}}$
$P = \frac{\pi imes 0.008 imes 3600 imes 10^{-4}}{10^{-3}} = \pi imes 0.008 imes 3600 imes 0.1 = \pi imes 0.8 imes 3.6 = 2.88\pi ext{ W}$ (Option B is correct).