A solid cone of maximum radius R and vertex angle is to rotate at an angular velocity ω . An oil of viscosity μ and thickness t fills the gap between the cone and the housing, Derive an expression for the torque required and the rate of heat dissipation in the bearing.

Step 1: Torque on an elemental area (\( dT \)):

Consider an elementary area \( dA \) at radius \( r \) of the cone:

\( dA = 2\pi r \, dr \sin\theta \)

Shear force acting on the area:

\( \text{Shear force} = \tau \cdot dA = \mu \frac{u}{t} \cdot dA = \mu \frac{\omega r}{t} \cdot 2\pi r \, dr \sin\theta \)

Torque on the elemental area:

\( dT = \text{Shear force} \cdot r = \mu \frac{\omega r}{t} \cdot 2\pi r \, dr \sin\theta \cdot r = \frac{2\pi\mu\omega}{t\sin\theta} r^3 \, dr \)

Step 2: Total torque (\( T \)):

Integrate over the radius of the cone from \( r = 0 \) to \( r = R \):

\( T = \int_0^R dT = \int_0^R \frac{2\pi\mu\omega}{t\sin\theta} r^3 \, dr \)

Evaluate the integral:

\( T = \frac{2\pi\mu\omega}{t\sin\theta} \int_0^R r^3 \, dr = \frac{2\pi\mu\omega}{t\sin\theta} \cdot \frac{R^4}{4} \)

Simplify:

\( T = \frac{\pi\mu\omega R^4}{2t\sin\theta} \)

Step 3: Rate of heat dissipation:

The rate of heat dissipation is equal to the power utilized in overcoming resistance:

\( \text{Power} = T \cdot \omega \)

Substitute the expression for \( T \):

\( \text{Power} = \frac{\pi\mu\omega R^4}{2t\sin\theta} \cdot \omega = \frac{\pi\mu\omega^2 R^4}{2t\sin\theta} \)