A vertical axle carrying a load P ends in a hemisphere

A vertical axle carrying a load $P$ ends in a hemisphere of radius $R$ that bears against a socket, and the axle rotates. Assuming the normal pressure is spread uniformly over the hemispherical contact surface and the coefficient of friction is $\mu$, find the work done by friction in one complete revolution.

Step 1: Uniform pressure

With total load $P$ spread over the hemisphere's area $2\pi R^2$, the pressure is $p=P/(2\pi R^2)$.

Step 2: Area and normal force elements

Using polar angle $\theta\in[0,\pi/2]$ from the rotation axis and azimuth $\phi\in[0,2\pi]$, the area element is $dA=R^2\sin\theta\,d\theta\,d\phi$, so the normal force on it is

\[dN=p\,dA=\frac{P\sin\theta}{2\pi}\,d\theta\,d\phi.\]

Step 3: Friction force and travel distance

Friction on the element is $dF_f=\mu\,dN$. A point at polar angle $\theta$ is at distance $R\sin\theta$ from the rotation axis, so in one revolution it travels a circle of length $2\pi R\sin\theta$.

Step 4: Elementary work

\[dW=dF_f\cdot(2\pi R\sin\theta)=\frac{\mu P\sin\theta}{2\pi}\,d\theta\,d\phi\cdot2\pi R\sin\theta=\mu PR\sin^2\theta\,d\theta\,d\phi.\]

Step 5: Integrate over the hemisphere

\[W=\mu PR\int_0^{2\pi}d\phi\int_0^{\pi/2}\sin^2\theta\,d\theta=2\pi\mu PR\int_0^{\pi/2}\sin^2\theta\,d\theta.\]

Using $\sin^2\theta=\tfrac12(1-\cos2\theta)$, $\displaystyle\int_0^{\pi/2}\sin^2\theta\,d\theta=\frac{\pi}{4}$.

Step 6: Final result

\[W=2\pi\mu PR\cdot\frac{\pi}{4}=\frac{\pi^2\mu PR}{2}.\]

\[\boxed{W=\frac{\pi^2\mu PR}{2}}\]