b) If the shaft in (a) is hollow with inside diameter equal to $3/4^{th}$ of the outside diameter of the shaft for the same condition.

Given

$P = 50\hspace{0.05cm}KW = 50\hspace{0.05cm}\times\hspace{0.05cm}10^3\hspace{0.05cm}Watt\\ N = 120\hspace{0.05cm}rpm$

Solution

$\omega = \frac{2\pi N}{60} = 12.50\hspace{0.05cm}rad/sec\\ \rho = \frac{2\pi NT}{60}\\ \hspace{0.25cm} = \omega .T\\ T = \frac{P}{\omega} = \frac{50\hspace{0.05cm}\times\hspace{0.05cm}10^3}{12.50} = 3.98\hspace{0.05cm}\times\hspace{0.05cm}10^3\hspace{0.05cm}Nm = 3.98\hspace{0.05cm}\times\hspace{0.05cm}10^6\hspace{0.05cm}Nmm$

If $\theta = 0.5^\circ = 8.72\hspace{0.05cm}\times\hspace{0.05cm}10^{-3}\hspace{0.05cm}rad$

A] For solid shaft, L = 1000 mm, $J = \frac{\pi}{32}D^4$ …