Question

A tube of inner radius 4 cm and outer radius 5 cm can carry a maximum torque of T. This tube is now replaced by a solid circular shaft of the same material. The minimum radius of the solid circular shaft (in cm, rounded off to two decimal places) to carry the same amount of torque \( T \) is \(\underline{\hspace{2cm}}\).

Hint:

To find the radius of the solid shaft, equate the polar moments of inertia for the tube and the solid shaft.

Answer 1

Correct Answer: 4.17

The torque \( T \) is related to the polar moment of inertia \( J \) by the formula:
\[ T = \tau \cdot J \] For a tube, the polar moment of inertia is given by:
\[ J_{\text{tube}} = \frac{\pi}{2} (r_o^4 - r_i^4) \] For a solid shaft, the polar moment of inertia is:
\[ J_{\text{shaft}} = \frac{\pi}{2} r^4 \] Equating the torques for both cases:
\[ \frac{\pi}{2} (r_o^4 - r_i^4) = \frac{\pi}{2} r^4 \] Substituting \( r_o = 5 \, \text{cm} \) and \( r_i = 4 \, \text{cm} \):
\[ (5^4 - 4^4) = r^4 \] \[ 625 - 256 = r^4 \] \[ r^4 = 369 \] \[ r = \sqrt[4]{369} \approx 4.17 \, \text{cm}. \] Thus, the minimum radius of the solid shaft is \( \boxed{4.17} \, \text{cm}. \)