Question

A thin solid disk of 1 kg is rotating along its diameter axis at the speed of 1800 rpm. By applying an external torque of \(25\ \pi\) Nm for 40s, the speed increases to 2100 rpm. The diameter of the disk is ____ m.

Answer 1

Answer is : 40

\(\tau\ dt =I\Delta w\)

\(\Rightarrow 25 \pi \times 40 = I(300) \times \frac{2\pi}{60}\)

\(I = \frac{25\times60\times 40}{300 \times 2} = 100 = \frac{MR^2}{4}\)

\(R^2 =400\)

\(R = 20\ m\)

\(D = 40\ m\)