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 $25 \pi ~\mathrm{Nm}$ for 40 s , the speed increases to 2100 rpm . The diameter of the disk is ___________ m.

<p><p>Mass of the disk, $ m = 1 \, \text{kg} $.</p></p> <p><p>Initial angular velocity, $ \omega_i = 1800 \, \text{rpm} $.</p></p> <p><p>Final angular velocity, $ \omega_f = 2100 \, \text{rpm} $.</p></p> <p><p>External torque, $ \tau_{\text{ext}} = 25\pi \, \text{Nm} $.</p></p> <p><p>Time, $ t = 40 \, \text{seconds} $.</p></p> <p>First, convert the rotational speeds from revolutions per minute (rpm) to radians per second (rad/s):</p> <p>$ \omega_i = 1800 \times \frac{2\pi}{60} = 60\pi \, \text{rad/s} $</p> <p>$ \omega_f = 2100 \times \frac{2\pi}{60} = 70\pi \, \text{rad/s} $</p> <p>Next, use the equation of motion for rotation to find the angular acceleration $\alpha$:</p> <p>$ \omega_f = \omega_i + \alpha t $</p> <p>$ 70\pi = 60\pi + \alpha(40) $</p> <p>Solving for $\alpha$:</p> <p>$ \alpha = \frac{70\pi - 60\pi}{40} = \frac{10\pi}{40} = \frac{\pi}{4} \, \text{rad/s}^2 $</p> <p>The torque and moment of inertia relationship is given by:</p> <p>$ \tau = I \alpha $</p> <p>For a thin solid disk rotating along its diameter, the moment of inertia $ I $ is $ \frac{mR^2}{4} $. Thus:</p> <p>$ 25\pi = \frac{1 \times R^2}{4} \times \frac{\pi}{4} $</p> <p>Solving for $ R $:</p> <p>$ 25\pi = \frac{\pi R^2}{16} $</p> <p>$ R^2 = \frac{25\pi \times 16}{\pi} = 400 $</p> <p>$ R = 20 \, \text{m} $</p> <p>The diameter of the disk is:</p> <p>$ \text{Diameter} = 2R = 2 \times 20 = 40 \, \text{m} $</p>