Question 391: Centre of Mass and Rotational Motion

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

Accepted Answer

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

Centre of Mass and Rotational Motion

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.

Integer / Numeric Answer Type