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Step-by-Step Solution
Step 1: List the Given Data
ā¢ Mass of the solid cylinder, $m = 2\text{ kg}$
ā¢ Radius of the cylinder, $r = 4\text{ cm} = 0.04\text{ m}$
ā¢ Initial rotational speed, $3\text{ rpm}$ (revolutions per minute)
ā¢ Number of revolutions before coming to rest, $2\pi \text{ revolutions}$ (interpreted as $2\pi \times 2\pi = 4\pi^2$ radians)
Step 2: Convert the Angular Speed to Radians per Second
1 revolution per minute (rpm) corresponds to $2\pi$ radians per 60 seconds. Hence,
$ \omega_i = 3 \times \frac{2\pi}{60} \; \text{rad/s} \;=\; \frac{6\pi}{60} \;=\; \frac{\pi}{10}\text{ rad/s}.$
Step 3: Moment of Inertia of the Solid Cylinder
For a solid cylinder of mass $m$ and radius $r$ rotating about its central axis, the moment of inertia is
$ I = \frac{1}{2} \, m r^2. $
Substituting the values:
$ I = \frac{1}{2} \times 2 \times (0.04)^2 \;=\; 1 \times 0.0016 \;=\; 0.0016 \text{ kgĀ·m}^2. $
Step 4: Apply the Work-Energy (Torque-Angle) Theorem
The work done by the torque (magnitude) in bringing the cylinder to rest equals the loss in its rotational kinetic energy. Since final angular speed is 0,
$ \tau \, \theta = \frac{1}{2} I \, \omega_i^2, $
where $ \theta $ is the total angle turned while stopping. We are told it stops after $2\pi$ revolutions, which the problem sets as $4 \pi^2 \text{ rad}.$ Thus,
$ \theta = 4\pi^2 \text{ rad}, $
so
$ \tau \times 4\pi^2 = \frac{1}{2} \times 0.0016 \times \left(\frac{\pi}{10}\right)^2. $
Step 5: Calculate the Left-Hand Side and Solve for $ \tau $
First, compute $ \frac{1}{2} I \, \omega_i^2 $:
$ \frac{1}{2} \, I \, \omega_i^2 \;=\; \frac{1}{2} \times 0.0016 \times \left(\frac{\pi}{10}\right)^2
\;=\; 0.0008 \times \frac{\pi^2}{100}
\;=\; 8 \times 10^{-6} \pi^2. $
Hence,
$ \tau \times 4\pi^2 \;=\; 8 \times 10^{-6} \pi^2. $
Dividing both sides by $4\pi^2$,
$ \tau = \frac{8 \times 10^{-6} \pi^2}{4 \pi^2}
\;=\; 2 \times 10^{-6} \text{ NĀ·m}. $
Final Answer
$ \boxed{2 \times 10^{-6}\text{ NĀ·m}} $
