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Step-by-Step Solution
Step 1: Identify the physical quantities given
ā¢ Radius of the disc, R = 20\,\text{cm} = 0.20\,\text{m}
ā¢ Mass of the disc, m = 10\,\text{kg}
ā¢ Initial angular velocity, given as 600 revolutions per minute (rpm).
ā¢ Time over which the disc comes to rest, \Delta t = 10\,\text{s}
ā¢ Final angular velocity, \omega_f = 0\,\text{rad/s} (since it comes to rest).
Step 2: Convert the initial angular velocity from rpm to rad/s
1 revolution = 2\pi radians.
600 rpm means 600 revolutions per minute. Converting minutes to seconds:
600 rpm = 600 \times \dfrac{2\pi\,\text{rad}}{1\,\text{rev}} \times \dfrac{1\,\text{min}}{60\,\text{s}}
= 600 \times \dfrac{2\pi}{60}\,\text{rad/s}
= 600 \times \dfrac{\pi}{30}\,\text{rad/s}
= 20\pi\,\text{rad/s}.
Hence, \omega_i = 20\pi\,\text{rad/s}.
Step 3: Write down the expression for the required retarding torque
The torque needed to stop the disc in time \Delta t is found using the relation
\tau = \dfrac{\Delta L}{\Delta t} = \dfrac{I(\omega_f - \omega_i)}{\Delta t},
where I is the moment of inertia of the disc about its central axis of rotation.
Step 4: Calculate the moment of inertia of the solid disc
For a solid disc of mass m and radius R , about an axis through its center and perpendicular to its plane,
I = \dfrac{1}{2} m R^2.
Substitute m = 10\,\text{kg} and R = 0.20\,\text{m} :
I = \dfrac{1}{2} \times 10 \times (0.20)^2 = 5 \times 0.04 = 0.20\,\text{kgĀ·m}^2.
Step 5: Substitute the values to find the torque
Using
\tau = \dfrac{I(\omega_f - \omega_i)}{\Delta t},
with \omega_f = 0 and \omega_i = 20\pi\,\text{rad/s} , \Delta t = 10\,\text{s} , and I = 0.20\,\text{kgĀ·m}^2 , we get
\tau = \dfrac{0.20\bigl(0 - 20\pi\bigr)}{10} = \dfrac{-0.20 \times 20\pi}{10} = -\,0.4\pi\,\text{Nm}.
The negative sign indicates a retarding (opposing) torque. In magnitude,
|\tau| = 0.4\pi\,\text{Nm}.
Step 6: Final answer
Therefore, the retarding torque required to bring the disc to rest in 10 seconds is
0.4\pi\,\text{Nm} = 4 \times 10^{-1}\,\pi\,\text{Nm} \approx 1.257\,\text{Nm}.
