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Step-by-Step Solution
Step 1: Identify the Given Quantities
• Moment of inertia of the wheel: $I = 2\,\mathrm{kg \, m^2}$
• Initial angular speed: $60\,\mathrm{rpm}$ (revolutions per minute)
• Time to stop the wheel: $t = 60\,\mathrm{s}$
Step 2: Convert Angular Speed to Radians per Second
1 revolution = $2\pi$ radians.
Thus, $60\,\mathrm{rpm}$ means $60$ revolutions in $60$ seconds, or $1$ rev/s.
So, the initial angular speed in rad/s is:
$$
\omega_i = 1 \,\mathrm{rev/s} \times 2\pi \,\mathrm{rad/rev} = 2\pi \,\mathrm{rad/s}.
$$
Step 3: Determine Angular Deceleration (Retardation)
We want the wheel to come to rest in $60\,\mathrm{s}$. Using the relation for constant angular acceleration (or deceleration):
$$
\omega_f = \omega_i + \alpha t,
$$
where $\omega_f = 0$ (final angular speed). Hence,
$$
0 = \omega_i + \alpha \times 60.
$$
Solving for $\alpha$:
$$
\alpha = -\,\frac{\omega_i}{60} = -\,\frac{2\pi}{60} = -\,\frac{\pi}{30}\,\mathrm{rad/s^2}.
$$
The negative sign indicates a deceleration.
Step 4: Calculate the Required Torque
Torque $\tau$ is related to angular acceleration $\alpha$ by
$$
\tau = I\,\alpha.
$$
Substitute $I = 2\,\mathrm{kg \, m^2}$ and $\alpha = -\,\frac{\pi}{30}\,\mathrm{rad/s^2}$:
$$
\tau = 2 \,\mathrm{kg \, m^2} \times \left(-\,\frac{\pi}{30}\right).
$$
Taking the magnitude,
$$
|\tau| = \frac{2\pi}{30} = \frac{\pi}{15}\,\mathrm{N\,m}.
$$
Final Answer
The torque required to stop the wheel in one minute is
$$
\tau = \frac{\pi}{15}\,\mathrm{N\,m}.
$$
