AIPMT PRELIMS 2004 Physics Torque MCQ Question

To solve the problem of determining the torque required to stop a rotating wheel in one minute, we need to apply the concepts of rotational motion, specifically the relationship between torque, moment of inertia, angular velocity, and angular acceleration.

Given Data:

• Moment of inertia, $I = 2 \, \text{kg} \cdot \text{m}^2$
• Initial angular velocity, $\omega_i = 60 \, \text{rpm}$
• Time to stop, $t = 1 \, \text{minute} = 60 \, \text{seconds}$

Step 1: Convert Angular Velocity to Radians per Second

The angular velocity in revolutions per minute (rpm) must be converted to radians per second (rad/s). We use the conversion factor:

$$1 \, \text{rpm} = \frac{2\pi \, \text{rad}}{60 \, \text{s}}$$

Thus,

$$\omega_i = 60 \, \text{rpm} = 60 \times \frac{2\pi \, \text{rad}}{60 \, \text{s}} = 2\pi \, \text{rad/s}$$

Step 2: Determine Angular Deceleration

To stop the wheel, we need to decelerate it from $\omega_i$ to $\omega_f = 0 \, \text{rad/s}$ over a time period of $t = 60 \, \text{s}$. The angular acceleration $\alpha$ can be calculated using the formula:

$$\alpha = \frac{\Delta \omega}{t} = \frac{\omega_f - \omega_i}{t}$$

Substituting the values:

$$\alpha = \frac{0 - 2\pi}{60} = -\frac{2\pi}{60} = -\frac{\pi}{30} \, \text{rad/s}^2$$

Step 3: Calculate Required Torque

The torque $\tau$ required to produce this angular acceleration can be calculated using Newton’s second law for rotation:

$$\tau = I \cdot \alpha$$

Substituting the known values:

$$\tau = 2 \, \text{kg} \cdot \text{m}^2 \cdot \left(-\frac{\pi}{30} \, \text{rad/s}^2\right) = -\frac{2\pi}{30} \, \text{N} \cdot \text{m}$$

Taking the magnitude, we find:

$$\tau = \frac{2\pi}{30} = \frac{\pi}{15} \, \text{N} \cdot \text{m}$$

Conclusion

Thus, the torque required to stop the wheel in one minute is:

$$\boxed{\frac{\pi}{15} \, \text{N} \cdot \text{m}}$$

Explanation of Options

• Option A: $\frac{\pi}{12} \, \text{N} \cdot \text{m}$: Incorrect, as it does not match the calculated torque.
• Option B: $\frac{\pi}{15} \, \text{N} \cdot \text{m}$: Correct, matches our calculation.
• Option C: $\frac{\pi}{18} \, \text{N} \cdot \text{m}$: Incorrect, as it is less than the required torque.
• Option D: $\frac{2\pi}{15} \, \text{N} \cdot \text{m}$: Incorrect, as it is greater than the required torque.

In conclusion, the torque needed to stop the wheel within one minute is indeed $\frac{\pi}{15} \, \text{N} \cdot \text{m}$, confirming that option B is the correct answer.