Question

Applying a constant torque the speed of a flywheel is increased from \(1800\) rpm to \(2400\) rpm in \(10\) seconds. The number of revolutions made by the flywheel during this time is:

• A. 350
• B. 900
• C. 700
• D. 2100

Hint:

When acceleration is constant, use average speed \( = \frac{u+v}{2} \) to find displacement directly.

Answer 1

The Correct Option is A

Step 1: Convert angular speeds to rps.
\[ \omega_1 = 1800\,\text{rpm} = \frac{1800}{60} = 30\,\text{rps} \]
\[ \omega_2 = 2400\,\text{rpm} = \frac{2400}{60} = 40\,\text{rps} \]

Step 2: Use uniformly accelerated rotation concept.
Since torque is constant, angular acceleration is constant.
Average angular speed is:
\[ \omega_{\text{avg}} = \frac{\omega_1 + \omega_2}{2} \]

Step 3: Calculate average angular speed.
\[ \omega_{\text{avg}} = \frac{30 + 40}{2} = 35\,\text{rps} \]

Step 4: Calculate number of revolutions.
\[ \text{Revolutions} = \omega_{\text{avg}} \times t \]
\[ = 35 \times 10 \]
\[ = 350 \]

Step 5: Interpretation.
Flywheel completes 350 rotations in 10 seconds.

Step 6: Concept used.
For uniform angular acceleration, displacement = average angular velocity × time.

Step 7: Final conclusion.
\[ \boxed{350} \]