Question

A wheel undergoes a constant acceleration starting from rest at \(t=0\). The angular velocity of the wheel is \(3.14\ \text{rad s}^{-1}\) when \(t=2\ \text{s}\). The acceleration is abruptly ceased at \(t=20\ \text{s}\). The number of revolutions the wheel makes in the interval \(t=0\) to \(t=40\ \text{s}\) is

• A. \(100\)
• B. \(175\)
• C. \(225\)
• D. \(150\)

Hint:

When angular acceleration stops, the angular velocity becomes constant. Split the motion into two parts: \[ \theta_1=\omega_0t+\frac{1}{2}\alpha t^2 \] for accelerated motion and \[ \theta_2=\omega t \] for uniform angular motion.

Answer 1

The Correct Option is D

Step 1: Find angular acceleration during 0 to 20 s.
\( \alpha = \frac{3.14 - 0}{2} = 1.57\text{ rad/s}^2 \). At \( t = 20\text{ s} \): \( \omega_{20} = 1.57\times 20 = 31.4\text{ rad/s} \).

Step 2: Count revolutions in two phases.
Phase 1 (0–20 s, accelerating): \( \theta_1 = \frac{1}{2}\alpha(20)^2 = \frac{1}{2}(1.57)(400) = 314\text{ rad} \).
Phase 2 (20–40 s, constant \( \omega \)): \( \theta_2 = 31.4\times 20 = 628\text{ rad} \).
Total \( \theta = 942\text{ rad} \Rightarrow N = \frac{942}{2\pi} = \frac{942}{6.28} = 150\text{ rev} \).

\[ \boxed{150\text{ revolutions}} \]