Question

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

• A. 100
• B. 175
• C. 225
• D. 150

Hint:

Equations for constant angular acceleration: \(\omega = \omega_0 + \alpha t\) \(\Delta\theta = \omega_0 t + \frac{1}{2}\alpha t^2\) \(\omega^2 = \omega_0^2 + 2\alpha \Delta\theta\)
For constant angular velocity: \(\Delta\theta = \omega t\).
1 revolution = \(2\pi\) radians.
If values like 3.14 appear, check if they relate to \(\pi\) or \(\pi/2\). (Here \(3.14 \approx \pi\), \(1.57 \approx \pi/2\)). Using \(\pi\) simplifies calculations.

Answer 1

The Correct Option is D

Phase 1: Constant angular acceleration (\(\alpha\)) from \(t=0\) to \(t=20\) s. Starts from rest, so initial angular velocity \(\omega_0 = 0\). At \(t_1=2\) s, angular velocity \(\omega_1 = 3.14\) rad/s. Using \(\omega = \omega_0 + \alpha t\): \(3.14 = 0 + \alpha(2) \Rightarrow \alpha = \frac{3.14}{2} = 1.57 \text{ rad/s}^2\). (Note: \(1.57 \approx \pi/2\), and \(3.14 \approx \pi\). So \(\alpha = \pi/2 \text{ rad/s}^2\)). Angular velocity at \(t_2=20\) s (when acceleration ceases): \(\omega_{20} = \omega_0 + \alpha t_2 = 0 + (1.57)(20) = 31.4 \text{ rad/s}\) (or \(\pi/2 \times 20 = 10\pi\) rad/s). Angular displacement (\(\Delta\theta_1\)) during this phase (\(0 \le t \le 20\) s): \(\Delta\theta_1 = \omega_0 t_2 + \frac{1}{2}\alpha t_2^2 = 0 + \frac{1}{2}(1.57)(20)^2 = \frac{1}{2}(1.57)(400) = 1.57 \times 200 = 314 \text{ rad}\). (Or \(\frac{1}{2}(\pi/2)(400) = 100\pi\) rad). Phase 2: Constant angular velocity from \(t=20\) s to \(t=40\) s. The acceleration ceases abruptly at \(t=20\)s, so the wheel continues to rotate with the angular velocity it had at \(t=20\)s. Angular velocity \(\omega_{const} = \omega_{20} = 31.4\) rad/s (or \(10\pi\) rad/s). Time duration for this phase \(\Delta t_2 = 40\text{s} - 20\text{s} = 20\text{s}\). Angular displacement (\(\Delta\theta_2\)) during this phase: \(\Delta\theta_2 = \omega_{const} \times \Delta t_2 = 31.4 \times 20 = 628 \text{ rad}\). (Or \(10\pi \times 20 = 200\pi\) rad). Total angular displacement from \(t=0\) to \(t=40\) s: \(\Delta\theta_{total} = \Delta\theta_1 + \Delta\theta_2 = 314 + 628 = 942 \text{ rad}\). (Or \(100\pi + 200\pi = 300\pi\) rad). Number of revolutions: 1 revolution = \(2\pi\) radians. Number of revolutions = \( \frac{\Delta\theta_{total}}{2\pi} \). Using \(\pi \approx 3.14\): \(\Delta\theta_{total} = 942 \text{ rad}\). \(2\pi \approx 2 \times 3.14 = 6.28\). Number of revolutions = \( \frac{942}{6.28} \). \(942 / 6.28 = (300 \times 3.14) / (2 \times 3.14) = 300/2 = 150\). If we use \(\Delta\theta_{total} = 300\pi\), then Number of revolutions = \( \frac{300\pi}{2\pi} = 150 \). This matches option (d). \[ \boxed{150} \]