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 the angular acceleration.
The wheel starts from rest, so \[ \omega_0=0. \] At \[ t=2\ \text{s}, \] the angular velocity is \[ \omega=3.14\ \text{rad s}^{-1}. \] Using \[ \omega=\omega_0+\alpha t, \] we get \[ 3.14=0+\alpha(2). \] Therefore, \[ \alpha=\frac{3.14}{2}. \] \[ \alpha=1.57\ \text{rad s}^{-2}. \]

Step 2: Find angular displacement from \(t=0\) to \(t=20\ \text{s}\).
Acceleration continues up to \[ t=20\ \text{s}. \] Angular displacement during uniformly accelerated motion is \[ \theta_1=\omega_0t+\frac{1}{2}\alpha t^2. \] Since \[ \omega_0=0, \] we get \[ \theta_1=\frac{1}{2}(1.57)(20)^2. \] \[ \theta_1=\frac{1}{2}\times 1.57 \times 400. \] \[ \theta_1=314\ \text{rad}. \]

Step 3: Find angular velocity at \(t=20\ \text{s}\).
\[ \omega_{20}=\omega_0+\alpha t. \] \[ \omega_{20}=0+1.57(20). \] \[ \omega_{20}=31.4\ \text{rad s}^{-1}. \]

Step 4: Find angular displacement from \(t=20\ \text{s}\) to \(t=40\ \text{s}\).
At \(t=20\ \text{s}\), acceleration is ceased.
So, from \(t=20\ \text{s}\) to \(t=40\ \text{s}\), the wheel moves with constant angular velocity \[ \omega=31.4\ \text{rad s}^{-1}. \] Time interval is \[ 20\ \text{s}. \] Thus, \[ \theta_2=\omega t. \] \[ \theta_2=31.4\times 20. \] \[ \theta_2=628\ \text{rad}. \]

Step 5: Find total angular displacement.
\[ \theta=\theta_1+\theta_2. \] \[ \theta=314+628. \] \[ \theta=942\ \text{rad}. \]

Step 6: Convert radians into revolutions.
One revolution is \[ 2\pi\ \text{rad}. \] Using \[ \pi=3.14, \] we get \[ 2\pi=6.28\ \text{rad}. \] Number of revolutions is \[ n=\frac{\theta}{2\pi}. \] \[ n=\frac{942}{6.28}. \] \[ n=150. \]

Step 7: Final conclusion.
Therefore, the number of revolutions made by the wheel is \[ \boxed{150} \] Hence, the correct option is \[ \boxed{(4)} \]