A wheel turning with angular speed of 30 rev / s is brought to rest with a constant acceleration. It turns 60 rev before it stops. The time that elapses before it stops is

Options

A. 2 s
B. 4s
C. 5 s
D. 6 s

Explanation

To solve the problem step by step, we need to find the time it takes for a wheel to stop given its initial angular speed, the distance it travels before stopping, and the fact that it decelerates with a constant angular acceleration.

### Step 1: Convert Angular Speed to Radians per Second
The initial angular speed \( \omega_0 \) is given as 30 revolutions per second. We convert this to radians per second using the conversion factor \( 2\pi \) radians per revolution.

\[ \omega_0 = 30 \, \text{rev/s} \times 2\pi \, \text{rad/rev} = 60\pi \, \text{rad/s} \]

### Step 2: Calculate Angular Displacement
The wheel turns 60 revolutions before it stops. We convert this to radians:

\[ \theta = 60 \, \text{rev} \times 2\pi \, \text{rad/rev} = 120\pi \, \text{rad} \]

### Step 3: Use the Equation of Motion for Rotational Motion
We can use the rotational equation of motion that relates angular displacement, initial angular velocity, final angular velocity, and angular acceleration:

\[ \omega^2 = \omega_0^2 + 2\alpha\theta \]

Here, the final angular velocity \( \omega \) is 0 (since the wheel comes to rest), \( \omega_0 \) is \( 60\pi \, \text{rad/s} \), and \( \theta \) is \( 120\pi \, \text{rad} \).

Substituting the values into the equation:

\[ 0 = (60\pi)^2 + 2\alpha(120\pi) \]

### Step 4: Solve for Angular Acceleration \( \alpha \)
Rearranging the equation to solve for \( \alpha \):

\[ (60\pi)^2 = -2\alpha(120\pi) \]

\[ \alpha = -\frac{(60\pi)^2}{2(120\pi)} = -\frac{3600\pi}{240} = -15 \, \text{rad/s}^2 \]

### Step 5: Use the First Equation of Motion to Find Time
Now that we have the angular acceleration, we can use the first equation of motion for rotational motion to find the time \( t \):

\[ \omega = \omega_0 + \alpha t \]

Setting \( \omega = 0 \):

\[ 0 = 60\pi + (-15)t \]

Rearranging gives:

\[ 15t = 60\pi \]

\[ t = \frac{60\pi}{15} = 4\pi \, \text{s} \]

### Step 6: Approximate Time
If we need a numerical approximation, we can use \( \pi \approx 3.14 \):

\[ t \approx 4 \times 3.14 \approx 12.56 \, \text{s} \]

### Final Answer
The time that elapses before the wheel stops is approximately \( 4\pi \) seconds or about \( 12.56 \) seconds.

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